T–GALOIS EXTENSIONS ON RINGSAND A SUBMODULE CORRESPONDENCE

Ricardo Alfaro1

University of Michigan–Flint, Flint, MI 48502

§0. Introduction

The theory of separable Galois extensions is well understood and in the case

of central extensions there are analogous results to those for Galois extensions of

commutative rings. G. Szeto in [SM] generalize the one–to–one correspondence given

by DeMeyer in [D] for Azumaya algebras whose center C is G–Galois over the fixed

ring CG, to those faithfully flat (over C) center–Galois extension. This article uses

some of the ideas of Szeto in [SM] to extend this correspondence to rings S which

have a G–stable subring T on which the action of G is G–Galois.

We show that there is a one–to–one correspondence between the set of

TG–submodules of T and the set of “T–closed” SG–submodules of S. Then we

show that if we take T to be Z(S), the center of S, the “T–closed” algebras are

exactly the central extensions of SG in S and we obtain as corollary the correspon-

1Partially supported by a grant from the Faculty Development Fund of the University ofMichigan–Flint and a fellowship from the C.R.M., Bellaterra, Spain.

dence in [SM]. We then see T–Galois extensions in the context of H–separability of

the skew group ring S∗G over S.

§1. T–Galois extensions

Let S be a ring with 1, let G be a finite group acting faithfully as automorphisms

of S and let R=SG be the fixed ring under G. Writing g(r)= gr, the skew group ring

S∗G is the free left S–module with basis the elements of G and multiplication given

by the rule gs= gsg for all s∈S and g∈G. Denote by π the element∑g∈G

g∈S∗G.

The action of G on S is said to be G–Galois if S is finitely generated projective right

R–module and the natural map φ : S∗G→ EndRS given by φ(rg)(x)=r gx is a ring

isomorphism, or equivalently SπS=S∗G. The “trace map”, tr : S → R is given by

tr(x)=∑g∈G

gx which is an R–module homomorphism. Let T be a G–stable subring

of S (that is gt∈T for all t∈T ), we say that S is a T–Galois extension of R if the

action of G on T is G–Galois. Let U =TG denote the fixed ring of T under G and

let A be a right R–module contained in S and that contains R (that is, R⊆A⊆S).

We say that A is T–closed over R if (A∩T )R=A.

Lemma 1.1 Let S be a T–Galois extension of R. Then S is a T–closed right

R–module (i.e. S=TR).

Proof. Since the action of G on T is G–Galois, there exist ai, bi∈T i= 1, ..., n

such that∑iaiπbi = 1. Let x ∈ S, then x= 1 · x= (

∑iaiπbi)x=

∑iaitr(bix) ∈ TR.

Thus S=TR=(S∩T )R.

Lemma 1.2 Let S be a T–Galois extension of R, then TR ∼= T ⊗UR as right

R–modules.

Proof. Define the map σ : T⊗UR → TR via multiplication. Thus σ is a right

R–module homomorphism and clearly an epimorphism. Now let σ(∑iti ⊗ ri) =∑

tiri = 0. As in the proof of lemma 1.1, ti =∑jajtr(bjti) with aj , bj ∈T for every

ti. Since bjti∈T and tr(bjti)∈U , we have∑iti ⊗ ri =

∑i

(∑jajtr(bjti) ⊗ ri) =

∑i

∑jaj ⊗ tr(bjti)ri =

∑i

∑jaj ⊗ tr(bjtiri) =∑

jaj ⊗ (

∑itr(bjtiri)) =

∑jaj ⊗ tr(bj

∑itiri) =

∑jaj ⊗ 0 = 0. Hence σ is also a

monomorphism and TR∼=T⊗UR.

Corollary 1.3 Let S be a T–Galois extension of R. Assume R is flat as left

U–module. Let B be a right U–module in T (that is U⊆B⊆T ), then BR∼=B⊗UR

as right R–modules.

Proof. Since B ⊆ T , the sequence 0 → B → T is exact as right U–modules.

