Let \(X\) be a surface with function field \(K\). Show that every valuation ring \(R\) of \(K / k\) is one of the three kinds described in (II, Ex. 4.12).
Hint: In case (3), let \(f \in R\). Use (Ex. 5.1) to show that for all \(i \gg 0\), \(f \in \mathcal{O}_{X_i}\), so in fact \(f \in R_0\).
Solution
Under the standing hypotheses of Chapter V, \(X\) is a nonsingular projective surface over the algebraically closed field \(k\). Put \[ K=k(X). \]
Exercise II.4.12 explicitly excludes the trivial valuation ring \(K\) itself from its three surface types. We therefore prove the assertion for \[ R\subsetneq K; \] the ring \(K\) is the separate trivial case.
1. The valuation ring \(R\) has a unique center\[ x_0\in X. \]
Since \(X\) is projective, it is proper over \(k\). The valuation-center criterion Exercise II.4.5 says that every valuation ring of \(K/k\) has a unique center on a proper model. Equivalently, the generic-point map \[ \Spec K\longrightarrow X \] extends uniquely to \[ \Spec R\longrightarrow X, \] and the image of the closed point is \(x_0\).
2. The center \(x_0\) is not the generic point of \(X\).
If \(x_0=\eta_X\), then domination gives \[ \OO_{X,\eta_X}=K\subseteq R. \] Since already \(R\subseteq K\), this would force \[ R=K, \] contrary to the nontriviality assumption.
Because \(\dim X=2\), step 2 leaves two possibilities: \(x_0\) is the generic point of an irreducible curve, or \(x_0\) is a closed point.
3. If \(x_0\) has codimension one, then\[ \boxed{R=\OO_{X,x_0},} \]so \(R\) is of type (1) in Exercise II.4.12.
Since \(X\) is nonsingular and \(x_0\) has codimension one, \[ \OO_{X,x_0} \] is a one-dimensional regular local domain, hence a discrete valuation ring with fraction field \(K\).
The assertion that \(x_0\) is the center means precisely that \(R\) dominates \(\OO_{X,x_0}\). A valuation ring dominating another valuation ring of the same fraction field is equal to it P-AGH2412VALEX. Hence \[ R=\OO_{X,x_0}. \] This is the first surface construction in Exercise II.4.12.
Assume from now on that \(x_0\) is closed.
4. As long as the center \(x_i\) of \(R\) on \(X_i\) is closed, define\[ X_{i+1}=\Bl_{x_i}X_i. \]Then \(R\) has a unique center\[ x_{i+1}\in X_{i+1}, \]and\[ \boxed{x_{i+1}\in E_{i+1},} \]where \(E_{i+1}\) is the exceptional curve of\[ X_{i+1}\longrightarrow X_i. \]
The blowup of a nonsingular projective surface at a closed point is again a nonsingular projective surface. Hence step 1 applies to every \(X_i\) and gives a unique center \(x_{i+1}\) on \(X_{i+1}\).
The morphism \[ X_{i+1}\longrightarrow X_i \] is proper and the induced maps \[ \Spec R\longrightarrow X_{i+1}\longrightarrow X_i \] and \[ \Spec R\longrightarrow X_i \] agree on the generic point. By uniqueness of the center on \(X_i\), the closed point \(x_{i+1}\) maps to \(x_i\). Therefore \[ x_{i+1}\in\pi_{i+1}^{-1}(x_i)=E_{i+1}. \]
5. If for some \(n\ge1\) the center \(x_n\) is the generic point of\(E_n\), then\[ \boxed{R=\OO_{X_n,x_n},} \]and \(R\) is of type (2) in Exercise II.4.12.
The surface \(X_n\) is nonsingular, so the local ring at the generic point of the irreducible curve \(E_n\) is a DVR: \[ \OO_{X_n,x_n}. \] The valuation ring \(R\) dominates this DVR because \(x_n\) is its center. As in step 3, domination between valuation rings of the same fraction field forces equality.
The composite birational morphism \[ X_n\longrightarrow X \] maps the exceptional curve \(E_n\) to the original closed point \(x_0\). Thus \(R\) is exactly the local ring at the generic point of a curve on a birational model which is contracted to a closed point of \(X\), the second construction of Exercise II.4.12.
6. If no center ever becomes divisorial, then every \(x_i\) is closed andthe construction gives an infinite chain\[ X=X_0 \longleftarrow X_1 \longleftarrow X_2 \longleftarrow \cdots \]of blowups at the successive centers of \(R\). Put\[ A_i=\OO_{X_i,x_i}, \qquad R_0=\bigcup_{i\ge0}A_i. \]Then\[ \boxed{R_0\subseteq R.} \]
By definition of center, \(R\) dominates every \(A_i\), so \[ A_i\subseteq R \] for all \(i\). Hence their union is contained in \(R\).
