Proper morphisms and the valuative criterion of properness

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI, GPT-6.1 Sol (OpenAI), in Codex at Ultra. Public domain (CC0).

A closed map sends closed subsets to closed subsets. Universal closedness asks for that behaviour after every change of base. Valuation rings turn this apparently unlimited topological test into an extension problem: can a point defined over a fraction field acquire a compatible limit over the valuation ring? Properness combines existence of these limits with uniqueness and finite type.

We use the specialization witnesses, domination theorem, value groups, and separatedness criteria proved in Valuation rings and separatedness, and the integral and finite morphisms of Affine and finite morphisms. Quasi-compactness and finite type have the conventions of Finiteness of morphisms. No separation or Noetherian hypothesis is implicit. A valuation ring may have arbitrary rank; its fraction field is always the field occurring in its valuation diagram.

1. Closedness after changing the base

A morphism \(f:X\to S\) is universally closed if, for every \(S'\to S\), the projection \(X\times_S S'\to S'\) is a closed map of underlying spaces. It need not be a closed immersion. For example, an integral morphism is universally closed, even when many points lie over a point of the base.

We write \(s'\leadsto s\) when \(s\) lies in the closure of \(\{s'\}\). We say that specializations lift along \(f\) if, for every such specialization and every specified \(x'\) over \(s'\), there is a specialization \(x'\leadsto x\) with \(f(x)=s\). Specifying \(x'\) matters: merely finding some point over \(s\) is a weaker assertion.

Lemma 1.1. A closed morphism lifts specializations. Conversely, a quasi-compact morphism that lifts specializations is closed.

Proof. If \(f\) is closed, the image of \(\overline{\{x'\}}\) is closed and contains \(s'\). It therefore contains \(s\), giving the required \(x\).

For the converse, let \(Z\subset X\) be any closed subset, endowed with its reduced closed subscheme structure. The map \(Z\to S\) is quasi-compact. Its image is stable under specialization: a point of \(Z\) can be specialized as required in \(X\), and every such specialization stays in \(Z\). The quasi-compact image criterion of Theorem 1.2 in Finiteness of morphisms says that this image is closed. This applies to every \(Z\). \(\square\)

A valuation diagram for \(f\) consists of compatible maps

where \(R\) is a valuation ring and compatibility means \(fu=a|_{\operatorname{Spec}K}\). An extension is an \(S\)-morphism \(\widetilde u:\operatorname{Spec}R\to X\) restricting to \(u\). The existence part asks for at least one extension in every diagram; the uniqueness part asks for at most one.

Theorem 1.2 (universal closedness). The existence part is equivalent to lifting specializations after every base change. In particular, if \(f\) is quasi-compact, then

Proof. First suppose specializations lift after every base change. For (1.1), replace the base by \(\operatorname{Spec}R\). The resulting generic point \(x'\) of the diagram in \(X_R\) specializes to a point \(x\) over the closed point of \(\operatorname{Spec}R\). The given generic map identifies \(\kappa(x')\) with a subfield of \(K\). We have maps

whose composite is the given inclusion \(R\subset K\). The first map is local. The image \(C\) of the middle local ring in \(K\) is a local subring containing and dominating \(R\): its maximal ideal contracts to the maximal ideal of \(R\). A valuation ring is maximal among local subrings of its fraction field that dominate it, by the domination characterization in the valuation lesson. Thus \(C=R\). The resulting local map \(\mathcal O_{X_R,x}\to R\) defines \(\operatorname{Spec}R\to X_R\). Its generic restriction is the originally specified map, since all maps in (1.3) used that same embedding into \(K\). Projection to \(X\) gives the extension.

