Let \(X\) be an integral scheme. Show that the local ring \(\OO_{\xi}\) of the generic point \(\xi\) of \(X\) is a field. It is called the function field of \(X\) and is denoted by \(K(X)\).
Show also that if \(U = \Spec A\) is any open affine subset of \(X\), then \(K(X)\) is isomorphic to \(\Frac(A)\).
Solution
1. Since \(X\) is integral, it is irreducible and has a unique generic point \(\xi\).
An integral scheme is, by definition, reduced and irreducible. Every irreducible scheme has a unique generic point.
2. Every nonempty open subset of \(X\) contains \(\xi\).
The closure of \(\{\xi\}\) is all of \(X\). If a nonempty open \(U\) did not contain \(\xi\), then the closed set \(X\setminus U\) would contain \(\xi\) and hence would contain its closure \(X\), contradicting \(U\ne\varnothing\).
3. Let\[ U=\Spec A\subseteq X \]be a nonempty affine open. Then \(A\) is a domain and the point \(\xi\in U\) corresponds to the prime ideal \((0)\subseteq A\).
Because \(X\) is reduced and irreducible, every open subscheme is reduced and irreducible. Thus the affine scheme \(U=\Spec A\) is integral, which is equivalent to \(A\) being a domain.
The generic point of \(\Spec A\) is the zero prime. By 2, the generic point \(\xi\) of \(X\) lies in \(U\); its closure inside \(U\) is all of \(U\), so it is the generic point of \(U\). Hence it corresponds to \((0)\).
4. The local ring at the generic point is\[ \boxed{ \mathcal O_{X,\xi} \cong A_{(0)} = \operatorname{Frac}(A). } \]
The local ring of the affine scheme \(\Spec A\) at the prime \((0)\) is \(A_{(0)}\). Since \(A\) is a domain, localizing at the complement of \((0)\) means inverting every nonzero element of \(A\), which gives its fraction field.
The stalk of the restricted structure sheaf on the open set \(U\) agrees with the stalk of \(\mathcal O_X\) at \(\xi\), so \[ \mathcal O_{X,\xi}=\mathcal O_{U,\xi}\cong A_{(0)}. \]
5. Therefore \(\mathcal O_{X,\xi}\) is a field, and for every nonempty affine open \(U=\Spec A\) there is a canonical isomorphism\[ \boxed{K(X)=\mathcal O_{X,\xi}\cong\operatorname{Frac}(A).} \]
Step 4 identifies the local ring with a fraction field, hence with a field. The identification is canonical because both sides are the same stalk computed using the affine neighborhood \(U\).
6. Q.E.D.
Steps <1>4–<1>5 prove both assertions.