Why is $\mathbb{C}[x,y]$ not isomorphic to $\mathbb{C}[x] \otimes _{\mathbb{Z}} \mathbb{C}[y]$ as rings? I would like to know why $\mathbb{C}[x,y]$ is not isomorphic to $\mathbb{C}[x] \otimes _{\mathbb{Z}} \mathbb{C}[y]$ as rings.
Thank you! 1
 A: To show that $1\otimes 1+i\otimes i$ is nonzero
in $\mathbb{C}[X]\otimes_{\mathbb{Z}}\mathbb{C}[X]$ note that it maps
to $1\otimes 1+i\otimes i$ in
$\mathbb{C}[X]\otimes_{\mathbb{R}}\mathbb{C}[X]$.
This is a tensor product over a field, so a basis as an
$\mathbb{R}$-vector space is got by tensoring togther bases
on each side. Now as the set of elements of the forms $X^n$
and $iX^n$ are bases of $\mathbb{C}[X]$ over $\mathbb{R}$
then $1\otimes 1$ and $i\otimes i$ are linearly independent
over $\mathbb{R}$.
As
$$(1\otimes 1+i\otimes i)(1\otimes 1-i\otimes i)=0$$
then $\mathbb{C}[X]\otimes_{\mathbb{Z}}\mathbb{C}[X]$
isn't an integral domain, unlike $\mathbb{C}[X]\otimes_{\mathbb{C}}\mathbb{C}[X]$.
A: Note that the two are not even isomorphic as $\mathbb{C}$-vector spaces, since the LHS has countable dimension, while for the RHS $\{1\otimes z|\;\bar{z}\in \mathbb{C}/\mathbb{Z}\}$ is an uncountable set that is linearly independent over $\mathbb{C}$.

Edit: another way to see that the rings are not isomorphic, which is just an addition to Matt's answer, is e.g. to note that there is a different number of elements of finite multiplicative order $n$ for any $n>2$. E.g. on the LHS, an element must clearly be a scalar to be of finite multiplicative order, so the only elements of order 4 are $i$ and $-i$. On the other hand, on the RHS, $i\otimes 1$, $-i\otimes 1$, $1\otimes i$ and $-1\otimes i = 1\otimes -i$ are all distinct elements of order 4. 
To prove that two elements of the tensor product are really distinct (this applies to any tensor products), you have to construct a $\mathbb{Z}$-bilinear map from $\mathbb{C}[x]\times\mathbb{C}[y]$ to some other $\mathbb{Z}$-module that takes the two corresponding elements (in this case ($i$,1) and (1,$i$), say) to different images. To understand, why that is a necessary and sufficient condition for them to be distinct in the tensor product, go back to the definition of tensor product via a universal property.
A: Because $\mathbb C \otimes_{\mathbb Z} \mathbb C$ is not isomorphic to
$\mathbb C$.  (There is a natural map from the first to the second, with an
absolutely enormous kernel.)
Added: In light of the discussion below, it might help to add the remark that
this kernel is full of non-trivial zero-divisors (see e.g. Robin's answer).
One way to think of this is that $\mathbb C \otimes_{\mathbb Z} \mathbb C = \mathbb C \otimes_{\mathbb Q} \mathbb C$ contains $\overline{\mathbb Q}\otimes_{\mathbb Q} \overline{\mathbb Q}$, and this latter ring already is full of zero divisors.  (Robin's answer gives one explicit example.  More generally,
$\overline{\mathbb Q}$ is the union of all finite Galois extensions $L$ of $\mathbb Q$ lying in $\mathbb C$, and for any such $L$, the tensor product
$L \otimes_{\mathbb Q} L$ is isomorphic to a product of $[L:\mathbb Q]$ copies of $L$.)
