Valuative criteria using discrete valuation rings

Suppose \(R\subseteq L\) is a valuation ring containing \(K\) and with \[ \mathfrak m_R\cap K=(0). \] Let \(v\) be its valuation. Then \(v(c)=0\) for every \(c\in K^\times\).

If \(0\ne\alpha\in L\) is algebraic over \(K\), write a polynomial relation \[ c_0+c_1\alpha+\cdots+c_n\alpha^n=0 \] with \(c_0c_n\ne0\). If \(v(\alpha)>0\), the term \(c_0\) has strictly smaller value than every other term, impossible in a vanishing sum. If \(v(\alpha)<0\), the term \(c_n\alpha^n\) has strictly smaller value than every other term, again impossible. Hence \[ v(\alpha)=0. \] Thus the valuation is trivial on all of \(L\), so \(R=L\) is a field rather than a DVR.

Choose a transcendence basis \[ t_1,\ldots,t_r \] of \(L/K\). Then \[ L/K(t_1,\ldots,t_r) \] is finite.

If \(r=0\), then \(L/K\) is already finite. Since we have excluded 1, \(\mathfrak m\ne0\).

If \(r>0\), put \[ \OO_1 = \OO[t_1,\ldots,t_r]_{\mathfrak n}, \qquad \mathfrak n = \mathfrak m\OO[t_1,\ldots,t_r]+(t_1,\ldots,t_r). \] Then \(\OO_1\) is a noetherian local domain, its fraction field is \[ K(t_1,\ldots,t_r), \] and its maximal ideal \(\mathfrak n\OO_1\) is nonzero. Moreover \(\OO_1\) dominates \(\OO\) because \[ \mathfrak n\cap\OO=\mathfrak m. \]

If a DVR of \(L\) dominates \(\OO_1\), it also dominates \(\OO\). Thus replacing \((\OO,K)\) by \((\OO_1,K(t_1,\ldots,t_r))\) reduces to a finite extension and a nonzero maximal ideal.

The maximal ideal is finitely generated because \(\OO\) is noetherian.

Hartshorne I.6.1A says that every local subring of a field is dominated by a valuation ring of that field. Hence there is a valuation ring \[ V\subseteq K \] dominating \(\OO\).

All \(x_i\) lie in the maximal ideal of \(V\), so their values are positive. Among the finite set of values choose a minimum and relabel the corresponding generator as \(x_1\).

By minimality of \(v(x_1)\), \[ v(x_i/x_1)\ge0 \] for every \(i\). Hence \[ \OO'\subseteq V. \]

Because \(V\) dominates \(\OO\) and \(x_1\in\mathfrak m\), one has \[ x_1\in\mathfrak m_V. \] If \((x_1)=\OO'\), then \(x_1\) would be a unit of \(\OO'\), hence a unit of \(V\), contradiction. Thus \((x_1)\) is not the unit ideal.

The ring \(\OO'\) is a finitely generated \(\OO\)-algebra, hence noetherian, and it is a subring of the field \(K\), hence a domain.

Krull’s principal ideal theorem gives \[ \operatorname{ht}\mathfrak p\le1. \] Since \(x_1\ne0\) and \(x_1\in\mathfrak p\), one has \(\mathfrak p\ne(0)\), so \[ \operatorname{ht}\mathfrak p=1. \] Therefore \[ \dim\OO'_{\mathfrak p}=1. \]

For every \(j\), \[ x_j=x_1(x_j/x_1)\in\mathfrak p. \] Thus \[ \mathfrak m=(x_1,\ldots,x_n) \subseteq \mathfrak p\cap\OO. \] The contraction is a proper ideal of the local ring \(\OO\), so maximality of \(\mathfrak m\) gives \[ \mathfrak p\cap\OO=\mathfrak m. \] Hence the local inclusion \[ \OO\subseteq\OO'_{\mathfrak p} \] is a domination.

Finally \(\OO'\) lies between \(\OO\) and its fraction field \(K\), so its fraction field, and that of its localization, is \(K\).

The ring \(\OO'_{\mathfrak p}\) is a one-dimensional noetherian domain. The Krull–Akizuki theorem says that its integral closure in any finite extension of its fraction field is a noetherian domain of dimension one. Thus \(C\) has the asserted properties.

The extension \[ \OO'_{\mathfrak p}\subseteq C \] is integral, so lying over provides a prime \(\mathfrak q\) over the maximal ideal of \(\OO'_{\mathfrak p}\). Since \(C\) is one-dimensional, such a nonzero prime is maximal.

The localization \(R=C_{\mathfrak q}\) is a noetherian local domain of dimension one. It is integrally closed because \(C\) is integrally closed and localization preserves integral closedness. Hence Hartshorne I.6.2A implies that \(R\) is a discrete valuation ring.

