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I am trying to show that if $F,H$ are abelian groups with $F$ free abelian, and if $a \in F$ and $h \in H$ are non-zero, then $a \otimes h \ne 0$ in $F \otimes H$.

This is specifically in a section describing the derived functor Tor. Of course, that doesn't mean the solution has to involve that, but there is probably a way. I know that $F$ free abelian means that $F$ is torsion free and hence $\mbox{Tor}(F,A)=0$.

I was trying to use a formulation of $\mbox{Tor}$ in terms of exact sequences. If:

$$0 \to R \stackrel{i}{\hookrightarrow} F \to A \to 0$$ is an exact sequence then $\mbox{Tor}(A,B) =\mbox{ker}(i \otimes 1_b)$

Seemed to me if I picked the right sequence I could get that $\mbox{Tor}=0$ implies that the kernel is trivial, which would give the result, but I can't get this to work

Edit It appears that this is false from the answers below. Here is a link to the question.

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Can you describe more about what you know of tensor products? For example, I think most people would give the solution " $F$ is a direct sum of copies of $\mathbb{Z}$ and tensor product distributes over direct sums, so..." – user641 Apr 13 '11 at 0:45
No, not exactly; see the two answers below. – user641 Apr 13 '11 at 0:56
@Steve, yes retracted after seeing the comment – Juan S Apr 13 '11 at 0:57
up vote 1 down vote accepted

Here is an answer using Tor:

Consider the exact sequence $$ 0\rightarrow\mathbb{Z}\rightarrow F\rightarrow K\rightarrow 0,$$

where the map from $\mathbb{Z}$ to $F$ sends the generator $1$ to $a$, and $K$ is the cokernel. Tensoring with $H$ we get the long exact sequence $$ \cdots Tor(F,H)\rightarrow Tor(K,H)\rightarrow H\rightarrow F\otimes H\rightarrow H\otimes K\rightarrow 0. $$

Now $Tor(F,H)=0$ since $F$ is torsion-free. So the kernel of $H\rightarrow F\otimes H$, given by sending $h$ to $a\otimes h$, is $Tor(K,H)$. This is not always non-zero, and the problem as stated is incorrect. Consider $H=\mathbb{Z}/2\mathbb{Z}$, $F=\mathbb{Z}$, $a=2$.

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thanks for that. I guess we need some additional hypothesis like $H$ is torsion free as well? – Juan S Apr 13 '11 at 1:03
Well actually you can say exactly what you need. If I can write $a=b^m$ for some $b\in F$, then we need $mh\neq 0$. Basically $a$ looks like $(n_1,n_2,\ldots,n_r)$ for some integer $r$ (ignoring the zero coordinates), and if $m=GCD(n_1,\ldots,n_r)$, we need $mh\neq 0$. – user641 Apr 13 '11 at 4:12

Suppose $F=\mathbb Z$ and $H=\mathbb Z/2\mathbb Z$. Let $a=2\in F$ and let $\xi\in H$ be the non-zero element. Then $$a\otimes\xi=(2\cdot1)\otimes\xi=1\otimes(2\cdot \xi)=1\otimes 0=0$$.

It follows that what you want to prove is false.

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