I had the following doubt:
Show that the element $2\otimes 1$ is $0$ in $\mathbb Z\otimes_{\mathbb Z} \mathbb Z/2\mathbb Z$ but not a zero in $2\mathbb Z\otimes_{\mathbb Z} \mathbb Z/2\mathbb Z$.
I was able to prove $2\otimes 1$ is $0$ in $\mathbb Z\otimes_{\mathbb Z}\mathbb Z/2\mathbb Z$ by shifting $2$ using tensor property . now I can't see how to prove it's non-zero in $2\mathbb Z\otimes_{\mathbb Z} \mathbb Z/2\mathbb Z$.
kindly help out.
Clearly $2\otimes 1$ generates $2\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/2\mathbb{Z}$ as a module over $\mathbb{Z}$. So it is enough to prove that $2\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/2\mathbb{Z}$ is nonzero. But what would it mean if it was zero? This would mean: any $\mathbb{Z}$-bilinear map from $2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ to any $\mathbb{Z}$-module $P$ is zero. But, for example, the map $(2z,x)\mapsto zx \in \mathbb{Z}/2\mathbb{Z}$ is nontrivial.
$\textbf{Edit}$: This is essentially the same approach John suggested, noting that for any ring $A$ and an $A$-module $M$ the isomorphism $A \otimes_A M \cong M$ is given by sending $a \otimes m$ to $am$.
There's a homomorphism from $2\mathbb Z\otimes_{\mathbb Z} \mathbb Z/2\mathbb Z$ to $\mathbb Z\otimes_{\mathbb Z} \mathbb Z/2\mathbb Z$ defined by $2s \otimes q \mapsto s \otimes q$. So now: can you prove that
$$ 1 \otimes 1 $$ is nonzero in the latter group?
