0

Let $(X,\mathcal{O}_X)$ be a locally ringed space, and $f\in\mathcal{O}_X(X)$. Show that $$D=\{x\in X\mid f(x)\neq0\in\kappa(x)\}\subset X$$ is an open subset. What is $D$ in the case of $X=\operatorname{Spec}(A)$?

What I have come up with so far is if $x\in D$ then locally near $D$ we will get a $g$ so that $fg=1$. So, locally $f(x)\neq0$.

Am I on the correct path? But I am struggling to use that idea to figure out a formal proof based on my idea.

Can somebody help me?

Anish Ray
  • 857
  • 1
    You're on the right track. Luckily for you, both parts of this question have already been answered on MSE - see the duplicate links above. – KReiser Dec 05 '21 at 20:22

1 Answers1

1

Question: "What I have come up with so far is if x∈D then locally near D we will get a g so that fg=1. So, locally f(x)≠0. Am I on the correct path? But I am struggling to use that idea to figure out a formal proof based on my idea. Can somebody help me?"

Answer: If $X:=Spec(A)$ and $\phi_x:A\rightarrow \kappa(x)$ is the canonical map from $A$ to the residue field at the prime ideal $\mathfrak{p}_x$ it follows an element $f\in A$ has $\phi_x(f)\neq 0$ iff $f \notin \mathfrak{p}_x$. Hence the set of prime ideals $\mathfrak{p}_x \subseteq A$ with this property equals the open set $D(f) \subseteq X$.

hm2020
  • 10,015