Tensor product of domains is a domain I'm reading Milne's Algebraic Geometry course notes, version 5.22, as a companion to an algebraic geometry course I'm taking now. Proposition 4.15 states:
Let $A$ and $B$ be $k$-algebras, which are also domains, with $k = \overline{k}$ and $A$ finitely generated. Then $A \otimes_k B$ is a domain.
For homework I've proved that the direct product of two irreducible algebraic sets is again an irreducible algebraic set. The teacher simply stated that this implies the above result when $A, B$ are finitely generated. So, my first question is, can someone explain this implication to me?
My second question is, is the above result from Milne true if both $A$ and $B$ are not finitely-generated? If not, can you provide a counterexample?
 A: 1) Given a field $k$ and two domains $A,B$ over $k$, their tensor product $A\otimes _k B$ will be a domain as soon as one of the domains is a  separable $k$-algebra  and one of them (maybe the same!) has a fraction field which is a primary extension of $k$.
This is probably the most general possible result and is proved in Bourbaki's Algebra, Chapter v, §17, Corollary to Proposition 1.
(You can also find it on page 203 of  Jacobson's Lectures in Abstract Algebra: III Theory of fields and Galois theory.)   
Here separable means universally reduced  and an extension of fields $K/k$ is said to be primary if the algebraic closure of $k$ in $K$ is purely inseparable over $k$.   
These are quite advanced concepts in field theory but the good news is that for an algebraically closed field $k$  every algebra is separable and every extension field is primary, so that indeed for an algebraically closed field $k$, the $k$-algebra $A\otimes_k B$ is a domain as soon as $A$ and $B$ are domains.  
2) For a general (non algebraically closed) field $k$, if $X,Y$ are the affine varieties associated to the $k$-algebras of finite type $A,B$ without zero-divisors, the irreducibility of $X\times Y$ does not imply that $A\otimes_k B$ is a domain.    
For example, suppose  $p$ is a prime integer. Then for  $k=\mathbb F_p(t)$ ($t$ an indeterminate) and $A=k(\sqrt [p]t)$, we have $A\otimes_k A=\frac {A[X]}{\langle X^p-t\rangle}=\frac {A[X]}{\langle (X-\sqrt [p]t)^p\rangle }$, a ring with non-zero nilpotent elements (for example the class of $X-\sqrt [p]t$) which is thus certainly not a domain although the corresponding "variety" (or rather scheme) is irreducible (since its underlying topological space has just one point!).
Edit (September 28th, 2014)
  As noticed by @lee in the comments, a non irreducible example follows from the isomorphism $\mathbb C\otimes_ \mathbb R \mathbb C= \mathbb C \times \mathbb C$ . 
So I don't think that your teacher's remark is correct: it is not clear that the algebra corresponding to $X\times Y$ is the ring $A\otimes_k B$ rather than its reduction $(A\otimes_k B)_{\mathrm{red}}$ (obtained by killing the nilpotents: $(A\otimes_k B)_{\mathrm{red}}=A\otimes_k B/\mathrm{Nil}(A\otimes_k B)$).
This tends to show that, despite what  your teacher claims, you cannot replace the hard algebra in 1) by elementary topological considerations of irreducibility: Milne himself proves the result (for an algebraically closed field) purely algebraically  although, believe me, he knows what irreducibility means!
A: 
Let $K$ be an algebraically closed field and $A$, $B$ two $K$-algebras which are integral domains. Then $A\otimes_K B$ is an integral domain.

Let $x,x'\in A\otimes_K B$ such that $xx'=0$. Write $x=\sum a_i\otimes b_i$ and $x'=\sum a_j'\otimes b_j'$. By taking minimal representations (as sums of monomial tensors) for $x$ and $y$ one can assume that $(a_i)$, $(a_j')$, $(b_i)$ and $(b_j')$ are linearly independent over $K$. Now considering $A'$ the $K$-subalgebra of $A$ generated by $(a_i)$ and $(a_j')$, and analogously $B'$ the $K$-subalgebra of $B$ generated by $(b_i)$ and $(b_j')$ we reduce the problem to the affine case that is proved in Milne's book. 
