Algebra in a Localic Topos with Applications to Ring Theory

u E~

in ~( ~, ~) such

and B E 0n(U )

)lu

lu

The latter inequality corresponds to the following diagram :

64

~ A[u

~ u

a~lu

n[

^]~lu where B exists by ler~na 6 applied to Sh(u+,

'~u+) .

In other words, definition

19 says that ~0 is a characteristic map as soon as for any B £ 0n(u), v ~ u and xl, ..., xn E A(v) ff

^ ~ v ( X i ) < %~(~(x . . . .

, Xn)).

i=1

This notion of characteristic map will be a useful tool when classifying the localizationsof Sh(~,]r).

The reason for the name "characteristicmap" will

be found in theorem 22. Some more remarks about this definition. convention ^ ~ n>~ I, A n

= id~] r and if n = O, ^ ~

If n = I, clearly we use the

= tl?.

We know (lerana 5) that for

and (A lu)n coincide, so there cannot be any confusion in the notations

in this case; in particular ~n u = (~°[u)n acts componentwise.

Now if n = O,

the diagram reduces to ii u

B

.

A1 u

t ]I'tu which is thus equivalent to the commutativity and again no confusion is possible. If we think of ~o as the map which characterizes those elements x in A(u) such that q0u(X) = I, we deduce from definition 19 that ~0n(Xl) = ... --~n(Xn) = ] ~ n ( B ( x

I ..... Xn)) = 1.

This formula thus expresses the fact that those x are stable under the operations of ]~ (= are a subobject in Sh( Iq, Jr)). But definition 19 conveys more than

65

this simple fact : Proposition 20. Let f : A + B be a morphism in S h ( H , ~ )

and ~ : B ÷ ~

a characteristic

Then ~ o f is again a characteristic map.

map in 6(~,TF).

Consider the following diagram for any n E N, u E ~

A nu fn u

A[u

>

t Bn u

Blu

B

~ u

and B E 0n(u ) :

t

i +tu ~;~l u •

^~

u

The first square is cor~utative beca<~e f is a morphism in Sh(~,]Y).

Therefore I

o f satisfies the conditions of definition 19. Proposition 21. Let A be an object in S h ( ~ , ~ ) , and B C 00(u).

~ :A + ~

a characteristic map, u E

Then ~u(B) = a F h u.

I

For we have seen that ~lu o 8 = t~I u. Theorem 22. Let A be some object in Sh( ~, ~). There is a bijection between (I) the subobjects R ~-~ A in Sh(ll,~).

(2) the characteristic maps ~ : A + ~ F

in 6( ~, TF).

Moreover, if R and %0 correspond to each other under this bijection, the following square is a pullback in the topos 6( ~, ~)

:

66

R

~

l

A

)


Consider a subobject R >--+ A in Sh( ~, ~). map<0 : A ÷

q~

in~(I{,~).

]

I

A,IF We shall define a characteristic

Byproposition I - 5, for a n y u in}{we define

~(u) : A(u) --'J (a F hu, A) -~ 9.IF(U) ~(u) (f) = f-1 (R). is a natural transformation since for any g in (a F hv, A) with u ~< v we have

~(u)Cglu) = g ~I
a-](f-](R))

= (f o ~ ) - I ( R ) .

So ~ is a morphism in ~ ( H , ~ ) . We prove now that , is a characteristic map (notations of definition 19). If n C ~ ,

B E 0n(U), v < u

and (fl ..... fn) C A n ~v) = An(v) = (a F hv, A) n,

we need to show that

B(f~ . . . . .

fn )-1 (a) />

nn f]] (R), i=1

which means, by lemma ] ] n

~
. . . . .

~)-](RIv)~> n

fi] (Rlv) •

i= ] This is true by le~ma 4 applied to Sh(u4, ~u).

(By lemma 5, it does not matter

where inverse images are computed). The definition of ~p shows that f C A(u) is sended to t ~r(,) in ~ ( u ) if and only if f-](R) = a F hu, i.e. if and only if f factors through R :

67

f - I (R)

,

R

,

A.

I

I

a Fh u f

By p r o p o s i t i o n I - 5, t h i s shows e x a c t l y t h a t

~u 1(1) = ( a F h u, R) = R(u). I n o t h e r words, t h e f o l l o w i n g square i s a p u l l b a c k i n ~( 14, ~ ) R

>

I

1

pb

A

, a T. ko

Conversely, consider a characteristic map ko : A + ~ IF in a( Iq, ~), with A in Sh( 14, I~). Define R >-+ A by the following pullback in a( 14, ~) R-

~ I

I

p.b.

[

tit

~ ~a~.

A

It remains to show that R is a subobject of A in Sh( lq, I~), i.e. is stable under the operations of ~.

Con~lder u C14, n Cl~, ¢ E On(u) and xl, "''' Xn E R(u).

Thus Wu(Xi) = a F h u. n

Wu(B(x l..... Xn)) >i ^ k0u(Xi) i=I

definition 19

n

=

^ aFh i=1 u

=aFh

and t h i s shows t h a t g ( x l ,

""'

u

Xn) i s i n R(u).

We must prove that we have defined a one to one correspondance between the st~bobjects of A in Sh( 14, ~) and the characteristic maps on A in ~( 14, IT).

If

we start with a subobject R ~-+ A in Sh(14,~), we construct a map ~0 : A ÷ ~3£ in

~( I~, 7) and finally the subobject ko-I (t~) ~

subobject is just R, as follows from the pullback

A in Sh( 14, I~). But this

68

tll ~ A

+gIF


Conversely consider a characteristic map qo : A ÷ ~ F

in ~( ~, ]~), the

corresponding subobject R =
~]F in ~(}{,TF).

We must prove the equality
Thus if

and f E (a F hu, A) ~ A(u) we have to show that
But


But <0(u)(f) E ~F(u)

is some subobject of a F h u : call it

We shall prove that S = (
= s

= f-1 (
and the proof will be complete. So we must verify that the following square is a pullback in ~(~{,]~) : S

aFh Thus for any v E •

u

A

f


~ a

7F

and g C (a F h v, a F hu) ~ a F hu(V) we must prove that

(
iff

V

g E (a F hv, S) ~ S(v).

But g C a F hu(V ) and by construction of the associated sheaf functor, there is a covering (vi)iC I of v (v =

v vi) such that for any i, g iEI

= a(gi) where Vi

gi E F hu(Vi) or, equivalently, gi " F hr. + F h u is a morphism in Pr( IH, IT). I By lemma 2, the statement above reduces to : V i C I

(qo f)(v)(g) vi = a F

hv v i

iff

g

vi

E S(vi)

or in other words V i £ I

(~ f)(v i)(agi) = a F hv. 1

iff

ag i E (a F hT , S). 1

69

<,u,

Suppose first that v i

By lelnma 8, ag i factors through ag i

aFh

~ aFh

V-

a

U

a Fh u

a F hr. 1

From this we deduce

(q) f) (v i) (ag i) V. =

(~0 f) (Vi)(6 ~ o ~)

= a -1

((q~ f) (v i) (svi))

f ~ ~ ( ~ , ~r)

= - I ((£o f)(vi)(id a F h u vi)) f 6 Sh(~)

= -1((q) f)(u)(id a F hu)) vi -I =

o~

Yoneda le~mna

(S vi)

= c f f 1 (6vi) -I (S)

=

(agi)-l(s).

Therefore

(~o f)(vi)(agi) = a F hr. ]

iff (agi)-](S) = a F hr, 1

iff ag i 6 (a F hv. , S) 1

and this concludes the proof with v i 4 u. Now suppose v i ~ u .

Then gi 6 F hu(Vi) = FO ~ S(vi).

On the other hand

propositions 20 and 21 imply that (~ f)(vi)(agi) = a F hv . Hence, the result is trivial, i

70

§ 5. UNIVERSAL CLOSURE OPERATIONS ON Sh(H, 1T) As before, suppose the theroy ]~ to be con~nutative.

We now proceed to inves-

tigate the localizations of Sh( lq, ]~), i.e. the full reflexive subcategories of Sh( Iq, IF) whose reflection is an exact functor.

If IF is the initial theory,

Sh( II, ]~) is the topes Sh(lq) and its localizations can be classified at the same time by Grothendieck topologies on Jd, by Lax~were - Tierney topologies on the fl-object of Sh(lq) and by universal closure operations on Sh(lq). chapter is devoted to generalize these results to Sh( lq, ]~).

This

In the present

paragraph, we characterize a localization of Sh( Iq, ]Y) by- a universal closure operation. Definition 23. A universal closure operation on Sh( I~, ]~) is dey~ned by specifying for each subobject R >-~ A in Sh( ~, ~') , a subobject R >--+ A in Sh( lq, ~r) such that, for any subobjects R, S of A and any morphism f : B + A

(c 1) R < ~ (c 2) R < S ~ < g (c 3) ~ =

(c 4) ~ ( R )

= f-l(-~).

Proposition 24. An equivalent system of axioms for a closure operation on Sh( lq, IT) is given by (with the notations of definition 23)

(c s) ~

=

A

(c6) R n s =

(c 3) ~

=

RnS

R

(c 4) f - l ( R ) = f - l ( ~ ) . Start from (c I) to (c 4).

By (c I), A ~ < A

and thus X = A, this is (c 5).

Now R N S ~
Now by (c I) the following square is a pullback RnS c ~ S RNS~

~ S.

71

This shows that the closure of R A S in R A S is R A S (by (c 4)); but this closure is also (R A S) A (R A S) (again by (c 4)). Thus

R n g = (Rns) n (Rng] w h i c h shows t h a t R n S ~
T h e r e f o r e , by (c 2) - (c 3) and t h i s

last

a p p l i e d t o R, S and g , R we o b t a i n : RN S = RA S~>RN

S>~RAg.

This implies (c 6). Conversely start from (c 3) to (c 6). R AR

By (c 4), intersect R, R with R :

is the closure of R N R in R n R, i.e. R n R = R b y

that R ~ R , which is (c I).

(c 5).

This proves

On the other hand

R~S

~

RNS=R R n S = R

by (c 6)

R~<S I

which is (c 2). Proposition 25. £ <

Let ¢

) Sh(~,l~)

be a localization of S h ( ~ , I T ) .

For any subobject

i r

R ~ - + A in S h ( ~ , ~ )

define R by the following pullback, where nA is the

universal morphism of the adjunction £ ~ i,

¥ A

nA

,

i£(A)

This is a universal closure operation on Sh( ~, 7).

First point out that by exactness of £, £(r) and thus ~ are monomorphisms. From the following commutative diagram

72

nR R~. ,

.

i/(R)

¥

if(r)

A

•

i£ (A)

nA

we deduce that R < R, which is (c 1). Now R < S

implies if(R) < if(S) and thus UA 1- (il R) <~nA]- (il S); this is

(c 2). Now i is full and faithful, thus ilig is isomorphic to il and if(hA) is an isomorphism.

Let us apply il to the diagram defining R; I beeing exact

the new diagram is again a pullback :

~I

p.b.

i

P.b.

,. iS(A)

A

I

=

i/(r)

iS(A)

nA

The composition of these pullbacks shows that ~ = R, which is (c 3). Finally if f : B ÷ A is any morphism, the following cube is con~nutative

X I

.

f

f-1 (a)

il R

[

/ "

P il f-1 (R)

A

/

il A nA

/ ilB

nB

il r

73

where x is the unique factorization through the pullback R. qA and qB are pullbacks; the face over i l f

The faces over

is a pullback by exactness of 1.

By [2]] - 7 - 8 - 4, the face over f is also a pullback.

This proves that

f-](R) = f-](R), which is (c 4).

§ 6. LAWVERE - TIERNEY ]r-TOPOLOGIES O N ~

Let E be a frame.

A localization of S h ( ~ ]

(i.e. a subtopos of S h ( E ) )

can be classified by a Law~ere - Tierney topology j : ~ ÷ ~ on the ~-object of the topos Sh(~).

Now, if IT is any co~utative theory in S h ( H ) ,

a locali-

zation of Sh(H,]r) will in a similar way be classified by a morphism j : ~

÷ 9~

in the topos ~ ( ~ , ~ ) .

This morphism j satisfies the three

well-known conditions for a Lawvere - Tierney topologies. call it a Lawvere - Tierney ]r-topology o n ~ .

For this reason, we

The present paragraph is devoted

to the construction of such a j from a universal closure operation on S h ( ~ , ~ ) .

Definition 26. Let ~ be a fram~ and ]r a com~m~tative T-topology

on H is a morphism j : ~

A Lawvere - Tierney

theory in Sh(I~). ÷

in the topos ~( ~, IF)

such that

(LTI)

j ot

=t

(L T 2) j o j = j (L T 3) j o ^ 7

= ^'F o (j × j),

Proposition 27. Let (R I-+R) be a universal closure operation on S]I(~,~). U in ~, define j(u) : ~ ( U ) morphism j : ~

÷ ~

~ ~ ]r(u) by j(u)(R) = R.

which is a Lawvere - Tierney

For any

This defines a

~F-topology on ~.

If v 4 u inI~, we know by (c4) that for any R in ~iT(u), ~

= R[v"

This says exactly that j is a nor~hism in Sh(I~).

Now for any f : a F h u +

÷ a F hu, (c 4)also implies that f-] (R) = f-I ([).

The latter equality simply

states that j is a morphism in the topos ~(I{,7).

Now (L T I - 2 - 3) follow

i~ediately from (cS - 3 - 6).

Proposition 28. With j given as in proposition

27, the image I >--+ ~ r

of j is such that

74

for any u in

I(u) = (Re a~(u) I R = ~ . By definition of j, the elements in I(u) have the form R = S with S E ~]y(u); thus by (c3),R = ~ =

S = R.

Conversely if R = R, then R = Ju(R)

and R is in I(u). Proposition 29. With j given as in proposition 27, if R >---+A is a subobject in Sh( II, ~) with characteristic map ~, then R has characteristic map j o ~.

Consider the following diagram, where the right hand square is cor~autative

£ On(U)

and u E]~, B

~0n u

An

jn

.......... ~ ~

U

U

An'~ u

B

........................ ~ ~l

AU

u

u u

)

^~ u ~lu

It shows that

"~r u

o (j o ~ ) n u

= A

=

J

n

U

o

jn u

IO^%u U

~<Jluo
o

o


u

~ o B

since j is order preserving (by (c 2)). So j o ~ is a characteristic map on A and the subobject of A classified by j o q0 is (j o ~g)-1 (t It) - (cfr. theorem 22).

Now if u E I~, by proposition I - 5 an element in A(u) is just a morphism

75

f : a F h u + A.

But ~(u) was defined by %0(u)(f) = f-1 (R), (cfr. theorem 22).

Therefore, by (c 4) (j o ~0)-I (t IF) (u) = {f : a F h u ÷ A

I

(j o ,,o)(u)(f)

={f:

I

j (u) (f-1 (R)) = a F h u}

= {f : aFhu÷A

{

f-J(R)

= a F h u}

= {f : a F h u ÷ A

I

f-l(~)

= aFh

= {f : a F h u ÷ A

I

f factors

aFh

U

+A

= a F h u}

u}

t h r o u g h R)

= ~(u).

An this shows that ~ = (j o ~0)-I (t ~), which concludes the proof. § 7. GABRIEL - GROTHENDIECK ~-TOPOLOGIES ON H For the topos Sh(~), the equivalence between Lawvere - Tierney topologies and Grothendieck topologies is well-known.

A Grothendieck topology on H is

given by specifying certain subobjects of the representable functers h u (u £ ~). Now if you want to investigate sheaves of abelian groups o n e , a Gabriel localizing system o n ~

is defined by specifying certain additive subobjects of the

additive representable f~mctors ~u (u E H) on the free additive category~ generated b y ~

(cfr. [6] or [20]).

Then the Gabriel localizing systems on H

classify exactly the localizations of the category of presheaves of abelian groups on~.

Now observe that the axioms for a Gabriel localizing system are

exactly the additive version of the axioms for a Grothendieck topology-. This brings us in this paragraph, to define the concept of Gabriel - Grothendieck TF-topology on I% for a commutative theory ]Y in Sh(~).

This is done by

specifying for each a F hu, (u E ~), in Sh(~,TF) a certain set of subobjects. These subobjects satisfy the three usual Gabriel - Grothendieck axioms plus an additional axiom which takes into consideration the fact that we are dealing with sheaves and not simply presheaves as in the Gabriel and Grothendieck cases. Definition 30. Let H be a frame and ~ a com~nutative theo~d in Sh(~I). Grothendieck

A Gabriel -

~F-topology on ~ is defined by specifying for each u in }],

a family J(u) of subobjects of a F h u in Sh( ~, ~) such that

76

(G G ]) a F h

E J(u)

U

(G G 2) (R E J(u) and f : a F h v ÷ a F hu) * (f-l(R) E J(v)) (G G 3 ) ( ; E

J(u);

v EII (G G 4)

R ~

(

S >--+

a Fh u

)

V f : a F hv ÷ R

a F hu , u =

v i c I

v u. i n l t iEI 1

R ui E J(ui)

f-l(R N S) C J(v)

1

~ (S E J(u))

(a ~ S(u)).

Proposition 31. Let j be a Lawyers

J(u) This defines

- Tierney

a C~riel

For any

u

in ~,

consider

= a F hu}.

~-topo;ogy

J on N,

Now consider a morphism f : a F h v ÷ a F h

u

By lemma 8, f can be factored into aF

Therefore

on ~.

I j(u)(R)

- Grothendieck

By (L T I), (G G l) holds. in Sh(II, ]I).

IY-topology

= {RE tilt(u)

hv

a

~aFh

v~

, aFh

u.

f o r any R i n J ( u )

f-l(a) = -t(atv). But j is a morphism in Sh(If), thus j(v)(R Iv) = j(u)(R)Iv = a F h u

= a F h v. V

Moreover, j is a morphism in &( I~, IF), thu~ j(v)(-1(Rlv)) =

-1(j (v) (Rl v) ) = - 1 ( a F hv) = a F h .

Finally we have shown that j(v)(f-l(R)) = a F h V

which proves that R is in J(v).

This in.lies (G G 2).

We consider now the assumptions of (G G 3) and we denote by r • R >-+ a F h u the inclusion of R.

Consider the morphism rf : a F h v ÷ a F h u.

I - 5, rf is some element in a F hu(V ).

By proposition

By the construction of the sheaf

associated to F hu, there is a covering v =

v v. in I~ such that iEI l

77

rf vi C F hu(Vi); in other words, for any i E I there is a morphism fi : F hvi

÷

F h u (proposition I - 4) such that rf vi = af i. if v i { u

Now by lemma 9,

(rf vi )-](ju(S)) = a F hvi.

Now if v i ~< u, then rf vi can be factored into (lemma 8) :

a F hvi

+ a F hvi~

~ a F h u.

Then, since j is a morphism in ~( ]~, IF), we obtain :

(rf vi

)-1

:

(Ju(S))

-l(Ju(S) vi )

= - 1 (Jvi(S vi ) = Jvi(a -1 (S vi ) = Jvi(rf vi )-1 (S) = Jv. (f-l(R n S)) 1

=aFh

V. 3_

Thus we have proved that for any i f-](R n Ju(S)) vi = (rf vi )-](ju(S)) = a F h v . 1

By le~ma 2, this proves that f-1(R n Ju(S)) : a F h v. Thus any morphism f : a F h v + R factors through R N Ju(S) >--+ R. tion I - 6, the a F hv'S determine a proper set of generators; thus R N ]u • (S ) = R.

Therefore by (L T 2 - 3) :

a F h u : Ju(R) = Ju(R) N Ju Ju (S) : a Fh un

Ju(S)

By proposi-

78

= Ju(S) which shows that S is in J(u).

This implies (G G 3).

Finally consider the assumptions of (G G 4).

As j is a morphism in

~( ~, ]r), we obtain again the following equalities : Ju(R) ui = Jui(R ui ) = a F hm'l Therefore by lemma 2, Ju(R) = a F h u and R is in J(u).

This implies (G G 4).

Proposition 32. With the notations of proposition 31, J becomes a subobject of ~F in ~( ~, ~) and the following diagram is a pullback in the topos ~( ~, ~) J

,

1

J If u 4 v in ~, the restriction mapping J(v) + J(u) is obtained by pulling back along the canonical inclusion a F h u + a F hv; this definition makes sense because of axiom (G G 2). Thus J is already a presheaf on 14. This presheaf is separated by lemma 2 applied to Sh(u¢, ]Iu¢ ) :if u = v u i and R, S C J(u) with R ui = S ui, then iel R = U R I and S = U S ui; so R = S. iEI ui iE I

Finally (G G 4) says exactly that J

I

is a sheaf. By definition, J is a subobject of ~]r in Sh(~).

But by (G G 2) J is

stable under action of M]r; so J is a subobject of ~]r in the topos Sh(~,]I). Now R E J(u) if and only if Ju(R) = a F h u = t]y(*).

This shows that J

is the inverse image of tit along j. Proposition 33. Consider a Gabriel - Grothendieck T-topology J on the frame~.

J(u) is a filter in the lattice of subobjects of a F h u.

Each

79

By (G G 1),a F h u is in J(u), so J(u) is not empty. in J(u); we shall prove that R n S is in J(u).

Now consider R, S

For any v in 14 and

f : a F h v + R, consider the following diagram where the squares are pullbacks f-1(R n S) >

~, R N S

I aFh

S

I v

*

Y

R>

f

,

aFh

u.

r

By (G G 2), f-1 (R n S) = (rf)-I (s) is in J(v) because S is in J(u).

Now R n S

is some subobject of a F h u and R is in J(u); moreover for any f : a F h v ÷ R, f-1(R N (R n S)) = f-1(R n S) is in J(v); so R n S is in J(u) by (G G 3).

Finally consider R in J(u) and S I>R.

For any v in14 and f : a F h v ~

R

f-1(R 0 S) = f-1(R) = (rf)-I(R). By (G G 2), f-1(R n S) is in J(v) because R is in J(u),

Therefore S is in

J(u) by (G G 3).

II

§ 8. LOCALIZING AT SOME ]r-TOPOLOGY

In §§ S - 6 - 7, we started from a localization of Sh( 14, ]r), where IF is a commutative theory in the topos of sheaves on the frame 14, and we constructed successively a universal closure operation on Sh( 14, ]r), a Lawvere - Tierney It-topology on I~ and a Gabriel - Grothendieck ]r-topology on 14.

In this para-

graph, we close the loop : from a Gabriel - Grothendieck ]r-topology on 14, we construct a localization of Sh( 14, IF).

The reader may be surprised by the terminology in this paragraph where the words "prelocalized" and "localized" are used instead of the usual terms "separated presheaf" and "sheaf". topologies.

In fact we are working with two different

The first topology is the canonical one on 14 : we use the termino-

logy "separated presheaf~' and "sheaf" in its usual sense when we refer to the canonical topology on 14.

But in this paragraph we consider also a ]r-topology

J on 14 and thus there will be a corresponding notion of "J-separated-object" in Sh( 14, ]r) and "J-sheaf-object" in Sh( 14, 3~).

To avoid any confusion, we

prefer in the latter case to use the words "prelocalized-object" and "localized object".

80

Thus throughout this paragraph J is a fixed Gabriel - Grothendieck ~-topology on ~.

Definition 34. An object A in Sh( I~|, ~) is called "prelocalized" (with respect to J) if for any u in I~, R in J(u) and f, g : a F h u -~ A we have fiR = glR

~

f=

g.

Definition 35. An object A in Sh(~,~r) is called "localized" (w~th respect to J) if for any u l n a ,

R in J(u) and f : R ~ A ,

there exists a unique

g : a F h u ~ A sUCh that g lR = f. R~

• a

hu

A.

We shall denote by L the full subcategory of Sh(~,Ir) whose objects are the localized ones. zed.

It is obvious that any localized object is also prelocali-

We shall prove that ~ is a localization of Sh( H, 7).

Definition 36. A monomorphism s : S ~-~ A in S h ( ~ , ~ )

is called "dense r' (with respect

to J) if for any u i n ~ and any f : a F h u ÷ A, f-l(S) is in J(u). Proposition 37. The class of dense monomorphisms is stable under inverse images. Consider the following diagramwhere s is dense:

g - I (f-1 (s))

f - l (S)

~

a F hu . . . . . . . .

g

~

A

f

÷

s

~

B

8~

For any u C]l and g : a F h u ÷ A, g-](f-J(S)) = (fg)-1(S) is in J(u); thus f-1 (S) is dense in A. Proposition 38. A monomorphism s : S >-~ a F h u is dense if and only if it is in J(u).

By (G G 2), any monomorphism in J(u) is a dense monomorphism. if S ~

Conversely

a F h u is dense, choose R = a F h u in (G G 3) : this implies exactly

that S is in J(u). Proposition 39. in Sh( H, 'IT)

Consider the composite of two m o n o m o ~ h i s m s r

s

R>--+S >-+A. Then s o r is dense i f and only i f

First suppose r and s are dense. g : a F h v ÷ f-] (S) are pullbacks:

For any u, v i n ~ and f : a F h u + A,

consider the following diagram where all the squares

(fg)-1 (R)

~, f-1(R)

I

- -

f-1 (R)

i

a F hv

a Fh

r and s are dense.

v

i

, f-1(S)

g

~.

f-

~ R

S)

I

f-1(S) ~

--

~.

aF

h

u

r

S

f

,

A

a F h u is in J(u). But f-J(S) >--* a F h u is in So by (G G 3) the problem reduces to show that

We must prove that f-1(R) ~ J(u) because s is dense.

(fg)-1 (R) >--+ a F h v is dense. (fg)-](R) ~

And that is true because r is dense and

a F h v is just (f'g)-1(r).

Conversely suppose s o r dense. f-1 (R) is dense.

This implies that for any f : a F h u + A ,

But f-] (S) > f-] (R) ; thus f-] (S) is also dense (proposition

33); therefore s is dense.

On the other hand, for any g : a F h v + S consider

the following diagram where the squares are pullbacks :

82

g-1 (R)

aFh

V

,

R

R

, S>

, A

g

s

s o r is dense, thus g-1 (R) = (sg)-l (s o r) is in J(v); this proves that r is dense. We now turn to the definition of the localizing functor £ : Sh(~,~F) ÷ ~. This will be realized in several steps. : Sh(}I,~)

+ Pr(~,~)

First we construct a functor

and we consider the composite alax where a is the

associated sheaf functor.

We prove that this composite functor takes values in

L and we define it to be I. For any A in Sh(l~,~r) and u inl{ define

X(A) (u) =

lira (R, A), a~J (u)

where the colimit is computed in Sets ~F(u)

This definition makes sense

because (cfr. theorem I I - 5) u! u* R ~ R

(R, A) ~ (u! u* R, A) ~ (u* R, u* A) ~ (u! u* R, u! u* A). Therefore (R, A) % (RIu, Alu ) is provided with the s t r u c t u r e o f a ~ ( u ) - a l g e b r a since lr is cor~nutative ( l e n a 6 applied to Sh(u+, Iru¢)). Note also t h a t this colimit is filtered (proposition 33). If v ~< u in 1~, we need to define a restriction mapping ~ (A) (u) + x (A) (v) which is a morphism of ~(u) algebras.

But ~ (A)(u) is defined as a colimit.

For any R 6 l(u), Rlv is in J(v) by (G G 2) and by composition with R l v c _ + R we obtain a ~(u)-homomorphism

v SRlv

(R, A) ÷ (Rlv , A)

........ X(A)(v),

where the second morphism is the canonical inclusion into the colimit x(A)(v). I f S ~< R in J ( u ) , the following diagram is c l e a r l y commutative:

83

(R, A)

P

(Rtv, A)

v

SRlv

X(A) (v)

(S, A)

-" (Sly, A)

SSj v

where the vertical arrows are the composition with the canonical inclusions S ~ R and S jv >--+ R] v" Therefore we have a cone and thus a unique factorization X(A)(u) ÷ ~(A)(v). It is obvious that this makes X(A) into a T~-presheaf.

Finally, consider a morphism f : A ÷ B in Sh( Id, IF).

We must define

Xf : hA ÷ XB, thus for any u in 14, a ~F(u)-mo~phism Xf(u) : hA(u) ÷ xB(u). But again xA(u) is defined by a colimit.

Consider R in J(u) and the following

morphism

(l, f) (R, A)

s~ , (R, B)

........... X(B) (u).

If S ~ R in J(u) and s : S >-~ R is the inclusion, the following diagram is con~nutative (I, f) I,

(R, A)

(s, I)

i

(R,B)

U

(s, 1)

(S, A)

-- (s, B) (1, f)

sS

Therefore there exist a unique factorization X(f) (u) : X(A) (u) + X(B) (u) through the colimit ~ (A)(u).

Clearly this makes x into a functor

: Sh(I~,]F) ÷ Pr(l~,~).

_Proposition 40. For any A in Sh( ~l, 7),

Consider u = x ui = y ui o

hA is a separated presheaf.

v u. inll and x, y in x(A)(u) such that, for any i in I, iEI i

We must prove that x = y.

~(A)(u) is a filtered colimit; there-

fore there exists some R in J(u) and x, y • R++ A such that x and y are repre-

84

sented by x, ~ in the colimit.

From

ui x

, we deduce that x = y ui

and ui

y ui

represent the same element in the colimit XA(ui); this means that there exists some R i £ J(ui) such that (~ ui ) Ri = (y ui ) Ri or, looking at R i as a subobject of R,

Ri

= y

Ri.

Denote by K the equalizer of x, y; from ~ Ri = ~ Ri we deduce that

R.~
R°

i~l 4NN~ K;

~

R

| A.

Thus for any i E I, K u i ~ > R i ui = R i and from R i ¢ J(u i) we deduce that K ui ¢ J(ui) (proposition 33).

By (G G 4) this implies that K E J(u).

But

and ~ coincide on K, so x = y in XA(u) by construction of filtered colimits in algebraic categories, (cfr. [21 ] - 18 - 3 - 6). For any object A in Sh( I-I,~F), we denote by XA : A + XA the morphism whose component at u ¢ I{ is given by (cfr. proposition I - 5)

A(u) ~" (a F hu, A)

, XA(u). U

SaFh

U

If P is some presheaf, we also denote by ap : P + aP the canonical morphism arising from the construction of the associated sheaf functor.

Proposition 41. For any u ¢ 14, R ¢ J(u), A E Sh( 14, 7), ~ : R ÷ A and x : a F h u + aXA,

the following square is commutative R>

r

~ aFh

I

U

A

~ XA~ XA

if and only ~f x

~ axA axA

determines (via proposition I - 5) an element in XA(u)

represented by -x E (R, A) in the filtered colimit.

85

By proposition 40, XA is separated and thus aXA is a monomorphism. any v ¢ ~

and f : a F h v + XA in P r ( ~ , ~ ) ,

f o a F h

: F hv+

a F hv÷

by propositions

Now for

I 4 - I 5

XA determines an element in XA(v) ~ aXA(v) and

v aXA o f : a F h v ÷ XA >--+ aXAdetermines the same element

an element in a~A(v); this is clearly

as the one determined in Sh( H, ~) by axA o f o a F h

: F h v ÷ a F b~ ÷ XA ~

aXA.

v

Suppose first that x = ax' where x' : F h u ÷ xA determines some element in xA(u) and moreover that this element is represented by ~ in the colimit XA(u). By theorem II - 5, R %

u! u* R = R]u and thus any morphism R ÷

factors through the canonical monomorphism Blu >--+ B.

B in Sh(~I,~)

Thus the commutativity

of the diagram is equivalent to the commutativity of its restriction at u. R Iv "

~

Rlv~

aFh

•

v

R >

r

~

aFh

P

a F hu

v

x,uZ

(axA ° XA) tu

A

• XA

kA "~

P, aXA ax A

But the diagram restricted at u is con~autative if and only if it is cor~autative when preceded by any morphism f : a F h v + R, with v < u applied to Sh(u+,~u+).

(proposition I - 6

Finally, composing again with the monomorphism

axA u >--+ a~A, the diagram we need to consider is co~utative

if and only if it

is co~nutative when preceded by any f : a F h v + R, with v ~ u .

So, consider v ~ u

in}~ and f : a F h v ÷ R.

aXA o XA o x o f o a F h

Via proposition I - 5,

determines some element in xA(v).

By definition of

v XA, this is the element represented by xf 6 (a F hv, A).

Thus a~A o XA o ~ o f

86

determines in fact an element in XA(u) represented by xf 6 (a F h v, A). other hand x is represented in XA(u) by x C (R, A).

On the

We must consider x o r o f.

But rf can be factored by some ~ : a F h v + a F h v through the canonical inclusion : a F h v c_+ a F h u (lenmm 8).

Since XA is a presheaf, the element of ~A(v)

determined by X]v. is represented by ~Iv+ 6 (Rlv , A). is compatible with the algebraic It-structure

Now the construction of ~A

and in particular with the action

of the 1-ary operation ~ (cfr+ lemma 7).

Therefore the element of XA(v) determi-

ned by Xlv o ~ is represented by Xlv o ~ 6

(R]v , A) where we still denote by

~ : RIv ÷ Rlv, the action of the 1-ary operation~

on Rlv.

So we Irrost prove that

f 6 (a F hv, A) and Xlv ° ~ 6 (RIv, A) represent the same element in the colimit xA(v). Consider the following conmutative diagram R Iv

R v

>

>

a Fh v

~ c ~

w,

R

~ aFh u

l

I

a Fh v

f

r

~ R.

Since r is a monomorphism, we obtain the comnutativity of the following diagram

RIv

RIv

/R

•A

aFh v o c~ and~ f coincide on Rlvwhich is in J(v). So'xlv o c~ and x f represent the same element in ~A(v). This concludes the proof of the

which shows that Xlv I

commutativity of the given diagram. Conversely suppose the diagram to be cor~autative,

x 6 (R, A) represents

some element in ~A(u) determined by a morphism x' : F h u ÷ ~A in Pr( If, TF) (pro-

87

position I - 4).

