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For every valuation ring $D$ with fraction field $K$, consider compatible morphisms of schemesAn extension is a morphism of schemes $\ell :\mathrm{Spec}D\to X$ satisfying $\ell {|}_{\mathrm{Spec}K}=u$ and $f\ell =v$. The valuative criterion for universal closedness statesEquivalently, among quasi-compact morphisms, the existence condition characterizes universal closedness. A universally closed morphism is automatically a quasi-compact morphism; the converse direction of the criterion needs this compactness condition. All valuation rings are allowed, with arbitrary fraction fields; there is no Noetherian ring hypothesis and one cannot replace them by discrete valuation rings without additional assumptions.^[ref][ref]

Let $X\stackrel{f}{\to }Y\stackrel{g}{\to }S$ be morphisms of schemes. Start with $u:\mathrm{Spec}K\to X$ and $v:\mathrm{Spec}D\to S$ satisfying $gfu=v{|}_{\mathrm{Spec}K}$.

For existence, first apply the extension condition for $g$ to $fu$ and $v$, obtaining $w:\mathrm{Spec}D\to Y$. Then apply the extension condition for $f$ to $u,w$, obtaining $\ell :\mathrm{Spec}D\to X$. This $\ell$ solves the problem for $gf$. If both $f,g$ are universally closed morphisms, both are quasi-compact morphisms; their composite is a quasi-compact morphism by the allowed stability fact. Part (a)(i) therefore proves that $gf$ is a universally closed morphism.

For uniqueness, suppose ${\ell }_1,{\ell }_2$ are extensions for $gf$. Uniqueness for $g$ gives $f{\ell }_1=f{\ell }_2=w$, and uniqueness for $f$, now over this same $w$, gives ${\ell }_1={\ell }_2$. If $f,g$ are separated morphisms, their composite is a quasi-separated morphism by the allowed stability fact. Part (a)(ii) proves that $gf$ is a separated morphism.

Finally, if $f,g$ are proper morphisms, these arguments give both existence and uniqueness. Stability under composition of quasi-compact morphisms, quasi-separated morphisms and locally of finite type gives the other hypotheses of (a)(iii). Consequently

Let $S^{\prime }\to S$ be a morphism of schemes and set $X^{\prime }=X{\times }_S{S}^{\prime }$, with projection $f^{\prime }:X^{\prime }\to S^{\prime }$. Given compatible morphisms of schemes $u^{\prime }:\mathrm{Spec}K\to X^{\prime }$ and $v^{\prime }:\mathrm{Spec}D\to S^{\prime }$, compose them with the projections to obtain a valuative problem for $f$. An extension $\ell :\mathrm{Spec}D\to X$ for that problem and the prescribed $v^{\prime }$ determine a unique morphism of schemesThe universal property of a fibre product shows that ${\ell }^{\prime }$ extends $u^{\prime }$ and lies over $v^{\prime }$. Conversely, every extension ${\ell }^{\prime }$ has such a first component $\ell$. Thus the sets of extensions in the two problems are in bijection.

If $f$ is a universally closed morphism, it is a quasi-compact morphism and satisfies existence. The allowed stability of quasi-compact morphisms under base change and (a)(i) prove that $f^{\prime }$ is a universally closed morphism. If $f$ is a separated morphism, it is a quasi-separated morphism and satisfies uniqueness. Stability of quasi-separated morphisms under base change and (a)(ii) prove that $f^{\prime }$ is a separated morphism.

If $f$ is a proper morphism, the same bijection supplies existence and uniqueness. The permitted stability under base change of quasi-compact morphisms, quasi-separated morphisms and locally of finite type supplies all remaining hypotheses of (a)(iii). Hence

We first justify surjectivity is preserved by base change for morphisms of schemes. Given $y^{\prime }$ over $y$ and a point $x$ over $y$, the fiber after the residue-field extension contains a point becauseIndeed, tensoring a nonzero vector space over a field with a field extension remains nonzero; a nonzero commutative ring has a prime ideal, giving the required point. Thus every base change of $f$ is surjective.

Now make an arbitrary base change $Z^{\prime }\to Z$. For a closed subset $C\subseteq Y^{\prime }=Y{\times }_Z{Z}^{\prime }$, its inverse image under $f^{\prime }:X^{\prime }\to Y^{\prime }$ is closed. The proper morphism $(gf{)}^{\prime }:X^{\prime }\to Z^{\prime }$ therefore has closed image on that inverse image. Surjectivity of $f^{\prime }$ givesso $g^{\prime }(C)$ is closed. This proves that $g$ is a universally closed morphism. Together with the given separatedness and finite type, it proves

A proper morphism of schemes is a separated morphism of finite type which is a universally closed morphism: after every base change, the resulting map of underlying topological spaces takes closed subsets to closed subsets.

