Prove that $2 \otimes 1 \neq 0$ in $2\Bbb Z \otimes {\Bbb Z/2\Bbb Z}$ the tensor is over $\Bbb Z$. Prove that $2 \otimes 1 $ is zero in $\Bbb Z \otimes {\Bbb Z/2\Bbb Z}$  but not a zero in $2\Bbb Z \otimes {\Bbb Z/2\Bbb Z}$ the tensor is over $\Bbb Z$.
It is easy to show the first part that $2 \otimes 1= 1 \otimes 2=1 \otimes 0=0$
How to prove that $2 \otimes 1 \neq 0 $ in $2\Bbb Z \otimes {\Bbb Z/2\Bbb Z}$
It is evident that $2 \otimes 1 \neq 1 \otimes 2 $ in $2\Bbb Z \otimes {\Bbb Z/2\Bbb Z}$ . [As $1 \otimes 2 \notin 2\Bbb Z \otimes {\Bbb Z/2\Bbb Z}$]
Moreover, $2 \otimes 1$ is the generator of $2\Bbb Z \otimes {\Bbb Z/2\Bbb Z}$. So the whole group would be zero if $2 \otimes 1 =0 $. Next what?
I was also thinking in the line of homological algebra "For any $K$-module $T$ there is a canonical isomorphism
$$
\operatorname{Hom}_K(A\otimes B,T)\simeq\operatorname{Bil}_K(A\times B,T).
$$
Thus in order to prove that $2\otimes0\neq0$ it is enough to produce a bilinear map $\phi:A\times B\rightarrow T$ such that $\phi(2,1)$ is not a zero map." But this is an intense proof.
So, is there any easy straight forward proof?
 A: This can possibly be better understood by considering functors. No need to be very formal, though.
Groups here are abelian.
The idea is that $\mathbb{Z}$ and $2\mathbb{Z}$ are isomorpic, with $1$ mapping to $2$. Therefore $2\mathbb{Z}\otimes\mathbb{Z}/2\mathbb{Z}$ is isomorphic to $\mathbb{Z}\otimes\mathbb{Z}/2\mathbb{Z}$, which in turn is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ and we can follow elements.
Suppose you have a group homomorphism $f\colon A\to B$; then, for every group $M$, this induces a homomorphism $f_M\colon A\otimes M\to B\otimes M$ by
$$
f(a\otimes x)=f(a)\otimes x
$$
This is a straightforward check with the universal property of tensor products. It follows that if we also have $g\colon B\to C$, then
$$
(g\circ f)_M=g_M\circ f_M
$$
by just checking on elementary tensors $a\otimes x$. In particular, if $f$ is an isomorphism, then also $f_M$ is an isomorphism, because if $i\colon A\to A$ is the identity, then also $i_M$ is the identity.
Finally, the map $j_M\colon \mathbb{Z}\otimes M\to M$, $j_M(n\otimes x)=nx$ is an isomorphism. In your case, letting $M=\mathbb{Z}/2\mathbb{Z}$, and considering $\mu\colon\mathbb{Z}\to 2\mathbb{Z}$, $\mu(n)=2n$, we have
$$
2\otimes 1=\mu_M(1\otimes1)=\mu_M(j_M^{-1}(1))\ne0
$$
because both $\mu_M$ and $j_M$ are isomorphisms.
A: Define a bilinear map $\newcommand{\Z}{\Bbb Z}2\Z\times\Z/2\Z\to\Z/2\Z$ by
$$B:(2m,n+2\Z)\mapsto mn+2\Z.$$
(You should check it's well-defined and bilinear). It induces a linear map
$$\beta:(2\Z)\otimes \Z/2\Z\to\Z/2\Z$$
with 
$$\beta(2m\otimes(n+2\Z))=mn+2\Z.$$
But $\beta(2\otimes(1+2\Z))=1\otimes 2\Z\ne0$ so that $2\otimes(1+2\Z)\ne0$
in $2\Z\otimes \Z/2\Z$.