So, by the first part of the proof, the following diagram is

commutative r

R

> .......

'

aFh

U

x

A

"~

XA

~

XA

~ a~A .

aXA

But on the other hand XA is separated (proposition 40) and thus by construction of the associated sheaf functor, there is a covering u =

v u. in H such that i6I x

for any i, x u. determines an element in XA(ui); thus the following diagram is commutative I i

r-

1 +

R ui

x ui

x ui .......~

XA

;

~A

ax'

U. l

[ A

where Xlu

aFh

~,

atA

a xA

represents some element in xA(ui), ax' is represented by x and thus

{sirepresented by x

.

If we can prove that x

. is also represented U1

Ui

Ui

by xl ui , then for any i, x ui = ax' ui and by le~mm 2, x = ax'.

Then we can

I

reduce the problem to the case where x represents some element in XA(u). such an element x, we must prove that x represents x in XA(u).

For

Consider

6 (R, A) which represents x in XA(u); there is no loss of generality to suppose R = ~, (if not, simply work on R N R which is still in J(u) by proposition 33). So, by the first part of the proof, we have two co~autative diagrams r

R

A

>

.......~ XA XA

and x represents

r

~ aFh

~

R

U

aXA

A

aXA x in

XA(u).

~

~ aFh

~~ A xA

It

suffices

to

prove

that

U

~ aXA aXA

x and x represent

the

88

N

same element in ~A(u), i.e. that x and x coincide on some K ~< R with K E J(u). Take K to be the equalizer of x and x. to prove that K E J(u).

x and x coincide on K and it suffices

We have the following situation k

r

K> with

r dense.

By p r o p o s i t i o n

~ R>

39,

~ aFh

r k will

u

be dense and thus K will

by proposition 38, as soon as k is a dense monomorphism. consider v £I~ and f : a F h v + R.

be in J(u)

To verify the latter,

We must prove that f-1 (K) is in J(v)

In

the following diagram Fh

V

I

.j f-] (K) ;

"

t

~-~ -" a F h

l'

,, "~

a Fh v

V

k

r

-"- R

KI

>

,

aFh

U

Ix A

•

A

• axA

we have :

XA

a~A o ~A ° ~ ° f -- x o r o f = a~A o ~A ° ~ ° f"

But ~A is a separated presheaf

aXA

(proposition 39); thus a~A is a monomorphism and ~A ° ~ ° f = ~A ° ~ ° f" Thus ~A ° ~ ° f ° aF h v and ~A o ~ o f o a F hv determine the same element of ~A(v).

By definition of ~A' this element is thus represented by x o f E (a Fhv,

By c o n s t r u c t i o n s : S ~

of the filtered

a F hv in J(v)

factors

through

Finally

s and t factor

colimit,

such that

the equalizer through

A) a n d ~

k of

this

o fC

(aF

means that

x o f o s = 7 o f o s. (x, x).

the pullback

Let t be the f - 1 (K).

h v, A). there

exists

some

But then f o s factorization

This shows that

But S is in J(v) and thus by proposition 33, f-] (K) is in J(v).

Proposition 42. For any A in Sh( H, ]Y), ah(A) is a prelocalized object.

morphism. S < f - 1 (K).

•

89

Consider u E ~ , that fr = gr.

r : R >--+ a F h u in J(u) and f, g : a F h u

We must prove that f = g.

aX(A) such

Now by proposition I - 5, f is an element

in a~A(u) and since XA is a separated presheaf (propasition 40), the construction of the associated sheaf functor shows that there is a covering u = such that for any i ¢ I, flu. ¢ xA(u.).

Thus f

i

fi £ (Ri' A) with R i in J(~i~.

IU.

v ui inn iEI

can be represented by some

A similar thing c~n be done about g and again

without any loss of generality we may assume that the covering working for g is the same as the one for f (eventually consider a common refinement of both coverings).

We may also assume that the corresponding subobjects R i agree (even-

tually consider the intersection of both subobjects).

Thus for any i C I,

g ui E XA(ui) is represented by gi E (Ri, A). Consider the following commutative diagram (by proposition 41) where aXA is the canonical morphism arising from the construction of the associated sheaf functor,

axA is a monomorphism because IA is separated (proposition 40).

~ e.

1

A

• a F h

>

-~ aXA

r.

f R>

a F h r

J~

U

aX(A).

g

From fr = gr, we deduce that xA o T i = hA o gi" Thus the following diagrams are co~autative ri R. ~i

ri P aFh ui

A

~ a~A XA

R. ~ i

-~ a F h ui

A

~

a~A

xA

and thus, by proposition 41, gi represents f ui in ~A(ui) and ~i represents g ui

90

in xA(ui).

But fi represents f ui and ~i represents g ui.

So f ui

g ui for

any i and thus f = g, since XA is a separated presheaf (proposition 40).

•

Proposition 43. For any prelocalized object A in Sh( I~, 0), the composite a~A o XA : A ÷ ÷ hA + ahA is a dense monomorphism. aXA is a monomorphism because xA is a separated presheaf (proposition 40). To verify that ~A is a monomorphism, consider f, g : B ~ A such that XA o f = kA ° g"

For any u Elq and

k : a F bu ÷ B, a~A o ~A o (f k) = a~A o ~A ° (g k); by proposition 41, this implies (consider the following diagram where a F h u E J(u)) aFh

aFh

U

fk

gk

I

U

aXA o XA o f o k = = axA ° ~A ° g ° k

A

~ kA XA

*

aXA

axA

that fk and gk represent the same element ~A fk = XA gk of XA(u).

By construc-

tion of a filtered colimit, this means that fk and gk coincide on some R E J(u). r R

>

fk ~ aFh

~A. u

gk

But A is prelocalized, thus from fkr = gkr, we deduce fk = gk. I - 6, we conclude that f = g.

By proposition

Thus ~A is a monomorphism.

Now we will establish that aXA o ~A is a dense monomorphism in Sh( lq, 0). (Note that the monomorphisms a~A and ~A are not in Sh( lq, O) but only in Pr(ll,~r)).

For any u Clq and f : a F h u ÷ a~A, we must prove that f M ( a ~ A o ~A )

is in J(u).

But f determines an element in akA(u) (proposition I - 5) and by

construction of the associated sheaf functor (via proposition 40), there is a covering u = iEIV uol in Id such that each f ui determines some element in ~A(ui). This in,plies that f u. is represented by some T i C (Ri, A) with R i E J(ui) and making the following ~iagram

91

R. 1

u -1

(A)

P

A XA

aFh

p.b.

XA aXA

aFh

where the square is a pullback,

axA

U

commute

(proposition 41).

R i ~ u-1(A) and since R i ¢ J(ui) we have u-1(A) ¢ J(ui),

This implies that (proposition 33).

Proposition 44.

If A is a prelocalized object in Sh( ~, ~),

Consider u ¢ E ,

RE

: a F h u + aXA of f. is prelocalized

J(u) and f : R + aXA.

axA is a localized object.

We need to find an extension

This extension will be necessarily unique because aXA

(proposition 42).

Consider the following diagram where the squares are pullbacks and ~ as defined below g

-I

(S) ;

~

a F h

S T

V

r

S~

,

S

A :

~

U

XA :

xA

aFh

R>

~

aX

aXA

Now aXA o XA is dense, thus s is dense (proposition 37), thus r o s is dense (propositions 38 - 39).

Therefore g ¢ (S, A) represents some element in ~A(u);

this ~lement is determined

(proposition 41) by a morphism ~ : a F h u ÷ axA such

that f o r o s = a~A o XA o g.

We must prove that ~ o r = f.

By proposition

I - 6, it suffices to show

N

that for any v ¢I~ and g : a,F h v ÷

R, f o r o g = f o g.

of g and s, we obtain ~o

r o g o s' = f o g o s'

FornLing the pullback

92

because ~ o r and f coincide on s. J(v) by proposition 38.

By proposition 37, s' is dense and thus in

But alA is prelocalized; so ~ o r o g = f o g.

I

Definition 45. If J is a g-topology on ~ and L is the full subcategory of localized objects of S h ( ~ , ~ ) , 1

the functor I : S h ( ~ , ~ )

÷ L is defined by

= aXa~.

By propositions 42 - 44, each aXahA is localized and thus this definition makes sense.

We shall prove that £ is a localization.

Proposition 46. Let f : A ÷ B be a morphism in S h ( ~ , ~ )

with B a localized object.

Then

f factors uniquely through hA.

We must define g in Pr( If, IT) such that the following diagram commutes XA A

........ ~

XA

For any u E l{, we must define g(u) • hA(u) ÷ B(u).

Consider some element

x E hA(u) represented by ~ E (R, A) with R C J(u).

From the following situation

r R~

1 1

~ aFh u /

A

f

/f

,"

Y

B

and the fact that B is localized, we get a unique extension y such that y o r = f o x.

Thus y determines some element in B(u) : this is g(u)(x).

This definition does not depend on the choice of x representing x.

If

x' E (R', A) is another element representing x, ~ and x' coincide on some R" C J(u) and thus the corresponding extensions y, y' coincide also on R" E J(u); because B is localized, this implies that y = y'.

93

Now if ~ E 0n(U ) is some operation and if x I, ..., X n E by the morphisms~1, xl' "'" ~

..., ~ ,

XA(u) are represented

without any loss of generality, we may suppose that

are defined on the same R C J(u) - (if not, take their intersection).

Denote by Yi the unique extension of f o xi to a F h u.

From the conmlutativity of

r R>

>

A

aFh u

+

f

B

and the fact that f is a lr-homomorphism, we deduce the commutativity of R:

; aFh u

I

"'

t

A

,

'"

B

(this makes sense because R and a F h u are in Sh(u+,]ru+)). g(u) is a ~(u)-homomorphism.

This shows that

Finally if v ~ < u inll and x E hA(u) is represented by x E (R, A) then Xlv E >.A(v) is represented by ~Iv.

Consider the following diagram

I

rlv

Rv>

.. a F h

v

Y R

A

+t

B

where y = g(u)(x)

~

-

and z = g(v)(Xlv).

T h i s d i a g r a m shows t h a t

I

on R I v ; b e c a u s e This proves

that

B is localized,

z = ytv or in other words,

Ylv a n d

z coincide

J

g(u)(X)tv

= g(v)(Xlv).

g : ),A ÷ B i s a raorphism i n P r ( N , Tr).

Now for an element x E A(u), ~A(U)(X) is represented by x £ (a F h u, A) and thus the unique extension y is necessarily f o x

94

aFh

aFh U

U

A

B.

This shows that g o XA = f and concludes the proof. Theorem 47. Let J be a Gabriel - Grothendieck £

F-topology o n ~ .

The situation

~c_~ Sh( ~, ~) is a localization of Sh( ~, ~). i

We must prove that £ is left exact and left adjoint to the canonical inclusion i.

But £ = axax

and a is left exact.

Moreover, x is also left

exact because finite limits are computed pointwise in Sh(~,]T) and Pr(~,]r) and xA(u) is defined by a filtered colimit in Sets TF(u) where filtered colimits commute with finite limits (cfr. [21] - 18 - 3 - 6).

Finally, £ is left exact.

Now consider the following situation with A, B in Sh( 14, ]~) and B localized. XA A

aXA > ~A

fv /

Xa~A

axa

> axA .1

~ ~aXA ~

,,,>aX aXA ~

i

There is a unique extension fl by proposition 46, a unique extension f2 because B is a sheaf, a unique extension f3 by proposition 46 and finally a unique extension f~ because B is a sheaf.

This shows that the morphism A + aXaXA has

the universal property making £ left adjoint to i.

§ 9.

•

CLASSIFICAT.ION OF THE LOCALIZATIONS OF S h ( . ~ , 1 ] ' )

In § 5 - 6 - 7 - 8, we have described correspondances between localizations of Sh(~,]~), universal closure operations on Sh(~,~F), Lawvere - Tierney ]Y-topologies on ~ and Gabriel - Grothendieck ~F-topologies on ~.

In this para-

graph, we show that all these correspondances are bijective and thus we get a

95

three-fold classification of the localizations of S h ( ~ , ~ ) . Proposition 48. £ Let L ~-~ Sh( 14, lY) be a localization of Sh( 14, IF) and f : A ÷ B a i morphism in Sh( ~, ~). Then f is carried by 1 to an isomorphism if and only if the image of f and the equalizer of the kernel pair of f are carried by 1 to isomorphisms.

This follows easily from the fact that £ is a right and left exact functor between regular categories.

•

Proposition 49. 1 Let L ~--~ Sh( ~, ~r) be a localization of Sh( 14, 7) and r : R >-+ A i a monomorphism in Sh( ~, 7), Then r is applied by 1 on an isomorphism if and only if for any u in14 and f : a F h u ÷ A, f-l(r) is sended by 1 to an isomorphism.

The dowlward implication is obvious since 1 cer~nutes with inverse images. Now suppose that for any u in 14 and f : a F h u ÷ A, /f-1(r) is an isomorphism. Consider the following composite

f:

.[j_ f - l ( R ) v¢14 a F hv ÷ A

q

>

f:

.lJ_ a F h v v¢tt aFh v ÷ A

P

~> A

where q acts by inverse image on each component and p is the canonical regular epi~rphism whose existence is implied by proposition I - 6.

Each monomorphism

f-l(R) >--+ a F h V

is sended by 1 to an isomorphism.

Thus /(q) is an isomorphism.

obviously factors through R via the morphism ing co~utative diagram in

f-I (R) ÷ R.

But p o q

So we have the follow-

96

IR

ZA

Now £p o £q is a regular epimorphism; so Ir is both a monomorphism and a regular epimorphism; thus it is an isomorphism, Proposition 50.

i

£

Consider a localization

L ~

Sh( H, IT) and the corresponding closure

i operation.

A monomorphism r : R >--+ A in Sh( ~, 7) is sended by £ to an

isomorphis m if and only if R = A.

is defined by the following pullback, where A -~ £A is the canonical morphism arising from the adjunction ,,,,

p.b.

1R

]"

A O b v i o u s l y i f £ r i s a n i s o m o r p h i s m , R = A. diagram.

Conversely, if R = A apply £ to this

S i n c e £ i s i d e m p o t e n t , we g e t g.A

tt

gA

£R

£A.

This shows that £r is a regular epimorphism and thus an isomorphism. Proposition 51. Consider a Gabriel - Grothendieck ~-topology J on ~ and the corresponding £ localization • ~-~ Sh( }I, ~) of Sh( }~, ~F). A monomorphism r : R ~ a F hu i is sended by £ to an isomorphism if and only if it is in J(u).

97

Suppose R in J(u).

For any localized object A, we have the natural isomor-

phisras (£R, A) ~ (R, A) ~ (a F h u, A) G (£ a F h u, A) which prove that £R ~ I a F h~u. Conversely suppose Ir to be an isomorphism.

Consider the following diagram

where the upper square is a pullback r

R~ -

t

s

S> .......

~ aFh

U

XR

XaFh u

p.b.

Xa F h u axa F h u

aXR

I

XaXr

~- aXa F h

>

aXr

Xaxa F h U

XaXR

Xa ~a F h

T

axaxR

U

I

I

£R

U

axa xa F h u

£ a F hu

£r By propositions 37, 38, 39, 42, 43, axaxR o Xax is dense and thus axr is dense; therefore s is dense.

Now the cor~nutativity of the diagram gives rise to a

monomorphic factorization t : R >--+ S and to prove that r = s o t is dense, it suffices to prove that t is dense. So consider v C14 and g : a F h v-~ S. dense.

We want to prove that g-1(R) is

But s g E (a F h v, a F hu) represents some element x in X a F hu(V) _

_~ a X a F hu(V); the commutativity of the diagram shows that this element x is in fact in a~R(v).

This implies the existence of a covering v =

v v. in II iCI i

such that for any i, x vi is in ~R(vi) and is thus represented by gi £ (Ri' R) with R i C J(vi).

Then the following diagram is commutative (proposition 41).

98

r.

R. >.........

1

l

R

.....

r

a F h

P

aFh

b XR

V. 1

~ aXR

i

¢~+

.

U

•

Xa F h

~ aXa F h

u

Xa F h u

U

axa F hu

Again by proposition 4], the following diagram is also comnutative aFh I vi

aFh

I vi

~+

aFh

a Fh v

V

l S

C.+

i a Fh u

xa F h u - - ~ Xa F h u

and t h e r e f o r e

s o g

Thus t h e y c o i n c i d e

X

aXa F h u

axa F h u

and r o g i r e p r e s e n t t h e same e l e m e n t i n xa F h u ( V i ) . Lvi on some S i 6 J ( v i ) . So, c o n s i d e r t h e f o l l o w i n g d i a g r a m

where the squares are punbacks

\r . ~ .

"'. g

(R) I v - "

a F hvi >

. go

*

J

(R)

,.

aFh v

g b aFh

U

All composites from S i to a F h u are equal and since s is a monomorphism, all composites from S i to S are equal.

So we get a monomorphic factorization

.

99

S i >---+ g-l(R) Ivi' which shows, because Si £ J(vi) , that each g-l(R)[vi' is in g(vi). i

But this implies that g-1(R) is in J(v) (by (G O 4)).This concludes the proof.

•

Propositio n 52. The correspondence sending a localization of S h ( ~ , ~ )

to the corresponding

universal closure operation is injective.

£

LetK~

Sh(H, IY) be a localization.

By [21] - 19 - 3 - I, f,

i K % Sh(~,~)

[Z-I] where z is the class of all morphisms in Sh(lq, T£) sended

by I to an isomorphism.

Thus a localization of Sh(~,~T)

is completely charac-

terized by the class z of morphisms f such that £(f) is an isomorphism.

But,

by proposition 48, z is itself completely characterized by its monomorphisms. And by proposition 50 the monomorphisms in Z are completely characterized by the corresponding universal closure operation.

Finally, the localization is

completely characterized by the corresponding universal closure operation. Proposition 53. The correspondence sending a universal closure operation on Sh(lq,~) to the corresponding Lawvere - Tierney T-topology on H is injective.

Consider two universal closure operations R b R and R b* ~ which give rise to the same Lawvere - Tiemey of each a F h u, u 6 ]q. any u 61~.

T-topology,

i.e. which coincide on the subobjects

Consider any monomoI~hism R >--+ A in Sh( lq, ~F) and

An element inR(u)

is represented by a morphism f : a F h u ÷

which factors through R; thus the following square is a pullback

aFh

f

I

t u

aFh u But

A.

f

a F h u = f M (~)

= f-l(R)

= ffl

('~).

A

•

100

This shows that f factors through R and thus f ff ~(u). same way, ~ R

and thus R = ~.

Finally R < ~ .

In the

This concludes the proof.

[]

Proposition 54. The correspondence G~briel

sending a Lawvere

- Grothendieck

~-topology

- Tierney

~-topology

on H to a

on ~ is injective.

Suppose that two La~were - Tierney IF-topologies j and j' give rise to the same Gabriel - Grothendieck

IF-topology.

Consider u E ~

and R E ~ ( u ) ;

we must

prove that j (u) (R) = j' (u) (R). Consider any v ff~ and any f : a F h v + j(u)(R), i.e. any element f E j(u)(R)(v),

f is thus an element in a F hu(V).

By construction of the asso-

ciated sheaf functor, there is a covering v = iEIV v.1 i n n such that each f vi is in F hu(Vi) , i.e. is a morphism F hv. ÷ F h u" pullbacks a F h

aFhv..-

aFhv.

, v

f

aFh

;

~

+ f

v

i

If v i ~ u ,

Now consider the following

j(u)(R)

aFh

. u

we know by lemma 9 that f-] (R) vi = (f vi)-1(R) = a F hvi

and in particular : J(vi)(f-l(R) vi ) = a F hvi. If v i < u, we know by lemma 8 that f Ivi can be factored in the following way : a F hv.

~

aFhv.

l

~

~

1

Since j is a morphism of presheaves, we have : J(vi)(R vi ) = j(u)(R) and since j is a morphism in ~( 14, IT)

Vi

aF

h u.

101

J(vi)(-l(Rvi))

= 5 -](j(u)(R) vi).

Hence, J(vi)(f-l(R)

vi)

= f-l (j (u) (e) ) vi

[

a F hv"1 =

Thus for any i, f-](R) Ivi E J(vi). Now look at j'.

In the same way, if v i ~ u , we know by lem~a 9 that f-] (j' (u) (R)

vi

= a F hvi

Now if v i ~ u, we have again by l e n a 8

f-1 (j, (u) (R)) vi = j' (vi) (f-] (R) vi) =aFh 1

the last equality holds because f-1 (R) vi is in J(vi).

But these relations

show that the two elements f-I (j'(u)(R))Iv and a F h v of 2.~(v) have the same restrictions in each ~]1,(vi). So, since 9IT is a sheaf, f-l(j'(u)(R))lv = = a F h v. In other words, f factors through j'(u)(R) and determines an element i

in j' (u) (R) (v). Finally, we have shown the inclusion j (u) (R) ~ j' (u) (R). Conversely, we have j' (u) (R) ~ j (u) (R) and finally j (u) (R) = j' (u) (R).

So

j = j' and the proposition is proved. Proposition 55. The correspondence which sends a Gabriel - grothendieck to the corresponding localization of S h ( ~ , ~ )

~-topology on

is injective.

Consider two Gabriel - Grothendieck topologies J and J' o n ~ which corres£ pond to the same l o c a l i z a t i o n L ~ Sh(~,~). For any u in H, J(u) and J'(u) f are equal to the set of those subobjects r : R >--+ a F h u such that £r is an isomorphism (proposition 51).

Thus J = J'.

Theorem 56. The results of § S - 6 - 7 - 8 describe one-to-one correspondences between

(I) the localizations of S h ( ~ , ~ ) , (2) the universal closure operations on Sh(~,~F),

102

(3) the La~vere

- Tierney

(4) the Gabriel

- grothendieck

By p r o p o s i t i o n s

52 - 53 - 54 - 55, we have injections

{Iocalizations}

T-topologies

~F-topologies

~

{closure operations}

on ]~,

t' B

on ~.

J~

{G - G

~F-topologies}

t'

{L - T 7 r - t o p o l o g i e s } .

Let us prove t h a t ~ o ~ o g o a = i d . To t h a t end, c o n s i d e r a l o c a l i z a t i o n £ L ~Sh(N, ll'). Via ~, ~, ~, t h e c o r r e s p o n d i n g G a b r i e l - Grothendieck i "II'-topology J i s g i v e n by J(u) = {Re a~(u)

I R = a F h u}

= {R£ aT(u ) I £R~£

a F hu};

t h e l a s t e q u a l i t y h o l d s by p r o p o s i t i o n 50. £, Now c o n s i d e r t h e l o c a l i z a t i o n L ' ~ Sh(N,~) g i v e n by ~ ( J ) . By p r o p o s i t i o n i' 51, a n~nomorphism r : R ~ a F h i s such t h a t £ ' r i s an isomorphism i f and U

only if r is in J(u). morphism.

So, £ ' r

i s a monomorphism i f and o n l y i f £ r i s a mono-

But we know a l r e a d y ( p r o o f o f p r o p o s i t i o n 52) t h a t two l o c a l i z a t i o n s

a r e e q u i v a l e n t when t h e y t r a n s f o r m the same monomorphisms i n t o isomorphisms. But t h e n , by p r o p o s i t i o n 49, t h e o n l y monomorphisms t o be c o n s i d e r e d a r e t h o s e w i t h codomain an o b j e c t o f the form a F h . 1 l' u L ~

Finally, both localizations

Sh(~,TF)

and ~' ~--+ S h ( ~ , ~ ' ) are equivalent b e c a u s e i' the same monomorphisms r : R >--+ a F h u into isomorphism.

they transform

i

We know already that ~ o ¥ o ~ o a = id. since ~ is injective,

~ o B o a o ~ = id.

Thus ~ o ¥ o 8 o a o 6 = ~ and

In the same way, ~ o B o a o ~ o ¥ = y

and thus B o ~ o 6 o ~ = id; B o ~ o ~ o ~ o B = B and thus ~ o 8 o y o B = id. This concludes

the proof.

[]

§ 10. THE CASE OF GROUPS A N D A B E L I A N GROUPS

In chapter 6, the results of the p r e s e n t chapter 5) will be p a r t i c u l a r i z e d In this paragraph,

chapter

(as w e l l as those of

to the case o f a theory o f modules on a ring.

w e shall treat the more p a r t i c u l a r

This investigation w i l l also p r o v i d e

some information

case o f a b e l i a n groups. about non abelian groups

103

and show why our results fail for the latter.

This justifies our assumption that

is commutative.

We consider the frame lq of open subsets of the singleton, frame {0, I} and the theory

~ of abelian groups.

i.e. the initial

Thus Sh(I~, ~r) is the category

Ab of abelian groups and Sh(~q) is the category Sets of sets. table algebras are a F h 0 = (o) and a F h I = (22, +).

The two represen-

The m o n o i d M ~

the monoid of 1-ary operations of IF, which is (TZ, x).

is thus

So the topos g( lq, IF)

is the topos of (77, x)-sets.

~

is the set of subgroups of ( G ,

+); it is thus isomorphic to IN.

The

action of (7z, x) on IN goes as follows : consider z £ 7z and n E IN; z corresponds to the homomorphism g : ( 7z, +) + (~, n corresponds to the subgroup

+) which is the multiplication by z;

(n 72, +).

Therefore the action (2Z, ×) × l q ~ is

defined

by the fact

that

z,n

z-l(n

IN; ( z , n ) ~ - ~

is

z,n

the generator

of~-l(n

7z) = { x E7z ] z x E n

_

n

I

n

But

7z}

= {x C77 [ n d i v i d e s

={x¢~

7z).

z x}

divides

x}

72

n^z where n ^ z denotes

the greatest

common d i v i s o r

of n and z.

Thus z * n -

N = a~r has an ordering corresponding to the inclusion n groups; thus n ~ m in IN = a T

if and only if m divides n.

is I and the intersection n A m corresponds

77. ~ m

n nA z

Z; of sub-

The greatest element

to the intersection n

7z n m

7z of

subgroups; thus n A m = n v m where n v m is the smallest contain multiple of n and m.

Now consider some subgroup R >-+ A in Ab.

There is a characteristic map

: A +IN. If a E A, denote ~ : Z; ÷ A the homomorphism~qlich sends I to a. ----7 a (R) is some subgroup of 7z, thus has the form n 7z; n is the value of %o(a). So ----] a (R) = {z £

77 [ a z EIR}

and thus ~(a)

= m i n { n £IN*

I

n a £ R}

104

if this set is non empty and w(a) = O if this set is empty. computed for the usual ordering on N

(! The minimum is

!).

If A is any abelian group, a map ~ : A + N

is a characteristic map if it is

a morphism in g(I{,]r) satisfying the conditions of definition 19. morphism in & ( N , ] ~ )

~ is a

if and only if for any a E A and z C A, ~(za) = z*~(a),

that is to say _

~(za) Now ~ must s a t i s f y

w(a)

z

^

~--caT"

a c o n d i t i o n f o r any o p e r a t i o n o f t h e t h e o r y ;

express these conditions for the basic operations O, +, -.

it suffices

to

From the remark

following definition 19, the condition on O is

~(O)

=

1.

The c o n d i t i o n on + means ~o(a + b) > (~o(a) a qo(b))

that is to say ~o(a + b) divides w(a) v w(b).

The c o n d i t i o n on - means ~(-a) that

~> q~(a)

is to say w ( - a ) d i v i d e s ~0(a)

which is already satisfied because ~(a)

%o(-a) = ( - 1 ) A ~o(a)

= ~(a).

To s u n m a r i z e , a mapping W : A ÷ l q i s t h e c h a r a c t e r i s t i c

mapping o f some

s u b g r o u p o f A i f and o n l y i f (1)

w(O)

(2) ~ ( z a )

= 1 =

~(a)

z 7~---~

z ¢ ZZ

(3)
A Lawvere - Tierney

]r-topology on l~ is a mapping j : I~ ÷ 1~ which is compa-

tible with the action of (Tz, x) and satisfies (LT -I - 2 - 3). •

j(n)

n

(1) j ( ~ - - X ~ ) = j ( n ) (2)

j (1)

= 1

(3) j j ( n )

.

^ z'

= j(n)

(4) j ( n v m) = j n v jm.

z eTZ,

This means

105

If p is some prime number, the ~mual localization process at the prime ideal p Z

is thus described by such a morphism j : IN ÷ I~. If n7Zc-+ 7z is some ideal I of 77, its localization is obtained by adding an element ~ for any a 6 n2zwhich cannot be divided by p.

These two ma ideals have the same localization if and only if each element of the form m--~ n a where p does not divide m b can be written in the form ~ where p does not n divide n b . But clearly, this is possible if and only if ~ cannot be divided by p.

Now consider

Thus the greatest ideal

nZ~ c

mZg, i.e. m divides n.

mTZ which has the same localization as n 2Z

obtained when m is the smallest integer such that n cannot be divided by p, i.e. m when m is the greatest power of p dividing n. But the greatest ideal which has the same localization as

n~

is the closure of

nTZ for the universal closure

operation associated to the localization (via proposition 50).

Thus the Lawvere -

Tierney ~-topology corresponding to the localization associated to some prime number p is given by

j :IN+IN j (n) = greatest power of p dividing n.

Consider now the theory ]r of arbitrary groups. frame.

Then Sh( I% ]~) is the category G r of groups,

functors are again a F h ° = (O) and a F h I = (77, +).

Again, let 14 be the initial ll~e two representable Thus the mono~dMlF of

l-ary operations, the topos g ( ~ , ~) and the object ~]r can be constructed in the same way and are exactly the same as in the commutative case.

But the prin-

cipal results of this chapter are no longer valid and this explains why we required ]r to be commutative.

We will show that theorems 22 and 56 fail in the

case of groups.

Consider the coproduct 7zJL Z~ in G_~r,where the two basic generators are denoted by x and y.

Consider the subgroup R generated by the words xx and yy.

The corresponding characteristic mapping ~0 : ~ IL Z~ + ~ is thus given by ~o(w) =

[ inf {n I n w 6 R} f o r 0 if this set is empty.

In particular, e(x) = z, qo(y) = z, e(xy) = O.

Thus the rule ~p(xy) divides

q~(x) v e(y) is not satisfied in this case.

In order to compare localizations and IT-topologies, observe that Ab and G r have the same representable objects and thus the same Gabriel - Grothendieck ]~-topologies; they have also the same topos &( IN, ]~) and the same object ~]r, thus the same Lawvere - Tierney ]r-topologies.

For ~ ,

these ]r-topologies are

106

in one-to-one correspondance with the localizations of Ab (theorem 56) and there are many non obvious such localizations (see [6], [20] or even our easier example of localizing at some prime ntunber p).

On the other hand, the category

Gr of groups has only the two trivial localizations (0) and Gr (proposition 57). Thus theorem 56 fails in the case of the category Gr of groups. Proposition

57.

(0) and Gr are the only two localizations of the category Gr of groups.

Consider a localization L~-~E~ Gr of Gr ~daich is not the identity on Gr. i Observe that propositions 48 - 49 - 50 - 52 do not depend on the commutativity of ~.

1 is not the identity thus 1 takes some non-isomorphic morphism into

an isomorphism (proof of proposition 52) and thus finally some s : n Z ~ + (n J 1) into an isomorphism (propositions 48 - 49).

Denote by f the morphism which sends I to the word x y C Z ~ and y are the two basic generators of Z]L Z.

Z, where x

Consider the following square

which is obviously a pullback (because n j 1).

(0)

n ZZ .[L n

s£s 7zJLZ

/(s) is an isomorphism, thus /(s/h s) is an isomorphism and /(r) is an isomorphism (1 is exact).

Thus l ( Z ) ~" l(O).

But 1 is exact and therefore

l(2Z) % (0). Now ZZ is a generator and any group can be reconstructed as a colimit of a diagram containing only ZZ; since 1 preserves colimits, /(G) Z (O) for any group G.

Thus the localization is just (0).

For the reader who is interested in the non commutative case, let us point 1 out that if IL ~-+ Sh( 11, IT) is a localization of Sh( ~i, 7F) for an arbitrary i theory IT, the sets defined by J(u) = {r : R >--+ a F h u I /(r) is iso} for any u E 11,describe a Gabriel - Grothendieck %-topology on 11 in the sense of definition 30.

Moreover, it is still true that this correspondence from a

l

107

localization to a q-topology is injective.

Thus any localization of Sh( 14, ~)

is completely characterized by a IF-topology on I~, but there are q-topologies on ~ which do not derive from a localization of Sh( ~, IT) - (counterexample of groups).

So the localizations of Sh( 14, ~) are exactly classified by

]/-topologies on II satisfying additional axioms.

To conclude, the notion of

q-topology does not seem to be the right one in the non coramutative case.

Or

maybe the usual notion of localization is not the right one in the non commutative case.

This problem is open.

CHAPTER 4 : INTEGRAL THEORIES AND CHARACTERIZATION THEOREM

L e t ~ be a frame and Sh(~) the topos of sheaves over~. t h a t ~ can be recovered from Sh(~).

It is well-known

H is exactly the frame of subobjects of I

in Sh(~).

Now consider a fixed external theory ~. can ~ be recovered from Sh(~,~r) ? provide a counterexample.

We ask an analogous question :

The answer is "no" in general : in § I, we

We consider the boolean ring 2 ~ (power set of ~)

and the corresponding (external) theory ~ of 2~-modules.

We prove that for any

non zero integers n, m, Sh(2n,~) is equivalent to Sh(2m,~).