Conversely, existence survives arbitrary base change. An extension into \(X\), together with the prescribed map to the new base, defines the extension into the fibre product. It is therefore enough to lift one specialization \(s'\leadsto s\) along \(f\). Given \(x'\) over \(s'\), apply the specialization-witness theorem of the valuation lesson with the specified field extension \(\kappa(x')/\kappa(s')\). It supplies a valuation ring \(R\subset K=\kappa(x')\) and a map \(\operatorname{Spec}R\to S\) whose generic point maps to \(s'\) and whose closed point maps to \(s\). The canonical map \(\operatorname{Spec}\kappa(x')\to X\) completes a valuation diagram. Any extension sends the closed point to the desired specialization of \(x'\).

If \(f\) is universally closed, Lemma 1.1 gives lifting after every base change, hence existence. If \(f\) is quasi-compact and satisfies existence, its base changes are quasi-compact and lift specializations, so Lemma 1.1 makes all of them closed. \(\square\)

These are Stacks, Tags 01KC, 01KE and 01KF. The domination argument explains why the extension uses the original valuation ring, rather than an unspecified larger valuation ring.

2. Properness and its permanence properties

A morphism is proper if it is separated, of finite type, and universally closed. Each requirement has a distinct role. Finite type controls how much algebra is needed; separatedness controls uniqueness; universal closedness controls existence. An infinite integral algebra can be universally closed and separated without being of finite type.

Theorem 2.1. Properness is local on the target for open covers and is stable under arbitrary base change and composition. Closed immersions and finite morphisms are proper. If

has \(gf\) proper and \(g\) separated, then \(f\) is proper. Consequently, every \(S\)-morphism from a proper \(S\)-scheme to a separated \(S\)-scheme is proper and has closed image.

Proof. Universal closedness is target-local: after any base change, the inverse images of a target open cover form an open cover, and the images of a closed subset are closed on all those opens exactly when its full image is closed. Separatedness and finite type are target-local by the first two lessons. Their conjunction is therefore target-local.

Base changes of universally closed maps remain universally closed by associativity of fibre products. A composite of universally closed maps is universally closed because every base change of the composite is a composite of closed maps. Separatedness and finite type have the same permanence properties, proving the corresponding assertions for properness.

A closed immersion is separated, of finite type, and remains a closed immersion after base change. It is therefore proper, even when its ideal is not finitely generated. A finite morphism is affine and hence separated; its finite module algebra is finitely generated as an algebra. It is integral, so universally closed by the integral-morphism theorem. This proves properness of finite morphisms without a Noetherian assumption.

For cancellation, factor \(f\) as

The graph \(\Gamma_f\) is a closed immersion because \(Y\to S\) is separated. The projection is the base change of the proper map \(X\to S\). Both arrows are proper, hence so is \(f\). The final assertion is this factorization applied to the structure maps. A proper map is closed, and the image of its entire source is thus closed. No finite-type hypothesis on \(Y\to S\) is needed. \(\square\)

The target-local, composition, base-change, closed-immersion, image, and finite assertions are Stacks, Tags 01W2–01W6 and Tag 01WN.

Corollary 2.2. If \(p:X'\to X\) is surjective and universally closed and \(fp:X'\to S\) is universally closed, then \(f\) is universally closed.

Proof. Every base change of a surjective scheme morphism is surjective: over a point, a pair of nonempty fibres gives a point of the fibre product because the tensor product of two field extensions of the residue field is nonzero. Given a closed subset \(Z\subset X_T\), its inverse image in \(X'_T\) is closed, and its image in \(T\) equals the image of \(Z\). That image is closed by universal closedness of \((fp)_T\). \(\square\)

In fact the proof only needs universal surjectivity of \(p\); the stated form is the one we will use.

3. The general valuative criterion

Theorem 3.1. Let \(f:X\to S\) be quasi-separated and of finite type. Then \(f\) is proper if and only if every valuation diagram has exactly one extension.

