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
\[ AB = \{ z \in X \times _ k Y \mid \text{pr}_ X(z) \in A, \ \text{pr}_ Y(z) \in B\} \subset X \times _ k Y \]
\[ \overline{A} \times _ k \overline{B} = \overline{AB} \]
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
which holds as $A$ and $B$ are affine. Details omitted. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #8748 by Michael on
Comment #9328 by Stacks project on