However, if we add some assumption on the theory ~, the answer becomes i'yes". The assumption is that each non constant 1-ary operation is an epimorphism in ~, where we regard ~ as a category with a denumerable set of objects. In the case of the theory ~ of modules on the ring R, this is equivalent to the fact that R is an integral domain.

Such theories are therefore called integral.

The notion of integral theory includes as examples the theories of sets, mono~ds, groups, rings, modules on an integral domain (in particular abelian groups and vector spaces), algebras on a field, boolean algebras, sets on which a group acts, and so on ....

Integral theories are studied in § 2.

In § 3, we prove the characterization theorem.

If ~ is an integral

theory, two frames~ andS' are isomorphic if and only if the categories Sh(~,~)

andS h(~',~)

are equivalent.

To prove this, we show that for an

integral theory ~, ~ is just the frame of formal initial segments of Sh(~,TF). In particular, ~ can be recovered from S h ( ~ , ~ ) . § 1. A COUNTEREXAI'-~LE

Consider an algebraic theory ~F in the usual sense.

For any frame H,

can be seen as a theory in Pr(14) (= the constant functor ~F) or in Sh(H) (= the corresponding associated sheaf functor). nally, Sh( ~, ~F) depends only on ~. characterizes ~ completely.

Thus when ~r is fixed exter-

So it makes sense to ask if Sh( 14, ~r)

In this paragraph, we exhibit a counterexample.

Consider the set I~ of integers (or more generally any infinite set E). Choose any bijection I~ ~ J ~

N and apply the power set functor, which takes coli-

109

mits into limits (it is represented by 2) 2N g 2N ~ N ~ 2N × 2N This isomorphism of 2 N and 2 N x 2 N is in fact an isomorphism of boolean rings (the addition is the s3q~netric difference A A B = (A U B) x (A N B) and the multiplication is the intersection A N B), (see [II] - II - 3 - 9).

From the isomorphism of rings 2 ~

2 ~ x v N , we deduce an equivalence

between the corresponding categories of modules

Mod %Mod 2 ~ Nod 2 x b~d 2 2N N x 2N -N -N (cfr.

[?7]).

But Mod N i s the c a t e g o r y o f sheaves o f 2 N-modules on t h e one-

p o i n t t o p o l o g i c a l spa~e and Mod2 N × M°d 2 N i s t h e c a t e g o r y o f sheaves o f 2 N - m o d u l e s on t h e d i s c r e t e

two p o i n t s s e t .

Thus we have proved t h e e q u i v a l e n c e

Sh(21 , g ) g Sh(2 2, lY) where % is the theory of 2 M-modules.

Iterating the process, we deduce that

for any two non zero integers n, m Sh(2 n, %) g Sh(2 m, ~F). This is clearly a counterexample to our problem since for n ~ m, 2n and 2m are not isomorphic. § 2. INTEGRAL THEORIES

If we consider a category Sh(H,~F), where 'F is a theory a n d ~ a frame, the objects a F h u generate Sh(~,]~), (proposition I - 6).

But any hom-set

in H is reduced either to a singleton or to the empty set : therefore the generators a F h u are constructed from the two free algebras FO and Ft.

This explains

why assun~tions on 1-ary operations (= elements of FI) and constants (= elements of FO) of ~ induce strong consequences on Sh( H, Tr), even without any assumption on n-ary operations with n ~ I.

In this paragraph, we describe such assumptions :

integrability of the theory with, as special cases, left or right simplicity.

As mentioned in the introduction of this chapter, a theory ~ i s

integral

if any non constant 1-ary operation is an epimorphism ill ~ or, equivalently, if any non constant ~-endomorphism of FI is injective.

This is in particular

the case when ~r is left simple, i.e. when FI has only the trivial subobjects

110

FO and FI, or when ~ is right simple, i.e. when FI has only the trivial quotients FO (if there are constants) and Ft.

Left simplicity is also equivalent to the

fact that any non constant 1-ary operation has a section. As already mentioned, for a theory ]r of modules on a ring R, integrability of ]r is equivalent to the fact that R is an integral domain.

Other examples

are given by sets, monoids, groups, rings, boolean algebras, and so on .... We start with two technical lemmas which make explicit and precise the form of some well-known isomorphisms. Len~aa I. For a theory

~, there is a contravariant homomorphism of mano¢ds

the usual c o ~ o s i t i o n

of arrows)

(for

:

]T(T I, T I) % Sets]~(F1, FI). Looking at an algebra as a product preserving functor IF ÷ Sets, FI is just ]~(T I, -), (cfr. [21] - 18). Sets~F(F1, F])

Thus by the Yoneda ler~na, we have

= Nat(]T(T I, -), ]T(T I, -)) % ]T(T I, T I) and the last isomorphism def

takes IT(a, -) to ~ : T I ÷ T I.

From the equality

we deduce that the isomorphism is contravariant. Lenm~a 2. For a theory

"IF, the bijections

]F(T I T I) ~ Sets IT(F] FI) ~ FI

are such that a(S) = So~ .

The first bijection is the one given by proposition I and the second comes from the adjunction F ~

U where U : Sets IF ÷ Sets is the forgetful functor :

Sets IT(FI, FI) ~ Sets(l, UFI) % UFI. FI, seen as a product preserving functor ~ + Sets is just]Y(T I, -) and

111

is just ~(~,

-), (cfr. proposition

1).

So for any integer n,

]T(T I, T n) ÷ ]~(T I, r n) is defined by Tr(~, T n)(Y) = 7 o ~.

]T(~, T n) :

In particular

]I(~, T I) : ]~(T I, T I) ÷ ]r(T ], Z I) is defined by ~r(a, TI)(p) = 6 o e. ~r(~, TI) ~

But

and thus ~(~) = 8 o ~.

Proposition 5.

For a theo~d ~, a l-ary operation a : T I ÷ T I is an epimorphism if and only if for any 1-ary operations ~, y : T I +÷ T I B o e = Y o a If ~ is an epimorphism,

~

~ =7-

the condition certainly holds.

Now suppose the

condition holds and consider 6, ~ : T I ~ T n such that ~ o ~ = ~ o ~.

For any

projection Pi : Tn ÷ TI' we then have Pi ~ ~ o ~ = Pi o ~ o ~ and since Pi o and Pi ° ~ are 1-ary operations,

our assumptions

imply

Pi ° 6 = Pi ° s"

This is

true for any i and thus ~ = E, which proves that ~ is an epimorphism.

•

Definition 4.

A theory~F is integral if and only if any non constant 1-ary operation of ~is

an epimorphism in~F.

Proposition 5.

A theory~F is integral if and only if for any non constant 1-ary operation, the corresponding endomorphism of FI is injective. By proposition 2, a 1 - a ~ for any 1-ary operations

operation ~ is an e p i m o ~ h i s m

if and only if

6, 7, S o a = 7 o ~

~

6=7.

If we denote by a, B, Y the corresponding e n d o m o ~ h i s m s o f is equivalent to : ~roposition ~

o

B

FI, this condition

I) =~

o

7

~

8 =y.

But FI is a generator in ~ t s ~', thus this last condition is equivalent to the injectivity of ~.

Proposition 6.

Let 72 be the theory of modules over a fixed ring R (with unit). integral if and only if R is an integral domain.

is

112

The mono~d of l-ary operations of ~

Thus the

is exactly the mono~d (R, ×).

theory is integral if and only if for any x ~ 0 and any y, z y x = z x

~

y = z

(y-z)x = 0

~

y - z = O.

or equivalently

Because y, z are arbitrary, this is equivalent to w x = 0 for x ~ 0 and an arbitrary w.

~

w = 0

In other words, the condition is equivalent to w x = 0

~

x = 0 or w = 0

which is the definition of an integral domain.

The free algebra functor F : Sets ÷ Sets IT preserves monomorphisms I - 1).

So the inclusion 0 ~

] is sended to an injection FO ~

(cfr.

FI in Sets

Definition 7.

A theory ~Fis called left simple if the only subobjects of FI are FO and F].

Proposition 8.

A theory ~ is left simple if and only if any non constant 1-ary operation has a section in ~F. Consider a left simple theory •

and a non constant l-ary operation 6.

corresponds to an endomorphism ~ : FI ÷ Fl and thus to the element ~(,) E FI where * is the universal generator of FI.

~ is non constant, thus

~(.) is not a constant and the subalgebra generated by ~(*) must be FI, because TF is left simple.

In particular, there is an operation ~ (necessarily a

1-ary operation) such that ~(~(.)) = *. • = ~(*))

But (with the notations of proposition 2)

= ~(~(*))

and by proposition I, this means ~ o B = id.

= ~)

= s o

Thus ~ has section 8.

Now suppose that any non constant 1-ary operation has a section in ~ . Take a subalgebra R ~-~ FI which contains a non-constant l-ary operation ~. Choose a 1-ary operation 8 such that ~ o B : id in IF. *=~

But ~ E R, thus ~ )

o B =~(~)

=~(~(*))

E R and • E R.

= s(~(*))

In FI, we have = ~).

This proves that R = FI.

113

Proposition 9.

Any left simple theory is integral. Trivial since a morphism with a section is an epimorphism.

For any constant @ g FO, there is a unique homomorphism g : FI ÷ FO such that g(*) = @. This homomorphism is surjective because I~D is an initial object 2r (F preserves colimits) and thus the composite

in Sets

FO > is the identity on FO.

~ FI .

~ FO

Thus ~ is a coequalizer (cfr. [21] - 18 - 8 - 8) and FO

is thus a quotient of FI via @.

Definition 10.

A theory "IF is called right simple if any quotient of FI is the identity on FI or some ~ : FI + FO. Proposition

11.

A right simple theory is integral. Consider a right simple theory It, a non constant 1-ary operation and two 1-ary operations B, y such that ~ o ~ = y o ~. ty B = ~.

We must prove the equali-

Via proposition I, we consider the following diagram where Q is the

coequalizer of 8, y q F1

....

~. F1

-~ Q

F1.

By proposition I, ~ o ~ = ~ o y and thus there is a unique factorization p : Q ÷ FI such that p o q = ~.

But the theory is right simple.

If Q = FO

and q = ~ with @ C FO, then = 5(*)

= p q(*)

= p(O)

= @

which contradicts the fact that ~ is non-constant. and ~ = y, thus ~ = y.

So q is the identity on FI

114

There are many examples of integral theories, Example 12 : Sets. FI is the singleton which has only two subobjects : FO = ~ and Ft.

So the

theroy is left simple and thus integral. Example 13 : Mono~ds. F1 is ( IN, +). tion by n.

For n E IN, the homomorphism n : I~ ÷ IN is just the multiplica-

Indeed,

~(0)

= 0

~(I)

= ~(,)

n(m)

= n(l

= n +

...

+

I)

If n # O, n is injective.

=n(l)

+

...

+ n(1)

= n

+

...

+ n

= mn.

Thus the theory of monoids is integral.

Example 14 : Groups. FI is (Tz, +).

For z E77, the homomorphism-z : 7z +ZZ is just the multipli-

cation by z, (same proof as in example 13, with the additional equality -ff(-m) = - -z(m)). If z # O, ~ is injective. Exile

Thus the theory of groups is integral.

15 : Abelian groups.

This theory is integral by proposition 6. Exile

16 : Rings.

We do not suppose con~autativity nor existence of a unit.

FI is the ring

of polynomials 1 an Xn + ... + a I X where a i is an integer.

If P(X) is any such polynomial, P : FI + FI takes

X = . to P(X) and thus Q(X) on Q(P(X)).

Now if

QI (P(X)) = Q2 (P (x)) where P(X) is not O, we must prove that Q1 and Q2 coincide.

Suppose P(X)

given by P(X) = a n Xn + . . .

+ a 1 X1

a

# O.

I f QI(x) h a s d e g r e e m, Q1 p ( x ) h a s d e g r e e n m1 and t h u s a l s o Q 2 ( P ( X ) ) ; so Q2 t o o h a s d e g r e e m.

Then, sup_pose Ql(X) and Q2(X) g i v e n b y

Q1 (x) = am xm + "'" + ~I Xl

115

Q2 (X) = Bm xm + "'" + ~i X l" We now cor~pare the polynomials QI(P(X)) = Q2(P(X)).

The terms of degree n m are

equal, thus am amn = ~m amn which implies am = $m' since an ~ O.

The terms of degree n(m-1) are equal, thus

m-1 + E(~m' an' "" ., al) = Bm-1 am-] + E(Sm, an, .... a l) ~m-1 an n where E(~m, an .... , a I) is the coefficient of X n(m-1) in ~m(P(X))m.

Because

am = Bm' we deduce E(~m, an, ..., al) = E(~m, an, ..., al) and finally am_ ] = 8m_], since an ~ O. We iterate the process with the terms of degree n k (k ~ m) and finally we prove the equality o k = 8k for all k. is injective.

Thus QI = Q2 and

So the theory of rings is integral.

Example 17 : Rings with unit. We do not assume commutativity.

Now F] is the ring ~ [X] and the proof

of the integrality of the theory is analogous to that of example ]6. Example 18 : Commutative rings. We do not assume the existence of a unit.

Here F] is the ring described

in example 16 and the proof of the integrality is the sam~ as in example 16. Example 19 : Commutative rings with unit. Again FI is ~ [ X] and the proof of the integrality is analogous to that of example ]6. Example 20 : Commutative algebras. If R is a commutative ring with unit which is an integral domain, the corresponding free colr~mtative algebra with unit FI is just R[X].

If we do not

assume the existence of a unit in the algebra, FI reduces to those polynomials of the form Xn + ... + a

a

n

X I. 1

The proof given in example ]6 transposes to the present case because R is an integral domain. Exemple 2] : Modules on an integral domain. By proposition 6, this theory is integral.

116

Example 22 : Vector spaces. Any (skew) field is an integral domain. integral by proposition 6.

Thus a theory of vector spaces is

The free algebra FI is the field K itself; it has

only the trivial subobjects and quotients : thus the theory is in fact left simple and right simple. Example 23 : Sets with base point(s). The free algebra FI is the set with a single element which is not a base point.

The only subobjects are FI itself and the subset of base points.

So the

theory is left simple and thus integral. Example 24 : G-sets for a group C. The free algebra FI is the group G itself where the action is the multiplication of the group.

If g E G, the homomorphism g : FI ÷ FI is thus the multi-

plication by g, which is injective because it has an inverse g

So the theory

is integral. Example 25 : Boolean algebras. The free algebra FI is 22 •

x.

C x

Any subalgebra contains O and I and if it contains x (resp. C x) it must contain its complement which is C x (resp. x).

So the theory of boolean algebras is

left simple and thus integral. § 3. THE CHARACTERIZATION THEOREM We now proceed to prove our characterization theorem for an external integral theory ~F.

For any frame ~, we orove that I~ is isomorphic to the frame of

formal initial segments of S h ( H , ~ ) .

We deduce that two frames ~ andll' are

isomorphic if and only if the categories Sh( I% ~) and Sh( i~', ~) are equivalent. Proposition 26. Let H be a frame and "IF an external integral theory.

The canonical inclusion

117

of ~ in the frame ~ of formal initial segments of Sh( ~, ~) is an isomorphism of frames. Up to now, we have described in the following way the internal theory associated to the external theory ~ presheaf ~

(external).

: it is the sheaf associated to the constant

But this constant presheaf is generally not a sepa-

rated presheaf : indeed for a separated presbeaf P, P(o) has at most one element (o is covered by the empty covering and thus two elements of P(o) have always the same restriction at all the (unexisting) elements of the empty covering; thus they must be equal).

In fact, this difficulty at o is the only reason why

a constant presheaf Q is generally not a sheaf. u =

Indeed, if u ~ o in ~ and

v ui, at least one u i is distinct from o; say uj ~ o. -EI and ~or any i C I, x u. = y u ' then in particular i i

Thus if x, y E Q(u),

x = x uj = y uj = y"

On the other hand, if A is a sheaf on H, then A(o) is exactly a singleton. A is separated, thus we know already that A(o) has at most an element.

Now the

empty family of elements is a compatible family for the empty covering of o; by the sheaf condition, this empty family can be glued into an element of A(o). Thus A(o) is non-empty and A(o) is a singleton.

Finally this shows that the

sheaf associated to a constant presheaf Q is the same as the sheaf associated to the presheaf Q' which coincides with Q at each u ~ o and which is such that Q' (o) is a singleton.

For all these reasons and for the simplicity of the proof which follows, we now change our conventions slightly.

~F is the external integral theory.

We associate with it a separated presheaf of theories which is TF for each u ~ o and the degenerate theory (a single operation in each dimension) when u = o.

The corresponding category of algebraic presheaves is denoted by Pr( ~, ~F).

Finally we consider the associated sheaf of theories and corresponding category Sh( ~, ~) of algebraic sheaves.

The previous remarks show that the sheaf

of algebraic theories is exactly the sheaf associated to the constant presheaf on ~; in particular Sh( ~, ~) is just the category defined previously (see I - 4).

Finally recall that we denote by F " Pr(H) ÷ Pr( ~, ~) the left adjoint to the forgetful functor Pr( ~, ~F) + Pr(~). and v C 14

Thus, if P is some presheaf on H

1t8

F P(v) =

I F(Pv)

if

v ~ o

[

if

v = o

{*}

where F(Pv) denotes the free 1!?-algebra on the set P(v).

F hu(V ) =

FI

if

FO

if

{*}

if

In particular if u 6 14 o ~ v~u

v~u o = v.

This is a separated presheaf because F hu(O ) = (*} and if o / v =

v v i in}{ i£1

and x, y are elements in F hu(V ) such that for any i, x vi = Y [ w " then x = y. I 1 Indeed if v < u, choose v i ~ o and the restriction to v i is just the identity on FI; now if v ~ u ,

choose v i ~ u

and the restriction to v i is just the identity

on FO; in both cases, we deduce x = y.

This separation property of F h u is the

reason why we introduced this slightly different presentation. Now consider a formal initial segment U of Sh( 14, ZF).

We must prove that

U is in fact the formal initial segment generated (via proposition II - 5) by some element u 6 14.

By proposition II - 13, it suffices to prove the equality U = u

in the larger frame Heyt(a F hi), that is to say

U! U*(a F h i ) ~ a F h uFirst, we need to construct u. 14 is a subframe of ~ and so we may define u by

u=v

{v 6141 v<~ u}.

For any v E Id, v < u, we have U, U*(a F hl)(v) ~ u !

u*(a F hl)(v) = a F hu(V) .

But F h I and F h u coincide for all v < u ;

thus a F h i and a F h u coincide for

all v < u; in particular a F hu(V) = a F h i (v) which proves finally that U! U*(a F hl)(v) = a F hi(v) Now consider the following pullback in Pr( 14, It)

P ;

I

F h1 >

~ U! U*(a F h i )

,

I

a F hi;

if v < u .

119

F h I is indeed a subobject of a F h I because F h I is a separated presheaf.

Apply

the associated sheaf functor to this diagram to obtain a P ;

~

Uf U * ( a F h )

a F h1 again this t o P.

a F hi;

i s a p u l l b a c k w h i c h shows t h a t U t U*(a F h i ) i s t h e s h e a f a s s o c i a t e d

Combining t h i s

last

result

w i t h t h e former, we o b t a i n t h a t

f o r v ~< u i n 1-I,

U! U*(a F h l ) ( v ) = a F h l ( v ) and t h u s

P(v)

Comparing

with

F1

if

o ~ v ~< u

(*}

if

o = v.

F hi(v)

the description

of F hu, we deduce

Fhu>

"Fh

a canonical

factorization

1

w h i c h , f o r v - ~ u, i s t h e u n i q u e morphism FO + P ( v ) .

In particular,

F hu is a

s u b o b j e c t o f P. We will prove that U! U*(a F hl) = a P is just a F h u.

Suppose a P ~ a F h u.

Then P cannot be equal to F h u and thus there exists some v £ H F hu(V) is not the whole P(v). that v ~ u.

such that

Now the descriptions of P(v) and F hu(V) show

Then F hu(V ) = FO and the inequality F hu(V) < P(v) shows the

existence of a 1-ary operation u 6 P(v) which is not a constant. <~> ~ FI the subalgebra of FI generated by ~.

Denote by

Define also the subpresheaf

<<~>> of P generated by ~, i.e. i <<~>> (w) =

<~> FO

if if

o ~ w ~ v w ~ v

{*}

if

o = w.

By proposition I - 4, there is a morphism f : F h

v

+ <>

which corresponds to the choice of ~ 6 F hv(V). (I) if o ~ w ~ < v

We can easily describe f :

f(w) : FI + <6> is the morphismwhich sends • to ~; by

definition of ~, this is a surjection.

But the composite

120

FI

f(w)~

<~>

~

> FI

is injective because a is non-constant (integrality of the theory; proposition 5); thus f(w) is injective.

Finally f(w) is an isomorphism.

(2) if w ~ v

f(w) : FO + FO is the identity on FO; it is an isomorphism.

(3) if w = o

f(w) : {*} ÷ {*} is again the identity and thus an isomorphism.

So, f is an isomorphism.

Finally, consider the following composite a F h

"~ a <<~>> v a(f_1)

>--+ a P = U! U*(a F hl).

By proposition II - 15, this implies v ~< U in Heyt(a F hl) and thus in ~'. definition of u, this implies v ~ u, which contradicts v ~g u. U~ U*(a F hl) = a P = a F h u and U = u.

Finally, ~ =

By

Thus

H w h i c h proves the proposi-

tion.

•

Theorem 27 (Characterization theorem).

Consider an integral theory ]~. TWo frames ]4 and ]-I' are isomorphic if and only if the categories Sh( ]4, 7) and Sh( ~', ~F) are equivalent. By proposition 26.

•

CHAPTER 5 : SPECTRUM OF A THEORY Recall that if R is a co~utative ring, the Grothendieck - spectrum of R is some topological space X constructed from the prime ideals of R. structure of X reflects some algebraic properties of R.

The topological

Moreover, R can be

presented as the ring of global sections of some sheaf of rings on X.

The inte-

rest of this representation lies in the fact that the stalks of the sheaf are local rings.

Thus the study of an arbitrary co~autative ring can be reduced to

the study of local rings if one accepts to replace a single ring by a sheaf of rings. There are many other notions of spectrum and sheaf representation.

For

example, Pierce's representation is based on the properties of idempotents of R. It is simpler than Grothendieck's, but on the other hand Pierce needs assumptions on R to get interesting properties of the stalks : for regular rings, the stalks are fields. In this chapter, we introduce, for an arbitrary algebraic theory 7, a notion of spectrum for ~ and a sheaf representation on this spectrum for any T-algebra. This is obtained from the general theory of formal initial segments.

In further

chapters, we shall particularize these constructions to the case of a ring R, via the theory of R-modules. If ~ is a frame and ~F is an algebraic theory in Sh(~), we saw in chapter 2 how to construct the frame ~ of formal initial segments of Sh( ~, ~). depends on I~ and on ~.

This frame

But if we fix ]I to be the initial frame {0, I), thus if

we work with sheaves on the one point topological space, the theory ~r is just a theory in the external sense and Sh( ~, 7F) is simply Sets ~.

In other words, to

any algebraic theory ~, we can associate a frame ~iILwhich depends only on ~ this is the frame of formal initial segments of Sets ~.

:

In this chapter, we

prove that this frame is spatial, i.e. is the frame of open subsets of some topological space called the spectrum of ~l~. The inclusion of frames 14 ~-~ ~ studied in chapter {O, I} ~

~

2 reduces here to

and the restriction functor r : Sh(~,~)+

Sh({O, I},~F)~Sets ~

is simply the "global sections" functor.

In proposition II - 14, we produced a

122

right inverse A to this functor.

In other words, for any ~-algebra A, there is

a sheaf ~A of T-algebras on the spectrum of IF, whose algebra of global sections is A.

This is a sheaf representation theorem for T-algebras. Considering the results of chapter 4, we deduce that for an integral theory

Tt, the spectrt~a of ~ is just the one point set and therefore the corresponding representation theory vanishes in that case.

Thus the theory developed in this

chapter will be useful in the case of non-integral theories. § I. THE PURE SPECTRUM OF AN ALGEBRAIC THEORY In this paragraph, we prove that the frame of formal initial segments of Sets

, where ~ is any theory, is the frame of open subsets of a compact space

(not necessarily Hausdorff). Definition I.

If ~ is any algebraic theory, ~

denotes the frame of ~Jrmal initial seg-

ments of Sets We still denote by

FX the free T-algebra on the set X.

Definition 2.

If ~ is any algebraic theory, a ~-ideal is a sub-IF-algebra of Fl. If R is a ring with unit and ~ is the theory of left R-modules, Fl is simply R and a left-submodule

of R is simply a left ideal of R.

This justifies our

terminology. Definition 3.

If IT is any algebraic theory, a ~r-ideal is pure if it is of the form

U[ U*(F1) for

some

U in ~']r"

We denote by p(IT) the set of pure Tt-ideals.

In chapter 6, we will see

that in the case of left modules on a ring R with unit, definition 3 reduces to the usual definition of a pure ideal. Proposition 4.

Any pure ~T-ideal is a Heyting subobject of FI.

123

By definition of a formal initial segment (II - 6). Proposition 5. The pure q-ideals constitute, for the usual operations of union and intersection of sub-T-algebras, a frame. Moreover, p(~) is isomorphic to ~ . By proposition II - 12, the inclusion ~ y ~ - + Heyt(F1) is the map sending the formal initial segment U to the pure IF-ideal U! U*(F]).

•

Definition 6. A pure TF-~deal is purely maximal if it is maximal among the proper pure ~F-ideals. Definition 7. A pure T-ideal J is purely prime if it is proper and for any pure q-ideals If, 12 11D

12 ___J

~

II _aJ or 12___J.

Proposition 8. Any purely maximal ~F-ideal is purely prime. Suppose J ~ FI is a purely maximal Xr-ideal and If, 12 are pure It-ideals such that 11 n 12_a J.

If I 1 _ J, then 11 U J (union as sub-q-algebras

of F])

is still a pure ideal (proposition 5) and is strictly larger than J; the maximality of J implies I l U J = Ft.

But then, by proposition 4

12 = 12 N F 1 = 12 N (11 U J) = (I 2 D Ii) U (I 2 n J)

cJuJ =j

which concludes the proof. Proposition 9. Any proper pure T-ideal is contained in a purely maximal Let I ~ FI be a pure IF-ideal, I # Ft. clusion

T-ideal.

Consider the set X, ordered by in-

124

X = {J I J pure F-ideal; I c J ~ FI}

X is not empty because I is some element in X. family of elements in X, k, I £ Jk and the union

If (Jk)k6K is a totally ordered

U Jk is again a pure ideal containing I. k£K

Now for any

U Jk is filtered : it is thus exactly the union of k6K

the underlying sets ([21] - 18 - 8 - 8) and I ~

U

J~.

So we can apply Zorn's

k6K ~. ~ lemma t o X and o b t a i n a p r o p e r G - i d e a l J which xs maxima± among t h e p r o p e r pure i d e a l s c o n t a i n i n g I.

But o b v i o u s l y t h i s i m p l i e s t h a t J i s p u r e l y maximal.

I

Proposition I0. If I is a pure ~-ideal and a 6 F] ~ I, there exists a purely prime T-ideal J such that I c J and a ~ J.

Consider the set, ordered by inclusion, X = {J I J pure T-ideal, I _c.7, a ~ J}. X is not empty because I 6 X.

Now if (Jk)kC K is a totally ordered family of

elements in X, just as in the proof of proposition 9, such that a ~ in X.

U Jk" k6K

U Jk is a pure ~-ideal k6K We may then apply Zorn's lemma and choose some J maximal

We assert that J is purely prime. with Ii ¢ J and 12 C J ;

By contraposition, take If, 12 pure

we must show that I l N 12 ¢ J.

From Ik ¢

J, we deduce

that Ik U J is a pure T-ideal containing strictly J; by definition of X, this implies a 6 Ik + J.

Finally a £ (11N I2) + J and a ~ J, thus I l n 12 C J . I

Proposition 11. Any proper pure F-ideal is the intersection of the purely prime T-ideals containing it.

By propositions 8 - 9, there are purely prime ideals containing a proper pure ideal I and by proposition 10, any element which is not in I is outside one of them, thus outside their intersection.

This proves the proposition.

We turn now to the construction of the pure spectrum of T.

We consider

first the set pp (IF) of purely prime T-ideals and its power set 2pp (T) shall define a topology on pp(~)

and the corresponding topological space

We

125

Spp(~) will be the pure spectrt~n of ~.

In order to do so, we define a mapping :

2pp ("It)

0 : p(~)

by 0(I) = {J E pp(Ze) I I _

J}.

Proposition 12. 0 is a morphism of fr~nes. (1) smallest element : O(FO) = {J E pp(]T)

FO ~_J} = %.

(2) greatest element : O(FI) = {J [ pp(TO

I F1 ~_ J} = pp(Xr).

(3) finite intersections : Consider Ii, 12 in p(]Y) and J in Spp(IT).

Since J is purely prime,

we have the equivalence I 1 fl I z ~_J*=~ 11 _ J

and 12_~J.

Therefore O(I i n I2) = {J E p p ( ~ )

[ I i N 12 _~ J}

= {J E pp(]Y) [ Ii ~ J and

12 _

J}

= {J [ pp(~r) IIi _¢ J} n {J E pp(~r) I 1 2 _ J} = O(Ii) N 0(I2).

(4) arbitrary unions • Let (Ik)kE K be a family of elements in p(~F).

o(u

ik) = {JCpp(~r)

kEK

I u

Then

ik¢_ J}

kEK = {J E pp(~r) I 3k [ K = =

O {J E pp(~) kEK U

kCK

I Ik {

Ik ~ J} J}

OUk).

Proposition 13. 0 is injective. Consider 11 ~ I2 in p(IF).

One of these ideals is not contained in the

other one; suppose I l ~ 12 and choose a C 11 ~ 12.

By proposition 10, we can

choose a purely prime T-ideal J such that 12 c J and a ¢ J; in particular I i ~ J.

But this implies J E O(Ii) and J ~ 0(I2).

is injective.

Thus O(Ii) ~ 0(I2) and 0

126

Proposition 14. The subsets 0(I), for I running through p(]T), constitute a topology on pp(]I). By proposition 12, this family of subsets contains 0, FI and is stable for finite intersections and arbitrary unions.

[]

Definition 15. The pure spectrum Spp(]~) of the theory ~F is the set pp(]F) of purely prime T-ideals equipped with the topology whose open subsets are the O(I) 's, for any pure ~r-ideal I. Proposition 16. The frame of open subsets of the pure spectrum Spp(~)

of ~ is isomorphic

to the frame p ( ~ ) of pure ~F-ideals. By propositions 12 - 13.

I

Theorem 17. The pure spectrum Spp(~) of a theory ~

is a cor~pact space.

Consider a family (Ik)k£ I of pure If-ideals such that By proposition 16, this is equivalent to algebras) the Ik'S. ak. 6 I k . , 1 1

is the sub-IT-algebra

o f F1 g e n e r a t e d by a l l

U Ik (union as ]~k6K

the elements of all

U I k h a s t h u s t h e form c~(akl, . . . , a k ) where k£I n n i s any i n t e g e r a n d c~ i s a n - a r y o p e r a t i o n . The e q u a l i t y

ak ). n

But t h i s

g e n e r a t o r . o f F1 i s o f t h e form

n U 0(Ik. ). i=I i

n U Ik. and t h u s a l s o any i=I i n Finally, FI = U Ik. and Spp(]T) = i=I I

implies that • is in

6(*) = 6 for any 1-ary operation 6 £ F]. = 0(FI) =

But

An e l e m e n t i n

U I k = F1 i m p l i e s t h a t t h e u n i v e r s a l k6K a(a k , ..., 1

U Ik = FI. k6K

U 0(Ik) = Spp(]r). k6K

I

This concludes the proof.

The spectrv~ Spp(~D

is generally not a Hausdorff space.

will be given in chapter 7 w h e n

A counterexample

IT is the theory of right modules on the ring R

of linear endom~rphisms of a vector space with infinite countable dimension.

127

§ 2. REPRESENTATION THEOREM FOR 'W-ALGEBRAS In this paragraph, we interpret some results of chapter

2 to present any

IF-algebra as the algebra of global sections of a sheaf of IT-algebras on the pure spectrtml of 'IT. Proposition 18. The '~lobal sections" functor

r : Sh(Spp(~), ~) ÷ Sets ~' has a right inverse A.

Sh(Spp(%), IF) is simply Sh(~/]~, ]Y) and Sets ~F is simply Sh({O, I}, °F). Moreover, ~Tf is the frame of formal initial segments of Sh({O, ]},IF).

The

"global sections" ftmctor r : Shgf%, l r ) ÷ Sh({O, 1}, ~r) is given by r (A) (0) = {*} r(A)(1) = A ( ] ) , Thus r (A) i s t h e r e s t r i c t i o n

o f A t o {0, 1}.

i m m e d i a t e l y from p r e p o s i t i o n

I I - 14.

Therefore the result

follows []

Theorem 19. For any IF-algebra A, A is isomorphic to the ~ - a ~ e b r a of global sections of the sheaf AA on the pure spectrum Spp(~) of IF.

This is another way of saying that A is right inverse to F.

[]

CHAPTER 6 : APPLICATIONS TO MODULE THEORY In this chapter, we apply our previous results to the particular case of module theory on a ring R. R is a ring with unit.

Thus ~ w i l l

The frame H w i l l

be the theory of right R-modules, where again be the initial frame {O, I} so

that all the sheaves we consider are sheaves on the singleton. frame of open subsets of the singleton).

({O, I} is the

Thus the category Sh(~,~F)

is simply

the category~4OdROf right R-modules. If R is cor~autative, the theory of R-modules is a commutative theory and the results of chapter 3 ~(H,~),

can be applied.

