Lemma 33.24.1. Let k be a field. Let X, Y be schemes over k, and let A \subset X, B \subset Y be subsets. Set
Then set theoretically we have
Some results on the relation between closure and products.
Lemma 33.24.1. Let k be a field. Let X, Y be schemes over k, and let A \subset X, B \subset Y be subsets. Set
Then set theoretically we have
Proof. The inclusion \overline{AB} \subset \overline{A} \times _ k \overline{B} is immediate. We may replace X and Y by the reduced closed subschemes \overline{A} and \overline{B}. Let W \subset X \times _ k Y be a nonempty open subset. By Morphisms, Lemma 29.23.4 the subset U = \text{pr}_ X(W) is nonempty open in X. Hence A \cap U is nonempty. Pick a \in A \cap U. Denote Y_{\kappa (a)} = \{ a\} \times _ k Y the fibre of \text{pr}_ X : X \times _ k Y \to X over a. By Morphisms, Lemma 29.23.4 again the morphism Y_ a \to Y is open as \mathop{\mathrm{Spec}}(\kappa (a)) \to \mathop{\mathrm{Spec}}(k) is universally open. Hence the nonempty open subset W_ a = W \times _{X \times _ k Y} Y_ a maps to a nonempty open subset of Y. We conclude there exists a b \in B in the image. Hence AB \cap W \not= \emptyset as desired. \square
Lemma 33.24.2. Let k be a field. Let f : A \to X, g : B \to Y be morphisms of schemes over k. Then set theoretically we have
Proof. This follows from Lemma 33.24.1 as the image of f \times g is f(A)g(B) in the notation of that lemma. \square
Lemma 33.24.3. Let k be a field. Let f : A \to X, g : B \to Y be quasi-compact morphisms of schemes over k. Let Z \subset X be the scheme theoretic image of f, see Morphisms, Definition 29.6.2. Similarly, let Z' \subset Y be the scheme theoretic image of g. Then Z \times _ k Z' is the scheme theoretic image of f \times g.
Proof. Recall that Z is the smallest closed subscheme of X through which f factors. Similarly for Z'. Let W \subset X \times _ k Y be the scheme theoretic image of f \times g. As f \times g factors through Z \times _ k Z' we see that W \subset Z \times _ k Z'.
To prove the other inclusion let U \subset X and V \subset Y be affine opens. By Morphisms, Lemma 29.6.3 the scheme Z \cap U is the scheme theoretic image of f|_{f^{-1}(U)} : f^{-1}(U) \to U, and similarly for Z' \cap V and W \cap U \times _ k V. Hence we may assume X and Y affine. As f and g are quasi-compact this implies that A = \bigcup U_ i is a finite union of affines and B = \bigcup V_ j is a finite union of affines. Then we may replace A by \coprod U_ i and B by \coprod V_ j, i.e., we may assume that A and B are affine as well. In this case Z is cut out by \mathop{\mathrm{Ker}}(\Gamma (X, \mathcal{O}_ X) \to \Gamma (A, \mathcal{O}_ A)) and similarly for Z' and W. Hence the result follows from the equality
which holds as A and B are affine. Details omitted. \square
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