Codimension 1 points

Question

Today I was reading a proof of the following Lemma from Liu's "Algebraic Geometry and Arithmetic Curves"

Recall: A a point $x \in X$ is called a codimension 1 point if $ \overline{ \{x \}}$ has codimension 1.

$\bf{Lemma(Sect\ 7.2, \ 2.5)}$: Let $X$ be an integral Noetherian scheme and let $ f \in K(X)^\times$, then there are finitely many codimension 1 points $ x \in X$ such that $ f \notin \mathcal{O}_{X, x}^\times$.

My question is about the first couple of lines of the proof.

We start out by taking an open affine scheme $U = Spec(A)$ of $X$ such that $ f = \frac{a}{b}$ for $ a, b \in \mathcal{O}_X (U)$ and then define a set:

$\begin{equation} Y:= V(a) \cup V(b) \cup (X-U). \end{equation}$

(where I assume that $V(a)$ (resp. $V(b)$) refers to what taking it inside the affine scheme $U$)

$\bf{Claim}$: For any codimension 1 point $ x \in X$ such that $ x \notin Y$, we must have $ f \in \mathcal{O}_{X, x}^\times$.

$\bf{Question}:$ I believe I have proof of this claim but I think "more" is true. Does the claim still hold if we modify the definition of $Y$ so that instead of $ Y: = V(b) \cup (X-U)$?

I feel like this must be the case since then $ X - Y = D(b) $ and because $X$ is integral, we have

$\begin{equation} f\in A_b = \Gamma(D(b), \mathcal{O}_X) = \bigcap_{z \in D(b)}\mathcal{O}_{X,z} \end{equation}$

and the claim follows. Am I missing something? It seems the whole proof actually goes through with this smaller $Y$. Thanks!

Answer 1

You are right to conclude that for $x\notin V(b)\cup(X-U)$, you will have $f\in\mathcal O_{X,x}$, but you do not have $f\in\mathcal{O}_{X,x}^\times$ - indeed, if $a$ vanishes at $x$ and $b$ does not, then $f(x)=0$ and you have $f\in\mathfrak m_x$, in particular $f$ is not a unit.