# Product of varieties

Why is a product of varieties over an algebraically closed field k a variety?

After some preliminary reductions this reduces to the question: Why is A ⊗ B a domain if A, B are domains over k (tensor product over k). To prove this suppose that (∑ a_i ⊗ b_i) (∑ c_j ⊗ d_j) = 0 in A ⊗ B with a_i, c_j ∈ A and b_i, d_j ∈ B. After recombining terms we may assume that b_1, …, b_n are k-linearly independent in B and also that d_1, …, d_m are k-linearly independent in B. Let A’ be the k-subalgebra of A generated by a_i, c_j. Unless all of the a_i and c_j are zero, we can find (after rearranging indices) a maximal ideal m ⊂ A which does not contain a_1 and c_1 (use that A’ is a domain). Denote a_i(m) and c_j(m) the congruence classes in A/m. By the Hilbert Nullstellensatz A/m = k and we can specialize the relation to get

(∑ a_i(m) b_i) (∑ c_j(m) d_j) = 0

in B! This is a contradiction with the assumption that B is a domain and we win.

This blog post is my atonement for having “forgotten” this argument. What are some standard texts which have this argument? (Ravi will add it to his notes soon he just told me…)

## 4 thoughts on “Product of varieties”

1. BCnrd on said:

Johan, here’s another proof via specialization, which doesn’t involve manipulating tensor expressions. By localizing A to its fraction field, it suffices to prove that if B is a domain finitely generated over an algebraically closed field k and if K/k is an extension field then B_K := K \otimes_k B is a domain. We may replace B with B[1/b] for any nonzero b in B, so since k is perfect we may (upon choosing a separating transcendence basis) arrange that B is a localization of = k[x_1,…,x_n,t]/(h) where h is irreducible in k[x_1,…,x_n,t]. It suffices to prove that h is irreducible when viewed in K[x_1,…,x_n,t]. Any hypothetical factorization only involves finitely many elements of K, and so by specializing at a well-chosen maximal ideal of the k-algebra they generate we contradict the irreducibility of h in k[x_1,…,x_n,t].

• Johan on said:

Thanks! Your argument also is a precursor to the proof that given X —> S of finite presentation, then the set of points in S where the fibres are geometrically irreducible is constructible.

On the other hand, it has the disadvantage that one needs a separating transcendence basis. I think finding a separating transcendence basis is nontrivial in char p, even for function fields of varieties. Or is there an easy proof in that case?

2. Kai on said:

Hi Johan: I don’t think you need algebraically closed to prove that the product of two irreducible algebraic sets is an irreducible algebraic set. So even if the ground field is not algebraically closed, the tensor product of two affine coordinate rings of two affine varieties is a domain. (The affine coordinate ring is here defined as the quotient of the polynomial ring by the ideal of the affine variety.)

• Dear Kai, this depends on what you call a variety. My advisor, like you apparantly, defines a variety to be a geometrically irreducible gadget. On the other hand, I define it to be just a separated, finite type, integral scheme over a field. In the stacks project we use the same definition: Definition Tag 020D.