Proof. A proper morphism is quasi-compact and universally closed, so Theorem 1.2 supplies existence. Its separatedness supplies uniqueness by the valuation lesson. Conversely, finite type includes quasi-compactness, so existence gives universal closedness. Quasi-separatedness and uniqueness give separatedness by the general separatedness criterion of that lesson. Finite type was already assumed. These are precisely the three properness conditions. \(\square\)

This is Stacks, Tag 0BX5. Notice where quasi-separatedness enters: it is needed for the converse uniqueness criterion. Notice also that the valuation test alone does not impose finite type.

For a concrete failure of the latter inference, let \(L/k\) be an infinite algebraic field extension. The map \(\operatorname{Spec}L\to\operatorname{Spec}k\) is integral and separated, hence satisfies existence and uniqueness. It is not of finite type: a field generated by finitely many algebraic elements has finite degree. Thus it is not proper.

4. Projective space supplies limits

Theorem 4.1. For every scheme \(S\) and every \(n\geq0\), the projection \(\mathbf P^n_S\to S\) is proper.

Proof. Separatedness was proved using chart intersections in the diagonal lesson. The standard charts are affine spaces over the base, and there are \(n+1\) of them, so the morphism is of finite type. It remains to prove existence.

After pulling back a valuation diagram, the base is \(\operatorname{Spec}R\). A \(K\)-point of \(\mathbf P^n_R\) has homogeneous coordinates \([a_0:\cdots:a_n]\), not all zero. Among the nonzero coordinates choose \(a_j\) of least value in the totally ordered value group of \(R\). Then

Zero coordinates cause no problem. These ratios define a map into the standard chart \(D_+(T_j)\cong\mathbf A^n_R\), giving the required extension. Separatedness gives its uniqueness. Theorem 3.1 now proves properness. Since the argument works on every base change, it proves the assertion for arbitrary \(S\). \(\square\)

In a DVR one often multiplies the coordinates by a power of a uniformizer. Formula (4.1) is the version that also works when the valuation is not discrete and has higher rank: only the minimum of a finite set of values is used.

The same mechanism works for positive weights, although one must compare values after taking their degrees into account.

Theorem 4.2 (finitely generated graded algebras). Let \(G=\bigoplus_{d\geq0}G_d\) be a graded ring generated by finitely many homogeneous elements of positive degree as a \(G_0\)-algebra. Then \(\operatorname{Proj}G\to\operatorname{Spec}G_0\) is proper. In particular it satisfies both parts of the valuative criterion, without assuming that the generators have degree one.

Proof. Choose generators \(x_1,\ldots,x_m\) of degrees \(d_1,\ldots,d_m>0\). If there are none, Proj is empty and the assertion holds. Otherwise the \(D_+(x_j)\) form a finite cover. Each degree-zero chart ring \(G_{(x_j)}\) is a finite-type \(G_0\)-algebra. To verify this last point, write a degree-zero monomial with powers of \(x_j\) cancelled from its numerator. For every other exponent write \(e_i=q_i d_j+r_i\), with \(0\leq r_i<d_j\). Peel off the powers of \(x_i^{d_j}/x_j^{d_i}\). The remaining numerator has bounded exponents \(r_i\); its degree is divisible by \(d_j\), so its denominator is the corresponding integral power of \(x_j\). There are only finitely many such residual monomials. They and the displayed powers generate the chart ring. Thus Proj is of finite type; it is separated by the diagonal lesson.

Consider a valuation diagram and write \(v:K^*\to\Gamma\) for its value group. The generic point lands in some \(D_+(h)\), where \(h\) is homogeneous of degree \(e>0\). Let \(\varphi:G_{(h)}\to K\) be its ring map and put

At least one \(q_i\) is nonzero, since the \(D_+(x_i)\) cover Proj. Set \(D=\operatorname{lcm}(d_1,\ldots,d_m)\). Among the nonzero \(q_i\), choose \(q_j\) minimizing

The point lies in \(D_+(x_j)\), so it determines a map \(\psi:G_{(x_j)}\to K\). Every generating degree-zero monomial has the form

