When can we say elements of tensor product are equal to $0$?

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 $x,y\neq 0$.

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 $2\otimes 3=2\cdot3\otimes 1=0\otimes 1=0$.

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 $0\cdot n = (1 + (-1))\cdot n = n - n = 0$.)