# Tensor-commutative abelian groups

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 $\mathbb Q$?

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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.

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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 $\mathbb Q$.

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

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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.

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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