Its fraction field is \(L\), and the contraction of its maximal ideal to \(\OO'_{\mathfrak p}\) is the maximal ideal there. Thus \(R\) dominates \(\OO'_{\mathfrak p}\), which by 5 dominates \(\OO\). Therefore \(R\) dominates \(\OO\).

Step 2 reduces to the finite-extension, nonfield case; steps <1>3–<1>7 construct the required DVR. Step 1 records the unique excluded case in the source statement.

These are immediate special cases of Hartshorne II.4.3 and II.4.7, whose valuative criteria are stated for arbitrary valuation rings.

Consider the diagonal \[ \Delta_f:X\longrightarrow X\times_YX. \] It is always an immersion. It is enough to show its image is closed.

Suppose it were not closed. Since \(X\times_YX\) is noetherian and the image of an immersion is locally closed, choose a point \[ z\in\overline{\Delta_f(X)}\setminus\Delta_f(X) \] and an irreducible closed subset \[ T\subseteq\overline{\Delta_f(X)} \] containing \(z\) whose generic point \(\xi\) lies in \(\Delta_f(X)\). Give \(T\) its reduced induced structure.

The local ring \[ \OO_{T,z} \] is a noetherian local domain. Since \(z\ne\xi\), it is not a field. Its fraction field is \[ \kappa(\xi). \] Part (a) supplies a DVR \[ R\subseteq\kappa(\xi) \] dominating \(\OO_{T,z}\).

The corresponding morphism \[ \Spec R\longrightarrow T\subseteq X\times_YX \] has two projections \[ u_1,u_2:\Spec R\rightrightarrows X. \] At the generic point these agree because \(\xi\) lies on the diagonal. They are morphisms over \(Y\) because they come from the fibre product.

If \(u_1=u_2\), the whole morphism \(\Spec R\to X\times_YX\) would factor through the diagonal, so its closed point \(z\) would lie in \(\Delta_f(X)\), contradiction. Thus the same DVR-valuative diagram has two distinct lifts, contrary to hypothesis.

Hence the image of the diagonal is closed. An immersion with closed image is a closed immersion, so \(f\) is separated.

Uniqueness in particular gives the ``at most one’’ condition. Apply 10.

Because \(Y\) is noetherian and \(f\) is of finite type, \(f\) is of finite presentation. In checking universal closedness, the standard noetherian-approximation step in the proof of the valuative criterion reduces arbitrary base changes to base changes \[ Y'\longrightarrow Y \] with \(Y'\) noetherian and of finite type over \(Y\). Universal closedness of a quasi-compact morphism may be tested after base changes locally of finite presentation (Stacks Project, Tag 05JX); restricting around a point witnessing failure of closedness then gives the noetherian affine case used here.

Fix such a base change and put \[ X'=X\times_YY'. \] Let \(Z\subseteq X'\) be closed. Suppose its image in \(Y'\) were not closed. Since \[ Z\longrightarrow Y' \] is of finite type between noetherian schemes, Chevalley’s theorem makes its image constructible. Choose \[ y\in\overline{p(Z)}\setminus p(Z). \] There is an irreducible closed subset \[ C\subseteq\overline{p(Z)} \] containing \(y\) whose generic point \[ \eta \] lies in \(p(Z)\); this follows from the generic-point characterization of dense constructible subsets. Give \(C\) its reduced induced scheme structure.

Choose a point \[ z\in Z \] over \(\eta\). The residue-field extension \[ L=\kappa(z) \quad/\quad K=\kappa(\eta) \] is finitely generated because \(Z\to Y'\) is of finite type.

The local ring \[ \OO_{C,y} \] is a noetherian local domain with fraction field \(K\). Since \(y\ne\eta\), it is not a field. Part (a) gives a DVR \[ R\subseteq L \] dominating \(\OO_{C,y}\).

The generic point gives a morphism \[ \Spec L\longrightarrow Z\subseteq X'=X\times_YY', \] and domination gives \[ \Spec R\longrightarrow C\subseteq Y'. \] Composing the former with \(X'\to X\) and the latter with \(Y'\to Y\) produces a DVR-valuative diagram for \(f\). By hypothesis it has a lift \[ \Spec R\longrightarrow X. \]

Together with the fixed morphism \(\Spec R\to Y'\), this lift gives by the fibre-product property a morphism \[ \Spec R\longrightarrow X'. \] Its generic point is the chosen point of \(Z\). Since \(Z\) is closed and the generic point of \(\Spec R\) is dense, the entire morphism factors through \(Z\). The closed point therefore maps to a point of \(Z\) lying over \(y\), contradicting \[ y\notin p(Z). \]

Thus every such base change is closed, and the noetherian-approximation reduction gives universal closedness.

Steps <1>1–<1>8 prove the corrected form of part (a), and steps <1>9–<1>14 prove part (b).