When can we say elements of tensor product are equal to $0$? I am learning about tensor products of modules, but there is a question which makes me very confused about it! 
If $E$ is a right $R$-module and $F$ is a left $R$-module, then suppose we have a balanced map (or bilinear map) $E\times F\to E\otimes F$. If some element $x\otimes y \in E\otimes F$ is $0$, then can we say $x$ or $y$ must be equal to $0$? I know if $x = 0$ or $y = 0$, then $x\otimes y$ is $0$. Are there other cases where $x\otimes y$ is $0$? Can someone give me a specific example? 
Really thank you!
 A: By the universal property of tensor product, an elementary tensor $x\otimes y$ equals zero if and only if for every $R$-bilinear map $B:E\times F\to M$, $B(x,y)=0$. While this may seem like a difficult thing to check, in practice it is usually not so bad.
As an example, we will show that $\mathbb{Z}/5\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q}=0$. Let $x\otimes y\in\mathbb{Z}/5\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q}$ be an elementary tensor. Then by the bilinearity of the canonical map $\mathbb{Z}/5\mathbb{Z}\times\mathbb{Q}\to\mathbb{Z}/5\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q}$, we have
$$
x\otimes y=x\otimes 5y/5=5(x\otimes y/5)=5x\otimes y/5=0\otimes y/5=0.
$$
This shows that all elementary tensors are zero, and thus since the tensor product is generated by elementary tensors, $\mathbb{Z}/5\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q}=0$.
We can show that an elementary tensor $x\otimes y$ is nonzero by giving an $R$-bilinear map $B:E\times F\to M$ such that $B(x,y)\neq 0$. As an example, consider $E=F=\mathbb{Z}$ as $\mathbb{Z}$-modules and the $\mathbb{Z}$-bilinear map $B:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ given by multiplication: $B(x,y)=xy$. Then if $x,y\neq 0$, $B(x,y)\neq 0$, so that $x\otimes y\neq 0$ in $\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}$ when $x,y\neq 0$.
A: I found this question while seeking to answer a related one. I think it's worth saying that $0\otimes n = (0\cdot 0)\otimes n = 0\cdot(0\otimes n) = 0\otimes(0\cdot 0) = 0\otimes 0 = 0\in M\otimes_A N$ where $M$ and $N$ are left $A$-modules (or right--I can never remember which is which).
(You may also note that $0\cdot n = (1 + (-1))\cdot n = n - n = 0$.)
A: It is not possible to find a nice characterization of when simple tensors are zero.
To give an example where $a, b$ are nonzero but $a\otimes b=0$, consider $\Bbb Z/6\Bbb Z\otimes_\Bbb Z \Bbb Z$ where $2\otimes 3=2\cdot3\otimes 1=0\otimes 1=0$.
A: To add to other good answers and comments: perhaps a one slightly abstracted version of the question can be construed as asking about the exactness of the tensor product functor(s) $M\to M\otimes N$. This is not an exact functor (in many interesting categories), and its failure to be exact is measured by "Tor" groups (which are also construable as derived functors, apart from specific explicit constructions in various categories). In particular, at the very least, ${\mathrm Tor}^i(M,N)$ is very often non-zero even for $i=1$.
In different words: it is a non-trivial problem, with no simple general formulaic solution, to find the collapsing/relations in tensor products. And there are many well-known examples with some at-first-possibly-surprising collapsing, as mentioned already, as $\mathbb Z/m\otimes_{\mathbb Z} \mathbb Z/n\approx \mathbb Z/\gcd(m,n)$.
