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Let $A$ be an abelian group (i.e., a $\mathbb Z$-module), and consider the tensor product $\mathbb Q\otimes_{\mathbb Z} A$. I want to show that the element $\frac{1}{d}\otimes n=0$ iff there's a positive integer $r$ such that $rn=0$.

I have shown that this tensor product is isomorphic to $U^{-1}A$, where $U$ is the set of nonzero integers, which allows me to extend the $\mathbb Z$-module $A$ to a $\mathbb Q$-module. The isomorphism is given by $\frac{a}{b}\otimes n\mapsto b^{-1}(an)$, and the inverse by $b^{-1}n\mapsto \frac{1}{b}\otimes n$.

Thus, if $\frac{1}{d}\otimes n=0$, then it follows that $d^{-1}n=0$. But this means that $n=0$, right? I'm a bit confused as to why I need to find an $r$...perhaps my confusion stems from not exactly understanding the equivalence relation on $U^{-1}A$? I think the relation is $u_1^{-1}n_1=u_2^{-1}n_2$ iff $u_2n_1=u_1n_2$. But this means that, since $0=1^{-1}\cdot 0$, $d^{-1}n=0$ iff $n=1$...do I have the wrong equivalence relation?

Nishant
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    The equivalence relation you've given is only the case for integral domains. The general equivalence relation is $(u_1, n_1) \sim (u_2, n_2)$ iff there's some nonzero $r \in U$ such that $r(u_1n_2-u_2n_1)=0$. –  May 20 '14 at 19:09
  • But $\mathbb Z$ is an integral domain...in addition, this problem comes from Dummit and Foote, and that problem also said that $R$ was an integral domain, and $U^{-1}N$ was the extension of the $R$-module $N$ to the field of fractions of $R$... – Nishant May 20 '14 at 19:20
  • Oh I see, you need that condition for the relation to be transitive, since you can't just "cancel" the nonzero elements, like you could if you were constructing the field of fractions. – Nishant May 20 '14 at 19:38
  • I made a stupid mistake, sorry. Nonetheless the second construction is the correct one. (I was thinking of $A$ as a ring for whatever reason, and in that case, the ring $A$ must have been an integral domain to use your construction.) –  May 20 '14 at 19:38
  • All right, thanks! – Nishant May 20 '14 at 19:41
  • To see why, by the way, try to construct $U^{-1}\Bbb Z_2$ with both ways - only one will actually be $\Bbb Q \otimes_{\Bbb Z} \Bbb Z_2$. –  May 20 '14 at 19:43
  • So the latter is just the trivial group, since you're tensoring a divisible group with a torsion group. However, the former contains the element $2^{-1}\cdot 4$, which is nor equivalent to $0$ since the numerator is not $0$. However, multiplying by $2$ makes the $u_1r_2-u_2r_1$ terms into $0$. – Nishant May 20 '14 at 19:50
  • Almost - your "equivalence relation" isn't even an equivalence relation here, which is the real problem: $(2,1)\sim (2,0)\sim (1,0)$ but $(2,1)\not\sim (1,0)$. (But the rest was spot on.) –  May 20 '14 at 20:16
  • More generally: $$ \text{$R$ commutative ring, $M$ torsion $R$-module, $N$ divisible $R$-module $\implies M \otimes_R N = 0$.} $$ – Watson Jan 23 '17 at 13:36

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