By flatness of R we have 0 → B⊗UR → T⊗UR is also exact. But by lemma 1.2

T⊗UR∼=TR, hence by restriction of the isomorphism we obtain B⊗UR∼=BR as right

R–modules.

We now have the following correspondence.

Theorem 1.4 Let S be a T–Galois extension of R. Assume R is faithfully flat

as left U–module. Then there is a one–to–one correspondence α between the set

of right U–modules in T and the set of T–closed right R–modules in S, given by

α(B)=BR.

Proof. Let B be a right U–module in T . We first prove that BR∩T =B. Since

B ⊆BR∩T , the sequence 0 → B → BR∩T → (BR∩T )/B → 0 is exact as right

U–modules. Then by flatness of R over U we have that

0→ B⊗UR→ (BR∩T )⊗UR→ (BR∩T/B)⊗UR→ 0 is also exact. By Corollary 1.3

we have B⊗UR∼=BR and (BR∩T )⊗UR∼=(BR∩T )R. But

(BR∩T )R⊆(BR)R⊆BR⊆(BR∩T )R, hence (BR∩T )R=BR (this is BR is T–closed).

Therefore B⊗UR∼=(BR∩T )⊗UR and hence (BR∩T/B)⊗UR=0. Since R is faithfully

flat as left U–module, then (BR∩T )/B=0 and hence BR∩T =B. Now define the map

α : { right U–modules in T} → {T–closed right R–modules in S} by α(B) =BR.

Since (BR∩T )R =BR , BR is T–closed and α is well defined. The left and right

inverse map is given by β(A) =A∩T because αβ(A) =α(A∩T ) = (A∩T )R=A for

every T–closed right R–module A in S and βα(B) =β(BR) =BR∩T =B for every

right U–module B in T . Therefore α is a bijection.

The hypothesis of the theorem can be changed to S being faithfully flat as left

T–module as the following technical lemma shows

Lemma 1.5 Let S be a T–Galois extension of R. Assume that T is also flat as

left U–module and there exists an element w∈T with tr(w)=1. Then:

(1) S is flat as left T–module if and only if R is flat as left U–module.

(2) S is faithfully flat as left T–module then R is faithfully flat as left

U–module.

Proof. (1)(⇒) U is a direct summand of T as U–module because the trace

map splits. But it is clear that S∼=T⊗UR also as left U–modules (lemma 1.1 and

lemma 1.2), hence R is a direct summand of S as left U–modules. Now T is flat as

left U–module so S is flat as left U–module and hence R being a direct summand

of S is also flat as left U–module.

(⇐) Since R is flat as left U–module and U⊆T , then T⊗UR is flat as left T–module.

But S∼=T⊗UR also as left T–modules by lemma 1.1 and lemma 1.2. Hence S is flat

as left T–module.

(2) Let M be a right U–module with M⊗UR=0. Then 0=M⊗UR⊗RS∼=M⊗US∼=

(M⊗U T ) ⊗T S but S is faithfully flat as left T–module, hence M⊗U T = 0. Since

the action of G on T is G–Galois and there is an element of trace 1, U is direct

summand of T as left U–module and thus M=0.

With this lemma we can rewrite theorem 1.4 as follows:

Theorem 1.6 Let S be a T–Galois extension of R. Assume S is faithfully flat

as left T–module, T is flat as left U–module, and there exists an element w ∈ T

with tr(w)=1. Then these is a one–to–one correspondence between the set of right

U–modules in T and the set of T–closed right R–modules in S.

§2. Central Extensions

Consider now the case T =Z(S), the center of S. Then a Z(S)–Galois extension

of R is a “central Galois extension” of R as in [SM]. In the same article, Szeto defines

a subring A of S containing R to be a “central extension” over R if A=Z(A)R where

Z(A) is the center of A. We now show that the Z(S)–closed right R–modules which

are subrings are exactly the central extensions of S over R.

Proposition 2.1 Let S be a central Galois extension of R. A subring A of S

containing R is a Z(S)–closed right R–module if and only if A is a central extension

over R.