If a factor with positive exponent has \(q_i=0\), its image under \(\psi\) is zero. Otherwise raising the monomial to the \(e\)-th power and using the degree equality gives

A totally ordered abelian group has no torsion and positive integer multiplication detects the sign. Hence \(v(\psi(M))\geq0\). The coefficients from \(G_0\) already map into \(R\), so the whole chart ring maps into \(R\). This defines an extension through \(D_+(x_j)\). Its generic restriction is the prescribed point, and uniqueness follows from separatedness. Theorem 3.1 finishes the proof. \(\square\)

The weighted valuative statement is Stacks, Tag 01MF. The value comparison avoids dividing in \(\Gamma\), where division by degrees need not exist. Finite homogeneous generation is essential to the finite-type conclusion; arbitrary graded algebras need not produce proper morphisms.

5. Why discrete valuation rings suffice over a Noetherian base

Theorem 5.1 (Noetherian valuative criterion). Suppose \(S\) is locally Noetherian and \(f:X\to S\) is of finite type. Then \(f\) is proper if and only if every valuation diagram with \(R\) a DVR has exactly one extension.

The additional ingredient is the Noetherian form of Chow's lemma: for a separated finite-type \(X\) over a Noetherian scheme \(S\), there are a proper surjection \(p:X'\to X\) and an immersion \(j:X'\to\mathbf P^N_S\), with \(p\) an isomorphism over a dense open of \(X\). Its exact full open proof is Stacks, Tag 0200. The construction is also proved in Theorem 4.1 of Projective morphisms and Chow's lemma, after the lessons on very ample and ample sheaves. It uses properness of projective space and properness permanence, already proved above; it does not depend on the DVR criterion being proved now.

Proof of Theorem 5.1. Properness implies existence and uniqueness for all valuation rings, hence for DVRs. Assume conversely the DVR test. Properness is target-local, so restrict to an affine open of \(S\). Its inverse image is Noetherian because \(f\) is of finite type. The Noetherian separatedness criterion in the valuation lesson shows first that \(f\) is separated: every DVR diagram has at most one extension. We may therefore apply the stated form of Chow's lemma.

We claim that the immersion \(j:X'\to\mathbf P^N_S\) is closed. Otherwise its image has a point \(y\) in its closure outside the image. The Noetherian space \(X'\) has finitely many irreducible components, so \(y\) lies in the closure of the image of one of them. Give that closure its reduced structure and call it \(Z\). It is an integral Noetherian scheme; the component's generic point has residue field \(K=K(Z)\), and \(y\) is not its generic point. The local domain \(\mathcal O_{Z,y}\) is not a field. By the discrete domination theorem used in the valuation lesson, there is a DVR \(R\subset K\), with fraction field exactly \(K\), dominating \(\mathcal O_{Z,y}\). It defines a map

whose closed point goes to \(y\), while its generic point has its specified lift \(u:\operatorname{Spec}K\to X'\).

Apply the assumed DVR existence test to \(pu:\operatorname{Spec}K\to X\) and the base map supplied by \(c\). It yields \(a:\operatorname{Spec}R\to X\). Since \(p\) is proper, the general existence criterion lifts this diagram to \(b:\operatorname{Spec}R\to X'\). The two \(S\)-maps \(jb\) and \(c\) into projective space agree on \(\operatorname{Spec}K\). Projective space is separated and \(\operatorname{Spec}R\) is integral, so the reduced-source uniqueness lemma gives \(jb=c\). Its closed point would then map both into \(j(X')\) and to \(y\), a contradiction.

An immersion with closed image is a closed immersion: near the image its closed-immersion descriptions give the ideal sheaf, and the open complement has empty inverse image. Thus \(j\) is closed. Theorem 4.1 and composition make \(X'\to S\) proper. Corollary 2.2 then makes \(X\to S\) universally closed. It is already separated and of finite type, hence proper. \(\square\)

The theorem and the boundary-point argument are Stacks, Tag 0208. Discrete domination uses the exact algebra proof Tag 00PH, with its hypotheses and dependencies explained in the valuation lesson. This proof tests the boundary of a finite-type projective model; it does not assert that an arbitrary higher-rank valuation can itself be replaced by a DVR.