The valuative criterion for properness, in the Noetherian setting used here, says that a finite type morphism $f:X\to Y$ is a proper morphism if and only if the following existence and uniqueness condition holds. For every valuation ring $V$, its fraction field $K$, and compatible morphisms of schemes $a:\mathrm{Spec}K\to X$ and $b:\mathrm{Spec}V\to Y$ with $f\circ a=b{|}_{\mathrm{Spec}K}$, there is exactly one morphism of schemes $h:\mathrm{Spec}V\to X$ such that $h{|}_{\mathrm{Spec}K}=a$ and $f\circ h=b$. All valuation rings are allowed; no discreteness assumption is needed in this formulation. The following arguments also use that each of the three defining conditions for a proper morphism is preserved by base change.

A morphism of schemes is proper when it is a finite type morphism, a separated morphism, and a universally closed morphism. The valuative criterion for properness says, under the usual finite-type and Noetherian hypotheses, that $f:X\to Y$ is proper exactly when every commutative squarewith $R$ a discrete valuation ring and $K=\mathrm{Frac}R$ has a unique diagonal lift $\mathrm{Spec}R\to X$.

For ${\mathbb{P}}_k^2\to \mathrm{Spec}k$, a $K$-point is $[x_0:x_1:x_2]$ with not all $x_i$ zero. If $\pi$ is a uniformizer, multiply all coordinates by one power of $\pi$ so that ${\min }_{i}v(x_i)=0$. The resulting coordinates lie in $R$ and at least one is a unit, so they define an $R$-point extending the given $K$-point. If two extensions exist, on a chart where one coordinate is a unit their affine coordinate ratios agree in $K$ and therefore in the integral domain $R$; hence the extensions agree. This verifies existence and uniqueness directly.

Take the projective line with a doubled point over $\mathbb{C}$, obtained by gluing two copies of ${\mathbb{P}}_{\mathbb{C}}^1$ along the complement of one point. After every base change of a morphism of schemes, a closed subset has closed image from each of the two projective-line charts, so its total image, the union of those two images, is closed. The structure morphism is therefore universally closed. The two doubled points have no disjoint neighborhoods, so the scheme is not separated and hence is not proper.

A morphism of schemes is proper when it is of finite type, separated, and universally closed.

Each property is local on the target. Finite type is local by its affine definition. The restrictions of the diagonal morphism of $\phi$ over the open sets $U_i$ are closed immersions; since being a closed subset is local on an open cover, the diagonal itself is a closed immersion, so $\phi$ is separated. Finally, after any base change $Y^{\prime }\to Y$, the inverse images $U_i^{\prime }=Y^{\prime }{\times }_Y{U}_i$ cover $Y^{\prime }$. For every closed $C\subseteq X{\times }_Y{Y}^{\prime }$, its image has closed intersection with every $U_i^{\prime }$ because the restricted base-changed morphism is closed. The image is therefore closed in $Y^{\prime }$. Thus $\phi$ is universally closed and hence proper. This proves that properness is local on the target.

The fibre product of schemes $X{\times }_{S}Y$ comes with projections to $X$ and $Y$ having equal composites to $S$, and is universal with this property. A morphism is universally closed when every base change is closed on underlying topological spaces.

The structural morphism ${\mathbb{A}}_k^1\to \mathrm{Spec}k$ is closed because its target has one point, but it is not universally closed. After base change by ${\mathbb{A}}_k^1$, it becomes the projection ${\mathbb{A}}_k^2\to {\mathbb{A}}_k^1$; the closed hyperbola $V(xy-1)$ has image $D(x)$, which is not closed.

For a finite-type universally closed nonseparated example, glue two copies of ${\mathbb{P}}_k^1$ along the complement of one rational point, producing a projective line with a doubled point. It is of finite type. After every base change, each of its two projective-line charts maps closedly to the base, so the image of any closed subset, being the union of the two closed images, is closed. Thus the structural morphism is universally closed. It is not separated because the two doubled points violate uniqueness in the valuative criterion for separatedness, or equivalently because its diagonal is not closed.

Among quasi-compact morphisms of schemes, universal closedness is equivalent to the existence part of the valuative criterion: for every valuation ring $D$ with fraction field $K$, every compatible pair $\mathrm{Spec}K\to X$, $\mathrm{Spec}D\to S$ extends to $\mathrm{Spec}D\to X$. All valuation rings must be tested without further hypotheses. A universally closed morphism is automatically a quasi-compact morphism.