Proving a Tensor Product of Modules is Nontrivial I'm just starting to learn about tensor products of modules, and I'm having trouble verifying the  basic fact that $\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z} \neq 0$. I could do this for example by showing that $1 \otimes [1]$ is nonzero, but I'm having some unexpected trouble with that. What's the overall strategy to prove something like this?
Related question: Can we guarantee in general that the tensor product of two nonzero modules won't collapse to the zero module?
 A: This can be again turned into question about the universal property: $\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z}$ being zero would mean that any bilinear map with domain $\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ is zero. Therefore, it is e.g. sufficient to find a bilinear map $$\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z}$$ that is nonzero. You can try to find one. Similarly, in case you are interested in the actual elementary tensor $1 \otimes [1]$ being nonzero, you should find a bilinear map coming from $\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ that does not send $(1, [1])$ to $0$.
Alternatively: You noted before that for any ring $R$ and any $R$-module $M$, $R \otimes_R M \simeq M$. Thus, in our case ($R=\mathbb{Z}, M=\mathbb{Z}/2\mathbb{Z}$), one gets $\mathbb{Z}\otimes \mathbb{Z}/2\mathbb{Z} \simeq \mathbb{Z}/2\mathbb{Z}.$
And no, in general, the tensor product of two nonzero modules may be $0$. You can try to show, using the bilinear relations, that the module $\mathbb{Z}/3\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z}$ is trivial.