They imply the existence of a topos

which is the topos of (R, ×)-sets, in which lives an object ~ .

This object is the set of right ideals of R. classified by a morphism ~ : M ÷ ~ sified by a "topology" j : ~ W +

Each submodule

B ~

A can be

and each localization of Mod R can be clas-

~IT"

Now if R is any ring with unit, the results of chapter 5 apply to the theory of right R-modules.

This produces a "pure spectrum of ~' and for any

R-module A, a sheaf representation of A on the pure spectrum of R.

This spec-

trum and this representation are thus obtained via the formal open subsets of MOdR , which are characterized by certain ideals of R; in chapter 5, we called them "pure T-ideals".

Here we show that in the case of R-modules, they are

exactly the pure ideals of the ring in the usual sense of module theory. R itself is a R-module. on its pure spectrum.

Thus it can be represented by a sheaf of R-modules

In fact this sheaf of R-modules is also a sheaf of rings.

Moreover, the sheaf representation of any R-module is a sheaf of modules on the sheaf of rings which represents R. § ]. THE CLASSIFYING OBJECT FOR MODULE THEORY Consider a commutative ring with unit R and the corresponding theory ~r of modules.

Consider the frame l~ = (0, I} of open subsets of the singleton.

Then Sh( lq, TF) is the category Mod R of R-modules. tions of ~ is the multiplicative monoid (R, ×). described in chapter

3

The monoid of 1-ary operaThus the topos &( ]4, ~)

is the topos Sets R of (R, ×)-sets.

129

The two representable algebras are a F h 0 = (O) and a F h I = R. Thus air is the set aR of submodules of R, i.e. the set of ideals of R. aR has the structure of a (R, ×)-set via ~?R×R

........ ~ R

(I, r)

!

a R i s o r d e r e d by i n c l u s i o n o f i d e a l s ,

, r-l(I)

= [I : r] .

has greatest

e l e m e n t R and f i n i t e

inter-

sections.

Now if B >-+A is the inclusion of some submodule, the results of chapter 3 tell us also how to construct the corresponding characteristic map ~ : A ÷ ~ %. If a £ A, consider the corresponding linear mapping a : R ÷ A such that a(1) = a.

Then ~0(a) is a-1(B),i.e. w(a) = {r E R I a r 6 B}.

In other words, the ideals of R are the truth-values of the theory of R-modules and for a 6 A, B_cA, the truth value of a 6 B is the ideal constituted by those r such that a r 6 B is true.

By theorem I I I - 56, there is a one-to-one correspondance between localizations of ModR and ]r-topologies on e ]r. mapping j : aR + eR in Sets W

A ]Y-topology on ~]r is a

satisfying three conditions.

More precisely, a

It-topology on ~R is a mapping j : ~ R - + ~R s u c h t h a t f o r I , J ¢ a R and r C R (1) j ( [ I

: r])

= [j(I)

: r]

(2) j (R) = R (5) j j ( I ) (4) j ( I

= jU)

a J) = j(I)

T h i s i s e x a c t l y what H. Simmons c a l l s

a j(J).

a "Iocalizor"

(cfr.

[22]).

If P is a prime ideal in R and R ÷ Rp is the Lksual localization of R at P, this morphism of rings induces a full and faithful functor : MOdRD ÷ Mod R between the corresponding categories of modules.

But Rp is

a flat R-module; thus the ftmctor -@ ~R~) : M o d o ÷ Mod R is exact. On the other R hand, -® Rp is left adjoint to ~, so that M o d ~ is a localization of Mod R. R

130

This localization is thus classified by a IF-topology jp : eR ÷ ~R"

We now

proceed to describe this topology. The localization Ip of an ideal I is the set of fractions

Ip= {~i I ic where

i' qi _ q,

if iq' = i'q.

and o n l y i f any q ' • P. j r E I.

I; qCP}

Therefore,

if I _cJ are two ideals,

Ip = Jp if

w i t h j E J , q g P can be w r i t t e n as a f r a c t i o n ~-r w i t h i £ I ,

T h i s i s e q u i v a l e n t t o t h e e x i s t e n c e o f an element r g P such t h a t Indeed, i f such an r e x i s t s

P i s prime.

'

J-" = j r w i t h j r C I and q r ¢ P, s i n c e q

r

C o n v e r s e l y i f j = ~-,i j q ' = i q E I and J q, q' = j with j q' E I q q q

and q q' ~ P, s i n c e P i s prime. Ip = Jp

Thus

~=~

¥ j 6 J

3 r ~ P

j r 6 I.

But the closure I of I for the universal closure operation associated to jp is jp(1) and this is the largest ideal J such that Ip = Jp. jp(I) = {j C R [ 3 r ~ P

Therefore

j r £ I}.

Indeed, the only thing we still need to prove is that jp(I) is an ideal.

If

j r 6 I and j' r' e I with r, r' ~ P, then j r r' C I and j' r' r 6 I with r r' ~ P (P is prime) and (j + j')(r r') £ I.

So jp(1) is stable under addition.

It is obviously stable under multiplication by some element of R.

So it is

an ideal. § 2. THE PURE IDEAL ASSOCIATED TO A FORMAL INITIAL SEGMENT Consider any ring with unit R (not necessarily commutative) and the category Mod R of right R-modules.

By proposition II - 12, any formal initial seg-

ment of Mod R can be completely characterized by some Heyting subobject of R in MOdR, i.e. by some (special) right ideal of R. segment, this ideal I is U T U* R.

If U is the formal initial

In this paragraph we will prove that I is

a 2-sided left-pure ideal of R. Definition ] (cfr. [5]). A submodule B >--~ A in the category of left R-modules is called r~$ght pure if, for any right R-module M the canonical morphism M©B

,

~

M®A

131

is injective.

Before stating the next proposition, observe that any morphism O ÷ A is injective in Mod R (cfr. definition II - 6).

Proposition

2.

If U is a formal initial segment of Mod R characterized by the ideal I = U! U* R, the objects of the full subcategory U are those modules A such that the canonical morphism A I + I is an isomorphism; moreover

U*(A) : A I. All modules and ideals are defined on the right. canonical inclusion.

U r : U~-+MOdR is the

U* : Mod R + U is its right adjoint and U, : U ÷ Mod R

is right adjoint to U*.

U! and U* are thus cocontinuous and U~ U* R = I.

We know that I R = I, since I is a right ideal.

Consider a free module F(K) on a set K of generators.

U, U* F(K) = U, U* F(J_[_ 1). K

F1) K .EL LI~ u* F1 K .~U! U*R K

= U, U*(]l.

11. I K (ik)k£K

ik £ I is zero except for a finite][ d nfimber of indexes

Now any module A is a quotient of the free module FA.

We then have the

following commutative diagram U i U* (FA) >

U! U*(z)

+

t

[ v

v

U! U*(A) >

÷

and U! U*(A) is the image of z U! U*(FA) >---~ FA where z is defined by :

FA

~> A

A t

132

FA =

(ra)aE A

Z((ra) a ~ )

I r a E R is zero except for a finite number of indexes

= z a . r . aC~ a

The description of U! U*(FA) implies that U~ U* A = ~ •

[a

z EA

=A.

So U* A = A

.

a . i

ia E I is zero except for a finite number a

of indexes

}

I.

I.

Now, a module A is in U if and only if U! U* A = A

(cfr. [23]), i.e. if and only if A . I = A.

This concludes the proof.

Corollary 3.

Let I characterize the formal initial segment U of Mod R-

Then I is

2-sided. I = U! U* R = R I (proposition 2)

Proposition 4.

Let I characterize the formal initial segment U of Mo~.

Then for every

finite family ( i k ) ] ~ n of elements in I, there exists ~ E I such that for all k we have ik s = ik. n J~ I generated by (il, ..., in). k= ]

Consider (ii, ..., in)R the submodule of Then

n

(i I .... , in)R . I = (i I .... , in)R N U! U*( ~ I) k=1 n

(i I .... , in)R N

J~ (U! U* I) k=] n

= (ii, ..., in)R N ( i[ I) k=1 = (il, ..., in)R. Therefore we can write (i l, ..., in ) = £Z (i~ .... , in)r £ j£

= (il, ..., in) . (~ r£ j£)

133

m

where r£ c R and j£ E I; thus E = z rZ j£ E I by corollary 3. L Proposition S.

Let I character~2ze the formal initial segment U of ModR.

Then I, as a

left module of R, is right pure. Consider any A C Mod R and the canonical morp_hism A ® I + A © R g A which sends a © i to a i.

We must prove it is injective (definition I).

Take

a k • i k E A ® I such that Z ak ik = O. By proposition 4, choose e in I such k k that for any k, ik . ~ = ik. This implies that z ak ~ i k = z (ak ® i k . ~) k k = z (a k

. i k © ~)

k = (z ak . i k ) ® E k =0©~ =

O.

This concludes the proof.

§ 3. FORMAL INITIAL SEGMENT ASSOCIATED TO A PURE IDEAL

Again we work in the category Mod R of right R-modules on the ring with unit R.

We fix a 2-sided ideal I of R which is right pure when we regard it

as a left submodule of R. MOdR, starting from I.

We will construct a formal initial segment U of

Since I is 2-sided, A ®

I is a right module for any

A E Mod R (cfr. [20]).

Proposition 6.

Let I be a 2-sided ideal of R. I is right pure if and only if, for any A E MOdR, the canonical morphism A © I ~ A I is an isomorphism. The morphism A ® I

÷ A ®R

= R has image A I; thus it is injective if and

only if A © I ÷ A I is bijective.

134

Proposition 7. Let I be a 2-sided right pure ideal of R.

Then for every finite family

(ik)1~c~n of elements in I, there exists ¢ 6 I such that for all k we have ik ~ = ik. Consider the cokernel (i I ..... in)R

~ Rn

Rn 7> ~i I ..... in)R

~ O.

Tensoring on the right by I preserves cokernels; by proposition 6, we then obtain (i I, ..., in)R . I

~ Rn . I

~ >

(ii, ..-, in--~.R

o

I

+0

and since I is 2-sided : (i I,

. in)I ---+ in .. ,

in 7> (ii, ..., i n ) R - - + O .

But (il, ..., in) 6 In is sended to 0 in the quotient; thus it comes from some element in (i I .... , in)I : (i I, ..., in ) = (il, ..., in). ¢ with ¢ 6 I. Proposition 8. Let I be a 2-sided right pure ideal of R. A I = {a i ] a E A ;

An element i n A

For any A £ Mod R we have

i 6 I; a i = a}.

I has the form z a k ik w i t h

a k £ A and ik 6 I.

By proposi-

tion 7 choose E 6 I such that for an~ k, ik s = ikkZ a k ik = ~ (ak ik c) = (Zk ak ik)s = a E, with ~ 6 I.

I

Proposition 9. Let I be a 2-sided right pure ideal of R.

The full subcategory U of blodR

whose objects are those modules M such that M = M I is a formal initial segment U of Mod R.

Moreover I = U! U* R.

We denote the inclusion by U! : U ¢-+ Mod R. U* = - ® I.

A ¢ mdR

We define U* : Mod R +

U by

This makes sense since I . I = I by proposition 7 and thus for

135

(A © I ) . by proposition 6. right module.

I =A

. I . I =A

. I =A©I

On the other hand, A © I is a right module since I is a

The adjunction U! ~

U* holds : indeed, consider A E ~ d R and

B £ U; we must prove there is a bijection ( B , A) ~

(B, A . I ) .

Any morphism B + A.I is in particular a morphism B ÷ A . I c_A.

On the other

hand, if f : B + A and b E B, then b C B = B . I and thus b = Z b k i k with ik E I; so f(b) = z f(bk) ik E A . I and f factors through A .kl. right adjoint to U k.

Thus U* is

But U* = -® I has itself a right adjoint U, = (I, -) because

I is 2-sided (cfr. [20]).

We still need to prove condition (F 3 - 4 - 5) of definition II - 6. (F 3) means that if B is in U, any submodule A of B is in U. a is in B = BI thus a = ai with i C I (proposition 8).

Indeed take a E A;

So a is in AI.

(F4)~s obvious : the canonical morphism A I + A is simply the inclusion, for any A £ MOdR.

It is also a Heyting sl~object (F4).

Indeed, (~3) is satisfied

since Mod R is an abelian category (cfr. [21] - 14 - 2 - 7).

Now if S and T

are any submodules of A, the inclusions (A I N S) U (A I N T) c A

I N (S U T)

(S N T) O (S N A I) ¢ S N (T U A I) certainly hold because they do for any subobjects.

Now take a £ A I N (S U T).

From a E A I, we deduce a = a i with i C I (proposition 8) and from a E S U T, we deduce a = s + t with s C S and t C T.

Thus

a = a i = (s + t)i = s i + t i. But s i E S because S is right sided and s i C I because I is left sided; so s i £ S N I and in the same way t i C T n I. This proves

(H3).

Finally a is in (S N I) U (T N I).

We verif~f (H4) in an analogous way : choose s in S n CT U AI).

From s E T U A I and proposition 8, we deduce s = t + a i; t E T; a E A; i E I. From proposition 7 choose e ~ I with i E = i.

So

s=t+ai = (t - t ~) + (t + a i)e = t(1

Now we have s ~ E S I c S N A

-

~) + s ~ .

I.

On the other hand, t(l - ~) E T since t E T

and t(] - E) = s - s ~ E S since s E S; so t(] - c) E S n T.

Therefore, s is in

136

(S n T) U (S n A I).

This proves (H4).

Finally U! U* I = I ® I

= I . I = I by propositions 6 - 7.

§ 4. PURE SPECTRA OF A RING

For the theory of R-modules, the results of §§ 2 - 3 imply that a pure ideal in the sense of definition V - 3 is simply a 2-sided right-pure ideal of R. This yields an easier description of the pure spectrt~ of the theory of right R-modules : we call it simply the right pure spectrum of R.

Proposition 10.

If R is a ring with unit and ~F is the theory of right R-modules, the pure ~F-ideals as in definition V - 5 are exactly the 2-sided right-pure ideals of R. A n y pure ]~-ideal is a 2-sided left-pure ideal by corollary 3 and proposition 5; the converse is true by proposition 9.

Theorem 11.

Let R be a ring with unit.

Consider the set r-pp(R) of 2-sided right-pure

ideals J such that, for any 2-sided right-pure ideals If, 12 11 n I 2 cJ~

I l _cJ

or 12 c J.

For any 2-sided right pure ideal I define 01 = {J 6 r-pp(R) I I ~_ J}.

The subsets 01 constitute a topology on r-pp(R).

This space is called

the right pure spectrum of R; it is compact. By propositions VI - 10, V - 14, V - 17.

Thus for a ring R, we have defined two different spectra : the right pure spectrum of R and dually (working with left-modules) the left pure spectrum of R.

These two spectra are generally not homeomorphic; a counterexample will be

given in § VII - 4.

137

§ 5. PURE REPRESENTATION OF A MODULE

If A is a right R-module, theorem V - 19 presents A as the module of global sections of some sheaf ~A of right R-modules. easy description AA.

The results of §§ 2 - 3 produce an

From this description of AA, it follows irm~ediately that

AR is in fact a sheaf of rings and that ~

is a sheaf of modules on the sheaf

of rings AR.

Theorem 12.

Let R be a ~ n g with unit and Mo h

the category of right R-modules.

For any 2-sidedright pure ideal I of R, define

AR(01) = MOdR(l, I). AR is a sheaf of rings on the right pure spectrum of R; R is isomorphic

to the ring of global sections of AR. If U is the formal initial segment of M o ~

generated by I, we have

MOdR(l, I ) ~ U(I, I) U(l, u* R)

ModR(I, U* R) U, U* R. as follows from the considerations of proposition 9.

T h ~ AR is exactly the

sheaf considered in theorem V - ]9 and proposition II - 14.

The composition

of linear endomorphisms makes ModR(I , I) into a ring and thus AR into a sheaf of rings.

Theorem 13.

Let R be a ring with unit and A a right R-module. For any 2-sided r~ght pure ideal I of R define AA(OI) = ModR(l, A).

AA is a sheaf of right modules on the sheaf of rings AR; A is isomorphic to the module of global sections of AA. If U is the formal initial segment of Mod R generated by I, we have

138

ModR(I, A) ~ Mo~(U! I, A) U(I, U* A) MOdR(I , U* A) % U, U* A as follows from the c o n s i d e r a t i o n s of p r o p o s i t i o n 9.

Thus AA i s e x a c t l y the

sheaf considered in theorem V - 19 and proposition II - 14.

Moreover

MOdR(I , A) is a right module on MOdR(I, I); the scalar multiplication is given by

(f, g) [

~ f o g.

•

CHAPTER 7 : PURE REPRESENTATION OF RINGS

In chapter 6, we obtained, from the general theory of formal initial segments, the description of the pure spectra of a ring R a n d the corresponding representation theorems for R a n d any R-module.

The object of this chapter is

twofold : we intend to study more deeply pure ideals and the representation theorems; on the other hand, we want to give a direct treatment of what has been done in chapter 6, i.e. a treatment independant of the theory of T-ideals. However we insist on the fact that all the results of chapter 6 have been discovered first from the general theory of formal initial segments; the direct algebraic treatment came later.

We work on an arbitrary ring with unit R, not necessarily com~mtative. When nothing is specified, '~aodule" and "ideal" always mean "right R-module" and "right R-ideal". R-linear mappings.

We denote by Mod R the c a t e g o ~ of right R-modules and If M and N are two modules, (?4, N) denotes the set of linear

mappings from M to N.

Several notions of "pure submodule" can be found in the litterature.

In

the case of a 2-sided ideal, they turn out to be equivalent; this is what we prove in § I.

The definition we adopt is the one wich appears to be most useful

in the proofs : a pure ideal of R is a 2-sided ideal I of R such that for every i E I, there exists an element ~ E I such that i . a = i. dual notion with ¢ "unit" on the left.

There is clearly a

In § I, we describe also some basic pro-

perties of pure ideals and in § 2, we give examples.

A spectrum of a ring R is some topological space associated to the ring R and whose topological properties reflect some aspects of the algebraic structure of R.

For example, Grothendieck's spectrum is constructed from the prime

ideals of R, Pierce's spectrum is constructed from the iden~otents of R, and so on ....

In § 3, we propose a spectrum of R - we call it the pure spectrum

of R - constructed from the pure ideals of R; it is always a compact (not necessarily Hausdorff) sober space.

As the notion of pure ideal can be defined on

the left and on the right, we obtain in fact two different pure spectra of R : a right one and a left one; they are generally not homeomorphic.

In § 4, we

give some examples and counterexamples.

If X is some topological space, a sheaf of rings on X can be regarded in two ways : to any open subset U of X, we assign a ring F(U) in such a way that

140

certain restriction and glueing conditions are satisfied; or we consider a local homeomorphism p : F ÷ X such that the family (p-1 (X))xC X is a continuous family of rings.

The correspondance between the two definitions comes from the fact

that F(U) is isomorphic to the ring of

continuous sections of p on the open

subset U.

The continuous sections of p on the total space X are called the global

sections.

The ring p-1(x), for x C X, is called the stalk of the sheaf at x.

When a spectrtm~has been defined for a ring R, one tries generally to construct a sheaf of rings on this spectrum in such a way that R is isomorphic to the ring of global sections of this sheaf.

This process is interesting as soon

as the stalks of the sheaf have additional properties : in Grothendieck's case, they are local rings; in Pierce's case, they are fields as soon as the ring is von Neumann regular.

Thus, for example, the study of a regular ring can be

reduced to the study of fields if one accepts to replace a singlc ring by a sheaf of rings. In §§ 5 - 6, we propose two different sheaf representations of a ring R on its pure spectrum.

The first representation is easily described as a mapping

on the open subsets of the spectrum via the rings of endomorphisms of the pure ideals.

The second representation has the advantage that the stalks of the sheaf

are quotients of the ring R. theorems for R-modules.

At the same time we give analogous representation

In chapter 8, we shall study the rings for which

these representations have nice properties : these are the Gelfand rings; in particular, both representations will coincide for Gelfand rings. § 7 is merely a counterexample.

We show that Pierce's method for construc-

ting a sheaf representation in terms of "espace ~talg" does not work in general when dealing with the pure spectrum.

In fact our representations of §§ 5 - 6

both coincide with that of Pierce in the case of regular rings (see chapter

8).

But in general the pure spectrum is richer than Pierce's spectrum and the representation theorem splits into two different results. Finally in §§ 8 - 9, we look at what happens to pure ideals and the pure spectrumwhenwe

let the ring R vary-. We find that finite products of rings

co~mute with the construction of pure spectra.

On the other hand, we need the

comnutativity of the ring to prove that a ring homomorphism induces a continuous mapping between the corresponding spectra.

141

§ I. PURE IDEALS OF A RING

Let R be a ring with unit. valent definitions.

We define a (right) pure ideal an give equi-

We prove some properties of pure ideals.

Definition t . A (right) pure ideal of R is a 2-sided ideal I o f R such that for every i 6 I, there exists an element s C I such that i c = i.

Again we use the convention that, when nothing is specified, "right pure". of i 6 R.

"pure" means

In the same way, Ann i = {r I i r = O} is the right annihilator

Several aspects of the following proposition are well known (cfr. [5]).

Proposition 2. The following conditions are equivalent for a 2-sided ideal I o f R

(]) I is pure (2) V i 6 I

3 s £ I

(3) V il, ..., in 6 1 (4) V A 6 M o d

R

(5) V A 6 Mod R (6) -~I

i = is 3 s 6 1

A®I

~A

V k

i k s = ik

. I

A ® I -~ A ® R is injective

is a left flat module

(7) for any ideal J~ J 0 I = J . I (8) V i 6 I

I +Ann

i = R.

(I) ~=~ (2) by definition I. (2) ~ (3) by induction on n.

Clearly,

(3) ~ (2) is obvious.

(3) is valid when n = 1 (by (2)).

(3) is valid for n and let i I, ..., in+ I be n+1 elements in I.

We will nrove Now suppose that Choose

s 6 1 such that in+ I s = in+l, 6 I such that for k = ], ..., n

(ik - ik s)~ = (ik - ik s).

This implies that in+ ] ( s + ~ - s e) = in+ ] s + in+ ] ~ - in+ 1 s ~p = In+ I + in+ I • - in+ I = in+ I

and for k = ], ..., n

142

ik (~ + ~ -

e ~) = ik a + ik ~ = ik ~ +

- ik ~

(i k -

ik e)~

= ik ~ + ik - ik a =

ik.

Thus E + ~ - ~ ~ E I satisfies (3) at the level n + I.

To prove (3) ~ (4), observe there is a canonical linear mapping A®I÷A

. I ; a®i

[-+a i

and (4) must be understood as the fact that this mapping is an isomorphism. It is clearly surjective as any z ak ik E A. I is the image of k z ak @ i k E A ® I. We will now show that this mapping is injective. If k z ak ® i k is sended to O, i.e. if z ak ik = O, choose ~ E I such that for any k k k, i k = i k

¢

Then, kZ ak ® i k = ~ (ak ® i k e)

(ak ik ® ~)

= z

k =

(z

ak ik

®

k

=

O.

This proves the injectivity and finally the isomorphism A ® 1 % A . I.

Conversely suppose (4) to be satisfied and for any i E I, consider the exact sequence of modules 0

~ i R

R ~ I-R

,R

~ O.

Tensoring with I, we obtain an exact sequence i R®I

'

'

" R®I

R

~ -i--~ @ I

÷0

or, using (4) iRI

~RI

R

~iR

I

70.

But I is 2-sided, thus we obtain iI

~I

I -

IR

~ O°

143

I Now i £ I is sended to O in ~--~, thus by exactness of the sequence, it is the image of some element z ii k E i I. k

So

i = z i ik = i(~ ik) k and z ik £ I. k

So I satisfies (2) and we proved the implication (4) ~ (2).

It suffices to consider the factori-

The equivalence (4) ~=~ (5) is easy. zation, valid for any module M, M®

I -+> M . I >

)M.R=M.

The first mapping is thus an isomorphism if and only if it - or equivalently the composite - is injective.

But this is the equivalence (4) ~=~ 5.

have already proved (I) ~=* (2) ~=~ (3) *=~ (4) *=~ (5).

Thus we

It should be pointed

out that (5) is simply Cohn's definition of a pure left-submodule I >-+ R (cfr. [5]).

We will now prove (2) ~ (7).

If J is right sided, J I E J and since I is

left sided, J I c_ I; thus J I ~ J such that i = i~.

N I.

Now take i E J N I and choose ~ £ I

We have i E J and E £ I, thus i ~ = i E I . J.

suppose (7) to be satisfied and choose i £ I. iR=iRNI

=iR.

Conversely

From the equality I =iI,

we deduce that i = i . I ~ i R is in i I, thus i = z i i k = i(~ ik) k with ik E I and thus z ik £ I; this is (2). k Now we must prove that (6) is equivalent to the other conditions. I is pure. J ® ~I

~I

Suppose

is a left flat module if for any right ideal J, the morphism

÷ R ® ~ I is injective. 0

Consider the exact sequence of left modules

~I

......~ R

~I

~0.

Tensoring by J, we obtain an exact sequence J®I

, J®R

~ JQT

R

'

~0

or equivalently 0

) J

N

I

....

,J

~ J®ll/~

/I

~ O.

144

This proves the isomorphism

~ J®Yl

J Jn

I"

Finally we need to show that J JnI

I'

is injective, which is obvious.

Now suppose that ~ I is a left flat module and choose i E I. Tensoring i the injection i R >--~R with the left flat module Y I ' we obtain an injection i R ®

YI -

R® %

Now any generator i r ® s o f i R ® y i By i n j e c t i v i t y , i R ® Y I = (0). 0

..... ~ I

YI.

i s sended to i r s £ I i n R/I, thus to O.

Consider the exact sequence ,R

~YI

~0.

i

Tensoring with i R we obtain an exact sequence i R@I

, i R®R

, i R®y

I

~, 0

> iR

>0.

or equivalently iR@l This proves that the mapping i R ® I

÷ i R is surjective; thus i = i.I E iR

is the image of z i r k ® ik; so k i = Z i r k i k = i (z r k ik) k k and Z r k ik E I. k

This proves

(2).

Finally we prove the equivalence

(2) ~=~ (8).

If I satisfies

(2) and i £ I,

the annihilator of i is the ideal defined by Ann i = {r E R I i r = O}. Choose ~ £ I such that i c = i; this implies i(I - ~) = O and thus I - e £ Ann i. Therefore we have I +Anni=R because c E I and I - e E Ann i, thus I = (I - ~) + E E I + A n n i . Conversely if

145

I +Anni=R then for i E I, we can write e+r=] where e C I and r £ Ann i.

Multiplying both sides by i, we get iE+ir=i.

But i r = 0 since i r E Ann i.

Thus i = i E.

Proposition 3.

Let I be a pure ideal of R and r C R. r E I

~=~

One implication is simply

Then

I +Annr

= R.

(8) in proposition

I +Annr

Now if

= R,

write I = e + ~0with ~ £ I ands0 £ Ann r. r = r ~ + r~o=

2.

Multiplying by r, we obtain

r ~ + O = r ~ £ I.

Proposition 4.

Any (right) pure ideal is a left flat module.

For any injection S >--+ A of (right) modules, we must prove the injectivity of S ® I ÷ A ® I; but this is simply the inclusion S I ~ - + A

I (proposition 4).

Proposition 5.

If R is a commutative ring with unit and I a pure ideal, then the ring (I, I) of linear endomorphisms of I is con~nutative.

Choose f, g two linear endomorphisms

of I.

For any i £ I, choose ~ E I

such that i ~ = i. (f o g)(i) = f(g(i ~)) = f(i g(E)) = g(c) f(i) (g o f)(i) = g(f(i ~)) = g(e f(i)) = g(E) f(i).

Proposition 6. Let I be a pure ideal and Jl, J2 two ideals. I + J1 = I + J2 % I NJI

=INJ

2

J

~ J1 = J2-

Then

146

Take a £ J1 ~ I + Jl = I + J2.

We can write a = i + j with i £ I, j £ J2.

Choose E such that i E = i.

a s = (i + j)e = i ~ + j ~ = i + j ~. Therefore i = a E - j c E I A J I = I A J2_cJ2. So a = i + j 6 J2 and thus Jz _c J2.

Proposition

Conversely J2 c_J1.

7.

(0) and R are pure ideals of R. Any sum and any finite intersection of pure ideals is a pure ideal.

0 is a

unit in (0) and I is a unit in R so (0) and R are trivially pure.

Let (Ik)k6 K be a family of pure ideals of R.

An element in + Ik has kEK n the form z iI where iI 6 Ix . We will show, by induction on n, that there /=I n n n exists some ~ 6 + Iko such that ( z i/)E = z i 1. If n = I, i I 6 Ikl and l=I ~ l=I l=I thus there is E 6 Ikl such that i I . ~ = i I. for n.

Now suppose the result is true

To prove it for n + I, choose ~ 6 IIkn+

such that in+ I . ~ = in+ I.

Consider also n Z

1 =1

n and choose ~ 6

+ /=I

n iI -

iI ~ 6

1+=i

I1

I 1 such that n ( ~

1=I

n iI -

i I ~)~ =

~

1=I

i I - i I E.

n+ 1

We have ~ + ~o - ~ ~0 6

+ I1 and i=I n+ I ( z i/) (~ + ~ - c ~ ) /=I

n =

( Z

/=I

n iI

-

i I ~)m +

Z

/=I

iZ c + in+ ] ~ + in+ I ~ - in+ I e

m

147

n = ( Z

i£ -

/=1 n+l =

N~,

Z i£. £=I

t~e

i ~ = i, i ~

n n Z= i£ ~ + Z i£ ¢ + in+ I + in+ 1 ~o - in+ I ~o 1 1 /=I

I, J p u r e i n R a n d i

= i.

~en

C I N J.

~oose E £ I andw£

J su~

that

i E ~ = i and ¢ ~ = I J = I N J.

•

Proposition 8.

Any ideal I contains a largest pure ideal.

We call it the pure part of

o

I; it is denoted by I. o

I is simply the sum of all pure ideals contained in I.

Such ideals exist

(at least (O)) and their sum is still a pure ideal (proposition 7); it is

I

obviously the largest pure ideal contained in I.

Proposition 9.

Let I, J be two ideals and (Ik)kE K a fcsnily of ideals. o

~

~--~

I nJ

Then

o

=

n j,

o

o

k£K Ik D + I k. ken

We have ~ ~ I and ~ J ,

thus

~n~c_InJ and ~ O ~ is pure by proposition 3.

This proves o

N o

cInJ. o

o

Conversely I N J c I N J c_I and I N J is pure; this prove I O J c

and in

o__

the same way, I O J c__ . Finally o o

INJ

o

cINJ.

o

Since k£I+ Ik is a pure ideal (proposition 7) contained in k£K+ Ik' the second

148

relation follows immediately from the definition of pure part.

An ideal is generally not the intersection of the maximal ideals above it, Moreover, the pure part of an arbitrary intersection of ideals is generally not the intersection of the pure parts of the ideals.

However, the following result

holds :

Proposition 10. Let I be a pure ideal.

Then o

I =nM=n

where the intersections are over all 2-sided (resp. right) maximal ideals M containing I.

The following proof works in both cases of 2-sided or right ideals. I _¢M implies I c N M and since I is pure, I c _ 6 1 . o

On the other hand, n M ~ M

o

ir~lies fl M c ~ M

and finally N M E N

. So it suffices to prove the inclusion

NMcI. o

If fl M_¢ I, choose a C n

x I.

From proposition 3, we deduce I +Anna~

R. o

Choose a maximal N containing I + Ann a, and thus I.

We have a E n M, thus

o

a E N and by proposition 3 N+Anna=R which is a contradiction since N and Ann a are in N. Proposition 11. Let A be a module and I a pure ideal. A I = {aCA

a

=

[ 3 c C I

Then a = a ~}.

Clearly each a c with a E A, ~ E I is in A I. Conversely consider n z ak ik £ A I and choose ~ E I such that for any k, ik ~ = ik. k=1

149

a = z a k i k = z (ak i k E) = (z a k ik)E = a c. k k k Proposition

12.

Let A and B be two modules and I a pure ideal.

Any linear mapping

A I ÷ B factors through B I.

Take f : A I ÷ B a linear mapping. that a = a E (proposition

II).

For any a E A I there

Therefore

is E £ I such

:

I

f(a) = f(a e) = f(a) E C B I.

Proposition

13.

Let A be a module,

( S k ) ~ K , S, T submodules of A and I a pure ideal.

Then

A I n ( + Sk) = + (A I n Sk) kEK kcK S

N (A I + T) = (S N A I) + (S N T)

A I + S

(S n T) = (A I + S) N (A I + T)

+ (A I R T) = (S + A I) n (S + T).

The inclusions + (A I n Ski c_A I n ( + Sk) kEK kEK (S N A I) + (S n T) c are obvious

and valid for any submodules.

S

N (AI+T)

We will now prove the converse

inclusions.

Take a E A I N ( +

Sk).

By proposition

"11, we can write

a = a s with

kcK ~£

= s I + ... + s n with s£ £ Sk£.

landa

a = as= wi~

s£~

£A

Therefore,

s I ~ + ... + s n

I n Sk£.

Now take s £ S n (A I + T). w i t h a C A, i E I, t E T.

Choose

s ~ =aie+t

By proposition

11, we can write

E £ I such that i s = i. E =ai+t

~

This

s = a i + t implies

150

and ~ s=s But s e E A

I N S andt-

t

E + t-

c = s -

t

c.

s c E S NT.

The last two relations can be formally deduced from the preceding ones, without going back to the definition of a pure ideal.

