About the ideal of a closed subscheme

Question

I have the following question concerning the structure of the ideal associated to a closed scheme. Let us suppose that $X$ and $Y$ are schemes over $Spec(k)$ being $k$ a field, and let us consider the fiber product scheme $X\times_k Y$. Let $x$ be a closed point of $X$. Is the subscheme $x\times Y\hookrightarrow X\times Y$ closed? What is the ideal sheaf of $x\times Y$?

Answer 1

Question: "Is the subscheme ${x}×Y↪X×Y$ closed? What is the ideal sheaf of ${x}×Y$?"

Answer: The affine situation. Let $X:=Spec(A), Y:=Spec(B)$ with $A,B$ commutative $k$-algebras and let $\mathfrak{m} \subseteq A$ be the maximal ideal of $x\in X$. There is an exact sequence

$$ 0 \rightarrow \mathfrak{m} \rightarrow A \rightarrow A/\mathfrak{m} \rightarrow 0$$

and tensoring with $B$ you get

$$\mathfrak{m}\otimes_k B \rightarrow^i A\otimes_k B \rightarrow A/\mathfrak{m} \otimes_k B\rightarrow 0$$

and there is an ismorphism $A/\mathfrak{m} \otimes_k B \cong A\otimes_k B /Im(i)$ hence locally the sheafification of the ideal $I(x):=Im(i) \subseteq A\otimes_k B$ is the "ideal sheaf" of the closed subscheme $\{x\} \times_k Y \subseteq X\times_k Y$.

Hence

$$\{x\}\times_k X :=Spec(A/\mathfrak{m}\otimes_k B) \cong Spec(A\otimes_k B/I(x)) \subseteq Spec(A\otimes_k B)$$

is a closed subscheme with ideal sheaf $\tilde{I(x)}$.

So: "Yes, the subscheme is closed" and "the sheafification of the ideal $I(x)$ is the ideal sheaf of $\{x\}\times_k X$".