FENG Jiangchao (), SHEN Ran ( )*, ZHANG Jiangang ()

1 College of Science, Donghua University, Shanghai 201620, China 2 Department of Mathematics, Shanghai Normal University, Shanghai 200234, China

Abstract： In algebra, the Jacobson-Bourbaki theorem is especially useful for generalizations of the Galois theory of finite, normal and separable field extensions. It was obtained by Jacobson for fields and extended to division rings by Jacobson and Cartan who credited the result to unpublished work by Bourbaki. In 2005, the Jacobson-Bourbaki correspondence theorem for commutative rings was formulated by Winter. And this theorem for augmented rings was formulated by Kadison in 2012. In this paper, we prove the Jacobson-Bourbaki theorem for noncommutative rings which is finitely generated over their centers. We establish a bijective correspondence between the set of subdivisions which are right finite codimension in A and the set of Galois rings of the additive endomorphisms End A of A which is finitely generated over its center.

Key words： noncommutative rings; Galois rings; simple module

Introduction

In mathematics, Galois theory provides a connection between field theory and group theory. Using Galois theory, certain problems in field theory can be reduced to group theory, which is in some sense simpler and better to be understood. The extension of Galois theory to normal extensions is called the Jacobson-Bourbaki correspondence. It is a correspondence between some subdivishion rings of a field and some subgroups of a Galois group by a correspondence between some subdivision rings of a division ring and some subalgebras of an algebra, instead of the correspondence between some subfields of a field and some subgroups of a Galois group. The Jacobson-Bourbaki theorem implies both the usual Galois correspondence for subfields of a Galois extension, and Jacobson’s Galois correspondence for subfields of a purely inseparable extension of exponent at most 1, see Refs.[4-9]. This theorem was successfully established by Nathan Jacobson and Henn Cartan which can be stated.

Theorem1[7]LetEbe a field,Φthe set of subfieldsFofEof finite codimension inE, Δ the set of subringsLof the ring of endomorphisms of the additive group ofEsuch that

(1)L?EIE, whereEIEis the set of multiplications inEby the elements ofE;

(2)Las left vector space overEis finite dimensional.

Auslanderetal. introduced the notion of Galois extension of a commutative ring[10-13]. Chaseetal. then adopted the Auslander-Goldman Galois extensions to generalize the classical Galois correspondence theorem from fields to commutative rings[13]. This theorem can be stated as follows.

Winter[14]generalized the Jacobson-Bourbaki theorem from fields to commutative rings and established the Galois rings correspondence for commutative rings. This theorem is as follows.

Theorem3[18]There exists a bijective correspondence

between the set of subdivision rings ofAof finite codimension and the set of Galois rings ofA.

Thoerem4[15]Let (A，D) be an augmented ring. There is a one-to-one correspondence between the set of division ringsFwithinA, whereFis a subring ofA, andAis a finite-dimensional right vector space asF, and the set of Galois subrings of EndA.

NotationThe notation used here is fairly standard except where otherwise stated. LetImdenote the identity matrix ofmdimension. Theδijdenotes the Kronecker symbol. LetAdenote a noncommutative ring. LetFdenote a subdivision ring ofF. LetAFdenote thatAis a right vector space overF. Let [A∶F]rdenote the dimension ofAwhich is a right vector space overF. Let EndAdenote the additive homomorphism ring ofA. EndAFdenote all of linear tranformation ofAwhich is a right vector space overF.

1 Preliminaries

In this section, we give some definitions and a preliminary theorem[14].

Definition1[14]LetAbe a noncommutative ring. An endomorphism ring ofAis a subringRof the ring EndAwhich containsAIA. Its centralizer is the subring

AR={b∈A|r(ab)=r(a)b,r∈R,a∈A}.

Definition2[14]LetAbe a noncommutative ring. An endomorphism ringRofAis irreducible ifAis a left simpleR-module.