Indeed, for the third

relation, we have (A I + S) N (A I + T) = ((A I + S) N A

I) + ((A I + S) N T)

= A I + ( ( A I N T) + (S N m))

=AI+

(SNT)

and for the last relation (S + A I) N (S + T) = (S N (S + T)) + (A I N (S + T)) = S + ((A I N S) + (A I N T))

= S + (A I N T). Proposition 14. Let I be a pure ideal and A a module

CAI

= {aEAI def

¥iEI

ai=O}

is the largest submodule of A whose intersection with A I is zero.

Consider the submodule S=

where all T are submodules of A.

+ AINT=

T (0)

Using proposition 13, we have

A I N S =A

I N ( + T) A I N T = (0)

=

+

(AI NT)

A I N T = (0) =

(o).

Thus S is the largest submodule of A whose intersection with A I is zero. We must prove the equality S = C A I.

151

Take s E S and i E I.

Then, s i ES

thus s i = O.

NA

Conversely take a £ C A

I = (O) I : we will show that

a R n A I = (0). If x £ a R n A I, by proposition

11,we can find e E I such that x = x e; but

we can also find r E R such that x = a r. r ¢ E I; so x = O.

Finally x = x e = a r e with

But this proves the inclusion a R c

S and thus a E S.

Finally S = C A I, which concludes the proof.

Proposition 15. Let I, J be two pure ideals, module. (I)

(Ik)k6 K a family of pure ideals and A a

Then,

ACI
(2) (I < J )

~ (CA

1 I>CA

J)

(3) C A ( + Ik) = N C (A Ik) k£K

kEK

(4) C A (I [I J) > C

A I + CAJ.

If r £ C I, then for any i £ I, we have r i = O and for any a £ A, we have a r i = O.

Hence, a r is in C A I.

This proves (]).

The ~secend relation is obvious. To prove the third relation, consider CA

( + Ik) = {a C A k6K

IV i £

= {a 6 A l V k = =

n {aEA kEK n

k£K

6 K

+ Ik, a i = O} k6K ¥ i C Ik,

kI V ,i ~ I

ai=

a i = O}

O}

C A 1 k.

Finally the second relation implies C A(I N J) >~C A I and C A ( I this implies the fourth relation.

Proposition 16. Let I be a pure ideal of R.

N J) >~C AI; •

152

C I is the left annihilator of I; it is a 2-sided ideal.

Indeed, by definition L-Ann I = {r 6 R j V i 6 1

r i = O} = C I.

By definition, C I is a right ideal (proposition 14) and L-Ann I is a left I

ideal (obvious); thus C I is a 2-sided ideal.

Prpposition 17. Let I be a pure ideal of R.

o

There is a largest pure ideal C I whose intersection with I is (0). o

From propositions 14 and 8, it follows that C I is simply the pure part of C l . Proposition ]8. The assignment o

o

I F--* C C I is a closure operation on the lattice of pure ideals of R. o

0

For any pure ideal I, we will denote C C I by I. a) R = R since C R = (0). o

b) I < T

since I N C I = (0) 0

0

o

o

c) T = T since C C C I = C I. 0

o

0

Indeed, by b) it suffices to prove the inclusion

o

o

0

o

o

C C C I ~
o

o

o

0

This l a s t e q u a l i t y holds since C C C I fl C C I = (o) and I c_ C C I. o

o

o

o

0

o

d) I < J * Y < 7 since I ~< J implies C J < C I and thus C C I ~< C C J. e) I n J = I A J.

Indeed, by d) we have I n J < T

and~<7.

So,

~< I A J andoit remains to show that the converse inclusion holds. By definition of C(I N J) we have o

C(l n J) n I n J = (0). o

0

0

From the definition of C J and C C J, we deduce o

o

C(I n J) fl I <~ C J o

C(I n d) A I n C

o

a < Cd NC

d = (0).

153

o

o

o

By definition of C I and C C I, this implies o

o

o

o

C(I n J) N C C J ~
.

C(I n J )

n o

Jn

.

CC

.

.

I~C

.

I N CC

I = (0).

o

Again by definition of C C (I N J), we obtain o

o

which is simply INJ~
Some

examples are quite trivial (an ideal with a unit), some are more typical (like continuous real functions which are zero at the neighbourhood of some point .).

In the non cor~autative case, we produce examples of ideals which

are pure on the left and not on the right and examples of ideals which are both left and right pure. Example 19.

A 2-sided direct sun.and is left pure. Suppose I is 2-sided and J is right-sided, with I + J = R and I N J = (0). Consider I = s + ~ w i t h

e E I a n d ~ E J.

For any i E I,

i=~i+~i; but <0 i E J I c J N I = (0); thus i = ~ i. Example 20.

A regular ideal (= generated by its central idempotents) is left and right pure. If (ek)kE K is a family of central idempotents in R, the corresponding generated ideal is n

I = {zi=1 ekl rk'l

n

I

rk'l E R} = {Zi=1 rkm ekl I rkl 6 R}. I

It is a 2-sided ideal.

Now for each k E K, ek = ekek; thus ek is a "unit"

154

for itself (on the left and on the right).

If ek, e I are two central

idempotents e k + eI - ek e I = ek + eI - eI ek C I is a "unit" on the left and on the right for e k and el; for example ek(e k + el - e k el) = e k e k + e k el - e k e k e l =

e k

+

=

ek .

ek eI

-

e k el

Iterating the process, each finite family of central idenmpotents e k , ..., e k has a "unit" on the right and on the left in I.

So if iz].= ekl. rkl. ~ I,

n

a "unit" on the right and on the left is obtained by choosing ~ which is "unit" on the right and on the left for ekl , ..., e k . n Example 21.

Let X be a normal topological space and C a closed subset of X. The continuons functions X + R w h i c h

are zero on a neighborhood of C

constitute a pure ideal in the ring C(X, IR) of continuous functions X ÷ R .

If f : X ÷ k

is zero on some n e i g h b o u r h o o d V of C and g : X ÷ ~

is zero

on some n e i g l ~ o u r h o o d W of V, f + g is zero on V N W and for any h : X ÷ ~ , hf is zero on V.

So the set I of continuous functions X ÷ ~ w h i c h

are zero

on a neighbourhood of C is an ideal.

If f : X ÷~R is zero on some neighborhood V of C, we may suppose V to be open.

As X is normal, we can find a continuous function ~ : X ÷ ~ such that

~(C) = 0 a n d ~ ( C V) = 1.

W = -1

([_ 2' +

]) is a closed neighborhood of C

Which is contained in V.

Choose ~ continuous such that ~(W) = 0 and ~(C v) = 1.

is in I and f c = f because ~(x) = ] as soon as f(x) ~ 0, (cfr. [8] for the results on normal spaces).

Exa~le

22.

Let V be an infinite dimensional vector space on some field K. be the ring of K-linear endomorphisms of V.

Let R

The set I of K-linear

endomor~hisms of V whose image is finite-dimensional is a left and right pure ideal in R.

155

If f : V ÷ V

and g : V +

V are two linear endomorphisms, the image of

f + g is contained in Im f + I m

g; thus it is finite-dimensional as soon as

Im f and Im g are finite-dimensional.

On the other hand, if h : V ÷ V is any

endomorphism and Im f is finite-dimensional,

Im(f o h) c Im f and

dim(Im(h o f))~ dim(Ira f) : so Im(f o h) and Im(h o f) are finite-dimensional. Finally I is a 2-sided ideal.

Now consider f : V + V with finite dimensional image. plementary subspace W of Im f.

Consider a sup-

The linear mapping ~ : V + V which is the

identity on Im f and zero on W has the same image as f and e o f = f; thus I is left pure.

On the other hand, f o ~ is generally different from f because

f o E(W) = (0) where E o f(W) is not necessarily reduced to (0).

Now consider the kernel "Ker f" of f and a supplementary subspace u of Ker f.

Each v E V can be uniquely written v = u + k with u E u, k £ Ker f.

Define <0 : V ÷ V by ~0(v) = u.

The image of ~ is U.

But f is injective on U

since Ker ~ O U = (O); thus dim U = dim f(U) ~
= f(,~(v))

which proves the equality f = f o q0.

=

(f

o ~)(v)

So I is right pure.

On the other hand,

<0 o f is generally different of f since f(v) could - for example - belong to Ker f.

Thus this is an example of an ideal which is left pure and right pure, but the "units" must be choosen differently on the left and on the right.

Example 23.

Let V be a vector space on some field K and v ~ 0 some vector in V. Let R be the ring of those linear endomorphisms eigenvector.

f : V ~ V having v as

Let I be the ideal of those endomor~hisms

v as eigenvector with eigenvalue O.

f : V ~ V having

I is a right pure ideal which is

not left pure.

If f(v) = ~v and g(v) = ~v, then (f o g)(v) = ~vv and (f + g)(v) = = (~ + ~)v.

So R is a ring.

Now if f(v) = O, g(v) = 0 and h(v) = ~v, one

deduces (f + g)(v) = 0 and (h o f)(v) = 0 = (f o h)(v).

So I is a 2-sided

156

ideal.

Let f : V ÷

V be a linear endomorphismwith

mentary subspace to the subspace spanned by v. and ~(w) = w if w E K.

f(v) = O.

Let W be a supple-

Define ~ : V ÷ V by ~(v) = 0

E is an element in I and for any x E V, x = kv + w

where k E K, w E W and f(x)

= f(kv)

+ f(w)

= f(w)

= (f o ~)(x).

So f = f o s and I is right pure.

On the other hand, I is generally not left pure. such that f(w) = v. (~o

and thus <0 o f 6 f.

Indeed, choose f

For any<0 E I,

f)(w) =~(v) = O ~ v - -

f(w)

So I is not left pure.

§ 3. PURE SPECTRUM OF A RING

In § ], we described the lattice of (right) pure ideals of a ring R. In proposition 7, we showed that an arbitrary sum and a finite intersection of pure ideals is again a pure ideal.

]"his property is analogous to the one

satisfied by the lattice of open subsets of a topological space.

In this para-

graph, we construct a compact topological space r-Spp(R), called the right pure spectrum of R, whose lattice of open subsets is isomorphic to the lattice of right pure ideals of R. constructed; § 4).

Dually a left pure spectrum £-Spp(R) of R can be

these two spectra are generally not homeomorphic

(see example 37,

Several proofs of this paragraph could be shortened using general lattice

theory.

Again, when nothing is specified, we work with right ideals and right

pure ideals.

Definition 24.

A pure ideal I of R is called "purely maximal" if it is maximal in the lattice of proper pure ideals. Definition 25.

A pure ideal I of R is called "purely prime" if it is proper and if for any pure ideals If, 12 : I t fl I2 m I ~ I 1 _c I or I2 _c I .

157

Proposition 26. Any purely maximal ideal is purely prime. Suppose I is purely maximal and Ii, 12 are pure with 11 N 12 c I.

If

Ii ~ I, then Il + I = R and by proposition 13 12 = 12 n R = 12 n (I I + I) = (I 2 N Ii) + (I 2 N I)

cl+l I.

=

Proposition 27. The pure part of any maximal ideal is purely prime. o

Let M be a maximal ideal and M its pure part.

Let 1 1 N

12 ~ M with

o

I1, I2 pure.

I f 11 ~M, then 11 ~ M and the r e s u l t holds.

I f 11 ~ M, then

I1 + M = R and therefore 12 = 12 n R = 12 n (I I + M) = (I 2 n Ii) + (I 2 N ~

~M+M cM which implies 12 c M, The pure part of a maximal ideal need not be purely maximal; a counterexample is given in § 4. Proposition 28. Any proper pure ideal is contained in a maximal pure ideal. Let I be a proper pure ideal of R.

Consider the set, ordered by inclusion

X = {J I J proper pure ideal; J ~ I}. I belongs to X and for any J 6 X, I ~ J.

In particular, any directed union of

elements in X is still in X and X is inductively ordered.

By Zorn's ler~na, we

158

can choose J maximal in X.

Any proper pure ideal containing J contains I

and is in X; by maximality of J in X, this ideal is simply J. maximal.

So J is purely

Proposition 29. If I is a pure ideal and a ~ I, there is a purely prime ideal J such that I c J and a ~ J. o

By propositiono]O , there is a maximal ideal M such that M ~ I and a ¢ M; []

by proposition 27, M is purely prime. Proposition 30. Any pure ideal I is the intersection of the purely prime ideals containing I.

[]

By propositions 10 and 27. Definition 31. We denote by p(R) the lattice of pure ideals of R and by pp(R) the set of purely prime ideals of R.

Theorem 32. For any pure ideal I of R define

01 = {J E pp(R) ] I ~ J}. The subsets Oi, with I pure, form a topology on the set pp(R).

Moreover,

the assignment I F-+O I is an isomorphism between the lattice p(R) of pure ideals of R and the lattice of open subsets of pp(R).

0(o)= (J E pp(R) [ (o) _~ J} = 4; thus the empty subset of pp(R) is simply

o (o)" O R = {J £ pp(R) [ R ~ J} = pp(R) since a purely prime ideal is proper.

So pp(R) is simply O R.

159

Consider two pure ideals 11 ,

12 .

0ii N 012 = {J £ pp(R) IIi ~_ J and I2 _¢ J} = {J 6 pp(R) I I1 N I 2 ~ J} = O l I N 12"

These e q u a l i t i e s hold because J i s purely prime. Consider now a family (Ik)k6 K of pure ideals

U kEK

= {J 6 pp(R) I 9 k E K, Ik ~ J}

Oik

= {J £ pp(R) I + Ik ~_ J} kCK =0

+

kCK and

Ik

+ Ik is pure (cfr. proposition 7). k£K Hence, the subsets 01 with I pure constitute a topology on pp(R) and the

proof above also shows that the assignment I ~--~01 preserves finite ^ and arbitrary v.

To conclude the proof, it suffices to show that this assignment

is a bijection between the pure ideals and the open subsets of pp(R).

Thus

we must prove the equivalence Ii = I2 ~=~ Oil = 012 for any pure ideals Ii, I z.

But 01i = 012 means that I i and I2 are

contained

exactly in the same purely prime ideals; thus they are equal by proposition 30. • Definition 33. The pure spectrum of a ring is the set pp(R) of purely prime ideals of R provided with the topolo~ 9iven in theorem 32; it is denoted by Spp(R). Again, when nothing is specified,everything

is defined on the right.

Dually a pure spectrum can be defined working with left pure ideals and left purely prime ideals.

When some confusion could arise, we will use the more

precise notations r-Spp(R) and l-Spp(R). not homeomorphic

(cfr. example 37).

In general, these two spectra are

160

Proposition 34. The pure spectrum of a ring is a compact (not necessarily Hausdorff) space.

Consider a family 01k of open subsets of Spp(R) such that kCKU 0ik = = Spp(R).

This means that

( t h e o r e m 32)

+

I k = R.

Thus 1C

kEK

+

I k a n d we

kCI

can choose kl, ..., kn E K and e i £ Ik. such that I = ~I + "'" + ~n E Ikl + 1 + ... + I k .

But this

implies

R = Ik

n =

01

kl

U

..

+ ...

+ Ik

1 "

U

and finally

Spp(R) =

n

I 0Ikn

§ 4. EXAMPLES OF PURE SPECTRA Having defined left and right pure spectra of a ring R, it is natural to ask whether the spectra are Hausforff and whether the left spectrum is always equal or homeomorphic to the right one.

The following examples will show that

in general, the answer to these questions is no.

In example 36, we also produce

a 2-sided maximal ideal whose left pure part is not pure maximal. Example 35. Example of a non-commutative ring whose left and right pure spectra are equal and homeomorphic to the Sierpinski space. Let V be a vector space (on any field K) with infinite sion.

Let R be the ring of K-linear endomorphisms of V.

those endomorphisms whose image is finite-dimensional. know that I is left and right pure.

countable dimen-

Let I be the set of By example 22, we

On the other hand, R has only three

2-sided ideals : (o), I and R (cfr. [4]). Thus the left pure spectrum of R and the right pure spectrum of R are equal and consist of (o), I and R.

R I purely maximal I (o) thus the spectrum is simply

(o), I purely prime

161

which is the Sieroinski space.

We now recall a proof of the fact that I is the only non trivial pure ideal of R.

Let J be a 2-sided ideal and o # f C J.

dimensional image.

on W and thus W is finite-dimensional. base en+1, ..., ek, ... of Ker g. some m.

Take g C R w i t h finite

Choose a supplementary subspace W of Ker g; g is injective Choose a base el, ..., en of W and a

From f # o, we deduce that f(em) ~ o for

We choose another base f(em), e~, ..., e!1, "'" .

For any. p E ~ * and

v £ V define two linear mappings rp and sv by rp(ep) = em rp(ei)

O, for i # p.

I Sv(f(em)) = v Sv(e~) = O, for i ~ 2. It follows immediately that for any i E Z*

g(ei) =

n Z s o f o rp(ei) p=1 g(ep)

which proves that n g = p=1 z

Sg (ep ) o f o r p

C J.

This proves that any 2-sided non-zero ideal J contains I.

Now suppose J 2-sided and containing strictly I. existence of f C J with infinite-dimensional image.

This implies the So the image of f has the

same dimension as the whole space V and we choose an isomorphism <0 : Im f ~

V.

Choose also a supplementary subspace W of Ker f : f induces

an isomorphism between W and Im f and thus W is also isomorphic to the whole space V.

Choose an isomorphism ~ : V ~-+ W and consider ~ o f o ~ E J.

composite ~ o f o ~ can be factorized in the following way

The

162

f V

I

t

thus it is an isomorphism.

, W

I

L

, Im f

I

I

~ V

So J contains an invertible element and J = R.

This concludes the proof.

This spectrum clearly is not Hausdorff.

It even fails to be T I.

Example 36. Example of a non commutative ring whose left and right pure spectra are not equal but are homeomorphic to the n t~ Sierpinski space.

Take some integer n > 2.

Consider the ring R of (upper) triangular

n × n matrices over some field K.

If A is a matrix, Aij is the matrix whose

(i, j)-entry is the (i, j)-entry of A, while all the other entries are 0. Iij is the matrix whose (i, j)-entry is 1, while all the other entries are 0.

Consider now a left pure ideal J of R and 0 # A £ J.

Supoose aij # 0.

We have I...

A

11

. I..

jj

= A..

6 J.

13

Since J is left pure, there is E 6 J with E . Aij = Aij.

Computing the

(i, j)-entry, we find eii • aij = aij , which implies eii = I.

I.-

ii

= E..

Ii

= I... ii

E

. I..

ii

We deduce

6 J.

Now if 1 ~< i Ill = I/i . Iii E J and using the same process as the one applied to Aij, we deduce Iz£ 6 J. i ~ I££. But if B 1=I is any matrix such that bk£ = 0 for k > i :

Finally each I££ with l ~ < i is in J and thus also Ii =

.B6J.

B=I. i

Finally, if J is a left pure ideal of R and A 6 J with aij ~ O, any matrix B such that bkz = 0 for k > i is also in J.

This proves that the only possible

left pure ideals of R are, for i = 0, I, ..., n L i = {B 6 R I the k t~ row of B is zero for k > i}.

163

In fact, it is obvious that each such L. is a 2-sided ideal; moreover each i L i is left pure since I i is a unit on the left for any matrix in L i. Thus we have described all the left pure ideals of R.

The right pure ideals of R are exactly the left pure ideals of the dual ring of R, i.e. the ring R* which has the same elements and the same addition as R but multiplication

is reversed : A,B

~ B,A. def

In fact, we shall describe a (covariant)

isomorphism between R and R* and this

will suffice to prove that the right pure spectrum of R (i.e. the left pure spectrum of R* g R) is isomorphic to the left pure spectrum of R.

If A is a n × n matrix, we define A T to be the matrix obtained from A by "transposing" around the second diagonal

:

A ~ = (an+1_j, n+]-i)i,j; for example a

b

c

o

d

o

o

T

f

e

C

e

o

d

b

f

0

0

a

]he mapping R*

;

A I

) AT

is such that A TT

= A

(A + B) T = A T + B ~ (A.B) ~ = B T. A T = A T * B T 0T = 0 IT=I. Thus, it is a (covariant)

isomm~phism between R and its dual R*.

the nature of the isomorphism At are exactly

R i = {B E R I the k t-h colum of B is zero for k ~< i} for i = O, 1, ..., n.

Moreover,

~ A T shows that the right pure ideals of R

164

Finally both spectra are isomorphic to the n t~ Sierpinski space but the only ideals which are both left and right pure are (O) and R.

Clearly this

spectrum is not Hausdorff nor TI. Consider also the 2-sided ideal RI, i.e. R l={B 6 R I b11 = 0}. This ideal is obviously left maximal, right maximal and 2-sided maximal.

But

the description of the left pure ideals then clearly shows that the left pure part of RI, i.e. the largest left pure ideal contained in RI, is just (O). This produces an example of a maximal ideal whose left pure part is not purely maximal. Example 37. Example of a non commutative ring whose left a~d right pure spectra are not homeomorphic. Let R be the ring of infinite (~dexed b y ~ ×~) triangular matrices on some field K, with usual addition and multiplication of matrices.

The multi-

plication makes sense because each c o l u ~ o f a triangular matrix has only a finite number of non-zero elements. For any integer i, define L i = {B 6 R I the k t~ row of B is zero for k > i} Rj = {B 6 R I the k t~ column of B is zero for k < j}. It is obvious that L i and Rj are 2-sided ideals. Moreover L i is left pure and R. is right pure. J

Indeed the matrix I E ~ = I if l ~ i

(Cl~)klwhere

~kl is a left unit for any matrix in L.. 1 ( ~ ) k Z ~lere

O otherwise

On the other hand, the matrix

~ = ] i f l ~ j O otherwise

is a right unit for any matrix in R i.

It should be pointed out that there are

J

other pure ideals than L i and R i : for example the matrices of finite rank J

constitute a left and r i g h t pure i d e a l of R (the proof given in example 35 works

165

here); as a consequence the intersection of this ideal with each R. is also 3 right pure. But we don't need these facts to produce our example.

L i is the ideal of those matrices which have non zero elements only on the first row; it is a left pure ideal.

We shall prove that L i is a minimal

left pure ideal and in fact is the unique minimal left pure ideal.

In other

words, we shall prove that any- non-zero left pure ideal contains L I . On the other hand, we shall prove that R does not have any minimal right pure ideal : this will conclude the proof that the left pure spectrum of R is not homeomorphic to its right pure spectrum.

Let J a non-zero left pure ideal of R a n d A 6 J with aij ~ O.

Denote by

Ik£ the matrix whose (k,£)-ent~Dz is I, while all the other entries are O. I

Then

Iii. A.Ijj = Ilj E J.

i] Because J is left pure, there is E E J such that E . Iij = Iij. the (I, j)-entry, we find ell . I = I.

Computing

But

Iii = Iii . E . Ill 6 J. Now for any matrix B E L i , B = lii . B E J. This proves that J contains L I .

On the other hand, if J is a non-zero right pure ideal and A £ J with aij ~ O, then J is not contained in the right pure ideal Rj.

This implies that

J is not minimal among the non-zero right pure ideals.

Finally the left pure spectruTa of R has a smallest non-empty open subset and the right pure spectrum of R has no minimal non-empty open subset.

Thus

these spectra are not homeomorphic.

§ 5. FIRST REPRESENTATION THEOREM

In this paragraph, we describe for a ring R~ a sheaf AR of rings on the pure spectrum of R; R is isomorphic to the ring of global sections of this sheaf AR.

For any- R-module A, we also describe a sheaf AA of modules on the

166

sheaf of rings AR and again A is isomorphic to the module of global sections of &A.

Again, everything is defined on the right.

Theorem 38. Let R be a ring and Spp(R) its pure spectrum.

For any pure ideal I,

the assignment o I ~-~ ( i ,

i ) s AR(I)

defines a sheaf AR of Y~ngs whose ring of global sections is isomorphic to R.

If R is commutative, AR is a sheaf of commutative rings.

In order to have a presheaf, we need to define a restriction map (I, I) ~ (J, J) when 0j ~ 01, i.e. when J ~ I.

By proposition 12, this is

just the usual restriction of a linear mapping f : I ÷ I to the submodule J. Now the linear endomorphisms of I form a ring and the restriction mapping (I, I) ÷ (J, J) is obviously a ring homomorphism.

So we have already defined

a presheaf AR of rings. This presheaf is separated.

Indeed, consider I =

f, g E (I, I) such that for all k, f ik = g ik.

+ Ik in p(R) and kCK

Any element i C I can be

written in the form i = i I + ... + in ; ik C Ik and thus f(i) = f(il) + ... + f(in) = g(il) + ... + g(in) = g(i).

This shows that f = g and AR is separated. Finally AR is a sheaf. fk £ (Ik' Ik) such that fk I1

Indeed, consider I = =

fl Ik"

+ Ik in p(R) and kCK

First we will show that there is no

loss of generality in assuming that the family (Ik)kE K of pure ideals is stable under finite sums or equivalently under binary sums.

Indeed, consider

fk : Ik ÷ Ik and fl : ll ÷ llWhich coincide on Ik N Ik. Ik + I1 can be expressed in the form

Any element i in

167

i = i k + iI ; ik £ Ik ; iI E I 1. Then, we simply define f : Ik + I1 + Ik + I1 by f(i) = fk(ik) + f/(i/). There is no ambiguity in the definition for if we have two such decompositions of i i = i k + iI = i t + i~ then we deduce

ik - it = il- i

n h

and therefore fk(ik - it) = f/(i~ - i/), or in other words fk(ik) + f/(i/) = fk(it) + f/(i~). So f is correctly defined and is obviously a linear extension of fk and f£. Moreover if Im is any pure ideal in the family (cfr. proposition 13). Im n (Ik + I£) = (Im n Ik) + (Im n I£). So, if i E I N (Ik + I£), i can be written in the form m i = ik + i£ ; i k E Im R Ik ; i£ C Im N I£ and therefore f(i) = fk(ik) + f/(i/) = fm(ik) + fm(i/) = fm(i). So we may assume the family (Ik)kE K to be stable under finite sums. Now, if i C

+ Ik, we can write kEK i = i I + ... + in ; i k E Ik.

Thus i is in a finite sum of Ik'S and therefore, by the first part of the proof, we can suppose that i is in some I1.

So we define f(i) = f/(i) and

there is no ambiguity in the definition of f :

+ Ik ÷ + Ik because kEK k£K

two different fl agree on i as soon as f/(i) makes sense.

Finally f is obvious-

ly a linear mapping extending each fk and AR is a sheaf of rings.

168

The ring of global sections of AR is just AR(R) ~ (R, R) which is isomorphic to R itself : the isomorphism sends on element r £ R to the left multiplication by r R÷R

; s ~-~ r s

which is however a linear mapping of right R-modules.

Conversely, each

linear mapping of right R-modules f : R ÷ R is simply the left multiplication by f(1) : f(s) = f(1 . s) = f(]) . s. Now if R is commutative, each ring (I, I) is commutative by proposition 5. If J is some purely prime ideal of R, the stalk of AR at J is the inductive limit of the AR(I) over all 01 containing J, i.e. (AR)j = lira (I, I).

We do not know what this stalk would look like in general, nor what its properties could be.

However, in the case of Gelfand rings (chapter 8), we

shall prove that the stalk (AR)j is just the quotient R/j. Theorem 39. Let R be a ring, Spp~) its pure spectrum and A some R-module. For any pure ideal I the assignment

011--~ (I, A) ~ AA(1) defines a sheaf AA of modules on the sheaf of rings AR.

A is isomorphic

to the module of global sections of AA.

The group of linear mappings from I to A can be made into a right R-module by the rule (f . r)(i) = f(r i) for f E (I, A), r E R and i ~ I. sided and thus r i C I.

This makes sense because I is also left

Moreover, (I, A) can be made into a (I, l)-module

by the multiplication

The assignment 01 F-+(I, A) defines a presheaf of modules on the sheaf

169

of rings AR.

Indeed, if J ( I, any linear mapping f : I + A restricts to a

linear mapping flJ : J ÷ A. ÷ (J, A).

This produces a restriction mapping (I, A) ÷

Now, if ~ E (I, I) is some element of AR(I),

which i m p l i e s t h a t the r e s t r i c t i o n mapping ( I , A) ÷ (J, A) i s l i n e a r . ~

Thus

i s a p r e s h e a f o f modules on AR. We may verify by methods similar to those of Theorem 38 that AA is a

sheaf.

The module of global sections of AA is just AA(R) = (R, A) which is isomorphic to A.

The isomorphism sends a C A to R+A

; r ~ar

and conversely each linear mapping f : R ÷ A is of this form with a = f(1) : f(r) = f(] . r) = f(]) . r.

§ 6. SECOND REPRESENTATION THEOREM

The topic of this paragraph is similar to that of paragraph 5 : we present a ring R as the ring of global sections of some sheaf VR on Spp(R) and we show that any R-module A can be presented as the module of global sections of some sheaf VA of modules on the sheaf of rings YR.

The sheaf representations

proposed in this paragraph differ generally of those described in paragraph 5; however in the case of Gelfand rings (chapter 8) they will coincide. Whereas the sheaf AR was easily described in terms of rings of endomorphisms, vR is defined in two steps : we construct first a presheaf by means of the pseudo complements C I of the pure ideals (proposition 14) and we define vR to be the associated sheaf.

But in the case of VR we are able to give some

information on the stalks : they are quotient rings of R. is defined on the right.

P r o p o s i t i o n 40. Let R be a ring.

For any pure ideal I the assignment

defines a presheaf of rings on the pure spectrum of R.

Again everything

170

In proposition 16, we verified that C I = {r £ R I v i E I, r i = O} is a 2-sided ideal of R.

In particular, ~ C

I is a ring.

If J c I are pure ideals, then clearly C J 2 C I and we deduce a ring homomorphism

This is the restriction map, wich completes the definition of the presheaf []

of rings. Proposition 41.

For a ring R, the presheaf defined by the assignment

°I ~-~ P)/C I is a separated presheaf. Consider I = kCI+ Ik in p(R). whose restrictions to any ~ C

We must show that for all [r], Is] E ~ C

Ik coincide, we have [r] = Is].

I

Or equivalently,

that Jr] g ~l c I is zero as soon as each of its restrictions to !y C Ik is zero. In other words, we must prove that for any r E R, we have (V k E K

r C C I k) ~ (r E C I ) .

This f o l l o w s i m m e d i a t e l y from p r o p o s i t i o n 15 : C I = C ( + I k) = n C I kkEK kEK

[]

Theorem 42.

The sheaf VR associated to the presheaf defined by

is a sheaf of rings on Spp(R); its ring of global sections is isomorphic to R. We recall the definition of the sheaf VR associated to the separated presheaf of proposition 40 (cfr. [9]). For a pure ideal I, consider all the hereditary pure coverings of I, i.e. all the families (Ik)kEKwhere Ik is pure, + Ik = I and such that with any Ik, the family contains any pure ideal smaller kEK

171

than Ik. Now if (Ik)k6K and (J£)£CL are two such hereditary pure coverings of I, the family (Ik N J£)(k,£) C K × L is again an hereditary pure covering of I and is contained in each of the families (Ik)kCK, (J£)£CL" Indeed, Ik N J£ is pure (proposition 7); now if I is pure and I ~ Ik N J£ then I ~ Ik and I ~ J £ and thus I = Ik, = J£, = Ik, N J£, for some k' 6 K and £' £ L. Finally, + (Ik n J£) = k + (+£ ( I k N J £ ) ) k,l = + (Ik n (+ J £ ) ) k £

=+IkNI k =+I k k =I.

Now if (Ik)k6K is an hereditary pure covering of I, a compatible family of elements in ~ C Ik is a family ([rk])kE K of elements [rk] E ~ C Ik such that for any k and £ in K, the restrictions of [rk] and [r£] coincide in ~C Two compatible families of elements on two arbitrary hereditary (Ik N I£)" pure coverings of I are called equivalent if they coincide on some common smaller hereditary pure covering. Now, VR(I) is just the quotient by this equivalence relation of the set of a] 1 compatible families as described above. If [[rk] 6 ~ C ik]k6K and [[s£] £ ~ C are equivalent to the families

II)£6 L are two compatible families, they

I[rk] 6 vC (Ik N I£))(k,£) 6 K x L

[ y ] and these two last families canbe added [s£] 6 C (Ik N I£) (k,£) E K × L or multiplied component by component.

This describes the ring structure of

VR(1).

Now if (Ik)k6K is an hereditary pure covering of I and J ~ I in p(R), we deduce by proposition 13 J = J N I = J N ( + Ik) = + (J N Ik) k6K k£K and from this, it follows inmediately that (J N Ik)k61 is an hereditary pure

172

covering of J.

Thus the assignment

[[rk] 6 ~ C

Ik]k6 K ~-+[[rk] £ ~ C

(J n Ik)]k6K

defines the restriction mapping VR(1) ÷ VR(J). There is a trivial ring homomorphism ~ C I + vR(I) ; [r] ~-+ ([r])p(i) where p(I) is just the family of all pure sub-ideals of I. By proposition 41, this mapping is injective. We must compute the ring of global sections of VR, i.e. the ring vR(R). Consider any element ([rk])k£K in vR(R); thus [rk] E ~ C

Ikwhere

+ Ik = R. kEK

This implies that ] = E

kl

+ "'" + ~k

n

; Ek. E I k

l

i

.

We shall consider the element r = rk

~k + "'" + rk ~k 6 R I i n n and we shall prove that [r] and [rk] coincide in each ~ C that the injection R =

~(o) = ~ C

R +

Ik"

This will prove

vR(R)

is also a surjection and thus an isomorphism. So, we need to prove that vk6

K

r - rk 6 C Ik.

Let us compute this difference in the following way r - rk = rkl

ekl

+ "'" + rkn ~kn - rk

= rkl Ekl + "'" + rkn ~kn - rk(~kl + "'" + ~ k ) n +

= (rkl - rk)ekl

.