6. Missing limits, duplicated limits, and closed subspaces

The affine line misses limits. In \(\mathbf A^1_k\times_k\mathbf A^1_k\), with base coordinate \(x\) and other coordinate \(y\), the closed subscheme \(V(xy-1)\) has coordinate ring \(k[x,x^{-1}]\). Its image on the \(x\)-line is \(D(x)\), which is not closed. Thus \(\mathbf A^1_k\to\operatorname{Spec}k\) is not universally closed. The map itself is closed: its target has one point. Universal closedness detects the failure only after a base change.

The same failure appears in a DVR diagram. Take \(R=k[[t]]\), \(K=k((t))\), and the point with affine coordinate \(y=t^{-1}\). An extension would require a ring map \(k[y]\to R\) sending \(y\) to \(t^{-1}\), which is impossible.

The doubled affine line duplicates some limits but still misses others. Glue two copies of \(\mathbf A^1_k\) along \(D(z)\) by the identity. The generic point \(z=t\) extends through either chart over \(k[[t]]\), with distinct closed-point images at the two origins. This proves failure of uniqueness. However, \(z=t^{-1}\) has no extension. Any proposed extension has a closed-point image in one affine chart. The inverse image of that chart is an open neighbourhood of the closed point of a local spectrum and hence the whole spectrum. It would supply the impossible ring map just described. Therefore the doubled affine line is not an example of existence in every diagram.

A doubled projective line has existence without uniqueness. Instead glue two copies of \(\mathbf P^1_k\) along \(\mathbf P^1_k\setminus\{0\}\) by the identity, retaining distinct origins \(0_1,0_2\). Every field-valued point lies in one copy of \(\mathbf P^1\), and its valuation diagram extends in that copy by Theorem 4.1. Thus existence holds for every valuation ring. The generic point with local coordinate \(z=t\) again has two extensions, one to each origin. The scheme is quasi-compact and of finite type, and so Theorem 1.2 makes its structure map universally closed. It fails separatedness and therefore fails properness.

Closed subspaces of projective space keep limits. If \(Z\hookrightarrow\mathbf P^n_S\) is any closed immersion, then \(Z\to S\) is proper by Theorem 2.1. Alternatively, extend a generic point in projective space. Every local equation of \(Z\) vanishes in \(K\), and hence in \(R\subset K\); the extension factors through \(Z\). This works even when the defining ideal is not of finite type.

7. Exercises with solutions

Exercise 7.1 (easy). Show that \(\mathbf A^1_k\to\operatorname{Spec}k\) is closed but not universally closed. Identify a specialization that fails to lift in a base change.

Solution. Every subset image in the one-point target is closed. Base change by \(\mathbf A^1_k\) gives the projection of the affine plane to the \(x\)-line. The closed hyperbola \(H=V(xy-1)\) maps to \(D(x)\). Its generic point maps to the generic point of the line, which specializes to the origin; \(H\) has no point over that origin. Equivalently, the image \(D(x)\) contains the generic point but omits one of its specializations. If the plane projection were closed, the image of \(H\) would be closed. Hence the original line map is not universally closed. This does not contradict the fact that specializations can lift along the plane projection itself: the obstruction is a specialization that must stay inside the specified closed subset \(H\).

Exercise 7.2 (easy). Prove that every closed subscheme \(Z\subset\mathbf P^n_S\) is proper over \(S\). Explain why no coherence assumption on its ideal is required.

Solution. Its inclusion is a closed immersion, hence separated, universally closed, and of finite type. The finite-type assertion follows on affine charts from the quotient map \(A\to A/I\): the target is generated as an \(A\)-algebra without adjoining any elements, regardless of the size of \(I\). The projective-space map is proper by Theorem 4.1. Composition proves the assertion. Finite presentation of the inclusion would require local finite generation of \(I\), but properness uses finite type, so that stronger requirement is unnecessary.