Moreover the transition maps are local inclusions \[ A_i\hookrightarrow A_{i+1}, \] because \(x_{i+1}\) lies above \(x_i\). Thus this is precisely the third construction of Exercise II.4.12, except that it remains to prove the assertion promised there: \[ R_0=R. \]
7. Let\[ f\in R. \]There is an index \(N\) such that, for every \(i\ge N\),\[ \boxed{f\in A_i=\OO_{X_i,x_i}.} \]
The case \(f=0\) is immediate, so assume \[ f\in K^\times. \] Regard \(f\) as the rational map \[ \varphi_f:X\dashrightarrow\PP^1, \qquad x\longmapsto[f(x):1]. \] Exercise V.5.1 P-AGH551RESOLVERATFN gives a finite sequence of point blowups \[ Y_m \xrightarrow{g_m} Y_{m-1} \longrightarrow \cdots \longrightarrow Y_1 \xrightarrow{g_1} Y_0=X \] such that \(\varphi_f\) extends to a morphism \[ \psi:Y_m\longrightarrow\PP^1. \]
For each \(j\), let \(y_j\) be the unique center of \(R\) on \(Y_j\). We compare the local rings at these centers with the sequence \[ A_0\subset A_1\subset A_2\subset\cdots. \]
Suppose the point blown up by \[ g_j:Y_j\longrightarrow Y_{j-1} \] is not \(y_{j-1}\). Then \(g_j\) is an isomorphism on a neighbourhood of \(y_{j-1}\), and therefore \[ \OO_{Y_j,y_j}\cong\OO_{Y_{j-1},y_{j-1}}. \]
Suppose instead that \(g_j\) blows up \(y_{j-1}\). Under any identification \[ \OO_{Y_{j-1},y_{j-1}}\cong A_i \] already obtained, this is the blowup at the center \(x_i\) of \(R\). The center \(y_j\) selected by \(R\) on that blowup is therefore the same infinitely near point as \(x_{i+1}\), and \[ \OO_{Y_j,y_j}\cong A_{i+1}. \]
Starting from \[ \OO_{Y_0,y_0}=\OO_{X,x_0}=A_0 \] and proceeding through the finite sequence of blowups, we obtain some \(N\le m\) with \[ \OO_{Y_m,y_m}\cong A_N \] inside the common function field \(K\).
It remains to see that \(f\) is regular at \(y_m\). Because \[ f\in R, \] the \(k\)-algebra map \[ k[t]\longrightarrow R, \qquad t\longmapsto f, \] defines a morphism \[ \Spec R\longrightarrow\AA^1\subset\PP^1 \] whose restriction to \(\Spec K\) is the rational map \([f:1]\).
On the other hand, the center morphism \[ \Spec R\longrightarrow Y_m \] followed by \(\psi\) gives another extension to \(\PP^1\) of the same generic-point map. Since \(\PP^1\) is separated, the two extensions agree. Consequently \[ \psi(y_m)\in\AA^1, \] and the affine coordinate \(t\) pulls back near \(y_m\) to the rational function \(f\). Hence \[ f\in\OO_{Y_m,y_m}=A_N. \] Since \[ A_N\subseteq A_{N+1}\subseteq A_{N+2}\subseteq\cdots, \] the same element belongs to \(A_i\) for every \(i\ge N\), exactly as asserted in the hint.
8. In the infinite closed-center case,\[ \boxed{R=R_0=\bigcup_{i\ge0}\OO_{X_i,x_i}.} \]Thus \(R\) is of type (3) in Exercise II.4.12.
Step 6 gives \[ R_0\subseteq R. \] Conversely, step 7 shows that every element \[ f\in R \] belongs to some \(A_N\), hence to \(R_0\). Therefore \[ R\subseteq R_0. \] The two inclusions give equality.
In particular, the local union \(R_0\) in the third construction is itself a valuation ring; no additional dominating valuation ring is needed. This is the assertion deferred from Exercise II.4.12 to the present exercise.
9. Every nontrivial valuation ring of \(K/k\) is one of the three kindsfrom Exercise II.4.12.
By steps <1>1–<1>2, its center on \(X\) is either divisorial or closed. The divisorial case is type (1) by step 3.
Starting from a closed center, step 4 gives successive centers on point blowups. If one of them becomes the generic point of the exceptional curve, step 5 gives type (2). If this never happens, every center remains closed and steps <1>6–<1>8 give type (3).
These alternatives exhaust all possibilities for points on the exceptional curve of a nonsingular surface.
10. Q.E.D.
Steps <1>1–<1>5 give the first two valuation types from their centers on successive smooth models. Steps <1>6–<1>8 prove the infinite-center case and the essential equality \(R=R_0\) using Exercise V.5.1. Step 9 assembles the exhaustive trichotomy, with the trivial valuation ring \(K\) separately excluded exactly as in Exercise II.4.12.