Closed subsets of an affine scheme

Question

Let $X=\mathsf {Spec} \: A$ be an affine scheme and $U\subseteq X$ an affine open subset. If $C=V (f)$, where $f\in \mathcal {O}_X (X)=A$ then is it true that $C\cap U=V (f_{|U})$?

Answer 1

Yes this is true. Let $U = \mathsf{Spec}~ B$ and $A \xrightarrow{\varphi} B$ the ring map corresponding to the open embedding $U \subset X$. Note that the inverse image $\varphi^{*}: \mathsf{Spec}~ B \to \mathsf{Spec}~ A$ is then said inclusion map, i.e. your statement is just

$V(\varphi(f)) = (\varphi^*)^{-1}(V(f)).$

This holds more generally for any ring map and for any ideal, i.e.

$V(\varphi(I)) = (\varphi^*)^{-1}(V(I)).$

This is nothing else but the proof that the map $\mathsf{Spec}~ B \to \mathsf{Spec}~ A$ arising from a ring map $A \to B$ is continuous.