What is $\mathbb{Z}/n\mathbb{Z}\otimes_\mathbb{Z} m\mathbb{Z}$? I would like to know what $\mathbb{Z}/n\mathbb{Z}\otimes_\mathbb{Z} m\mathbb{Z}$ is isomorphic to, where $n,m\in\mathbb{N}$. Of course there will likely be cases depending on coprimeness and whatnot; could someone demonstrate how to compute these for me? This is not homework.
 A: $\newcommand{\ZZ}{\mathbb{Z}}$
Since $m\ZZ$ and $\ZZ$ are isomorphic as $\ZZ$-modules, we can create an isomorphism of $\ZZ$-modules between $\ZZ/n\ZZ \otimes_\ZZ m\ZZ$ and $\ZZ/n\ZZ \otimes_\ZZ \ZZ$.
Let $f : \ZZ/n\ZZ \times m\ZZ \to \ZZ/n\ZZ \otimes_\ZZ \ZZ$ be given by $f(\overline{a}, mb) = \overline{a} \otimes b$. It is well-defined iff $m \ne 0$, and bilinear:
$$ f(\overline{a}_1 + \overline{a}_2, mb) = (\overline{a}_1 + \overline{a}_2) \otimes b = \overline{a}_1 \otimes b + \overline{a}_2 \otimes b = f(\overline{a}_1, mb) + f(\overline{a}_2, mb) $$
$$ f(\overline{a}, mb_1 + mb_2) = \overline{a} \otimes (b_1 + b_2) = \overline{a} \otimes b_1 + \overline{a} \otimes b_2 = f(\overline{a}, mb_1) + f(\overline{a}, mb_2) $$
$$ f(n \cdot \overline{a}, mb) = f(\overline{na}, mb) = \overline{na} \otimes b = n \cdot (\overline{a} \otimes b) = \overline{a} \otimes nb = f(\overline{a}, n \cdot mb) $$
So the corresponding map $\alpha : \ZZ/n\ZZ \otimes_\ZZ m\ZZ \to \ZZ/n\ZZ \otimes_\ZZ \ZZ; \ (\overline{a} \otimes mb) = \overline{a} \otimes b$  is a well-defined $\ZZ$-module homomorphism. It is clearly surjective, and so we only need show that it has trivial kernel. Thankfully, all elements in the domain are simple:
$$\sum (\overline{a}_i \otimes mb_i) = \sum a_i b_i (\overline{1} \otimes m) = \left( \sum a_i b_i \right) (\overline{1} \otimes m) $$
If $\alpha(n(\overline{1} \otimes m)) = 0$, then:
$$0 = \alpha(n(\overline{1} \otimes m)) = n(\overline{1} \otimes 1) = (\overline{n} \otimes 1) \implies \overline{n} = 0 \implies n(\overline{1} \otimes m) = 0 $$
So $\ZZ/n\ZZ \otimes_\ZZ m\ZZ$ and $\ZZ/n\ZZ \otimes_\ZZ \ZZ$ are isomorphic. And the latter is just $\ZZ/n\ZZ$. So $\ZZ/n\ZZ \otimes_\ZZ m\ZZ \cong \ZZ/n\ZZ$.
There's probably a prettier way to do this that exploits more universal properties, but I couldn't quite pin it down.
A: Hint: For $A$ commutative ring, $I$ ideal in $A$ and $E$ $A$-module, we have:
$$ A/I\otimes_A E \cong E/IE,$$
via $A$-module morphisms $\hat{a}\otimes x \to \overline{ax}$ and $\overline{x} \to \hat{1}\otimes x$.
For clear, detailed proofs, see this paper's Theorem 4.5 or Serge Lang's Algebra, Chapter XVI.  
In particular, for an abelian group $G$:
$$\mathbb{Z}/n\mathbb{Z}\otimes_{\mathbb{Z}} G \cong G/nG.$$
A: There are two ways to think of this tensor product: as a module, and as a ring (without unity).
As Henry has pointed out, $m\mathbb{Z} \cong \mathbb{Z}$ as $\mathbb{Z}$-modules.  We are assuming that $m\neq 0$; otherwise, $m\mathbb{Z}=0$ and the tensor product is $0$.
It is true, in general, that $M\otimes_R R \cong M$.  There are a number of ways to prove this, but it should appear imemdiately in any book that covers tensor products.  So we have, as $\mathbb{Z}$-modules (i.e. abelian groups), $\mathbb{Z}/n\mathbb{Z} \otimes_\mathbb{Z} m\mathbb{Z} \cong \mathbb{Z}/n\mathbb{Z}$.
As rings without unity, we need to be careful, as we do not have $m\mathbb{Z} \cong \mathbb{Z}$.  Since we already know that the product is $\mathbb{Z}/n\mathbb{Z}$ as an abelian group, we just need to determine the multiplication.  Because we have identified $m\in m\mathbb{Z}$ with $1\in\mathbb{Z}$, we have forced the multiplication to work as follows: $a\bullet b = mab$.
There is a nicer way to look at this, however.  The following general fact holds: if $R$ is a ring, $I$ is an ideal, and $S$ is an $R$-algebra, then $R/I \otimes_R S \cong S/IS$ as $R-$algebras.  Applying this here, $\mathbb{Z}/n\mathbb{Z} \otimes_\mathbb{Z} m\mathbb{Z} \cong m\mathbb{Z}/mn\mathbb{Z}$ as rings.
More concretely (if this abstract business is outside of your comfort zone), we are identifying the simple tensor $\overline{r}\otimes ms$ with the equivalence class $\overline{rms}$ modulo $mn$.  Note that making this rigorous requires something like Henry's proof.
