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!

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