Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
For the first part think about what the regular functions are for a product of algebraic sets. –  Alex Youcis Feb 12 '13 at 5:14
    
@AlexYoucis Thank you, I've figured that part out now. –  Julien Clancy Feb 12 '13 at 5:46
    
math.stackexchange.com/questions/152056/… –  user26857 Feb 12 '13 at 10:54
add comment

2 Answers 2

up vote 12 down vote accepted

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 a perfect field $k$ (in particular for an algebrically closed field) every algebra is separable and every extension field is primary, so that indeed for a perfect 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 perfect) 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!)

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 $(A\otimes_k B)_{red}$ (obtained by killing the nilpotents: $(A\otimes_k B)_{red}=A\otimes_k B/Nil(A\otimes_k B)$).
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!

share|improve this answer
    
Dear Georges, i would like to ask you: if $k$ is algebraically closed, then the difficult algebra of 1) can indeed be replaced by elementary topological considerations of irreducibility, right? For example, by problem I.3.15(a) in Hartshorne we can conclude that $A(X) \otimes_k A(Y)$ is an integral domain, simply because there is 1-1 one correspondence of irreducible algebraic sets and prime ideals. –  Manos Jun 11 at 18:11
    
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. –  Georges Elencwajg Jun 11 at 19:08
    
But since it is shown that $X \times Y$ is an irreducible algebraic set with vanishing ideal $A(X) \otimes_k A(Y)$, and $k$ is closed, we must have that $A(X) \otimes_k A(Y)$ is an integral domain by Corollary I.1.4 in Hartshorne! –  Manos Jun 12 at 19:47
    
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. –  Georges Elencwajg Jun 12 at 21:53
    
I am sorry, i meant coordinate affine ring, not vanishing ideal. I was thinking of taking the quotient with the vanishing ideal. I hope you feel less frustrated now. –  Manos Jun 13 at 22:23
show 1 more comment

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.

share|improve this answer
    
Can you use this trick to reduce both $A$ and $B$ to the finitely-generated case, or just one at a time? –  Julien Clancy Feb 12 '13 at 17:45
    
Also, can you explain the justification for the reduction? My algebraic foundations are not the strongest. –  Julien Clancy Feb 12 '13 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 Feb 12 '13 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. –  Julien Clancy Feb 13 '13 at 0:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.