Intuitions about tensor products My Question:

Prove, that as modules, $\mathbb{Z}/(10) \otimes_\mathbb{Z}\mathbb{Z}/(12) \cong \mathbb{Z}/(2)$.

Now I have no idea how to tackle this. I am guessing that in general $\mathbb{Z}/(a) \otimes_\mathbb{Z}\mathbb{Z}/(b) \cong \mathbb{Z}/(\gcd(a, b))$, but this is counter intuitive. Why isn't the amount of elements in a tensor product bigger than the number of elements in each module?
I would like a proof for my question and also intuitions and ways to tackle other similar problems.
 A: In $\mathbf{Z}/a$ you have $a = 0$. In $\mathbf{Z}/a \otimes \mathbf{Z}/b$ you have $a$ and $b = 0$. Therefore you have $ax + by = 0$ for all $x$ and $y$ which implies that $\gcd(a,b) = 0$.
Now because $\mathbf{Z} \cong \mathbf{Z} \otimes \mathbf{Z}$ we get a map $\mathbf{Z} \twoheadrightarrow \mathbf{Z}/a \otimes \mathbf{Z}/b$ by taking the tensor product of the two projections $\mathbf{Z} \to \mathbf{Z}/n$, $n = a, b$. Thus $\mathbf{Z}/a \otimes \mathbf{Z}/b$ is a quotient of $\mathbf{Z}$ where $\gcd(a,b) = 0$.
You can show that $\gcd(a,b)$ is the characteristic of $\mathbf{Z}/a \otimes \mathbf{Z}/b$ from which it follows that $\mathbf{Z}/a \otimes \mathbf{Z}/b \cong \mathbf{Z}/\gcd(a,b)$.

In general, tensor products of torsion-modules behaves much differently than you would expect from how they behave for vector spaces or free modules. There is a connection between flat and torsion-free. Every flat module is torsion-free which means every torsion module is not flat. Over sufficiently nice rings (e.g. a PID), every torsion-free module is flat. So for the integers, it is really the torsion that gets you.
A module $M$ is flat if given any module homomorphism $N \to N'$ one has
$$ \ker(M \otimes N \to M \otimes N') \cong \ker(N \to N'). $$
Which is sort of related to things not getting smaller.