Definition3[14]LetAbe a noncommutative ring. An endomorphism ringRofAis a Galois ring ifRis an irreducible endomorphism ring which is a finitely generated leftAmodule.

The following theorem is a very important preliminary theorem which is called the Jacobson-Chevalley density theorem.

Theorem5[7-8,17-18]LetRbe a ring. LetMbe a left simpleR-module,Dd=EndRM. If {x1,x2, …，xm} ?Mis a rightD-linearly independence set and {y1,y2, …,ym}?M. Then there existsr∈Rsuch thatr(x1)=y1,r(x2)=y2, …,r(xm)=ym.

It is easy to see the following facts from the above theorem.

(1)D=EndRMis a division ring by Schur’s lemma.

(2)Mis a rightD-module by (rx)f=r(x)ffor allx∈M,r∈R,f∈D.

(3) IfMis a ring, then, obviously,MRis equal toD=EndRM.

2 Main Results

Firstly, we give two important lemmas.

Lemma1LetBbe a commutative subring of noncommutative ringA, andAis finitely generated as a leftB-module.Ris an endomorphism ring such thatR=As1+As2+…+Asn. Supposing thatRcontainsr1r2…rmandAcontainse1e2…emsuch thatri(ek)=δik(1≤i,k≤m). Then there is a positive integer numberMsuch thatm≤M.

ProofAccording to the conditions, we can set up

R=Bs1+Bs2+…+Bsn.

(1)

Letsj(ek)=djk,

By Eq. (1), we have

CD=Im.

Definedφby

φ(x1,x2, …,xm)=

(x1,x2, …,xm)C,(x1,x2, …,xm∈A)

Obviously,φis injective.

Next, we prove thatnis a upper level ofm. If not, we suppose thatm>n, without loss of generality, we can assumem=n+1.

If martixChas a nonzeron×nminor, without loss of generality, we assume that the nonzeron×nminor is in the top-left ofC. Then, forl=1, 2, …,n, we have

whereAijis the algebraic cofactor of the element in theith row and thejth column.

Hence,

This is a contradiction to the injectivity ofφsince

(A1 n+1,A2 n+1, …,An n+1,An+1 n+1)≠(0, 0, …, 0, 0).

So alln×nminors ofCare zero.

SinceCis not zero, letdbe the maximal positive integer number such that all of the (d+1)×(d+1) minors ofCare zero but there exists a nonzerod×dminor ofC.

Without loss of generality, we assume that the nonzerod×dminor is in the top-left ofC.

Then, forl=1, 2,…,d, we have

Obviously,

This is also a contradiction to the injectivity ofφsince

(A1 d+1,A2 d+1, …,Ad d+1，Ad+1 d+1, 0, …, 0)≠(0, …, 0).

Hencem×n. Letn=M, then the Lemma holds.

Lemma2LetBbe a commutative subring of noncommutative ringAandAis finitely generated as a rightB-module.Ris a endomorphism ring such that

R=As1+As2+…+Asn.

Supposing thatRcontainsr1r2…rmandAcontainse1e2…emsuch thatri(ek)=δik(1≤i,k≤m). Then there is a positive integerMsuch thatm≤M.

ProofWriting

and applying both sides toekleads to

(2)

Letsj(ek)=djk,

By Eq.(2), we have

CD=Im.

Definedφ:Bm→Anbyφ(X)=DX,X∈Bm.

Obviously,φis a rightBmodule injective homomorphism.

Letα1,α2, …，αmbe the standard base of rightBmoduleBm, andβ1,β2, …,βMthe generators of rightBmoduleAnsinceAis a finitely generated rightBmodule.

Let

andf:Bm→Bndefined by

f(X)=TX,X∈Bm.

IfTX=0, then

φ((α1,α2, …,αm)X)=(β1,β2, …,βM)TX=0.

Sinceφis injective,α1,α2, …,αmis the standard base of rightBmoduleBm.

