# Can the tensor product of two non-free abelian groups be non-zero free?

It's pretty easy to construct an ($$R$$-$$S$$) bi-module $$M$$ and a left $$S$$-module $$N$$ such that neither $$M_S$$ nor $$N$$ is a projective $$S$$-module, but the tensor product $$M_S \otimes N$$ is a non-zero projective $$R$$-module.

However, taking $$R=S= \mathbb{Z}$$ defies my zoo of examples. Here projective=free is closed under direct sums and summands, and so it seems like $$M$$ and $$N$$ can be chosen indecomposable. The indecomposable abelian groups I know don't seem to fit. I don't know how to "uninvert" using a tensor product.

Can the tensor product of two non-free abelian groups be non-zero free?

Inspired by an AoPS question and finals week.

The tensor product $$M\otimes N$$ of abelian groups cannot be non-zero and free unless both $$M$$ and $$N$$ are free. This follows from the following facts.

1. Any subgroup of a free abelian group is free. This holds even in the infinitely generated case. Wikipedia has a proof in its article on free abelian groups.
2. Under certain conditions on $$M$$ and $$N$$, for non-zero $$n\in N$$ the homomorphism \begin{align} &M\to M\otimes N,\\ &m\mapsto m\otimes n, \end{align} is injective. It is easy to see that this holds when $$N$$ is free and $$n$$ is a basis element. It also holds when $$M$$ and $$N$$ are both torsion-free. You could prove this by extension of scalars to reduce it to the case of vector spaces over the rationals (and vector spaces are always free). This does not hold if you merely assume that $$N$$ is torsion-free, as pointed out by Jack Schmidt in a comment (and I must apologize for making the mistake of assuming this in my original answer).
3. Tensor products are right exact. That is, if $$0\to A\to B\to C\to 0$$ is an exact sequence of abelian groups, then $$M\otimes A\to M\otimes B\to M\otimes C\to 0$$ is exact. In particular, if $$M\otimes A$$ maps to zero, this shows that $$M\otimes B$$ and $$M\otimes C$$ are isomorphic, which I will use to quotient out the torsion subgroups before applying 2.

Now, suppose that $$M\otimes N$$ is non-zero and free. Let $$T$$ be the torsion subgroup of $$N$$, so that $$N/T$$ is torsion-free and $$0\to T\to N\to N/T\to 0$$ is an exact sequence. By 3, $$M\otimes T\to M\otimes N\to M\otimes(N/T)\to 0$$ is exact. However, as $$T$$ is torsion, $$M\otimes T$$ will also be torsion, so its image in the torsion-free group $$M\otimes N$$ is zero. This gives an isomorphism $$M\otimes N\cong M\otimes(N/T)$$. Applying the same argument with the torsion subgroup $$S$$ of $$M$$ shows that $$(M/S)\otimes(N/T)$$ is isomorphic to the free group $$M\otimes N$$. Then, as $$M/S$$, $$N/T$$ are torsion-free, picking any non-zero $$n\in N/T$$ and using 2 gives an injection \begin{align} &M/S\to (M/S)\otimes(N/T),\\ & m\mapsto m\otimes n. \end{align} So $$M/S$$ is isomorphic to a subgroup of the free group $$(M/S)\otimes(N/T)$$ which, as stated in 1, means that $$M/S$$ is free. Similarly, $$N/T$$ is free. Applying 2 again, the homomorphism \begin{align} &M\to M\otimes(N/T),\\ & m\mapsto m\otimes n \end{align} is injective for any basis element $$n\in N/T$$. So $$M$$ is isomorphic to a subgroup of the free abelian group $$M\otimes(N/T)\cong M\otimes N$$ and, using 1 once more, must be free. Similarly, $$N$$ is free.

• M need not inject into M⊗N, even when N is torsion-free. Take M=Z/2Z, N=Q. – Jack Schmidt Dec 16 '10 at 23:31
• Whoops! of course. That's a pretty bad mistake. Latlet me think some more about this – George Lowther Dec 16 '10 at 23:34
• No problem. Tensor products are weird. A funny example: Let M = { a/b in Q : b is square-free nonzero, a,b in Z } and N = Q / { a/b in Q : b is odd }, then M⊗N ≅ Z/2Z, I think. The "invertiness" of M kills most of the torsion of N, but not all. – Jack Schmidt Dec 16 '10 at 23:43
• Fixed the argument! (still blushing over the previous mistake though). This proof is a bit longer, but I'll try and clean it up. Hopefully it is correct now... – George Lowther Dec 17 '10 at 0:11
• Cool, and this works for the AoPS problem with Dedekind domains and even hereditary rings. I wonder if there is an example with R=Z[2i] or something silly like that. – Jack Schmidt Dec 17 '10 at 0:22

Any example must arise from infinitely generated groups. Suppose $A$ and $B$ are finitely generated, non-zero, non-free, abelian groups. Being non-free means there is torsion (since we're in the finitely generated case). If every element of $B$ has finite order, then every element of the tensor product has finite order. So, assume $b \in B$ has infinite order, and let $a \neq 0$ be an element of order $n$. Then $a \otimes b$ is non-zero and has order dividing $n$. Hence, $A \otimes B$ has torsion and, consequently, it is non-free.

• Exactly. In the torsion-free, finite rank case you get something similar, though less clear. Suppose a in A can be divided by n. Then a⊗b can be divided by n too. So if A is too divisible, then A⊗B stays too divisible to be free. You can get rid of the problem with torsion, but torsion sticks around. The only way to keep it, is to send everything to 0. Hence I am stuck. – Jack Schmidt Dec 15 '10 at 21:52