Proof.(⇐) Let A be a central extension over R, then A is a right R–module in S

with A=Z(A)R. Let x∈Z(A), so xr=rx for all r∈R,but S=Z(S)R, (lemma 1.1)

so x∈Z(S) and Z(A)⊆Z(S). Since R⊆A we have U =Z(S)G ⊆Z(A) and hence

Z(A) is a right U–module in Z(S). Then by proof of theorema 1.4, Z(A)R = A is

a Z(S)–closed right R–module.

(⇒) Let A be a subring of S containing R which is Z(S)–closed as right R–module,

then (A∩Z(S))R=A. We have Z(A)R⊆A=(A∩Z(S))R⊆Z(A)R since A∩Z(S)⊆

Z(A) in general. So A=Z(A)R and A is a central extension of R.

On the other side of the picture, it is clear that every right U–module in Z(S)

which is an algebra, is a U–subalgebra of Z(S) and viceversa. Also if A is a subring

of S containing R, A ∩ Z(S) is a U–subalgebra of Z(S) and if B is a U–subalgebra

of Z(S) then BR is a subring of S containing R.

Furthermore, since Z(S) is commutative, there is an element w ∈ Z(S) with

tr(w)=1, see [DI], and hence by restricting the map α in the proof of theorem 1.4

to these right U–modules of Z(S) which are algebras, we obtain theorem 3.4 of [SM]

as the following corollary:

Corollary 2.2 Let S be a central Galois extension of R. Assume S is faithfully

flat over Z(S), then there exists a one–to–one correspondence between the set of

U–subalgebras of Z(S) and the set of central extensions of R in S.

§3. H–separable extensions

In [H1] Hirata defined H–separable extensions of rings as generalizations of Azu-

maya algebras. A ring extension B ⊂ A is said to be H–separable if A⊗BA is

isomorphic to a direct summand of An as A − A–bimodules for some n. Let X

be a subset of A, let CA(X) = {a ∈ A/ax = xa ∀x ∈ X} be the centralizer of X

in A and let Z(A) be the center of A. There is a natural A − A–bimodule map

ϕ : A⊗BA → HomZ(A)(CA(B), A) defined by ϕ(a ⊗ b)(d) = adb. Sugano [S2] has

shown that A is H–separable over B if and only if CA(B) is f. g. projective as

Z(A)–module and the natural map ϕ is an isomorphism.

Now consider S,G and R as in §1. The action of G on S induces an action of G

on S∗G by conjugation, i.e. gα=gαg−1 ∀α∈S∗G.

Let ∆=CS∗G(S) , C=Z(S∗G) , N={g∈G / gα=α ∀α∈∆} and define

φg = {r ∈ S / r gx= xr ∀x ∈ S}. If φg 6= 0, g is called an ω–inner automorphism.

If φg = 0 ∀g 6= 1, G is called ω–outer . It is easy to see that ∆ is G–stable since

s gα= sgαg−1 = g g−1s α g−1 = g α g−1

s g−1 = g αg−1s= gαs ∀s∈S , α∈∆ , g ∈G and

the group G=G/N acts faithfully on ∆ by gα= gα. We want to relate S∗G being

H–separable over S with S∗G being a ∆–Galois extension.

Lemma 3.1

1) ∆=∑g∈G

φgg

2) ∆G=∆G=C

Proof. 1) Let α =∑g∈G

rgg ∈∆, then sx = xs ∀s ∈ S and therefore∑g∈G

srgg =∑s∈G

rggs g. Thus rg

gs=srg ∀g∈G ∀s∈S, so rg∈φg ∀g∈G. The other inclusion is

proven in a similar way.

2) The first equality is clear. Now let α∈∆G, so gα=gαg−1=α and αg=gα ∀g∈G,

that is α commutes with every element g ∈G, but by definition α also commutes

with every s∈S; hence α∈C. Conversely if α∈C, then αg= gα ∀g ∈G and then

gα=α ; ∀g∈G, so α∈∆G.

Now let’s assumme S ∗G is H–separable over S so ∆ is f.g. projective as

C–module. We are going to show that the action of G on ∆ is G–Galois.