Exercise 7.3 (medium). Let \(R\) be any valuation ring with fraction field \(K\). Extend a point \([a_0:\cdots:a_n]\in\mathbf P^n(K)\) to \(\mathbf P^n(R)\), and prove uniqueness. Do not assume that \(R\) has a uniformizer.

Solution. Choose a nonzero coordinate \(a_j\) of least value among the finitely many nonzero coordinates. For each \(i\), the ratio \(b_i=a_i/a_j\) belongs to \(R\), and \(b_j=1\). The tuple defines a surjection \(R^{n+1}\to R\) and, concretely, the chart map \(R[T_i/T_j]_{i\ne j}\to R\) with ratios \(b_i\). It gives the desired extension over \(\operatorname{Spec}R\). Choosing a different minimum coordinate produces a tuple differing by a unit, because the ratio of two equal-value coordinates is a unit of \(R\); hence the corresponding chart maps glue to the same point. More generally any two extensions agree on the generic point. Their equalizer is closed because projective space is separated, and it contains the schematically dense generic point of the integral scheme \(\operatorname{Spec}R\). They are therefore equal, by the reduced-source lemma. This also covers possible extensions described in different charts.

Exercise 7.4 (medium). Let \(X\xrightarrow fY\xrightarrow gS\) have \(gf\) proper and \(g\) separated. Prove that \(f\) is proper, without assuming that \(g\) is of finite type.

Solution. The graph is the pullback of \(\Delta_g:Y\to Y\times_S Y\) along the map \(X\times_S Y\to Y\times_S Y\) given by \((x,y)\mapsto(f(x),y)\). Thus \(\Gamma_f:X\to X\times_S Y\) is a closed immersion. The projection to \(Y\) is the base change of \(gf:X\to S\), hence proper. Their composite is \(f\). Closed immersions and compositions are proper, so the assertion follows. This proof establishes finite type of \(f\) as part of properness and does not try to infer finite type of \(g\).

Exercise 7.5 (medium). If \(X\to S\) is proper and \(Y\to S\) is separated, prove that every \(S\)-morphism \(h:X\to Y\) is proper and has closed image. Show also that the image remains closed after arbitrary base change on \(Y\).

Solution. Apply Exercise 7.4 to \(X\xrightarrow hY\to S\). The composite is the proper structure map of \(X\), and the second map is separated. Hence \(h\) is proper, in particular universally closed. For any \(Y'\to Y\), the base change \(X\times_Y Y'\to Y'\) is proper and closed. Its full source is a closed subset of itself, so its image is closed. Taking \(Y'=Y\) gives the original assertion. Moreover this image is the inverse image of \(h(X)\) as a set: a point over \(h(x)\) has a nonempty fibre after residue-field extension, by the nonzero field tensor-product argument in Corollary 2.2.

References and proof providers

The Stacks project, read in the AI Integrated Stacks Project edition, supplies the exact tagged treatments of universal closedness, properness permanence, general valuative criteria, weighted Proj, and the Noetherian DVR criterion cited above. The openly licensed full Chow proof at Tag 0200 and the internal construction in Theorem 4.1 of Projective morphisms and Chow's lemma supply that ingredient in Theorem 5.1. Discrete domination is the open algebra proof at Tag 00PH, already explained in the valuation lesson. These linked works retain GNU FDL 1.2. No source text is reproduced here.

Ravi Vakil, The Rising Sea: Foundations of Algebraic Geometry, draft of 27 July 2024, §§11.4 and 13.7, was consulted for alternative exposition of properness and valuation diagrams. Its expression is not reproduced. All four assigned result groups and all five exercises above have their proofs or exact open proof ingredients identified; no assigned theorem is left to a paid or merely free-to-read book.