.. + (rkn

rk)~k n"

Choose any s E Ik. We have (r - rk)s = (rkl

-

rk)Skl s

+

...

+

(rkn - rk)~kn s.

173

For any £ = ki, ..., kn, the element el s is in I1 Ik = I1 N Ik.

Now the

family (rl)/¢ K is a compatible family, thus rI and rk coincide in Y C in other words, rI - rk E C (I/ O Ik). (rl - rk)Els

(l/ N Ik);

So each term of the stem is of the form

with rl - rk E C (If N Ik) and a£ s E I1 N Ik-

Thus each of these terms is zero and (r - rk) s = O, which implies that I

r - rk is in C Ik. Corollary 43. Any local section of the sheaf VR is locally the restriction of a global section. By definition of vR, for any element x E VR(1), there is an hereditary pure covering I = [rk] in y C

Ik"

+ Ik such that the restriction xk of x to Ik Js an element kEK But an element [rk] in ~ C Ik is just the restriction of the

global element rk E R ~ VR(R). I

This is exactly what the corollary means. Proposition 44. For any purely prime ideal J of R, the stalk of VR at J is the quotient ring of R by the 2-sided ideal U C I, where the union is over all pure ideals I not contained in J. By proposition 16, C I and thus + C I are 2-sided ideals. *~+ C I is a ring.

So the quotient

The stalk at J is defined to be the inductive limit lim VR(1) JEO I =

lim

TR(1).

In other words, the stalk is just the ring U VR(1) where t h e e q u i v a l e n c e r e l a t i o n restrictions.

identifies

a n e l e m e n t i n v R ( I ) w i t h any o f i t s

174

Let x £ VR(I) with I ~ J.

By definition of vR, there is an hereditary

pure covering I = + Ik such that x restricted to Ik is some [rk] in ~ C But from I = + Ik ~

Ik"

J, we deduce the existence of some k such that Ik ~ J .

Thus this particular Ik appears in the union mentioned above and x is equivalent to [rk]. R ~ J.

But [rk] E ~ C

Ik is itself the restriction of rk E R and certainly

Finally, each element in vR(I) is equivalent in the union to some

element in R = VR(R).

This implies that the stalk is just ~ N where the equi-

valence relation identifies an element r to zero if there is some pure ideal I ~ J such that [r] is zero in ~ C

I' i.e. r £ C I.

Finally, define

= {r I ~ I pure ; I ~ J ; r E C I} =

U

I_~-3

CI=

I pure

+

I¢:3

CI.

I pure

We have shown that the stalk of vR at J is just the ring ~

~.

Proposition 45.

If d is a purely maximal ideal of ~

the 2-sided ideal ~ defined in propo-

sition 44 is contained in J. Let I be a pure ideal such that I _~ J.

Since J is purely maximal, J + I = R.

Therefore, by proposition 13,

Cl=

(C I) n R = (CI)

n (I +s)

= (CI

n I) + (CznJ)

= (C I )

n J.

This proves the inclusion C I c J and thus ~ c J. Theorem 46.

Let R be the disjoint union of the rings ~ , all J E Spp(R).

Spp(R) ~

for any

r E R.

where the union ranges over

We provide R with the final topology for all the mappings

~ ; s ~

[rl E a / ~

The mapping p : R ÷ Spp(R) ; [r] C ~

~-+ J

is a local homeomorphism and R is isomorphic to the ring of continuous sections of p.

175

For any pure ideal I and any element r £ R, the composite 01 ~-+Spp(R) ÷ R ; J ~-+[r] 6 ~ is continuous by definition of the topology on R. x 6 vR(I); there is a pure covering I =

Now consider some element

+ Ik such that for any k, k6K

x ik = [rk] 6 R/C Ik" l Consider the mapping 01 + R ; J ~-+[x] £ ~ . For any k, the restriction of this mapping to 0ik Olk~

01 -~ R

; J

I--*[x]

=

is

[rk] £ ~

and we have already proved the continuity of this mapping.

But the 01 's

form an open covering of 01 (theorem 32); thus the mapping 01 ÷ R is c~ntinuous. Finally the topology on R is also the final topology for all the mappings 01 ~ R ; J ~+ [x] E y ~ for any pure ideal I and any x 6 vR(1). pace ~tal~" version of

Therefore theorem 46 is just the "es-

our theorem 42 (cfr. [9]). The ring structure on the

set of continuous sections of p is defined pointwise from the ring structure of each stalk y~. We proceed now to construct an analogous sheaf representation VA for any R-module A. Proposition 47. Let A be a R-module.

For any pure ideal I the assignment

°I ~+ A/C AI defines a presb~af of R-modules on the pure spectrum of R.

If J_c I in p(R), then C A J 2 mapping

CAI and thus there is a quotient linear

-yc A,- -'- y c which produces the structure of presheaf.

176

Proposition 48. For a R-module A, the presheaf

0I ~

~/C AI

defined in proposition 47 is a separated presheaf.

Consider I = + Ik in p(R) and a C A such that for any k, a £ C AI k. By proposition 15 kEK

a~

n CAt k = C ( + CAI k) = C I

kCK which proves the separation.

kEK

Theorem 49. The sheaf VA associated to the presheaf

0I }'-+ y C AI of proposition 47 is, on the pure spectrum of R, a sheaf of modules on the sheaf of rings VR; its module of global sections is isomorphic to A.

_I

An element in VA(1) is thus represented by a compatible family

[ak] £ y C AIkkE K1

of elements for an hereditary vure~ covering (Ik)k£ K of I;

two such families are equivalent if they coincide on some co~aon smaller hereditary pure covering of I. Now an element in VR(I) is represented by a compatible family

[ [r/] E

Y C IlIIEL for an hereditary pure covering (I/)/EL of I. We

define an action of the ring VR(I) on the R-module VA(I) by [([ak])kCK]. [([rg])£EL] = [([am.rm])m~.l] where (Im)mEM is a smaller hereditary pure covering con~aon to (Ik)kEK and (I£)£EL (for example that described in the beginning of the proof of theorem 42). This definition is clearly compatible with the equivalence relations defining VA(I) and VR(I) but we still need to prove that it gives a structure of vR(I)-module on VA(I). The only thing to prove is that for any m E M [am] E ~ C A I m and [rm] E y C

Im ~ [am'rm] E ~ C A I m"

Or in other words a I £ C AI m or rm E C Im ~ am rm £ C AIm • The implication

177

am E C A I m ~ am rm C C A I m holds because C AI m is a right R-module. rm E C Im

The implication

~ a m rm £ C A I m

holds because A . C Im ~ C A.I m (proposition 15). Consider the R-linear mapping A/C AI + VA(I)

which sends

[a] on ([a])p(i) where p(I) is the set of pure subideals

By preposition

48, this mapping is injective.

In particular

of I.

we have an injec-

tion A = y O = ~ C AR + vA(R). We s h a l l prove t h a t t h i s i s a l s o a s u r j e c t i o n and thus an isomorphism.

Consider

an element i n VA(R) r e p r e s e n t e d by a family ([ak])kEKwhere (Ik)kE K i s an h e r e d i t a r y pure covering o f R and [a k] E ~ C AI k" ]

=

ek 1

+

.o.

+ sk

We can w r i t e

; Sk. 6 Ik . n

i

I

Consider the element a

=

+

ak 1 Sk 1

"'" + akn ekn C A,

and f o r any k E K, compute a-

ak = akl Skl + " "" + akn Sk n - ak = akl Skl + "'" + akn ~kn - ak(Skl + "'" + ekn ) +

(a- ak)Skl

...

+ (akn

ak)Skn"

For any s 6 Ik and any £ = kl, ..., kn sg s £ Il Ik = Il N Ik (cfr. proposition 2).

The compatibility of the family ([ak])kCK implies that

[ak] and [al] coincide in A/c A(ll N Ik) ; in other words, a£ - ak E C A ( I £ N Ik). Finally we have for any £ (a£ - ak)E £ s 6 A(I£ N Ik) N C A(I£ N Ik) = O.

178

So for any s E Ik w e have (a - ak) s = 0 and thus a - ak £ C AI k.

So [ak]

is the restriction at Ik of a E A and thus the mapping A ÷ VA(R) is surjective and thus an isomorphism. Corollary 50. In the sheaf VA, any local section is locally the restriction of a global section.

Any element [a] in A/C AI is the restriction of a E A = VA(R) and by definition of VA, any element of vA(I) is locally in some 7 C

AI k.

Proposition 51. For any purely prime ideal J of R and any module A, the stalk of vA at J is the quotient module of A by the submodule + C AI where the sum ranges over all pure ideals I not contained ~n J.

The stalk at J is the inductive limit U

vA(I)/~

_wJ where the equivalence relation identifies an element in vA(I) with each of its restrictions.

But an element in vA(I) is represented by a compatible family

[ [ak] E 7 C A I k ] k E K I =

for an hereditary pure covering (Ik)kc K of I.

+ Ik~_ J, we deduce that for some k £ K, Ik~_ J. kCK

vA(I) is equivalent to [ak] £ 7 C J is just ~ N w h e r e ideal I _ J.

From

So the element in

AI k and finally to ak £ A.

So the stalk at

a E A is equivalent to zero if it is in C AI for some pure

Thus the stalk at J is the quotient ~ J :

U

CAI

I pure

=

+

where

CAI

I pure.

Proposition 52.

^

If J is a purely maximal ideal of R and A a R-module, the submodule J of A defined in proposition SJ is contained in AJ.

If I is pure and I _~ J, we have I + J = R b y maximality of J.

Therefore

179

by proposition 13 C AI = C AI n A = C AI n A(I + J) = C AI N (AI + A J) = (C AI n AI) + (C AI N AJ)

=CAI

nAJ.

^

So C AI c AJ and thus J c AJ. Theorem 53. Let A be a R-module.

Let A be the disjoint union of the R-modules A/~,

where the union ranges over all J E Spp(R).

We provide A with the final

topology for all the mappings Spp(R) ~ A ; J b-*[a] E A / ] for any a £ A .

The mapping p : A + Spp(R) ; [a] E ~

b* J

is a local homeomorphism and A is isomorphic to the module of continuous sections of p.

For any pure ideal I and any element a E A the composite 0 I ~ - * Spp(R) ~ A ; J ~-+ [a] E A / ] i s c o n t i n u o u s by d e f i n i t i o n

o f t h e t o p o l o g y on g.

x C VA(I); t h e r e i s a pure c o v e r i n g I =

Now c o n s i d e r some e l e m e n t

+ I k such t h a t f o r any k, k£K

x ik = [a k] C A/C AI k"

Consider the mapping 01 ÷ A ; J P. Ix] ~ A/~. For any k, the restriction of this mapping to 0ik is Oik~-*0 I ÷ A ; J ~+ [x] = [ak] E A/~, and we have already proved the continuity of this mapping.

But the 01 's k form an open covering of 01 (theorem 32); thus the mapping 01 + R is continuous. Finally the topology on R is also the final topology for all the mappings 01 ÷ A ; J ~+[x] E C J

180

for any pure ideal I and any x 6 vA(1).

Therefore theorem 53 is just the

"espace ~tal~" version of theorem 49 (cfr. [9]). The module structure on the set of continuous sections of p is defined pointwise from the module structure of each stalk in A/t. /J § 7. A COUNTEREXAMPLE FOR PURE SHEAF REPRESENTATIONS In § 6, we have described a sheaf representation vR of a ring R. of "espace ~tal~", the stalk at a purely prime ideal J is ~

In terms

(proposition 46).

It is natural to ask whether there exists some sheaf representation of R on its pure spectrum whose stalk at J is just ~ j is of this kind).

(Pierce's representation

We will show that this cannot be true in general.

Consider a ring R with a single non trivial pure ideal I.

For example,

the ring of linear endomorphisms of an infinite countable dimensional vector space (example 35) or the ring of triangular 2 × 2 laatrices on some field (example 36).

For such a ring, the pure spectrum is the Sierpienski space

Now, c o n s i d e r a l o c a l homeomorphism p : R ÷ Spp(R) where p-1 (0) = t}/0 = R and p-1 ( I ) = R//I. condition Spp(R).

Take r 6 R i n t h e s t a l k

a t O; t h e l o c a l homeomorphism

i m p l i e s t h a t r i s an open p o i n t i n R b e c a u s e p ( r ) Thus t h e s t a l k

Consider [r] 6 ~ I "

= 0 i s open i n

at 0 is discrete.

The local homeomorphism condition implies the exis-

tence of some S[r ] 6 R in the stalk at 0 such that the pair {[r], S[r ] } is open in R.

Moreover, for any [r] 6 ~ I '

{[r], s} is open.

there is a unique s such that

Indeed if {[r], s} and {[r], s'} are open with s # s',

the intersection of these two open subsets is just [r] which is thus an open point.

But p([r]) = I is not open in Spp(R) and this contradicts the fact

that p is a local homeomorphism.

So the uniqueness of S[r ] is established

and we have in fact a mapping s : R/I ÷ R ; Jr] ~-+S[r ].

181

The open subsets described above form a base for the topology of R. Indeed, the intersection of two such open subsets is {r} n {s} =

~

{r} if r = s

t

otherwise. Sir ] if S[r ] = s

{[r], S[r ]} N {s} = otherwise. I

{[r], S[r ]} if [r] = [r']

{[r], S[r ]} N {[r'], S[r,]}=

{S[r ]} if [r] # [r'] and S[r ] = Sir. ] otherwise.

Clearly, the open subsets of R are just the unions of these basic open subsets.

In other words, U c R is open if [r]

E U =,

S[r ]

E U.

We are now able to compute the continuous sections of p.

First observe

that for any [r] C ~ I ' the mapping O[r ] : Spp(R) + R ; I ~-+ [r] ; 0 ~-+S[r ] is continuous.

Indeed, for any open subset u -I

(

°[r](U) = 1

~

if [r] ~ U, S[r ] ~ U

{0} if [r] ~ U, S[r ] E U {O,I} if [r] E U and thus S[r ] E W.

Now consider a continuous section : Spp(R) ÷ R. We know that {o(1), s (i)} is open in R and thus -I {o(I), s (i)} =

is open in Spp(R).

(

{O,I} if o(0) = s (i)

[

{I}

if ~(0) ~ s (i)

As I is net an open point in Spp(R), this implies

a(O) = s (1) and thus o = °o(I)" °[r] for any [r] in ~ I "

So the continuous sections of p are just the

182

Now, if p : R + Spp(R) is a sheaf representation of R, the set of continuous sections of p must be a ring for pointwise operations. exactly that s : Y I

+ R must be a ring homomorphism.

This means

Now a continuous section

C[r ] is exactly determined by [r] and the ring of global sections is just the ring ~ I "

As p : R + Spp(R) is a sheaf representation of R, this ring of

continuous sections is isomorphic to R. R/I are isomorphic.

So we conclude that the rings R and

This conclusion depends only on the fact that Spp(R) is

the Sierpinski space, not on the precise form of R. In the case of the ring R of example 35, R has a single non trivial 2-sided ideal which is precisely I.

This implies that ~ I

is a simple ring,

i.e. a ring with only the two trivial 2-sided ideals. Indeed, if J _~ y I is a 2-sided ideal its inverse image q -I (J) along the quotient map q : R ÷ y l is a 2-sided ideal in R and thus J = q(q-1(j)) is (0) or ~ I "

This proves

that in the specific case of example 35, R is not isomorphic to ~ I "

So in

that case, there cannot be a sheaf representation p : R + Spp(R) of R with -I p-1 (I) = y l and (0) = R.

p

In the case of example 36, the 2 × 2 triangular matrices on some field K, compute the "espaces ~tal~s" corresponding to the sheaves AR and yR.

The pure

ideals not contained in (0) are I and R and the only pure ideal not contained in I is R. Be

stalk of AR at I is the i~uctive l i ~ t of (R, R), thus it is

(R, R) ~ R.

The stalk of AR at (0) is the inductive limit of (R, R) ÷ (I, I),

thus it is (I, I).

Recall that I is the ideal of the ~trices of the form

f 00 ~ere

are c~onical

inclusio~

of rings

K ~ ~ere

ba 1 .

~ R ~---+ K 2 × 2

the first inclusion sends a E K to the diagonal matrix [a

tO

0 1

a

.

Any R - l i n e a r endomorphism o f I i s ~ K - l i n e a r . So ( I , t) i s contained in the r i n g o f K - l i n e a r endomorphis~ o f 1% K2 ~ i c h i s j u s t K2 x 2 So a w R - l i n e a r endomorphism o f I has the f o ~

183

O

b

1

~

0 Io a1

Y

6

O

b

.

But any such mapping is in fact R-linear; indeed, the R-linearity means that [ ~y

~B ]

[[ O0

ba ]

[ Xo

Y]]z = [[ ~

6~ ]

[ Oo

ba ]]

oX

y] Z

which is obvious. Thus (I, I) is just the ring K 2 x 2 and this is the stalk of AR at (O). The peculiar form of the space Spp(R) implies that any covering of a non-empty open subset must contain this open subset. Therefore, the two conditions defining a sheaf vanish in the case of a covering of a non-empty open subset. So the sheaf condition reduces to the condition on the empty open subset : the separation condition means that for a sheaf F, there is at most one element in F(~) and the glueing condition means that F(~) has at least one element. Finally a sheaf F on Spp(R) is just a presheaf F such that F(~) is a singleton. All this implies that the presheaf 0j V-+ R/c j ; J pure in R is already a sheaf and thus equal to yR. Clearly C R = (O). On the other hand, C I is a 2-sided ideal (proposition ]6) whose intersection with I is zero. If C I contains a non-zero matrix A, the condition A ~ I imolies that aii ~ O. But in this case I ] 0

O ] 0

[O A

0

1] 0

{O =

0

all ] E C i 0

which is a contradiction since this last matrix is also in I. Finally C I = (O). The stalk of vR at I is the inductive limit of ~(O)' thus it is Ro The stalk of VR at (0) in the inductive limit of ~ O + ~O" thus it is also R. Finally, when we compose both "espaces ~tal6s", we conclude that the stalks at (0) are different. For AR, it is K2 x 2 and for vR, it is just the subring R of K 2 x 2

184

§ 8. PURE IDEALS IN PRODUCTS OF RINGS

So far in this chapter, the ring R was fixed.

In these last two paragraphs,

we let R vary and investigate what happens to the pure ideals and the pure spectra.

In this paragraph, we consider the case of a product

× Rk of rings. Any kEK product of pure ideals is pure but the converse is generally not true; however it is true when K is finite.

Again when K is finite, we are able to describe

the purely prime ideals of

× Rk and we conclude that the pure spectrum of kCK × kCK Rk is just the disjoint union of the spectra of the rings Rk.

Proposition 54. Let (Rk)kE K be a family of r~ngs and for any kEK, In this case,

Ik a pure ideal in Rk.

× Ik is a pure ideal of the ring × Rk. kEK kEK

Clearly, ideal.

× Ik is a 2-sided ideal. Let (ik)kE K be some element in this kEK Then for any k, choose ~k E Ik such that ik ~k = ik" The equality (ik)k6 K • (Sk)kEK = (ik)kC K

holds in

× Ik. kEK

A pure ideal I in a product of pure ideals Ik in the Rk'S.

× R k of rings is generally not a product keK For example, if K is an infinite set, take I

to be the ideal of those families (rk)kC K such that all but a finite number of its elements rk are zero.

I is clearly 2-sided, left and right pure : the

unit (Ek)kC K of (rk)kE K can be choosen to be I when rk ~ O and 0 otherwise. I is not a product of pure ideals in the Rk'S as soon as infinitely many of the Rk'S are not the zero ring.

But if K is finite, any pure ideal in

is of the form × Ik with Ik pure in Rk. lemma, kEK

Le~

× Rk kCK To prove this, we need the following

55. Let f : R ÷ S be a surjective homomorphism of rings. I in R, f(I) is a pure ideal of S.

For any pure ideal

185

Clearly f(I) is a subgroup

of S.

Now for any s C S and f(i) E f(I), we

can write s = f(r) and therefore s . f(i) = f(r) . f(i) = f(r i) 6 f(I). f(i) . s = f(i) . f(r) = f(i r) C f(I) and f(I) is a 2-sided ideal. i ~ = i.

Moreover if f(i) E f(I), choose E E I such that

We have f(~) E f(1) and fCi)

. f(~)

= f(i

~) = fCi);

so f(1) is pure.

Proposition 56. Let (Rk)kE K be a ~ n i t e ~ l y × R k. kEK I =

of ~ n g s and I a pure ideal in

For any k E K there is a pure ideal Ik of R k such t ~ t

× Ik. kEK

For any k £ K, consider the canonical projection

Pk :

x kEK

h

By lemma 55, each I k = Pk(I) is a pure ideal in R k. I =

× I k. k£K

Obvio~ly

we have the i n c l ~ i o n

We shall Drove that

I m × Ik. - kEK

Now consider

(ik)kE K an element in kEK Ikk For any k, ik is in Pk(1); .£ t h ~ there is some element (ik)£C K in I with ik = i k. Choose (Sk)kE K in I, .£ a unit ~ r all (ik)£C K. T h ~ for any k and any £, we have •l l ik • ~£ = i k. In particular,

for any k

So we have t h e e q u a l i ~ (ik)kE K • (~k)kEK = (ik)kC I with (gk)kE K in I.

This implies that (ik)kC K is also an element in I.

From proposition 56, it follows easily that the lattice of pure ideals of

× R k is i s o m o ~ h i c k£K

to the product of the lattices of pure ideals of the

186

Rk'S.

Taking the associated Stone spaces, we deduce that the pure spectrum

of k×K Rk is the disjoint union of the pure spectra of the Rk'S. We propose a mo~e direct proof of this fact. This proof requires some ler~nas. Lengna 57. Let f : R ÷ S be a surjective homomorphism of rings and I, J two pure ideals of R.

Then, f(I N J) = f(I) N f(J).

Clearly f(l N J) ~ f(1) N f(J).

Now if f(i) = f(j) with i E I, j e J,

choose ~ E I such that i E = i f(i) = f(i s) = f(i) f(E) = f(j) f(s) = f(j ~) and j s E J I = J N I (proposition 2).

Thus f(1) N f(J) ~ f(I n J).

Lemma 58. Let f : R ÷ S be a surjective homomorphism of rings and J a purely prime o

ideal in S.

The pure part f-~(J) of f-J (J) is a purely prime ideal in R.

j = f f-](j) is proper, thus f-1(j) is a proper ideal. in R such that 11 N 12 _c ~I(~).

Take If, 12 pure

This implies, by len~na 57, that

f(I1) N f(I2) = f(I1on 12)

c f f-l(j) = j. By lemma 55, f(I1) and f(I2) are pure and since J is purely prime, f(Ii) _c J or f(12) c J. This implies ir~nediately that 11 _c f-1(j) or 12 _c f-1(J) and since II, 12 are pure

o

O

11 C ~--](~)or 12

I

Proposition 59. Let (Rk)kC K be a ~ n i t e family of rings.

The pure spectrum of

is the disjoint union of the pure spectra of the rings Rk.

xR

kEK

k

187

For any purely prime ideal J£ in R£, we obviously have p~1(j£) =

× Jk where Jk = Rk for k # £. kCK

.As p£](J£) is not the whole space

× Rk and as it is pure (proposition 54), kCK we deduce by ler~na 58 t h a t p~l (J/) is purely prime in × Rk. We w i l l prove kCK t h a t these ideals p~l~(j£) are the only purely prime ideals o f × ~.RI~" kEK Consider J purely prime in x Rk. By proposition 56, J = x Jk with kCK kEK Jk pure in Rk. As J is proper, at l e a s t one Jk is a proper ideal in Rk. In fact, exactly one Jk is a proper ideal in Rk. Indeed, if J£ and Jm were proper respectively in Rl and Rm with 1 # m, consider the two pure ideals

I/ =

×

Ik£ where

kEI

im =

× m where kCI Ik

I

II = R1

(

I~ = Jk if k # £.

i

im =

m

Rm

m Ik = Jk if k # m. From their definition, it follows immediately that Il N Im = J and ll_~ J, Im _~ J. Thus J is not purely prime.

Finally the purely prime ideals of × Rk are exactly the p£ I (J/) where J1 is purely prime in R1.

kCK So the assignment J1 ~-+ p~1 (j/)

describes a bijection between the disjoint union of the Spp(Rk)'S and Spp( × Rk). We need to show that this is an homeomerphism. kEK An open subset in

× Rk is just a pure ideal I = × Ik (proposition 56). kCK kCK This open subset contains the point J = × Jk C Spp( × Rk), where J1 is kEK kEK purely prime in R1, i f and only i f I _~ J. But for k # l , I k c Rk = Jk; thus I l - ~ Jl" F i n a l l y , an open neighbourhood of J = p~l (jr) in Spp( xK Rk) is j u s t a family (Ik)kE K of open subsets in each Spp(Rk) in such a way ~ a t the point J1 belongs to the open subset I 1. But this is exactly the d i s j o i n t union o f

188

I

the topological spaces Spp(Rk).

§ 9. CHANGE OF BASE RING

In this last paragraph, we study the action of a ring homomorphism f : R ÷ S on pure ideals and the pure spectra.

This problem is not easy at all and is still open in the general case. The difficulty comes from the fact that the image of an ideal is generally not an ideal and the inverse image of a pure ideal is generally not a pure ideal.

For example, consider the ring inclusion

but not in ~.

On the other hand, consider

in Z/n2Z but its inverse image n Z

n¢O,

Z ÷

2Z~-~; 2Z/nTZ;

Z is an ideal in (O) is a pure ideal

is not a pure ideal in Z as soon as

n # 1. To overcome these difficulties,

constructions

it seems reasonable to use natural

like "ideal generated by a subset" or "pure part of an ideal".

This approach of the problem produces a continuous mapping Spp(f)

: Spp(S) ÷

÷ Spp(R) in two particular cases " when f is surjective or when S is c o ~ u t a t i ve.

Moreover,

if R f S ~ T is a composite of two ring homomorphisms

each of them is either surjective or with codomain commutative,

such that

then

Spp(f) o Spp(g) = Spp(g o f).

In § 8, we described how pure ideals and purely prime ideals can be transformed by a surjective ring homomorphism

(lemmas 55, 57, 58).

We start

with analogous lemmas in the case of a ring homomorphism f : R ÷ S with S commutative.

Lemma 60. Let f : R ÷ S be a ring homomorphism with S commutative.

For any pure

ideal I in R, the ideal f(I) . S generated by f(1) in S is pure.

Take i E I and s C S, thus f(i) . s E f(1) i ~ = i.

. S.

Choose ~ in I such that

We have f(i)

.

s

and f(s) [ f(I) . S.

=

f(i~)

. s

=

f(i)

.

f(~)

Thus f(1) . S is pure.

. s

:

f(i)

. s

.

f(~)

I

t89

Lerfma 6 1 .

Let f : R ÷ S be a ring homomorphism with S co~nutative. two pure ideals in R.

Then

f(I N J) . S = (f(1)

. S) N (f(S)

Clearly f(I N J) . S_c (f(I) . S) N (f(J) . S). f(i)

Let I, J be

. S).

Now consider

. s = f(j) . s' ; i £ I ; j £ J ; s, s' £ S

some element in (f(I) . S) N (f(J) . S). f(i)

Choose a £ I such that i s = i.

. s = f(i a) . s = f(i)

. f(s)

=

f(i)

.

s

.

=

f(j)

. s'

=

f(j)

. f(~)

=

f(j

c)

. s f(~)

. :E(~) . s'

. S'

£ f(I N J) • S

(proposition 2).

Thus the equality holds.

Len~na 62.

Let f : R ÷ S be a ring homomorphism with S commutative.

Let J be a

o

~urelw prime ideal in S.

The pure part f:. ~'F", " (J) of f- 1 (J) is a purely

prime ideal in R. o

Take Ii, 12 pure in R such that 11 N 12 _c f'-](~).

This implies by

le~na 61 : (f(I1)

. S) N (f(I2)

. S) = f(I 1 N I2) . S o

c ~(f-/~O)) • s c f(f-] (J))

cJ C

By lermm

60,

f(I1)

. S snd f(Ii)

This implies

f(I2)

. S are pure

. S m J

or

f(12)

. S

. S J.

and

since

. S_m J.

J is purely

prime

190

f(ll) _c J

or

f-] f(ll) _c f-] (J)

f(12) _c J

or f-1 f(12) _c f-1 (j)

I i _c f-](j)

or

12 _c f-](j),

and finally, since I 1 and 12 are pure

o

o

I I _c f'~(J)

12 _m f-~(J).

or o

We still need to prove that f-'~(J) is a proper ideal of R.

But if

o

~ )

= R, then f-](J) = R and thus J m f(R).

So ] -- f(]) E J and J is not

proper, which is a contradiction. Proposition 63. Let f : R + S be a ring homomorphism which is either surjective or with codomain com3nutative.

Define a mapping o A

Spp(f) : Spp(S) ÷ Spp(R); S ~-+ ~-I(]). This mapping is continuous.

The mapping Spp(f) is well defined by le~nas 58 and 62.

To prove the

continuity, choose 01 an open subset of Spp(R), i.e. a pure ideal I of R. The points of 01 are the purely prime ideals J' of R such that I ~ J'. prove that Spp(f)-1(0I ) is open in Spp(S).

We must

In fact we shall prove that

Spp(f) -](OI) = O where is the Z-sided ideal generated by f(1) in S; is pure by lemmas 55 and 60. Consider a purely prime ideal J of S. J E Spp(f) -] (0i)

~=~ Spp(f)(J) 6 @I o

~=~ I ~ f-1 (j)

fl~fd ~ J 4=~ d 60

(because I is pure)

191

This proves the continuity of Spp(f). Proposition 64. Let f : R ~

S and g : S + T

be two ring homomorphisms such that each

of them is either surjective or with codomain commutative.

Then g o f

is either surjective or with codomain co.~tative and

Spp(g o f) = Spp(f) o Spp(g). Suppose T is not commutative; then g is surjective. quotient ring of S; thus S cannot be commutative. jective and finally g o f is surjective. Take J a purely prime ideal in T.

But then T is a

This implies that f is sur-

So Spp(g o f) is defined. We have o

Spp(g)(S) = go~(s) Spp(f) o Spp(g)(s) = o

Spp(g o fD(j) = ~ ) . o

But f ~ is a pure ideal contained in f-] g-1 (j), thus the following inclusion holds o

o

To prove the converse inclusion, consider the following diagrams in R and S

192

o

A=

E= o o

o

c

=

~ )

c; = g-1 (j).

D = f-1 g-](j) Again the brackets

notation

< > denotes

A, B, C, D, E, F, G are B sided ideals; are purely prime ideals is pure,

the generated

B-sided

A, C, E, F are pure ideals andA,

(lenmms 55, 58 and 60).

c = ~

= B.

Moreover

o

)

But C is pure, C _ D A

C, E

E is the pure part of G and F

F m G; this implies F m E and thus f-1(F) _m f-1(E) o

ideal.

cf-1 andA

f

( ~ ).

f-1(F) = B.

c

is the pure part of B, thus A = C.

This is' just I

what we needed to prove.

C~o~mterexample

65.

We conclude this paragraph with an example of a ring homomorphism f : R ÷ S which does not produce a continuous mapping Spp(S) ÷ Spp(R) when we apply to it the constructions of proposition 63.

Consider a field K, the c6~uutative ring R = K 2 and the non-co~utative ring S of triangular B x 2 matrices on K.

f : R--

S

;

(a, b)

Take f to be the inclusion

I

a

O

O

b

'

Clearly f is not surjective and S is not commutative.

The pure ideals of R

are (o), (o) x K, K x (o), K x K because a ~ = a in K implies a = o or E = I, since K is a field. and K x (o).

In particular, the purely prime ideals of R are (o) x K

On the other hand (example 36) the pure ideals of S are S, the

ideal J of matrices with first coluu~ zero and (o).

We have

o

f-~(j) = f-1(j) = (o)× K o

f-~(o) = f-1(o) = (o). (0) × K is purely prime in R but (o) is not purely prime in R.

193

This example also yields a situation where the pure part construction does not con~nute with the inverse image.

Let I be the 2-sided (and left pure)

ideal of S of those matrices with second row zero (see example 36). part of I is just (o) and thus f-1 (i) ° = (o).

The pure

On the other hand, f-1 (I) =

K × (o) which is pure in S. Moreover, consider the ring homomorphism

g : S ÷

R

a

c

a

b

;

F----+

which is surjective and has a cor~nutative image.

(a,b)

The composite g o f is just

the identity and therefore it induces the identity mapping on Spp(R) ; in particular Spp(g o f)(K x (o)) = K x (o).

On the other hand, g-1(K x (o)) = I and

the pure part of I is (o); thus Spp(g)(K × (o)) = (o).

If we apply the cons-

truction of proposition 63 to Spp(g)(K × (o)), we find (o) in R, which is not purely prime and which is not equal to Spp(g ~ f) (K × (o)).

CHAPTER 8 : GELFAND RINGS This last chapter develops the results of chapter 7 in the special context of Gelfand rings.

A detailed study of the structure of maximal and pure ideals

in Gelfand rings allows us to generalize to arbitrary Gelfand rings the results of Bkouche (cfr. [3]) on commutative Gelfand rings.

As a consequence, our

theory reduces to that of Pierce in the case of von Neumann regular rings (cfr. [19]). A ring is right Gelfand if its right maximal ideals satisfy a separation condition (cfr. [3], [16]).

Mulvey has proved (cfr. [16]) that right Gelfand

is equivalent to left Gelfand.

Here we explain the reason for this : in a

Gelfand ring, the left maximal ideals are exactly the right maximal ideals (§§

I -

3).

