Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Say that an abelian group $A$ is tensor-commutative if the equality $x\otimes y=y\otimes x$ holds in $A\otimes_{\mathbb Z}A$ for all $x,y$ in $A$.

The first question is somewhat vague:

Question 1. Can one characterize the tensor-commutative abelian groups?

It is easy to see that subquotients of $\mathbb Q$ are tensor-commutative, but I haven't been able to come up with other examples, whence

Question 2. Are there tensor-commutative abelian groups which are not subquotients of $\mathbb Q$?

[This is (I think) a natural continuation of Paul Slevin's question Are bimodules over a commutative ring always modules? More precisely, Paul asks implicitly which are the commutative rings $R$ such that any $R$-module admits only the boring $(R,R)$-bimodule structure. These are exactly the tensor-commutative rings. (In particular this condition depends only on the additive group of $R$.)]

EDIT. As explained here, the answer to Question 2 is yes. So, it seems natural to ask

Question 3. Is there a tensor-commutative ring which is not a subquotient of $\mathbb Q$?

share|improve this question
add comment

3 Answers 3

up vote 7 down vote accepted

If $A$ is a tensor-commutative abelian group, its localization $A_{\mathbb Q}=A\otimes_{\mathbb Z}\mathbb Q$ is at most one dimensional as a rational vector space. Indeed, one easily sees that $A_{\mathbb Q}$ is a «tensor-commutative $\mathbb Q$-vector space» and this is only possible if $A_{\mathbb Q}$ is at most one dimensional.

This tells us that the rank of $A$ is at most $1$. Since torsion-free groups of rank $1$ are isomorphic to subgroups of $\mathbb Q$, this deals with the torsion-free case.

On the other end of the spectrum, suppose $A$ is torsion and tensor-commutative. It is the direct sum of its $p$-primary components for all $p$, and each of these is tensor-commutative. Conversely, if the $p$-primary components are tensor-commutative, $A$ is (because the tensor product of the components corresponding to different primes is simply zero) We need only look at $p$-torsion tensor-commutative groups.

Let $A$ be such a thing. Then $A/pA$ is a $\mathbb Z/p\mathbb Z$ vector space which is tensor-commutative, so it must be of dimension at most $1$, that is, zero or a cyclic group of order $p$. Maybe one can keep this going and show that $A$ must be a Prüfer $p$-group in this case? To do this one wants to know if $pA$ is also tensor-commutative.

share|improve this answer
add comment

The answer to Question 2 is yes. Here is an example of a tensor-commutative abelian group which is not a subquotient of $\mathbb Q$: $$ M:=\mathbb Q\oplus\mathbb Q/\mathbb Z. $$ Indeed, $M$ is tensor-commutative because the morphism mapping $(a,b)\otimes(c,d)$ (with $(a,b),(c,d)\in M$) to $ac$ is an isomorphism from $M\otimes M$ onto $\mathbb Q$, and $M$ is not a subquotient $S$ of $\mathbb Q$ because any such $S$ is torsion or torsion-free (more precisely, $S$ is a submodule of $\mathbb Q$ or a subquotient of $\mathbb Q/\mathbb Z$).

EDIT. The answer to Question 3 is no. Here is a proof.

Let $A$ be a tensor-commutative commutative ring, and let us show that $A$ is a subquotient of $\mathbb Q$.

We'll freely use Mariano's answer.

Assume $A\neq0$. Let $R$ be the prime ring of $A$.

Case 1: $R\simeq\mathbb Z/(n),\ n\ge2$. Claim: $A=R$.

We can assume that $n$ is a power of a prime. Then $A$ is an $R$-algebra containing $R$.

Suppose by contradiction that there is an $a$ in $A$ which is not in $R$. For any subgroup $G$ of $A$ containing $1$ and $a$, form the commutator $$ c_G:=1\otimes a-a\otimes1\in G\otimes_RG=G\otimes_{\mathbb Z}G. $$ It suffices to show $c_A\neq0$. It is even enough to check that $c_G$ vanishes for every finitely generated subgroup $G$ of $A$ containing $1$ and $a$. But this is clear because $R$ is a direct summand of $G$.

[Details: $G=R\oplus H;\ 1=(1,0);\ a=(r,h);\ h\neq0$.]

Case 2: $R=\mathbb Z$. Claim: $A\subset\mathbb Q$.

As explained by Mariano, it suffices to prove that $A$ is torsion-free.

The rest of the argument is very similar to the previous one:

Suppose by contradiction that there is a nonzero torsion element $t$ in $A$. For any subgroup $G$ of $A$ containing $1$ and $t$, form the commutator $$ c_G:=1\otimes t-t\otimes1\in G\otimes_\mathbb ZG. $$ It suffices to show $c_A\neq0$. It is even enough to check that $c_G$ vanishes for any finitely generated subgroup $G$ of $A$ containing $1$ and $t$.

Let $G$ be such a subgroup. Then $G$ is isomorphic to $\mathbb Z\oplus T$, where $T$ is the torsion subgroup of $G$, and the result is clear.

[Details: $G=\mathbb Z\oplus T;\ 1=(z,t_1);\ z\neq0;\ 0\neq t=(0,t)$.]

share|improve this answer
add comment

Q1. I have asked the same question on mathoverflow. Will Sawin has found a classification of tensor-commutative (I call them symtrivial) modules over a Dedekind domain.

Q3. Do you mean subquotient after forgetting the ring structure? Tensor-commutative rings are precisely the epimorphisms of commutative rings with domain $\mathbb{Z}$. They can be classified, as rings. See here.

share|improve this answer
    
Dear Martin: Thanks a lot for your nice answer (+1). In Q3, $\mathbb Q$ is viewed as an abelian group. In the whole question, I only considered abelian groups. It was certainly a good idea of yours to consider a more general setting. –  Pierre-Yves Gaillard Feb 20 '13 at 14:25
    
Perhaps I will use "$\otimes$-commutative" instead of "symtrivial" in my thesis ... I quite like your terminology. –  Martin Brandenburg Feb 21 '13 at 1:18
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.