I'm trying to understand yet again another proof in Q. Liu's book. Specifically I refer to the proof of Lemma 2.5, chapter 7, section 7.2, page 268.
So, the statement is the following:
For X a noetherian integral scheme, and $f \in K(X)$, where $K(X)$ is the function field of $X$ and $f \neq 0$,
then the number of points $x \in X$ of codimension 1 (meaning that the irreducible subset $\overline{\{x\}}$ has codimension one) such that $f \notin \mathcal{O}^*_{X,x}$, is finite
(where $A^*$ denotes the group of units of a ring $A$).
Here is my proof:
If $f \in K(X)$ then by definition $f \in \operatorname{Frac}\mathcal{O}(V)$, where $V=\operatorname{Spec}A$ is any affine open. As $X$ is integral hence irreducible, $V$ must be dense in $X$ because it is open. If $x$ is such that $f \notin \mathcal{O}^*_{X,x}$, this means that $f = a/b$ with $a,b \in A$, $a,b \neq 0$ and $a_x,b_x$ cannot be both invertible in $\mathcal{O}_{X,x},$ whence $x \notin D(a)\cap D(b)$, i.e. $x \in V(a)\cup V(b)$. Moreover, we are assuming $x \in V$ for this to make sense.
Hence we are left to show that:
there are just a finite number of points of codimension 1 in $V(a)\cup V(b) \cup X - V$
As $X$ is noetherian, we can take $A$ noetherian, and there only a finite number of irreducible components of $V$ whence there are only a finite number of irreducible subsets in $V(a) \cup V(b)$, on the other hand X has only one irreducible component, given by the closure of the generic point, hence $X-V$ has only one irreducible component, and we conclude.
questions:
1 is the proof correct?
2 If so, I don't see where I'm using the fact that $\overline{\{x\}}$ has codimension $1$, and I don't see why he as to mention that $V$ is dense.
The proof is almost correct, but could be improved a bit. Everything through "Hence we are left to show that..." is correct and good work. What you're showing is that the only places $f$ could potentially fail to be in $\mathcal{O}_{X,x}$ are the complement of $V$ as well as the sets $V(a)\subset V$ and $V(b)\subset V$. It remains to explain why there are only finitely many codimension-one points in the union of these sets.
First, it suffices to consider the closure of these sets: if there are only finitely many codimension-one points in the closures of these sets, then there can only be finitely many codimension-one points in the original sets. This lets us reduce the claim to the following: if $X$ is an irreducible noetherian scheme and $Z\subset X$ is a proper closed subset, then $Z$ has finitely many points of codimension one in $X$.
To prove this claim, we note that any codimension one point of $X$ in $Z$ must be a generic point of an irreducible component of $Z$: if $\alpha$ is a codimension-one point but $\overline{\{\alpha\}}$ is not an irreducible component of $Z$, then $\overline{\{\alpha\}}$ must be contained in a larger irreducible closed set. But the only larger irreducible closed subset is $X$, which cannot be a subset of $Z$ since $Z\subset X$ is a proper inclusion. Now as a noetherian topological space has only finitely many irreducible components, we see that $Z$ can only have finitely many points of codimension one.
$V$ dense is potentially superfluous to note- the point there is that $V$ will have the same field of fractions as $X$, which is a consequence of the fact that $X$ is integral and $V$ is a nonempty open, which is the same information that shows that $V$ is dense in $X$. Whether you need to mention it or not may depend on what material you've already developed around integral schemes and fraction fields.