We have,

(α1,α2, …,αm)X=0,X=0.

Hence,f:Bm→Bnis a rightBmodule injective homomorphism. Similar to the proof of Lemma 1, we have a positive integerMsuch thatm≤M.

In the following theorem, we generalize Galois ring correspondence theorem from commutative rings to noncommutative rings which is finitely generated over their centers.

Theorem6LetBbe a commutative subring of noncommutative ringA, andAis finitely generated as a left (right)B-module.Ris a Galois ring ofA. Then,F=ARis a subdivision ring ofAof right finite codimension andR=EndAF.

ProofSinceRis a Galois ring ofA, EndRA=AR=Fis a division ring. Lete1,e2, …,emis a linearly independent team. SinceAis a leftRsimple module andD=EndRAis a division ring, so by the Jacobson-Chevalley density theorem, there existr1,r2, …,rm∈Rsuch thatriek=δik(1≤i,k≤m) andr1,r2, …,rmis leftAlinearly independent.

By Lemma 1 and Lemma 2, we havem≤Mfor some positive integer number, which means thatAFhave a maximum linearly independent team, may as well still be set ase1,e2, …,em. HenceAFis a finite dimension right vector space overFande1,e2, …,emis a maximum linearly independent team ofAF.

Obviously,

Ar1?Ar2?…?Arm?R.

Next, for everyf∈EndAF,fis defined byf(e1)=a1,f(e2)=a2, …,f(em)=am. Hence,

f=a1r1+a2r2+…+amrm.

So, we have

EndAF?Ar1?Ar2?…?Arm?R.

It follows that

R=EndAF.

Corollary1LetBbe a commutative subring of noncommutative ringAandAis a finitely generated left (right)B-module.Ris a Galois ring ofA. Then,Ris a free leftAmodule of finite rank.

ProofObviously, by Theorem 6, we haveR=EndAF. SoRis a free leftAmodule of finite rank.

The following theorem is the inverse of the Theorem 6.

Theorem7LetBbe a commutative subring of noncommutative ringAandAis a finitely generated left (right)B-module.F=ARis a subdivision ring ofAof right finite codimension. ThenF=AEnd AF.

ProofLetR=EndAFandF′=AR. Obviously,

F?F′.

SinceAis finite dimension right vector space overF, for everyx∈A,y∈A, there existsf∈Rsuch that

y=f(x).

Hence,Ris a Galois ring ofA. So, by Theorem 6, we have

R=EndAF′=EndAF.

IfFF′, then

F′=F?b2F?…?btF,

and

A=F?b2F?…?btF?…?bmF
(t≥2,m≥2,b2∈F′,b2, …,bm∈AF).

Then, by the Jacobson-Chevalley density theorem, there exists an elementr∈Rsuch that

r(1)=1,r(b2)=…=r(bm)=0.

Then, because ofR=EndAF′, we have

0=r(b2)=r(1)b2=b2.

This is a contradiction. Hence,

F=F′=AEnd AF.

Theorem8LetAbe a noncommutative ring which is finitely generated over its center. Then there exists a bijective correspondence

F→R≡EndAF,R→F≡AR

between the set of subdivision rings ofAof right finite codimension and the set of Galois ringsRofA.

ProofLetBbe the center ofA. Then, by Theorem 6 and Theorem 7, we have a bijective correspondence

F→R≡EndAF,R→F≡AR

between the set of subdivision rings ofAof right finite codimension and the set of Galois rings ofA.

3 Conclusions

In this paper, we prove the Jacobson-Bourbaki theorem for noncommutative rings which is finitely generated over their centers. We establish a bijective correspondence between the set of subdivisions which are right finite codimension inA, and the set of Galois rings of the additive endomorphismsAofAwhich is finitely generated over its center. We generalize the Jacobson-Bourbaki theorem to some extent. But it is an open problem for the general noncommutative rings yet.