Lemma 3.2 Let S∗G be H–separable over S. Then

1) Z(∆)=Z(S)

2) C=Z(S)∩R

Proof. 1) Since S∗G∼=∑g∈G

⊕(S ⊗ g) as S–S–bimodules, CS∗G(∆) =S (proposi-

tion 1.3 [S1]). Hence Z(∆)=CS∗G(∆)∩∆=S∩∆=S∩CS∗G(S)=Z(S).

2) Since C⊆∆ , C⊆Z(∆)=Z(S) by part 1). But also by lemma 3.1, C=∆G hence

C⊆Z(S)∩R. The other inclusion is clear. So C=Z(S)∩R.

Proposition 3.3 Let S ∗G be H–separable over S. Then G is ω–outer and

∆=Z(S).

Proof. By lemma 3.1 and 3.2, ∆G = ∆G = C = Z(S)∩R⊆ S. Let rg ∈ φg, so

x=rgg∈∆, and hence trG(x)= trG(rgg)=∑h∈G

hrggh−1=

∑h∈G

hrghgh−1∈S. But then

hrg =0 if hgh−1 6=1, this is, hrg =0 if g 6=1 and so rg =0 if g 6=1. Therefore since rg

was arbitrary, φg = 0 if g 6= 1, thus G is ω–outer and ∆ =φ1 · 1 by lemma 3.1. But

φ1=Z(S), so ∆=Z(S).

Remark. We can show furthermore that N is trivial. For, assume gx=x for all

x∈Z(S)=∆, then gx=xg and thus g∈CS∗G(∆)=S. Therefore g=1 and G = G.

Hence if S∗G is H–separable over S ,∆ is a commutative ring and thus to show

that the action of G on ∆ is G–Galois we can use [DI, proposition III. 1.2],which

states that a commutative B–algebra A is G–Galois over B with G a finite group of

automorphisms of A if and only if 1) AG=B, 2) A is a separable B–algebra, and

3) for each non–zero idempotent e∈A and each pair g 6=h∈G there is an element

x∈A such that gxe 6=hxe.

Theorem 3.4 Let S∗G be H–separable over S. Then the action of G on ∆ is

G–Galois over ∆G=C.

Proof. By lemma 3.1 ∆G = C, and by [S1, proposition 1.3] ∆ is a separable

C–algebra. Now assumme that there exists a non zero idempotent e∈∆ and a pair

h 6= g ∈G such that gxe= hxe for all x ∈∆. If we let e′ = g−1e, we have e′ 6= 0 and

xe′= g−1hxe′=e′ g−1hx since ∆ is commutative. But G is ω–outer, hence g−1h=1,

thus g=h, a contradiction. Therefore ∆ is G–Galois over C by [DI, proposition III.

1.2].

Corollary 3.5 Let S be a commutative ring. S∗G is H–separable over S if and

only if S is G–Galois over R.

Proof. (⇒) Since S is commutative, ∆ = Z(S) = S and C =R by lemma 3.2

and proposition 3.3. The result follows from theorem 3.4.

(⇐) This is the first part of [MS, theorem 3.3] of which this corollary is an improve-

ment.

Consider again the action of G on S ∗G by conjugation. It follows that the

centralizer of G in S ∗G is precisely equal to the fixed ring (S ∗G)G = I, which in

the language of C∗–algebras is called the algebra of G–central functions, (see [OP]).

Hence we obtain in the context of §1 the following

Theorem 3.6 Let S∗G be H–separable over S. Then S∗G is a Z(S)–Galois

extension of I and therefore S∗G=Z(S)I.

Proof. The first statement follows from Theorem 3.4 and the definition of T–

Galois extensions in §1. Moreover by lemma 1.1, S∗G=Z(S)I.

Acknowledgements. First I would like to express my gratitude to the Centre de

Reserca Matematica in Barcelona, Spain, where this work was done, for inviting me

to spent this year with them. I would also wish to thank the referee for his or her

helpfull comments.

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