## 33.24 Closures of products

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

$AB = \{ z \in X \times _ k Y \mid \text{pr}_ X(z) \in A, \ \text{pr}_ Y(z) \in B\} \subset X \times _ k Y$

Then set theoretically we have

$\overline{A} \times _ k \overline{B} = \overline{AB}$

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

$\overline{f(A)} \times _ k \overline{g(B)} = \overline{(f \times g)(A \times _ k B)}$

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

$\Gamma (A \times _ k B, \mathcal{O}_{A \times _ k B}) = \Gamma (A, \mathcal{O}_ A) \otimes _ k \Gamma (B, \mathcal{O}_ B)$

which holds as $A$ and $B$ are affine. Details omitted. $\square$

Comment #8748 by Michael on

Why the image of $f\times g$ is $f(A)g(B)$ in Lemma 04Q0 ?

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