Why is a product of varieties over an algebraically closed field k a variety?<\/p>\n
After some preliminary reductions this reduces to the question: Why is A \u2297 B a domain if A, B are domains over k (tensor product over k). To prove this suppose that (\u2211 a_i \u2297 b_i) (\u2211 c_j \u2297 d_j) = 0 in A \u2297 B with a_i, c_j \u2208 A and b_i, d_j \u2208 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 \u2282 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<\/p>\n
(\u2211 a_i(m) b_i) (\u2211 c_j(m) d_j) = 0<\/p>\n
in B! This is a contradiction with the assumption that B is a domain and we win.<\/p>\n
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…)<\/p>\n","protected":false},"excerpt":{"rendered":"
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 \u2297 B a domain if A, B are domains over k (tensor product over … Continue reading