The Jacobson Radical’s Role in Isomorphism Theorems for Mixed

Modules

Mary FlaggUniversity of Houston

Isomorphism Theorems

• A class of groups (modules over a ring R)is said to satisfy an isomorphism theorem if

a ring (R-algebra) isomorphism between the endomorphism rings of two groups (modules) implies the groups (modules) are isomorphic

Isomorphism Theorems

• Baer-Kaplansky – torsion groups

• Wolfson – torsion-free modules over a complete discrete valuation domain

• May and his colleagues – several classes of mixed modules over a complete discrete valuation domain

The Jacobson Radical

• The Jacobson radical of a ring is the intersection of all maximal ideals.

• It contains no non-zero idempotents• The structure of J(E) is not fully known

except in the simplest cases.• The most useful characterization is that

J(E) is the union of all quasi-regular ideals of E. (f in J(E) implies 1+f is invertible)

Question

• For classes of groups and modules that satisfy an isomorphism theorem, how much of the module’s structure is contained in the Jacobson radical?

Setting

• R a discrete valuation domain with prime p and field of quotients Q

• M and N – left R-modules • E(M) the endomorphism ring of M, and

J(E(M)) = the Jacobson radical of E(M)• All Jacobson radical isomorphisms are R-

algebra isomorphisms in the natural way.• T(M) the torsion submodule of a module M

Inspiration

• Torsion Groups –Hausen, Praeger and Schultz –Two abelian p-groups with unbounded basic

subgroups are isomorphic if and only if the Jacobson radicals of their endomorphism rings are isomorphic as rings.

This result carries over naturally to torsion modules over a discrete valuation domain

Torsion – free modules

• (Flagg, 2008) If M and N are torsion-free non-divisible modules over a complete discrete valuation domain, then M and N are isomorphic if and only if their endomorphism rings are isomorphic as R- algebras

Mixed Modules

• (May 1983) Mixed modules of torsion-free rank one with totally projective torsion submodules satisfy an isomorphism theorem.

• (May 1990) Mixed modules of at most countable torsion-free rank and a totally projective torsion submodule satisfy an isomorphism theorem.

The first step for mixed modules – determining the torsion submodule

• If we assume that the torsion submodules are non-trivial, a straightforward application of the proof of the Baer-Kaplansky Theorem shows that the torsion submodules of two mixed modules are isomorphic if their endomorphism rings are isomorphic R-algebras

• Furthermore, an isomorphism between the torsion submodules is constructed that induces the endomorphism R-algebra isomorphism on the torsion.

Problem

• The Baer-Kaplansky Theorem relies on the identification of primitive idempotents in the endomorphism algebra

• We know (Hausen and Schultz) that if a torsion group (R-module) is bounded modulo its divisible submodule, the Jacobson radical determines the structure of the group in many cases, but the isomorphism is not induced.

Determining the Torsion

• R need not be complete in the p-adic topology

• M/T(M) no longer needs to be divisible• All modules are assumed to have a torsion

submodule with an unbounded basic submodule

• Improvements over Flagg(2009, Comm. Alg.)

Theorem

• Let M be an R-module with torsion submodule which contains an unbounded basi submodule. Let N be arbitrary.

If there exists an isomorphism

Then, there exists an isomorphism : ( ( )) ( ( ))J E M J E N

: ( ) ( ) that induces on the torsionT M T N

Rank One Mixed ModulesTheorem:Let M and N are mixed modules of torsion-free rank one

with torsion submodules that contain an unbounded basic submodule.

If M is split and its torsion-free summand is isomorphic to R, and there exists an isomorphism

Then there exists

Note that totally projective torsion is not required for this.

: ( ( )) ( ( ))J E M J E N

: that induces .M N

M/T(M) Divisible

• If M/T(M) is isomorphic to Q, then even the split case is hard, so we need at this point to assume that T(M) is totally projective (which includes the condition that it is reduced)

• The split case is hard since J(E(Q))=0

The Technique

• Assume M and N are mixed modules of torsion-free rank one with unbounded, totally projective torsion submodules.

• Assume there exists an isomorphism

• Then there exists an isomorphism

• Call this common torsion submodule T

: ( ( )) ( ( ))J E M J E N

: ( ) ( ) that induces on the torsionT M T N

The Split Case

• If M is torsion-free rank one and split and it’s torsion-free summand is isomorphic to Q, then

• J(E(M)) is isomorphic to J(E(T))• If the torsion is totally projective, then a

split module is the only case in which the Jacobson radical of E(M) is isomorphic to J(E(T)).

• This determines splitting.

If M is not split• Embed M and N into T*, the cotorsion hull

or T, as pure submodules containing T.• Embed J(E(M)) and J(E(N)) in E(T*) using

the natural embedding.• Use the fact that pE(M) is contained in

J(E(M)) and pE(N) is contained in J(E(N)) to show that under this embedding

E(M) = E(N)By May (1990), M is isomorphic to N and the

isomorphism induces the Jacobson radical isomorphism.

Warfield Modules

• Recall that a module is called a Warfield module if it is a direct summand of a simply presented module.

• Assume that all modules are reduced, the torsion-free rank is finite, and the torsion is unbounded.

• Also assume R is complete in the p-adic topology.

Background for Warfield Modules

• Let M be a torsion-free finite rank Warfield module as described in the previous slide.

• Then, using the decomposition bases for M, we can show that pE(M) is contained in J(E(M))

• The modules M may each be decomposed

With F free of finite rank and M’/T(M) divisible and nonsplit.

'M F M

• Note that the ideals allow us to “separate this problem”

• Useful ideals

The ideal I is isomorphic to J(E(F))The ideal D is isomorphic to J(E(M’))

: 0/

: 0/ /

H tJ tJK J HL J JD J L I L H

tt

Theorem

• Let M and N be Warfield modules of finite torsion-free rank and assume that there exists an isomorphism

• Then M is isomorphic to N.• I am not yet sure if it is induced.

: ( ( )) ( ( ))J E M J E N

Final Notes• If a torsion-free module has countable rank, pE

is not contained in the Jacobson radical. So, the restriction to finite rank Warfield modules may be as good as this gets.

• For mixed modules we have the torsion ideals of the nilradical, Jacobson radical and Pierce radical equal, so the results for determining the torsion also hold if there is an R-algebra isomorphism between the Pierce radicals or the nilradicals of the endomorphism rings of two modules

Thank You

Mary Flaggmflagg@math.uh.eduUniversity of Houston