But a Gelfand ring can equivalently be characterized by properties of its pure ideals (§ 3).

The properties of pure ideals in a Gelfand ring are deduced

from an interesting formula describing the pure part of an ideal I (§ 2) : ~ = {a C R 1 3 ~ E I

a ~ = a}.

From this formula, we deduce characterizations of Gelfa~d rings in terms of pure ideals (theorem 31) and we deduce also the important fact that in a Gelfand ring, the left pure ideals are exactly the right pure ideals. This coincidence between left and right pure ideals implies that for a Gelfand ring, the left pure spectrum is exactly the right pure spectrum; moreover this pure spectrum is compact Hausdorff and homeomorphic to the usual maximal spectrum (§ 4).

Moreover, the sheaf representations VR and AR coincide

in the case of Gelfand rings and their stalks are local rings (§ 5).

This

extends a result of Bkouche for commutative Gelfand rings (cfr. [3]). The properties of § VII - 9 on the change of base ring also extend to Gelfand rings (§ 6). The last paragraph is devoted to examples of Gelfand rings. and yon Neumann regular rings are such examples.

Local rings

A very characteristic

example of a Gelfand ring which is generally not regular is given by the ring C(X, ~) of real valued functions on some topological space X.

From this example,

195

using rings of matrices or quaternion rings, we construct some examples of non c~autative Gelfand rings (§ 7).

Several results of §§ ] - 2 - 3 were obtained :first for con~nutative Gelfand rings.

Their translation into the non con~autative case was made possi-

ble by a fruitful collaboration with Harold Si~nons.

Several other interesting

results on Gelfand rings and pure representations will be found in further papers by him.

§ 1. GELFAND RINGS

We define the notion of right-Gelfand ring and we will show that it is equivalent to the notion of left Gelfand ring.

This result was known by Mulvey

(cfr. [16]) from an abstract categorical argument; here we make the reason of this fact very explicit : we show that in a Gelfand ring, an ideal is right maximal if and only if it is left maximal. easy lemmas on maximal ideals.

We start with some well-known

R is an arbitrary ring with unit; when nothing

is specified, all ideals are right ideals. Lemma 1.

Let M be a maximal ideal and ~ in M.

the greatest 2-sided ideal contained

Then NA =

[M:

r] = {xcR ~f~

Ma i s o b v i o u s l y be a 2-sided

a 2-sided

ideal,

s o I c Ma a n d M~ i s

I cM

N iN: rCR t rxcM}.

={xeRlVreR ideal;

r xCM}.

choosing

and x ¢ I;

the greatest

r]

r = 1, we d e d u c e Ma ~ M.

f o r a n y r C R, r x E I ¢ M

2-sided

ideal

Let I

thus x C 5¢;

i n M.

Lermna 2. Let M be a maximal ideal.

By lermna I, ~¢ = Then

Then ~

N [M : r]. rCR a g ~ 3

Therefore r a R + N = R and

is a prime idaal.

Now take a, b E R such that a R b c M a.

r ¢ R

r a gM,

[]

196

I = ras+m Rb=

;

sER

;

mCM.

rasRb+mRbcM. []

By len~aa I, we deduce b E ~ .

Lepta 3.

Let M be a maximal ideal and r ~ M.

Then [M : r] is a maximal ideal.

[M : r] is proper because I E [M : r] implies r C M. thus r s { M.

Take s £ [M : r],

We have r s R + M = R, thus r=

rs

t+m

;

tCR

;

mCM;

r(1 - s t) = m £ M. This implies ] - s t E [M : r] and thus I C s t + [M : r] c s R + [M : r]. Thus if we adjoin to [M : r] any element s £ [M : r], we generate the whole ring.

This says exactly that [M : r] is a maximal ideal.

I

Definition 4.

A ring R is a right Gelfand ring if for any distinct maximal ideals M ~ N, there exist r E R and s E R such that r ~ M, s ~ N and r R s = O.

Proposition 5.

In a @elfand ring, a 2-sided prime ideal is contained in a unique maximal ideal.

Consider P a 2-sided prime ideal and M, N two maximal ideals, P ~ M, P c N, M # N.

Choose r £ M

s £ N

r R s = O c P (definition 4).

Since P is prime, r E P or s E P and thus r C M N N or s C M N N, which is []

a contradiction.

Proposition 6.

Let M be a maximal ideal in a Gelfand ring and r ~ M.

We know that M ~ = is maximal

(len~na 3).

(proposition 5).

Then M = [M : r].

N [M : s] (lemma I); thus M A c [M : r] and [M : r] sCR But M ~ is prime (lemma 2) a n d M ~ _ C M , thus M = [M : r] []

197

Proposition 7.

fn a Gelfand ring, any maximal ideal is 2-sided.

Let M b e

a maximal ideal and r E R.

If r E M and m E M, r m E H.

If

I" ~ M and m C M, M = [M : r] (proposition 6) and thus mE

Proposition

[M : r] ~ r m E M .

M~mE

8.

In a Gelfand ring, a maraimal ideal is completely prime. Suppose M is maximal and r s C M; we must prove r E M or s E M.

If

r ~ M, by proposition 6, M = [M : r] and s E [M : r] because r s = o E M.

Proposition 9.

In a right Gelfand ring, any right maximal ideal is also left maximal. Let M be a right maximal ideal; M :is also a left ideal (proposition 7). Consider

a

left ideal I such that Hc

I.

Choose r £ I "- M; thus r ~ M and r R + M = R. 1 = r s + m

;

Consider

s C R

;

m E M.

We compute (I

-

since M is 2-sided.

s

r)s

s -

s

r

s

=

s(l

-

r

s)

=

s m

£

M,

But s ~ M because s C M implies I = r s + m £ M.

position 8, I - s r E M c So I C I and I = R.

=

I.

By pro-

But s r E i since r E I and I is left sided.

This proves that M is a maximal left ideal.

In order to conclude that in a Gelfand ring, right maximal ideals coincide with left maximal ideals, we still need to prove that in a Gelfand ring, any left maximal ideal is 2-sided.

This will be deduced from the study of pure

ideals in Gelfand rings (§ 2).

§ 2. PURE PART OF AN IDEAL IN A GELFAND RING

This paragraph is devoted to the computation of the pure part of an ideal in a Gelfand ring.

This pure part will be the set of all those elements of the

198

ring which have a unit in the ideal.

TI~ proofs are generally given first

for maximal ideals and then generalized to arbitrary ideals. is specified, everything is specified on the right.

Again, when nothing

We start with a definition.

Definition IO.

Let I be an ideal in a ring R. The "unit part" of I is the set e(I)

= {a E R I V r E R

3 ~ E I

a r = a r ~}.

Proposition 11.

The "unit part" of an ideal is a 2-sided ideal.

(I) is obviously stable by multiplication on the left and on the right. We must prove that it is also stable by addition. Choose s in I such that a r = a r s.

Take a, b in g(I) and r E R.

Consider b C s(I) and r(1 - s) E R;

choose ~ C I such that b r(1 - ~)~ = b r(] - ~). Finally, ~ + ~ - s ~ E I and (a + b)r (s + ~ - E ~o) = a r e + a r ~ - a r ~ + b r s + b r ~ - b r e ~ = a r +

a r~o-

a

r~+

b

r ~ + b

r(1

-

s)~

= a r + b r e + b r(1 - s) =ar+br =

(a + b) r.

Proposition 12.

For a maximal ideal N, ~(M) is contained in M.

Take a E ~(M) and r = ] E R; choose s C M such that a = a ~.

Because ~4

is 2-sided (proposition 7) and ~ E M, we deduce a E M.

The proof of proposition 12 does not work for an arbitrary ideal I (which is not left-sided).

However, the result is still valid for an arbitrary ideal,

but to prove it, we need some more lemmas.

Le~

13.

Let R be a @elfand ring, M a maximal ideal, and I a proper ideal.

Then

199

~(M) c I ~ I c M .

I is proper;

choose N maximal

such that I c N.

If N ~ M, choose

a ~ N, b ~ M with a R b = O. By maximality

of M, M + b R = R and thus I =m+br

This implies,

;mEM;

r£R.

for any s £ R, as=asm+asbr=asm.

Thus a E E(M) c I c N ,

Corollary

which is a contradiction.

Finally N = M.

14.

Let R be a Oelfand ring, M a maximal ideal and I an ideal. I + M=

I + ~(M) =

~(M); if I + e(M)

R~I

Then

+ E(~4) = R.

is proper,

then by lemma 13, I + E ( ~

cM

and I + M = M, which is a contradiction.

Lenma

15.

Let R be a Gelfand ring, M a maximal ideal, I an ideal and r E R. [I : r] c M ~ I

If I _~M,

cM.

I + M = R and thus I + ~(M) = R (corollary r = a + i ; a E ~(M)

Choose

14).

Write

; i E I.

~ E M such that a = a E.

r(1 - ~) = a(1 - ~) + i(I - ~) = i(I - ~) £ I. This

implies

I - E E[I

: r] and thus I - E E M.

I = (I - ~) + ~ E M, which

Proposition

is a contradiction.

But E E M and thus So I c M .

16.

Let I be any ideal in a Gelfand ring.

Let a £ ~(I) and choose

Then ~(1) is contained in I.

~ E I such that a = a ~. J = [I : a] + (I - ~)R.

Consider

the ideal

200

If J ~ R, consider a maximal E £ I cM.

ideal M D

J.

By lemma ]5, I c M

But I - e £ M and thus ] = (I - ~) + ~ C M w h i c h

and thus is a contradiction.

Thus J = R and I = r + (1 - ~)s

; a r E I ; s E R.

I

a = a r + a(1 - s)s = a r E I.

W e are n o w going to p r o d u c e an easier description

Proposition

of a(I).

First of all

:

17.

L e t I be an i d e a l i n a G e l f a n d r i n g . ~(I) = a {~(M)

Let J be the intersection ~(I) ~ E(Y0 a n d thus e(I) ~ J.

Then I M maximal;

on the right hand side.

I oN}.

From I ~ M, we deduce

Conversely

a ff J * ~ * 3 M _ m I; M m a x i m a l ;

a ff ~(M)

3 M _D I; M maxinkal 3 rCR

Ws

CM

arCarE

*~* 3 M _m I; M m a x i m a l 3 r C R

W ~ E M

] - ~ ~Ann

a r

* ~ 3 M 2 I; M maximal 3 r C R

M+

Ann a r ~ R

*~* ~ M ~ I; M maximal 3 r C R

Ann arcH

*~ 3 r E R

3 Mmaximal

*~-3 r C R

I + Anna

Hm

I + Ann a r

r ¢ R,

Finally a C l*~*V

r E R

I + Ann a r = R

*~*W r E R

3 ~ C I

I - ~ £Ann

*~V

3 ~ C I

a r(1 - e) = 0

3s£

ar=ar~

r C R

*~Wr£R

I

a r

I

*=~ a E ~(i).

Proposition

18

Let M be a maximal ideal in a Gelfand ring c(M) = {a E R [ 3 ~ £ M

Consider a E R a n d E £ M w i t h

a = a ~.

a = a ~}.

Then I - ~ E A n n a and I - ~ ~ M,

201

thus M + ~ m

a = R and by corollary I = ~+

r ; ~E

a = a~0+ by proposition

Proposition

Write

14, ~(M) + Ann a = R. ~)

; a r = o.

a r = a~E

E(M),

11.

20.

Let I be an ideal in a gelfand ring g(I)

By proposition ag

= {a E R

a=a~}.

[ 3 ¢ E I

17 e(I) ~=* 3 M D I; M m a x i m a l ;

a ~ ~)

3 M ~ I; M maximal W m E M

a f a m (proposition

18)

~=~ 3 M = I; M maximal VmEM ~=~ 3 M ~

I -mgAnn

a

I; M maximal; M + Ann a ~ R.

~=, 3 M D I; M maximal; M D Ann a ~=-3 M maximal; M ~ ~=~I + A n n

a # R

~=~I + A n n

a = R

I + Ann a

and finally a E a(I)

~3

a E I

I - ~ EAnn

3 E E I

a(1

~3cEI

The next step

Proposition

a

- ~) : o

a=a~.

will be to prove that ~(I) is a pure ideal.

21.

Let M be a maximal ideal in a Gelfand ring.

~ M ) is just the pure part

of M.

~) Hence,

is pure.

Indeed, consider a E ~CM) and ~ E M such that a = a E.

I - ~ E Ann a and thus M + Ann a = R.

and thus a(M) is pure

By corollary

14,

~(M) + Ann a = R

(proposition VII - 2). O

Conversely,

theoPUre part M of M is obviously contained in ~(M)

tion 18); so ~(M) = M.

(proposig

202

We now propose to generalize proposition

21 to an arbitrary ideal I.

To do this, we require some left-right syrmnetry properties of Gelfand rings.

Lemma 22.

Let M, N be two maximal ideals in a Gelfand ring. M = N*=* a(M] = c(N).

Immediate from lenmm 13.

Lenmm 23.

For any maximal ideal M in a Gelfand ring R and a, b E R, we have 1 -

a b ¢. c ( M )

*=* 1 -

b a ¢ ~(M).

Suppose I - a b £ e(M); we shall prove that e(M) + b R = R.

Indeed,

if

N is a maximal ideal such that ~(M) + b R c N

then e(M) c N

~ R

and thus by lenmm 22, M = N and so b ¢ M.

But then a b E M

and I - a b ¢ s(M) _cM, thus I ¢ M which is a contradiction. = R and we can choose r ¢ R such that I - b r ¢ a(M). {

I - b r E E(M) I - a b E c(M)

~

{

[b] . [r] = I [a] . [b] = I

Thus e(M) + b R

Thus

in~c(M

)

[r] = [a][b][r] = [a] in--~c(M% r-

a C E (M).

Therefore, I - b a = (1 - b r) + (b r - b a) =

(1 - b r) + b ( r

- a) E c ( M ) .

Lemma 24.

For any ideal I in a Oelfand ring R and a, b E R I - a b ¢ ~(I) ~ * I - b a ¢ ~(1).

By proposition I - ab

17 and lemma 23, we have

E~(1)~=*¥MD

I, M m a x i m a I ,

*=*¥ M = I, M m a x i m a l , *=* I - b a ¢ ~(1).

I - abe

c(M)

I - b a E ~(M) I

203

We know that M and e(M) are 2-sided So it makes sense to consider

ideals

the equivalents

for any maximal

ideal M.

of lerm~a 13 and corollary

14

for a left ideal I.

Lena

25.

Let R be a right Gelfand ring, M a right maximal ideal and I a proper left ideal.

Then E(~

Consider

a £ I.

cI~IcM.

If E(Y0 + a R = R, choose r £ R such that I - a r E ~ ) .

By lemma 23, I - r a E ~(M) c I and on the other hand, sided.

Thus

I £ I which

is a contradiction.

is not the whole

ring R.

e(M) + a R ~

From ~(M) ~ N ,

Corollary

N.

Consider

r a C I since I is left

Thus we deduce that E(M) + a R

a right maximal

ideal N such that

we deduce M = N (lemma 22) and thus a E M.

26.

Let R be a right Gelfand ring, M a right maximal ideal and I a left ideal.

Then, I + M=

R~

I + c(~) = R.

If I + e(M) ~ R, by lemma 25, I + e(M) 2 ~ ( ~

~ I + e(M) ~ M

and thus

I + M = M, which is a contradiction.

Lenmm 27.

Let R be a right Gelfand ring and M a right maximal ideal. (VaE

M is proper, of a.

Then,

position

R) (V e CM)

th~s I - ~ ~ M.

I - ~ C L-Ann

9).

Hence

(a = e a = ~ a E

e(M)).

Consider

(a), the left annihilator

L-Ann

(a) thus M + L-Ann

~(M) + L-Ann I = r + s ; rE

Then

(a) = R by ma×imality

(a) = R (corollary ~(M)

of M

(pro-

27); choose

; s a = Oo

Now, we obtain

I

a = r a + s a = r a C ~(~).

Le~

28.

Let R be a right Gelfand ring and I a right ideal.

(va~R)

(v~cl)

(a= ~ a ~ a c

~(I)).

204

By proposition 17 and len~aa 27, we have (a = E a, ~ E I) ~ V M m

I, M maximal, a = s a, E E M

V M_~ I, M maximal, a E ¢ff4) a e e(I).

Proposition 29.

Let R be a Gelfand ring, I an ideal and M a maximal ideal. I aM~=~(I)

aM.

From I _c M, we deduce s(I) _~ s(M) _~ M, by proposition 12. if ~(1) _~M and I C M ,

Then,

Conversely,

the maximality of M implies I + M = R and thus

I + ~(N0 = R (corollal 7 14).

Choose

I = i+

a ; iE

a=a~

I ; aE

¢(M)

; sEM.

Now we obtain l

-

~ =

(i

+ a)

(1

-

~)

=i-ia+a-ae =

From i 6 I and l e m a

i(I

-

E).

28, we have I - ~ C ¢(I) c M

which is a contradiction since e E M.

Proposition 30.

Let R be a Gelfand ring and I an ideal.

Take a £ E(1) and consider c(I) + Ann a.

Then E(1) is the pure part of I.

If ~(I) + A n n

in some maximal M, I is contained in M by proposition that a = a ¢; ~ E I a M

and l - ¢ E Ann a c M

29.

a is contained

Consider ¢ C I such

: this is a contradiction.

~(I) + Ann a = R and we conclude by proposition VII - 2 that ~(1) is pure.

On the other hand, the pure part of I is obviously contained in ~(1); thus ~(I) is the pure part of I.

Finally, by propositions

20 and 30, the pure part of an ideal I in a

Gelfand ring R is ~ = (a E R [ 3 ~ E I

a ~ = a).

Thus

205

§ 3. CHARACTERIZATIONS

OF GELFAND RINGS

We are now in a position to prove the equivalence between the notions of right Gelfand rings and left Gelfand rings : in fact, in a Gelfand ring, right maximal ideals coincide with left maximal ideals.

But a Gelfand ring

can also be characterized in terms of pure ideals : a ring is Gelfand if and only if the mapping which sends an ideal to its pure part is a continuous homomorphism on the lattice of ideals. consequences

From this theorem, we deduce several

: two of them are worth to be mentioned here.

In a Gelfand ring,

the "pure part" morphism and the "Jacobson radical" morphism determine a Galois connection on the lattice of ideals.

Moreover,

in a Gelfand ring, left pure

ideals coincide with right pure ideals.

Theorem 31.

For a ring R, the following conditions are equivalent : (RI) R is a right Gelfand ring. (R2) For any right maximal ideals M ~ N 3a~M

3b~N

aRb=o.

(R3) For any right ideals I, J o

I + J = R~ (R4) For any right ideals I, J,

I + J = R. (Ik)kE K

o o

o

InJ=InJ o o

+ I k = + I k. kEK kEK (L1) -

(L2) -

(L3) -

(L4)

: dual conditions o f (R1) -

(R23 -

(R3) -

(R43.

o

(R1) ~ (R2) i s j u s t

definition

4.

Let us p r o v e

(R2) ~ (R3).

If

I + J { R,

o

consider

a maximal i d e a l

M such that

I + J c M.

By p r o p o s i t i o n

29,

o

I = e(I) c M

and thus I c M .

This implies I + J c M w h i c h

Let us prove (R3) ~ (R4).

is a contradiction.

The condition on finite intersections

proposition VII - 9; by the same proposition, we have the inclusion o o

+ I c + I k. kEK k - kEK

is just

206

To prove the converse inclusion, consider a £ a e = a.

+ Ik and ~ £ + Ik such that k6K k6K

We can write e = el + "'" + en with ek 6 Ik.

Since I - e £ Ann a,

we can write 11 + ... + I n + A n n a =

R

and an iterated application of (R3) yields o

Ii + ... +

+Ann

a = R.

n o

But then there exists %0 £ 11 + ... +

such that I - %0 6 Ann a.

Thus

n o

a = a ~0 £ I l

o

o

+ ... + I c + Ik. n - k6K

To prove (R4) =~ (RI), consider two maximal ideals M # N. have M + N = R.

By (R4), we

So we can write o

1 = m+

n

; mEHandnE

.

Then there exists e £ N such that n e = n and I - ~ £ Ann n. o

Therefore

o

N+Annn=R

(~6N,

1 -

~6Annn).

By (R4), we obtain o

N+Annn=R x+

y = I

o

; y£~--"n

As Ann n is two-sided, we obtain V r 6 R Andn

Ry

= o with n £ M a n d y

o._z_. r y6AnnncAnnn.

~ N.

By the left-right duality, the equivalences are proved.

andy~N.

(LI) ~=~ (L2) *=* (L3) ~=~ (L4)

To conclude the proof, it suffices to show (RI) ~ (LI).

This will

be done if we prove that in a right Gelfand ring, any left maximal ideal is right maximal.

Thus let R be a right Gelfand rin~ and N a maximal left ~de~1. o

We shall prove the existence of a right maximal ideal M such that M _~ N. By lemma 26, this will imply N c M and thus M = N since M is also left maximal (proposition 9).

So we must prove that a left maximal ideal N contains the pure part of some right maximal ideal M. o

M_CN

Suppose that for every right maximal ideal M,

o

and choose aM £ M ~ N .

Consider :

207

I = + {Ann aM I M maximal}. If I # R, fix M maximal such that I m M and choose ¢ ¢ M such that aM • ¢ = aM.

NOW

1

- ~ ¢ oArm % i c

I =M

sEMcM

which yields a contradiction.

Thus I = R and

] = E l + ... + En ,• ~ k C Ann % where M k is maximal.

But for each maximal ideal M, ] { M and thus there exists

some index k(M) such that ek(M) ~ M. so we can write

The maximality of M gives }4 + ek(M)R = R;

1 = m~+ ~k(M) • rM ; rome M ; rM¢ R. ak(bl) = a k ~

" mM + ak(M)

= ak(M)

"

" ¢k(bl)

" rM

%1" o

This implies that for any maximal ideal M

By proposition 18, ak(M) E M.

o

%RaM 1

R ...

RaM

2

_M n

and finally by proposition VII - 10 applied to (o) : o

aM I R a M 2 R ... R a M n _ ~ a M

= (o).

But the 2-sided part of N is a prime ideal (lemma 2, valid for an arbitrary ring and thus also for left ideals). 2-sided partof N and thus in N.

This implies that some %

is in the

This contradiCtSo the choice of % ,

so there

must be some right maximal ideal M such that M m N. Corollary 32. In a Gelfand ring, the left maximal ideals are exactly the right maximal ideals. By (RI) ~=~ (Ll) and proposition 9. Proposition 33. In a Gelfand ring~ the left pure ideals are exactly the right pure ideals. Consider a 2-sided ideal I in the Gelfand ring R and its left pure part I£.

By definition of a left pure ideal

208

V a6

I£

3 g 6 I£

~ a = a.

By !en~na 29, this implies that any a 6 I£ belongs to the right pure part I r of I.

So I£ c I r.

But the equivalence (RI) *=~ (LI) in theorem 31 implies

dually I r c I~ and finally Ir = I£.

Now if I is left pure, I is 2-sided and I = I£ = I r which implies that I is right pure.

Again by theorem 31, any" right pure ideal

is left pure.

Thus in a Gelfand ring, we can speak of pure ideals without any specification of left or right.

We conclude this paragraph with a description of the

relation between pure ideals and Jacobson radicals.

Proposition 34. Let R be a Gelfand ring and r(R) the lattice of right ideals of R. The mappings : r(R)

÷ r(R)

;

I b-*

0 : r(R)

-* r ( R )

;

I

b-+ t a d

I

describe a Galois connection. A dual result holds for left ideals.

By rad I, we denote the radical of I : rad I = n {M I Mmaximal; M_m I}. We must prove that for any two ideals I, J o

I cJ

*=* I c rad J. o

Suppose first

_c J.

For any maximal ideal M ~ J, the inclusion M 2 I implies

M D I (proposition 29); thus I c rad J. --

--

Conversely, suppose I c rad J. O

o

--

For any maximal ideal M _m J, I _c M implies I _c M and finally O

O

I _c n {M I M maximal; M 2

J}

o

= J

(proposition VII - 10)

cJ. Corollary 35. In a Gelfand ring, any pure idaal is the pure part of its radical.

209

Let I be a pure ideal.

From proposition 34, we deduce :

o

rad I c I

~=~

rad I c rad I.

o

So r a d I c I . - -

Finally

On t h e o t h e r

hand,

I is pure and I c rad I,

o

thus

I c rad I.

- -

- -

I = rad I.

§ 4. PURE SPECTRUM OF A GELFAND RING

In chapter 7, we defined the pure spectra of a ring : the left pure spectrum and the right pure spectrum.

For a Gelfand ring, both spectra coincide.

More-

over, the points of this pure spectrum are just the purely maximal ideals. As a consequence, for Gelfand rings, the pure spectrum is homeomorphic to the usual maximal spectrum.

Proposition 36.

For a Gelfand ring, the left pure spectrum coincides with the right pure spectrum. By proposition 33 and the definition of a pure spectrum.

Proposition 37.

In a Oelfand ring, any purely prime ideal is purely maximal. Let J be a purely prime ideal in the Gelfand ring R. some purely prime maximal ideal M (proposition VII - 28). By theorem 31 o

M=

+ Ra aEM

Choose a 6 M such that R a_~J.

~

M=

+ Ra. a£M

By proposition VII - 2 M+Anna=R

and by theorem 31 M+Ann But by the dual of proposition VII - 16,

a=

R.

J is contained in Suppose J ~ M.

210

o

o

0

RanAnna=

o

~ n ~ - ~ a o

o

cRanAnnRa cRanAnnRa = o

(o)

c

J.

o

Since J is purely prime, R a c J or Ann a c J.

The choice of a implies ° Ann a c J c M which yields a contradiction with M + Ann a = R.

Proposition 38.

In a Gelfand ring, the t~ure part" operation induces a one-to-one correspondance between maximal ideals and purely maximal ideals. The pure part of a maximal ideal is purely prime (proposition VII - 27) and thus purely maximal (proposition 37). injective (lemma 22).

Moreover this correspondance is

To prove the surjectivity, consider ~ purely maxim~l

ideal J; it is contained in a maximal ideal M and thus J c M. o

But J and M

--

are purely maximal (first part of the proof), thus J = ~4.

Theorem 39.

The pure spectr&snof a Gelfand ring is compact Hausdorf~. By proposition VII - 34, it suffices to prove that the spectrumis Hausdorff.

Consider two distinct purely maximal ideals J1, J2 (proposition 37)

which are the pure parts of two distinct maximal ideals M~, M 2 (proposition 38). Since the ring is Gelfand 3 a~M

I

3 b~

M2

a Rb

= o.

We deduce, by theorem 31 and proposition 38, o

MI + a R = M2 + R b

R

~

Jl + a R =

= R

~

J2 + R b

R. = R.

In other words, a R _ ~ J I and R b _ ~ J 2 with aRM by proposition VII - 2.

Rb

= aR

RbcaR

Rb

= aRb

= o.

By theorem 31, this means exactly :

211

J1C O~

," J2 E 0

aR

," 0 o N O R~

= ~,

a~

and thus Spp(R) is Hausdorff.

Proposition 40.

The pure spectrum of a Gelfand ring is homeomo~hic to its usual maximal spectrum. The points of the maximal spectrum are the maximal ideals; the topology is generated by O r = {M I M maximal; r ~ M} for any r E R.

This is in fact a base for the topology since each maximal

ideal is completely prime (proposition 8) : O r N 0 s = {M [ Mmaxin~l; r ~ M; s ~ M} = {M [ M maximal; r s ~ M} =0

rs

Then proposition 38 describes a bijection between the pure spectrum and the maximal spectrum. and M a maximal ideal.

Let us prove it is an homeomorphism.

Consider r E R

By theorem 31 : ME0

~=~ r ~ H r

~=~M+ rR=R o

o

~=~H+

~"R= R

*~IC

0 rR

Thus any ftmdamental open subset of the maximal spectrum corresponds to an. open subset of the pure spectrum. Conversely let I be a pure ideal in R.

Again by theorem 31 o

I =

+

rEI Therefore by theorem VII - 32

r R

~

I =

+

rCI

r R.

212

01=

u 0 o ' ~EI ~-~ rR

and the f i r s t p a r t o f the proof shows t h a t each 0 o corresponds to a fundamental open subset i n the m a x i l ~ l spectrum.

rR This concludes the proof.

§ 5. PURE REPRESENTATION OF A GELFAND RING

In chapter 7, we described two different sheaf representations of a ring R on each of its pure spectra. coincide.

For a Gelfand ring, these four representations

Moreover the stalk of the representation at some point J C Spp(R)

is just the quotient ~ j

which turns out to be a local ring and the localiza-

tion of R at the unique maximal ideal containing J. Proposition 41.

For a Gelfand ring R, the four sheaf representations AR and VR (on the right and on the left) are isomorphic. By proposition 38, these four representations are defined on the same topological space.

Now if I is any pure ideal in R (left pure and right pure

by proposition 33), the greatest left-sided ideal whose intersection wi~l I is zero and the greatest right-sided ideal whose intersection with I is zero are 2-sided (propositions VII - 14 and 16) : thus these ideals coincide. As a consequence both sheaves VR, defined on the left or on the right, coincide.

To conclude the proof, it suffices to show that AR and VR, defined on the right, coincide.

Let I be a pure ideal in R.

The left purity of I (pro-

position 33) implies I =

+

{i E I I ¢ i = i}

and each subset I is obviously a right ideal.

E

= {i £ I I e i = i} By theorem 3] I =

+ ~ . eEI c

Now, if f : I ÷ I is a right-linear endomo1~phism, for any e E I and i E f(i) = f(E i) = f(E)i.

£

213

o

Thus the restriction of f at each I E is just the left multiplication by f(e). For a pure ideal I, we are now able to define a linear mapping ~R(I) ÷ VoR(l).

Take some f E AR(I) = (I, I); consider the pure covering

I = ~EI + Ic and for each ~ E I, take [f(~)] £ Y C

~ " This is a compatible C

family; indeedo choose el' ~2 E I : weo mUStoPrOVe that f(El) - f(E2) E E C(I i N Ic2 ).

Indeed for any i E I I N I 2

(f(~1) - f(E2))i = f~el )i - f(~2 )i = f(el i) - f(~2 i) =

f(i)

-

f(i)

To this compatible family [f(a)] 6 y

o

O.

=

corresponds a unique element in

CI

VR(I); this produces a mapping 61 : AR(I) + vR(1), which obviously respects the additive structure of the rings. obvious that ~I respects multiplication too. AR(I).

It is less

Consider f, g two elements in

To show that 6i(f o g) =o~!(f).~l(g), we will show that the restric-

tions of these elements to each Ic, with ~ E I, coincide i.e. o

[f(g(~))] = [f(~)][g(~)] in R/ o . Consider any element i 6 I . Then by / C I~ right-linearity of f and g and by definition of IE, we obtain : (f(g(a)) - f(e).g(~)).i = f(g(~).i) - f(e).g(e i) = f
The last equality follows from the fact that, since i 6 ~ o

there exists £

e' 6 I

such that i e' = i and thus E

g(i) = g(i ~') = g(i)~' 6 i So f(g(~))

c I .

- f(~)gC~) c C i • o

In a simSlar way, we can show that ] - ~ £ C I EvR(I).

Therefore ~I (idI) = ] in

Thus ~I is a ring homomorphism.

In fact, we have described a homomorphism of sheaves : : AR + YR. This means that for any pure ideals J c I, the following square

214

AR(I)

~ VR(I)

AR(J]

, VR(J) ~j

commutes, where the vertical arrows are the canonical restriction morphisms. Let f be an element in ~R(1). Again to prove the equality

~J(fb) : ~I
a

with ~ 6 J :

=

[f(E) ] ]

= tf< )l Yc ]

£

= if j(~)]

= ~j(f j)

5"

Thus ~ is an homomorphism of sheaves. We shall prove that each ~I is an isomorphism and this will conclude the proof.

Consider f 6 AR(1) = (I, I) such that ~i(f) = 0 6 vR(I).

anYi£ I~°6 I, [f(a)] = 0 in y C ~ or in other words, f(~) 6 C I~.

Then for Thus for any

f(i) = f(E i) = f(~).i = O. o

So f is zero on each ~a and I = + I ; hence f is zero on I and ~I is injectire. a6I To prove that ~I is surjective, consider some x E VR(I).

Then there is

some pure covering (Ik)k6K of I such that x can be represented by a compatible family of elements [xk] E Y C Ik"

But each pure ideal Ik itself has a pure

covering given by Ik = ~(k)EIk+ I(k).° Clearly each [Xk] also gives rise to a compatible family of elements [XE(k)] 6 ~

o

. Combining these facts, we

C I~(k)

215

see that x can be represented by a compatible family of elements for all ~ E I.

[x ] E R/ E

Now, for any E in I, define f C

E

C

E

;i

C

x.i. o

Then f

i s a r i g h t - l i n e a r mapping defined on ~E and thus takes values in IE;

hence fE E 4R(I ) .

Moreover, the family (fe)eE I i s compatible. o

For i f El , E2

o

are two elements in I , then for a ] l i E I El N I e2, we obtain : • fel(i) - f~2 (I) : xel

.i

-

x

~2 . i

-x =

(XC i

)i S2

=0 o

o

since xEI - xc2 E C(I E

N I 1

) by the compatibility of the family ([xE])eE I. E2

Now the fact that the family (fe)~El is compatible,

implies the existence of

some f E AR(1) whose restriction to ~E is just fE"

But f I° (i) = f(E)i.

o

So,

C

for all i E I , we have ( x

- f(E))i = O and [x E] = [f(~)] E ~

.

Hence

~l(f) = x and ~I is surjective.

In order to compute the stalks of the representation AR = VR, we need the following lenmm :

Lemma 42.

