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I think I'm missing something important about tensor products as I look over this question from an old qualifying exam:

Let $R$ be a principal ideal domain and let $A$ and $B$ be finitely generated $R$-modules. Show that if $a\in A$ and $b\in B$ are not torsion elements, then $a\otimes b \neq 0$ in $A\otimes_R B$.

I tried using the definition of tensor products, but did not have any luck. Any suggestions to solve this problem would be appreciated!

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Quite untrue, if you take $A=B=R=\mathbb{Z}$.This is because $A \otimes B$ is trivial, right? Or am I missing something? – Ravi Dec 30 '12 at 20:09
@FortuonPaendrag in your case $A \otimes_R B = \mathbb Z \otimes_\mathbb{Z} \mathbb Z = \mathbb Z$ but this doesn't seem trivial to me and more important for each pair of elements $a,b \in \mathbb Z$ you get that $a \otimes b = ab$ which is not $0$. – Giorgio Mossa Dec 30 '12 at 20:11
of course we suppose that $a,b$ as above must be not torsion element, i.e. both non null. :) – Giorgio Mossa Dec 30 '12 at 20:22
up vote 0 down vote accepted

Note. I've changed $A$ by $M$ and $B$ by $N$.

If $M$ and $N$ are free, we are done.

Now we reduce the general case to the free case. We have the following isomorphism: $$M/t(M)\otimes_R N/t(N)\simeq M\otimes_RN/S,$$ where $S$ is the submodule of $M\otimes_R N$ generated by the tensor monomials $x\otimes y$ with $x\in t(M)$ and $y\in t(N)$. (This isomorphism is canonical and sends $\bar x\otimes \bar y\in M/t(M)\otimes_R N/t(N)$ to $\widehat{x\otimes y}\in M\otimes_RN/S$.) If $a\otimes b=0$, then its residue class in $M\otimes_RN/S$ is also $0$. This shows that $\bar a\otimes\bar b=0$ in $M/t(M)\otimes_R N/t(N)$. But $\bar a\neq 0$ and $\bar b\neq 0$ and $M/t(M)$ and $N/t(N)$ are free, so...

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Your answer makes sense. I didn't occur to me to look at the isomorphism you described. Thanks! – phenomenalwoman4 Dec 31 '12 at 15:35

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