# Relation of Function Field of a scheme to the Local Ring of its Prime Divisor

Refer to p. 130 in Hartshorne: Let $X$ be a noetherian, integral separated scheme, regular in codimension 1, and let $Y$ be a prime divisor of $X$, with generic point $\eta$. Let $\xi$ be the generic point of $X$ and $K=\mathcal{O}_{X,\xi}$ is the function field of $X$. I can see that $\mathcal{O}_{X,\eta}$ is an integral domain and that it can be injected into $K$. But why is $K$ the quotient field of $\mathcal{O}_{X,\eta}$?

-

## 1 Answer

Pick an affine open $U=\mathrm{Spec}(A)$ containing $\eta$. Then $\mathscr{O}_{X,\eta}$ is the localization of $A$ at the prime ideal $\mathfrak{p}\in\mathrm{Spec}(A)$ corresponding to $\eta$. Also, since the generic point $\xi$ is in $U$, and necessarily corresponds to the generic point of $\mathrm{Spec}(A)$, i.e., the zero ideal, the local ring at the generic point is $A$ localized at the zero ideal, i.e., the field of fractions of $A$. Now you just have to prove that any localization of $A$ has the same field of fractions as $A$. Or more precisely, the canonical map $S^{-1}A\rightarrow\mathrm{Frac}(A)$ identifies the target as the field of fractions of the source for any multiplicative set $S\subseteq A$.

In general, since local rings can always be computed in affine opens containing the relevant point, for an integral scheme, the function field can be computed on any affine open. So my answer above does not use regularity in codimension one or the Noetherian hypothesis, or separatedness. It only uses integrality.

-
Great answer, you made it look simple :-) One question: what do you mean when you say that "the function field can be computed on any affine open"? –  Manos Oct 22 '12 at 16:34
If $X$ is any scheme, $x\in X$, and $U\subseteq X$ is open with $x\in U$, then the canonical open immersion $U\hookrightarrow X$ induces an isomorphism $\mathscr{O}_{X,x}\cong\mathscr{O}_{U,x}$. So we can identify the local ring of $X$ and $x$ and the local ring of $U$ at $x$. Now assume $X$ is integral with generic point $\xi$ and that $U$ is affine open (as long as $U\neq\emptyset$ it is automatic that $\xi\in U$). Now $U=\mathrm{Spec}(A)$ is an integral affine scheme, meaning $A$ is a domain, and $\xi\in U$ is its generic point, which means that it corresponds to the zero ideal. –  Keenan Kidwell Oct 22 '12 at 18:04
The definition of the structure sheaf of $\mathrm{Spec}(A)$ is such that at a prime $\mathfrak{p}$, the local ring is $A_\mathfrak{p}$. In a particular, taking $\mathfrak{p}=(0)$, the zero ideal, the local ring at the generic point is $A_{(0)}=\mathrm{Frac}(A)$ (by definition). So the function field of $X$ can be identified with the field of fractions of the ring of sections of any non-empty affine open. –  Keenan Kidwell Oct 22 '12 at 18:05