Let R be a Gelfand ring and J a purely maxin~al ideal.

Then the unique

maximal ideal M containing J is given by M=

{aE = {aE

R I V s E R, a s R I vs

E R, s a -

- I ~ J} I ~J}.

By proposition 38, we know that J is the pure part of some maximal ideal N.

Now lemma 23 shows that both formulae define the same subset M of R~

will first prove that M is a 2-sided ideal.

Take a, b in M and consider a + b. (a+ b)s-

For any s C R,

I = a s + b s - I = a s + (b s - I).

If (a + b)s - I E J, then a s ~ J since b s - I ~ J and similarly b s ~ J. Consider the maximal ideal N containing J. a s E N

and

If a s E N, then

a s + b s - I E JcN

We

216

imply b s - I E N and thus b s ~ N.

So, under the assumption (a + b)s - I E J,

we obtain a s ~N

or

b s ~N.

Now take the case a s ~ N (the case b s ~ N can be treated similarly). a~

Then

N and N+aR=R.

But J is the pure part of N.

So, from theorem 31, we deduce J+aR=R.

In particular, there is some r E R such that a r - I E J, i.e. a ~ M.

This is

So M is stable under

a contradiction and finally (a + b)s - I ~ J for any s. addition.

The relation M=

{aE

R I V s E R

as-

I ~ J}

shows that M is stable under right multiplication and the relation M=

{a£

R I Y s E R

s a-

shows that H is stable under left multiplication.

I ~ J) Finally, M is a 2-sided

ideal.

M contains J because a £ J and I ~ J imply a s - I ~ J. proof, it suffices to show that M is a maximal ideal.

To conclude the

Take a ~ M; we must

prove the equality M + a R = R. But a ~ M implies the existence of some s E R w i t h

a s - I E J.

Therefore

J + a R = R and since M contains J M+aR=R.

Theorem 43.

Let R be a Gelfand ring.

The stalk of the sheaf representation AR = VR

at some point J E Spp(R) is just ~j;

this stalk is the localization of

R at the maximal ideal containing J. By proposition VII - 44, we know that the stalk of vR at J is just ~'~

217

where = u {C I I I pure; I ~ J } . So we need to prove t h a t J equals ~.

Since J i s p u r e l y maximal ( p r o p o s i t i o n 37)

we know a l r e a d y t h a t ~ i s contained in J ( p r o p o s i t i o n VII - 45). prove the inclusion J c ~.

Thus we must

Again we define J

= {j £ J I ~ j = j}" E

J

is a right ideal and by theorem 31 avld the left purity of J J=

+ J ~£j ~

~

J=

+ ~ . e£j o

If J is not contained in ~, choose c such that J~ is not contained in ~. o

o

For any j E J , (I - a)j = 0 and thus I - e £ C J~.

This implies that

o

J+gJ

=R E

and by theorem 31 E

in other words, C ~

~ J and thus by definition of ~ and proposition VII - 18 C

-°

J

_=

_c

which c o n t r a d i c t s the choice o f ~.

J ~ _c ~

C

Thus we have a l r e a d y proved t h a t the

s t a l k at g i s j u s t ~ j . Now we shall prove that ~ j

is a local ring obtained by inverting all

the elements which are not in the maximal ideal M containing J. 38, J is the pure part of M.

By proposition

Consider the set N of non invertible elements

in ~ j . N = {[a] I a C R

V s E R

= {[a] ] a £

R

V s C R

[a][s] ~ I

a s - I ~ J or

or

= {[a] I a £

R

V s E R

a s - I ~ J}

where the latter equality holds by len~na 24.

[s][a] ~ I} s a-

l ~ J}

Thus by len~aa 42, the inverse

image of N in R is jt~st the maximal ideal M containing J. ideal in ~ j

Therefore N is an

and R/j is a local ring; moreover any element which is not in M

has an image which is not in N and thus is invertible in ~ j .

Thus the elements

which are not in M are exactly those which become invertible in ~ j .

Finally, we show that ~ j

is universal for the property of inverting the

218

elements of R which are not in M.

Consider a ring homomorphism f : R ÷ S

such that for any r ~ M, f(r) is invertible in S. E J EM

such that j E = j.

f(1 - s) is invertible in S.

For any j E J, choose

From e E M, we deduce I - e ~ M and thus But

f(j)f(] - ~) = f(j(] - s)) = f(j - j c) = f(j - j) = f(O) = O and since f(1 - ~) is invertible,

f(j) = O.

Finally f(J) = O and f factors

uniquely through t ~ j .

Theorem 45 extends a r e s u l t o f Bkouche ([3]) f o r corr~utative Gelfand rings.

It implies in particular the following corollaries

:

Corollary 44.

Let R be a Gelfand ring and M a maximal ideal.

The problem of localizing

R at the prime ideal M has a universal solution; it is the quotient of R by the pure part of M. Corollary 45.

Let R be a con~nutative Gelfand ring.

The sheaf representation AR = VR is

isomorphic to that given by Bkouche in [3].

By theorems 43 and VII - 46.

Bkouche's representation

reduces to Pierce's representation in the case

of conm~tative regular rings (cfr. [19]). our representation.

By corol]ary 45, the same holds for

But we shall prove a more precise result in our appendix.

§ 6. CHANGE OF BASE RING

In § 2 - 3, we proved some left-right duality conditions for Gelfand rings. They provide an effective means of transposing to Gelfand rings several results previously obtained for co~utative

rings.

In this paragraph, we use them to

establish the change of base-ring properties.

Proposition 46.

Let f : R + S be a ring homomorphism between two @elfand rings and I a pure ideal in R.

The right sided ideal f(1).S generated by f(1) in S

is pure and coincides with the left sided ideal S.f(I).

219

Consider the right sided ideal f(1).S. E £ I such that ~ i = i (proposition 33).

For any i 6 I and s 6 S, choose The equality

f(s) f(i)s = f(i)s implies that f(i)s is in the pure part of f(I).S. equals its pure part and thus is pure. contains S.f(I).

As a consequence f(I).S

Hence, it is also left sided and

A similar argument applied to the latter finally shows that

both ideals coincide.

This proposition 46 is the analogue of le~na VII - 60 for Gelfand rings. Let tm prove the analogues of lemmas 61 and 62.

Lenmm 47. Let f : R + S be a ring homomorphism between two Gelfand rings and I, J two pure ideals in R.

In that case

f(I n J).S = f(I).S N f(J).S.

Clearly f(I N J).S c f(1).S N f(J).S. f(i).s = f(j).t

;

Choose e E I such that ~ i = i.

i E I

Now consider

;

j 6 J

;

s, t 6 S.

Then

f(i).s = f(E i)s = f(E) .f(i) .s = f(~).f(j) .t = f(¢

j)t

E f ( I N J).S. Lermna 48. Let f : R ÷ S be a ring homomorphism between two Gelfand r~Jngs and M a maximal ideal in S.

The following equality holds o

0

f-1 and this is a purely maximal ideal in R. o

o

C l e a r l y f-1 (N) c _

o

f-1 (M).

Conversely take r in f-t(M).

o

t '-I(N) such t h a t r e = r .

We deduce :

f(r) f(s) = f(r)

with

f(r) E M, f(g) 6 M.

Choose g in

220

o

By proposition 30, f(r) is in M and thus r is in f-1(~). o

Thus

o

f-1 (g) = f-1 ( ~ . o

M is purely prime (proposition VII - 26) and the proof of lemma 62 o

works to show that f-1 (N) is purely prime (references to len~nas 60 and 61 are replacedobY references to p r o p o s i t i o n 46 and l e ~ a 47). Therefore f - l ( ~ ) is purely maximal (proposition 37). Proposition 49. Let f : R + S be a ring homomorphism between two Gelfand rings. a mapping

Define

o

Spp(f) : Spp(S) + S~p(R) ; J b+ f-] (S). This mapping is continuous. Moreover if g : S + T is another ring homomorphism between Gelfand rings

Spp(f) o Spp(g) = Spp(g o f). The proofs of propositions 63 and 64 work if the references to le~as 55, 57, 58, 60, 61, 62 are replaced by references to proposition 46 and ler~aas 47-48. § 7. EXAMPLES OF GELFAND RINGS We produce examples of co~utative and non commutative Gelfand rings. Example 50. The following conditions are equivalent for a ring R.

(]) R has a single left maximal ideal. (2) R has a single right maximal ideal. A ring which satisfies these conditions is a Gelfand ring; in particular any local ring is a Gelfand ring.

A ring which satisfies (I) or (2) satisfies obviously conditions (L 2) or (R 2) in theorem 31; thus it is a Gelfand ring. are equivalent by corollary 32.

Therefore (1) and (2)

A ring is local if the non invertible elements

form a 2-sided ideal; as a proper ideal never contains an invertible element,

221

this ideal of non invertible elements is necessarily the largest proper ideal.

Example 51.

Any commutative Von Neumann regular ring is a Gelfand ring. Let R be a co~utative ring.

An ideal I is regular if it is generated

by its idempotent elements (cfr. example VII - 20). regular if any ideal in R is regular (cfr. []9]).

The ring R is Von Neumann In particular, any ideal

is pure (example VII - 20) and thus any ideal is equal to its pure part. So condition (R 3) of theorem 3] is obviously satisfied.

It should be pointed out that if any ideal of a commutative ring R is pure, R is Von Neumann regular.

Indeed, for any a [ R] the ideal a R is

pure and so there is some r 6 R such that a . a r = a.

But a r is idempotent :

(a r)2 = a 2 r . r = a r , Thus any principal ideal is regular and consequently any ideal is regular.

Exar~le 52.

Let X be a topological space a n d ~

the real line.

The ring C(X, ]R)

of continuous functions is a commutative Gelfand ring; its pure spectrum is the Stone - Cech compactification of X. This is a result of Bkouche (cfr. [3]); we recall a proof below.

Before

this, observe that an idempotent element in C(X, ~) is a function which takes only the values 0 and I.

So if X is connected, the only two idempotent

elements in C(X, ~) are 0 and I and Pierce's spectrum of C(X, I~) reduces to a singleton (cfr. [19]).

On the other hand, the pure spectrum of C(X, I~)

is the Stone - Cech compactification of X (see below), which retains a lot of relevant information on X and thus on C(X, IR). Example VII - 21 describes some pure ideals in C(X, IR).

We will now outline a proof of the fact that C(X, ~{) is a Gelfand ring; we refer largely to [8]. First of all, there is always a completely regular space Y such that C(X, l~) is isomorphic to C(Y, I~) ([8] - 3 - 8). f £ C(Y, JR) define Zf = {y [ Y I f(Y) = O}.

Now for any

222

The subsets Zf form a lattice 7 :

zfnZg=Zlf I+ Igl Zf U Zg = Zfg. The maximal ideals of C(Y, R) are in one-to-one correspondance with the l-ultrafilters ([8] - 2 - S) ; a maximal ideal M _~ C(Y, IR) and the corresponding l-ultrafilter U are related by the formulae

M = { f £ C(X, IR) j Zf C U}

u = {zf I f cM}. The S t o n e - Cech c o m p a c t i f i c a t i o n o f Y (which i s a l s o t h a t o f X) i s t h e s e t

of all l-ultrafilters provided with the topology admitting the subsets Cf as a base for the closed subsets ([8] - 6 - 5). Cf = {U I g 7-ultrafilter; Zf 6 U}. In terms of maximal ideals and open subsets, the Stone - C~ech compactification of X and Y is the set of maximal ideals in C(Y, R) provided with the topology admitting the subsets 0f as a base 0f = {M I Mmaximal ideal in C(Y, IR); f ~ M}. But this is exactly the description of the maximal spectrum of C(Y, R). This space is Hausdorff ([8] - 6 - 5), and thus if M, N are distinct maximal ideals, there exists f, g 6 C(Y, IR) such that M 6 Of

; N 60g

Of n Og # ~.

Because any maximal ideal is prime Of n Og = Ofg. On the other hand (cfr. [8]). Ofg = % ~=~ ¥ M maximal ideal fg £ M ~=~ fg 6 rad(O) B n

(fg)n = 0

~=~ fg = O. Finally the Hausdorff condition on the maximal spectrum can be written under the form 3 f~M

3 g ~N

fg = 0

which is just the definition of a Gelfand ring.

223

Example 53. Let R be a commutative Gelfand ring. Let R be the ring of n x n triangular matrices on R such that all the elements on the diagonal are equal. R is a Gelfand ring with the same pure spectrum as R.

Thus the elements in R have the form a a.

a..

ij

"• a

The addition or multiplication of matrices in R is performed pointwise on the diagonal elements.

Therefore if I is an ideal in R, the subset

a(1) = {a 6 R I a appears on the diagonal of some matrix of I} is an ideal in R.

This ideal is necessarily 2-sided since R is cor~nutative.

(~ the other hand, if I is an ideal in R, the subset B(I) = {A 6 R I the diagonal element of A is in I} is a 2-sided ideal in R.

Moreover, if I is maximal in R, B(1) is maximal

in R : indeed the only way to add a matrix to B(I) is to introduce new diagonal elements and I, the ideal of diagonal elements, is already maximal• Now take M, N two distinct maximal ideals in R; because R is Gelfand : 3 a ~ M

3 b ~ N

:

a b = O.

Now consider a

b

0 a

0 .

-a =

O

0

a

b

For any matrix A in R a ¢ Bffl)

~ ¢ B(N)

Thus t o p r o v e t h a t R i s a G e l f a n d r i n g , i t

a A b = O. suffices

t o p r o v e t h a t any

maximal i d e a l i n R has t h e form 8(M) w i t h M maximal i n R.

T h i s w i l l be done

i f we p r o v e t h a t any p r o p e r i d e a l I o f R i s c o n t a i n e d i n some B(M) w i t h M maximal in R.

224

Take a proper right ideal I in R.

Obviously we h a v e / ~ B(~(1)); if

~(I) is proper in R, choose M maxirmal in R such that ~(I) = M. I E B(M).

This implies

So the problem reduces to prove that ~(I) is a proper ideal.

If

it is not, I E ~(I) and thus there is some matrix A £ I with I as a diagonal element.

Now denote by Iij the matrix whose (i,j)-entry is I, while all the

other entries are O.

For any j > I, we have I]j = A . I]j C I.

We deduce that n

B=Aand blj = 0 for j ~ 1.

Z llj . a]j £ I j=2

Therefore, for any j > 2

I2j = B . I2j C I . Iterating the process, we find that any matrix I.. ij wit~ j > i is in I.

There-

fore the identity matrix is also in I : I =A-

z

i<j thus I is not proper.

I..

,

a..

1J

1J

Hence, ~(1) must be proper, which shows that R is a

Gelfand ring. The pure spectrum of a Gelfand ring is just its maximal spectrum (proposition 40), thus B describes a bijection between Spp(R) and Spp(R). Let us prove it is an homeomorphism.

For any a C R and any maximal ideal

M in R

Thus any :fundamental open subset 0 a of Spp(R) corresponds to an open subset in Spp(R).

Conversely take A C R with diagonal element a; for any maximal

ideal M in R

A¢

B(M) , - = ' , a C M .

Thus any fundamental open subset in Spp(R) corresponds to an open subset in R.

Finally we have shown that Spp(R) and Spp(R) are homeomorphic. We will conclude this example with a remark.

In a Gelfand ring, all

maximal ideals are 2-sided (proposition 7) but an a~itrary ideal need not be 2-sided.

Consider in R the matrices with some fixed column zero (and thus

225

diagonal zero) : this is obviously a left ideal which is not right sided. Conversely the matrices with some fixed row zero (and thus diagonal zero) form a right ideal which is not left sided.

Example 54.

Let R be a com~nutative Gelfand ring such that each element of the form 1

+ (a i)2 +

+ (an)2

...

is invertible; the rings C(X, IN) of continuous functions satisfy this condition. In that case, the ring of complexes of R and the ring of quaternions of R are Gelfand rings with the same pure spectrum as R. A continuous function f c C(X, IR) is invertible if and only if for any x £ X, f(x) ~ O.

Thus I + (fl)2 + ... + (fn)2 is always invertible.

If R is a commutative ring, the ring of complexes of R is the ring ¢(R) of matrices of the form

Ia -b

with a, b in R.

a

This ring is obviously conmutative and the transposition is

a ring involution on ~(R).

The ring Iq(R) of quaternions on R is the ring

of matrices of the form

i

_B t

At

where A, B are in ¢(R); this ring is generally not cormmtative. us point out that the condition on R : "I + (al)2 + ... + (an)2 is invertible", implies the following condition on ¢(R) : "I + A

i

~

I

+ ... + A

n

~-- is invertible". n

Indeed ak

bk

ak

-b k

-b k

ak

bk

ak

0

Ak Ak = 0

Finally let

226

and therefore 1 + z a~ + z b~ k k

1 + Z Ak Ak

k

1 +z

0

k

a ~ + ~ b~

which is invertible since the element on the diagonal is invertible.

So the two steps in the construction of If(R) fit into the con~non following context : "Let R be a coranutative ring and (-) : R ÷ R a ring involution such that any element of the form 1 + a I ~I + "'" + an ~n is invertible in R.

From this

we construct the ring R of matrices of the form

bt

-b

a

with a, b in R".

Choosing the identity as involution on R, we obtain E(R); choosing transposition as involution on E(R), we obtain If(R). We will show later on that both involutions respect maximal ideals, i.e. for every maximal ideal M and m E M, we have m C M.

To prove the statements of our example, it then

suffices to show that in the situation described above, the following implication holds : "R is a Gelfand ring and the involution respects maximal ideals ~ R is a Gelfand ring with same pure spectrum as R". Let us prove the implication.

Let I be a right ideal in R. a(1) = { a E R I ~ b E

(I) is obviously an ideal in R. b

-a

a

Then define : R

I a

b

E I

E I};

I

Moreover the relation

bI o

shows that in fact I a

bt

a E c~(I) and b C c~(I).

227

Conversely if I is an ideal in R, define [ a

b]

8(I) = { 8(I) is obviously a right

C R I aC I

; b E I} ;

ideal in R and if I is proper so is 8(I).

If M is a maximal ideal in R, 6(M) is necessarily a maximal ideal in R. Indeed, consider a right ideal I D 6(M) and A E I x B(M)

A =

Is b] ;

agM

or

b ffbl.

According to the relation pointed out before, it suffices to consider the case a g M.

Since M is maximal, we then obtain M+

aR=R,

and h e n c e

mO] blr m+

1

br

ar = I ; m C H ; r C R. a

0 c

+

1

1

O]

0

1

o

o

I =

1+b~r7

I

¥

-br 1+b~rr E I

-~r

br

1

1+b~r7 and thus I = R.

I 1+b~rY

So 8(M) is maximal.

If I is a proper ideal in R, u(1) is necessarily a proper ideal. if ~(I) = R, there is some matrix I

b

CI -g and thus

I

Indeed

228

1

0

1

I

b

l+bg

b I

1+b}5

C I. 0

g

t

1

l+bb

l+bb j

This contradicts the fact that I is proper. is contained in some maximal ideal M.

So a(1) is a proper ideal and

But then I is contained in 8(M).

This shows that the maximal ideals of R are exactly the 8(F0 with M maximal in R. Take M and N two distinct maximal

Let us prove that R is a Gelfand ring. ideals in R.

Because R is Gelfand : ~ a~

M 3 b ¢ N

ab = O.

Now a ~ M, aa g M, thus aa

O

¢ 8 (N)

¢ B(M) 0

0

~a

and for every A E R, we have aa

OllbA 0

~a

= O.

0

Thus R is a Gelfand ring. Now let us compute the pure spectrt~a of R.

As R and R are Gelfand

rings, their pure spectrum is their maximal spectrum (proposition &O).

The

mapping B describes a one-to-one correspondance between the maximal ideals in R and the maximal ideals in R.

Let us prove that it is an homeomorphism.

For any" a E R and any maximal ideal ~ in R a ~ M

*=*

~ B(M). 0

Thus any fundamental open subset in Spp(R) corresponds to an open subset in Spp(R).

Conversely if a, b are in R and M is maximal in R,

I a A =

b } ~ ~(M)

~-~

a ~ H

or

b ~ H.

Thus the fundamental open subset of Spp (R) generated by the element A C R

229

corresponds to the union of the open subsets of Spp(R) generated by a and b : this is an open subset in Spp(R).

Finally B describes an homeomorphism

between Spp(R) and Spp(R). All that remains to be shown now is that the two particular involutions giving rise to ~(R) and ~(R) respect maximal ideals. does so and ~(R) is a Gelfand ring.

Clearly the identity

Now any maximal ideal in ~(R) is of the

form B(M) with M maximal in R (observe that the proof of this fact is done entirely without the additional assumption on the involution). I- ~ t

--

ba I be an element of ~(M)in ~(R)" thus a E M and b C M.

Let But then

J

- b C M and

b

a

This example 54 should be compared with example 2 - 20 in [18]. Orzech and Small prove that under the weaker condition that 2 is invertible in the co~autative ring R, the lattice of ideals of R is isomorphic to the lattice of 2-sided ideals of the quaternion ring ~(R) on R.

APPENDIX

: NOTE ON PIERCE'S REPRESENTATION THEOREM

Using central idempotents,

Pierce proposes a sheaf representation

theorem for any ring R (cfr. [19]).

In this appendix, we show that the fLmc-

torial description of this representation the spectrum)

(= mapving on the open subsets of

is just the sheaf of linear endomorphisms

of pure ideals.

particular when pure ideals are exactly the regular ideals

In

(for example in

the case of von Neumann regular rings or in the case of principal rings) Pierce's representation coincides with our representation AR.

Definition ].

Let R be a ring. Vi6I

An ideal I is regular if and only if 3 ~ 6 I N center R

~ = e2

i =

~

i.

Proposition 2. The definition of a regular ideal is left-right symmetric; in particular any regular ideal is 2-sided.

With the notations of definition I, i = e i = i e because E is central. In particular,

for any r E R and i 6 I • ri=rie

= e tic

I.

Proposition 3.

Any regular ideal is pure.

The converse holds for yon Neumann regular

rings and com~mtative principal rings.

A regular ideal is pure by definition.

A ring is said to be regular if

any ideal is regular and thus pure : hence, in that case all ideals are pure and regular.

If the ring R is commutative and principal,

choose a pure ideal

a R ; by purity of aR ~rER

a=a.ar

and ar is necessarily an idempotent element

:

(ar) 2 = aZr . r = at. Thus aR is regular.

Proposition 4. Let R be a ring.

The set B of central idempotents in R is a boolean

23t

algebra for the binary operations ^ and v given by

E

A

e I

=

EE

V

E I

=

~

l +

E ~

-

EE

I .

The points of the corresponding Stone space X are the maximal regular ideals of R and the open subsets are the 01 given by 01 = {M I M m a x ~ l for any regular ideal I.

regular ideal in R; I ~_ M} X is called the Pierce's spectrum of R.

B is a boolean algebra and the points of the corresponding Stone space are the maximal ideals of B; a base of its topology is given by O

= {N I N maximal ideal in B; E g N}

for any central idempotent ~ (cfr. []9]). in

We recall that J ~- B is an ideal

B if E C J andE' E J

=

c v E' E J

E £ J and ~' E B

=

aE' E J.

There is an obvious isomorphism between the lattice of regular ideals I in R and the lattice of ideals J in B : to I we associate the set J of its central idempotents; to J we associate the ideal it generates in R.

So the

space X can equivalently be defined as the set of maximal regular ideals in R. The fundamental open subsets are now 0

= { M I M m a x i m a l regular ideal; E ~ M} E

for any central idempotent ~.

If (~k)kEK is a family of central idempotents

and I is the regular ideal generated by this family U 0 = {M I M £ kEK ak

=

X ; 3 k C K

~k~M}

01 .

Thus any open subset in X has the fo~m 01 . Conversely if I is any regular ideal and (Ek)kE K is the family of central idempotents in I, the same equalities show that 01 is open in X. of X.

Finally the subsets 01 are exactly the open subsets •

232

Theorem 5 (Pierce's representation theorem - cfr. [19]). Let R be a ring and X its Pierce's spectrum. disjoint union of the quotient rings 7 M M.

Consider the set R,

for any maximal regular ideal

Provide R with the final topology for all the mappings X+

R

; M

b-+[r] E 7 M

for any r g. R . The canonical projection p : R + X is a local homeomorphism and presents R as a sheaf of rings on X.

The ring R is isomorphic to the ring of

continuous sections of p. Theorem 6. Let R be a ring and X its Pierce's spectrum.

The mapping defined by

01 ~-+ (I, I) for any regular ideal I, is a sheaf of rings on X.

The corresponding

"espace ~tal{" is just Pierce's spectrum as defined in theorem 5. The mapping 01 b~ (I, I) is a sheaf of rings, indeed the proof of our theorem VII - 38 is valid if we replace the word "pure" by the word "regular".

In particular, if ~ is some

central idempotent, a linear mapping f : ~R ÷ ~R is just the left multiplication by f(g) f(Er) = f(ccr) :

f(~)

. ~r.

Therefore we have an isomorphism of rings (oR, ~R) ~ ER. Now i f

e'

such that

is another central

idempotent such that

e' = e r and t h e r e s t r i c t i o n

o f f E (eR, ER) t o e ' R i s

f(~'s) = f(E'~'s) = f(Eer)~'s = ~'f(~)E's. In other words, via the isomorphisms

E'R c eR, t h e r e e x i s t s

r E R

233

(ER, ER) ~ ~R the restriction mappings ~re just ~R ÷ ~'R

;

~r ~+ ~'~r.

Now let us consider the sheaf of rings defined in theorem 5; we shall compute the continuous sections of p on some fundamental open subset 0 for a central idempotent c E R.

Any mapping

0 ~-~X÷R for any r C R is continuotm.

o:0 is of this form.

of X

" M ~+ [r] C ~ ' M

Conversely any continuous section of p : +R

Indeed E has a complement ]-c in B and thus the complement

of 0~ in X is just 0]_~ (cfr. [19]); if we define ~ to be 0 on 01_ E, we obtain a continuous extension of 7:X÷R. By theorem 5, ~ has the form

for some r E R.

Finally any continuous section on 0 0

But different

elements r,

E

÷ R ; M ~-~ It] E ~

has the form

M"

s E R c o u l d i n d u c e t h e same c o n t i n u o u s

section.

This is the case if V M E 0

,

r-s C M

or in other words VM£

X

(~ ¢ ~ 1 ~ r - s

E M).

But we know (cfr. [11]) that V ~ E B

(~ ~ M ~

I-~ e M).

Thus finally r, s E R induce the same continuous section on 0 V M ~ X

(1-s £ M ~

if and only if

r-s E M),

or in other words r-s E n {M I M E X; I-EEM}. But in a boolean algebra, any ideal is the intersection of the maximal ideals containing it (cfr. [11]); thus the principal ideal (I-~)R in R is the inter-

234

section of the maximal regular ideals containing it.

Finally two elements

r, s 6 R generate the same continuous section of p on 0 £ if and only if

r - s 6 (I-E)R. Therefore the ring of continuous

sections of p on 0

is

R//(I_~)R ~ ~R. Finally if E' is another central idempotent and c' = Er for some r 6 R, we have obviously 0E, c 0 . --

Let us consider a continuous section of p on 0 ;

E

E

we have seen it has the form 0 for some r £ R.

+ R ; M F-+[er] 6 R/

Its restriction on 0 , is just 0 , + R ; M ~+[~'Er]

6 R/ I M

simply because ~r-

E'Er = (l-~')sr 6 (l-s')R.

So, up to the isomorphism described above the resl-riction mapping is just ~R + ~'R ; ~r F-+a'~r.

We are now able to conclude.

The sheaves given by theorems 5 and 6

coincide on a base of the topology of X : thus they are isomorphic.

Finally we have shown that Pierce's representation of rings is just the analogue with regular ideals of our representation AR using pure ideals. Let us conclude by pointing out that Pierce's representation

is also the

analogue of our representation vR.

Corollary 7.

Let R be a ring and X its Pierce's spectrum.

Pierce's sheaf is the

sheaf associated to the separated presheaf

OI F-+YC I for any regular ideal I in R. Replacing the word "pure" by "regular", VII - 40 - 41 show that

the proofs of our propositions

235

is a separated presheaf on X,

If c is some central idempotent in R,

aR @ ( ] - e ) R = R and thus C ¢R = (]-~)R.

In o t h e r words

Yc

Yo- >

and the r e s t r i c t i o n s are induced by the q u o t i e n t maps.

Going back to the

p r o o f o f theorem 5, we conclude t h a t t h i s s e p a r a t e d p r e s h e a f c o i n c i d e s with P i e r c e ' s sheaf on a base of the topology of X.

Thus the s h e a f a s s o c i a t e d t o

this separated presheaf is j u s t P i e r c e ' s sheaf. Corollary 8. Let R be a ring in which any pure ideal is regular (cfr. proposition 3). In that case, Pierce's spectrum of R coincides with the pure spectrum of R and Pierce's representation is isomorphic to our representations AR and VR.

INDEX algebra (~-) algebraic theory characteristic map

63

classifying object

60

co~utative theory

2

complete lattice completely prime ideal complex ring dense monomorphism distributive lattice

2 197 225 80 2

external theory formal initial segment frame Gabriel-Grothendieck IF-topology

25 2

75

Galois connection

208

Gelfand ring

196

global sections

168

Heyting algebra

2

Heyting subobject ideal (~r-)

14

initial segment

122 13

integral theory

111

internal theory

4

lattice

2

Lawvere-Tierney ~-topology left simple theory

73 112

localization

70

localized object

80

local section

173

237

80

prelocalized object

3

presheaf prime ideal

195

pure ideal of a ring

141

purely maximal ideal

156

purely maximal

123

T-ideal

purely prime ideal

156

purely prime qr-ideal

123

pure part of an ideal

147

pure spectrum of a ring

136,

159

pure spectrum of a theory

126

pure submodule

130

pure Ir-ideal

122

quaternion ring

225

radical of an ideal regular ideal right simple theory separated presheaf

208 153, 221, 230 113 3

sheaf

3

stalk

173

Stone-C6ch compactification

221

Stone space

231

unit part of an ideal

198

universal closure operation yon Neumann regular ring

70 221

NOTATIONS 3

a

Anni

141

CAI

150

CI

151 152

(R) C(X, ~) ~(~,s) F

Sh(u+) Sh(u+, ~u+ ) Spp (R) Spp (/r)

13 159 126 I

221 6O

t /r tu~,

62 13

u+

13

2 U

20, 25

U,

20, 25

U 1

20, 25

UT

22

44

H(R)

13

225

1,4,5

a{

4

Sh( lq, IT)

225 122

hu

6

U=U[

U

22

147

(i ,A)

168

(I,I)

145 174 178

£

8O

[M : r]

0

XA

84

pp (R)

158

PP (]Y)

124

p (R)

158 3 4 121 7O

rad(I) Sets ~F

166

195

4

Pr(l~) Pr( 1-t, ~) p(Tr)

137, 168 137,

195

125, 158, 211, 231

n

46, 127 49, 127, 137

¢(I)

59

M1r 0

Zf F A AA AR

221

208 1

Sets

I

Sh(I~)

4

198

~(A)

82

f?2F

6O

0 VR

20

^~I"

170 62 2, 18

BIBLIOGRAPHY [I] ARTIN - GROTHEN~3IECK, Cohomologie 6tale des sch6mas, S6m. de g6om. alg. 4, 1963/64, IHES Paris. [2] BARR, Exact categories, Lecture notes in math. 236, Springer (1970), 1-120. [3]

BKOUCHE, Puret@, mollesse et paracomgacit~, C.R. Acad. Sci. Paris 270, (1970), A 1653-A 1655.

[4]

DIVINSKI, Rings and radicals, Mathematical expositions 14, Univ. of Toronto Press.

[5]

FAITH, Algebra II : Ring theory, Grund. der Math. Wiss. 191, Springer (I 976).

[6]

GABRIEL, Des cat6gories ab@liennes, Bull. Soc. Math. de France 90, (1962), 323-448.

[7]

GABRIEL

-

ULMER, Lokal pr~sentierbare Kategorien, Lect. Notes in Math.

221, Springer (1971). [8]

GILLMAN

-

JERISON, Rings of continuous functions, Grad. Texts in Math.

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GODEMENT, Th6orie des faisceaux, Hermann (1958).

[10] GOLAN, Localization of non commutative rings, Marcel Dekker (1975). [11] GRATZER, General lattice theory, BirkhaQser (1978). [12] JOHNSTONE, Topos theory, Academic Press (1977). [13] LAMBECK, On the representation of modules by sheaves of factor modules, Can. Math. Bull. 14-3 (1971), 359-368. [14] IAWVERE, Functorial semantics of algebraic theories, Proc. Nat. Ac. Sci. 50 (1963), 869-872. [15] LINTON, Autonomous categories and duality of functors, J. of alg. 2 (1965), 315-341. [16] ~JLVEY, A generalization of Swan's theorem, Math. Zeit. 151 (1976), 57-70. [17] MULVEY, Representations of rings and modules, Lect. Notes in Math. 753, Springer (1979). [18] ORZECH - SMALL, The Brauer group of conrautative rings, Marcel Dekker (1975).

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[19] PIERCE, Modules over con~atative regular rings, Morn. Am. Math. Soc. 70 (1967). [20] POPgSCU, Abelian categories with applications to rings and modules, Academic Press (1973). [21] SCHUBERT, Categories, Springer (1972). [22] SIMNDNS, The frame of localizations of a ring, Dep. of Math., Univ. of Aberdeen (I981). ADDITIONAL REFERENCE JOHNSTONE-WRAITH~ Algebraic Theories in Toposes, in Indexed Categories and their applications, Lect. Notes in Math. 661, Springer (1978).