I've been using Qing Liu's Algebraic Geometry for Arithmetic Curves as a supplementary reference in this chapter, but I'm a little bit confused as to what is going on at the end of this problem (Vakil 14.2.I):
If $X$ is Noetherian and factorial, show that for any Weil divisor $D$, $\mathscr{O}(D)$ is an invertible sheaf. (Hint: It suffices to deal with the case where $D$ is irreducible, say $D = [Y]$, and to cover $X$ by open sets so that on each open set $U$ there is a function whose divisor is $[Y \cap U]$. One open set will be $X − Y$. Next, we find an open set $U$ containing an arbitrary $p \in Y$, and a function on $U$. As $\mathscr{O}_{X,p}$ is a unique factorization domain, the prime corresponding to $Y$ is codimension $1$ and hence principal by Lemma 11.1.6. Let $f$ be a generator of this prime ideal, interpreted as an element of $K(X)$. It is regular at $p$, it has a finite number of zeros and poles, and through $p$, $[Y]$ is the “only zero” (the only component of the divisor of zeros). Let $U$ be $X$ minus all the other zeros and poles.)
So the hint basically sets everything up for us. Since $X$ factorial, we may use the same logic as the preamble of this answer by KReiser) to restrict our attention to $D = [Y]$ for some $Y \subset X$ regular of codimension 1. By a few of the prior exercises in this section, $X$ Noetherian and normal implies that $\mathscr{O}(D)$ is invertible iff $D$ is locally principle, so as suggested by the hint we really only want to cover $Y$ by open sets on which $D\vert_U = \operatorname{div} f \vert_U$ for some $ f \in K(X)$ (since $\mathscr{O}(D)$ is already isomorphic to $\mathscr{O}$ away from $Y = \operatorname{Supp}D$ by what Vakil calls Algebraic Hartog's Lemma).
If we basically follow word-for-word the next sentences, we have some $f \in K(X)$ such that, when considered as a germ in $\mathscr{O}_{X, p}$, it
generates the height 1 prime ideal $\mathfrak{q}$ corresponding to $Y$. However, I'm not quite sure what's going on with the $[Y]$ being the "only zero" part; I looked at Liu's text for this, which makes precise what is meant by the divisor of zeros (as expected, its just the effective divisor obtained by throwing away all the terms with negative coefficients ) but am still a little confused by what Vakil is implying. Does this mean that there is no other codimension one $Y' \subset X$ containing $p$ such that $f$ has a non-zero order of vanishing along $Y'$? In this case it seems like it should be true then that if we let $U$ be the open subset obtained by removing the finitely many (by Noetherian assumption) closed points corresponding to the zeros and poles of $f$, then $Y' \cap U$ can only have non-zero order of vanishing when $Y' = Y$ so that $\operatorname{div} f\vert_U = [Y \cap U] = D\vert_U$ — but this only really gives me an answer assumeing the line above is the correct reasoning.
Any help would be appreciated!
Best Answer
Yes, that's correct.
This requires some adjustment. We need to remove the closed subvarieties that are zeroes and poles of $f$, but they need not be closed points - think about $X=\Bbb A^2_k$, $p=(0,0)$, and $f=x(y+1)$ - then the divisor of $f$ is $[V(x)]+[V(y+1)]$. Other than that, you're correct.