Does tensoring the integers $\Bbb Z$ with any field $F$ give that field itself? How to prove that? Does tensoring the integers $\Bbb Z$ with any field $F$ give that field itself? How to prove that?
I know for $\Bbb Q$ it is true.
 A: You have to be careful over what ring you're tensoring. What this means is that we typically say, if $M$ is a right $R$ module and $N$ is a left $R$ module, then the tensor product $M\otimes_R N$ is an abelian group such that $mr\otimes n=m\otimes rn$ for all $r\in R$ (there is the description in terms of universal constructions such that if $f:M\times N\to V$ is a balanced homomorphism of abelian groups (i.e. $f(mr,n)=f(m,rn)$ for all $m\in M, n\in N, r\in R$), then there exists a unique map $\hat{f}:M\otimes_R N\to V$ such that $\hat{f}\circ i=f$, for $i:M\times N\to M\otimes_R N$.
All that's (very broadly) abstract nonsense. Let's take $\mathbb{R}$ as our field, and consider the ring $R=\mathbb{Z}$. Then $\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{R}$ is an abelian group consisting of elements which are sums of simple elements $a\otimes b$. You can show that this is just $\mathbb{R}$, since simple elements are $a\otimes b=1\otimes ab=1\otimes c$ for all $c\in\mathbb{R}$ since $a,b$ were arbitrary. 
What can go wrong? Let's take $\mathbb{Z}_4$ and $\mathbb{R}$ and consider $\mathbb{Z}_4\otimes_{\mathbb{Z}}\mathbb{R}$. Take an arbitrary simple element $a\otimes b$. We can write $b=4(b/4)$ for all $b\in\mathbb{R}$. Then $a\otimes b=a\otimes 4(b/4)=a4\otimes b/4=0\otimes b/4=0\otimes0b/4=0\otimes0=0$. So $\mathbb{Z}_4\otimes_{\mathbb{Z}}\mathbb{R}=0$. 
There are also issues when we take different rings $R$ to tensor over.
A: Consider the scalar multiplication map $m:\mathbb Z\times F\to F$ given by $m(n,x)=nx$. This is bilinear and hence induces a homomorphism $\mathbb Z\otimes F\to F$ that is surjective because $m$ is.  The kernel consists of sums $\sum a_i\otimes x_i=1\otimes \sum a_ix_i$ that map to 0, so
$$1( \sum a_ix_i)=0$$
Thus $ \sum a_ix_i=0$, so the element was $0$ in the first place, hence we have an isomorphism.
