# 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?

• For the first part think about what the regular functions are for a product of algebraic sets. Commented Feb 12, 2013 at 5:14
• @AlexYoucis Thank you, I've figured that part out now. Commented Feb 12, 2013 at 5:46
• math.stackexchange.com/questions/152056/…
– user26857
Commented Feb 12, 2013 at 10:54

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!

• Dear @Manos: no, I don't think that simple considerations of irreducibility suffice, even if $k$ is algebraically closed. I stand by my last two sections ( starting with "So I don't think...") and I don't know what solution Hartshorne had in mind for his exercise. Commented Jun 11, 2014 at 19:08
• Dear @Manos, $A(X)\otimes_k A(Y)$ is not the vanishing ideal of $X\times Y$. As far as I am concerned, I'll stop this discussion here. Commented Jun 12, 2014 at 21:53
• No problem, @Manos. Commented Jun 14, 2014 at 2:56
• According to your cited theorem, C tensor C over R is a domain(as C is a domain and R is perfect). But it's commutative 4-dim R algebra, hence is not a domain. Is there something wrong?
– lee
Commented Sep 28, 2014 at 13:45
• Dear @lee, you are perfectly right: it is not enough to suppose that $k$ is perfect. I corrected this false assertion by now demanding that $k$ be algebraically closed: thanks a lot for catching my former too weak hypothesis. It is quite ironical that I wrote that, since I am a great fan of the isomorphism $\mathbb C \otimes _\mathbb R \mathbb C=\mathbb C\times \mathbb C$ , as demonstrated at the end of the edit here ! Commented Sep 28, 2014 at 21:30

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.

• Can you use this trick to reduce both $A$ and $B$ to the finitely-generated case, or just one at a time? Commented Feb 12, 2013 at 17:45
• Also, can you explain the justification for the reduction? My algebraic foundations are not the strongest. Commented Feb 12, 2013 at 19:40
• @JulienClancy One can reduce to the affine case simultaneously. After reduction $x,x'\in A'\otimes_KB'$ and $xx'=0$; furthermore, $A'$ and $B'$ are integral domains as subrings of integral domains. (In fact, $A'\otimes_KB'$ is a subring of $A\otimes_KB$.)
– user26857
Commented Feb 12, 2013 at 21:25
• Great, thank you. I was confused by a statement in Atiyah/MacDonald that said that $x \otimes y = 0$ in a tensor product does not imply that it is equal to zero in a submodule, but clearly I just misread it for actual multiplication. Commented Feb 13, 2013 at 0:05