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Given $f:X\to Y$ and $g:Y^{\prime }\to Y$, the base change of $f$ along $g$ is the projectionProperties such as being proper and flat are stable under arbitrary base change.

For every scheme $X$, projection $p:X\times {\mathbb{P}}^n\to X$ is a closed map, and remains so after every base change. Hence projective space is universally closed over its base. The assertion concerns the underlying scheme points, not just rational points over a non-algebraically closed field.^[ref]

The existence part of the valuative criterion explains the role of projective space. Given a valuation ring $V$ with fraction field $K$, a $K$-point $[a_0:\cdots :a_n]$ can be rescaled so that every coordinate belongs to $V$ and one coordinate is a unit. Indeed, among the finitely many nonzero coordinates choose $a_j$ whose principal fractional ideal contains all the others; then $a_i/a_j\in V$ and $a_j/a_j=1$. These coordinates define the required morphism of schemes $\mathrm{Spec}V\to {\mathbb{P}}^n$. Since projective space is also separated and of finite type, this proves that projection is a proper morphism.

On a Noetherian scheme, one can instead use finitely many homogeneous equations and the Projective Nullstellensatz to express the image as closed rank-defect conditions on graded multiplication map for a projective fiber. An arbitrary affine base need not give a finitely generated defining ideal, so that finite-equation hypothesis must not be assumed in the general statement.

For an irreducible source, regard the morphism of algebraic varieties as dominant onto the closure of its image, with generic point $\eta$. The fiber dimension theorem gives the displayed formula. A proper morphism has closed image, so the closure can then be omitted.

Let $f:X\to Y$ be a proper morphism of irreducible algebraic varieties with equal dimension. If some fibre has an isolated point, its corresponding irreducible component has dimension zero. Apply the fiber dimension theorem to $X\to f(X)$: $0\ge \dim X-\dim f(X)$. Thus the closed image has dimension $\dim Y$ and equals $Y$. The incidence proof that a cubic surface contains a line is an application; the entire witness fibre need not be finite.

Let $F$ be a globally generated vector bundle of rank $r\ge 2$ on a projective curve over an algebraically closed field $k$. Then some global section of $F$ vanishes nowhere and defines a line subbundle ${\mathcal{O}}_C\subset F$ with locally free quotient.

Put $W=H^0(C,F)$. The incidence pairs $(p,s)$ satisfying $s(p)=0$ form the total space of the kernel of the surjective evaluation vector bundle morphism $W\otimes {\mathcal{O}}_C\to F$. Their dimension is $\dim W+1-r<\dim W$. Projection to $W$ is closed because projection from the product with $C$ is a proper morphism, so the sections vanishing somewhere lie in a proper closed subset of $W$. Choose a point outside it. Its fiberwise nonzero values extend locally to frames, giving the asserted line subbundle and locally free quotient.

The same rational-point choice works over an infinite field, after forming the bad locus geometrically. Over a finite field, a proper closed subset of an affine affine space can contain every rational point; the dimension argument alone does not provide a $k$-defined nowhere-vanishing global section.

With the same compatible pair $u,v$, the valuative criterion for properness isEquivalently, one can replace finite type by being a quasi-compact morphism and locally of finite type. Necessity follows from the definition of a proper morphism and the existence and uniqueness criteria in (i) and (ii). Conversely, finite type supplies the compactness needed in (i), while being a quasi-separated morphism supplies the compactness of the diagonal morphism needed in (ii). Thus the extension condition gives both universal closedness and separatedness, and these together with finite type give a proper morphism. These hypotheses are part of the criteria, even though the question is phrased for unrestricted morphisms of schemes.^[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

Write $X=\mathrm{Spec}B$, $S=\mathrm{Spec}A$ with $A,B$ integral domains, let $\alpha :A\to B$ be the ring homomorphism, and put $K=\mathrm{Frac}(B)$. We do not assume that $\alpha$ is injective. Write $A^{\prime }=\alpha (A)\subseteq B\subseteq K$.

Let $D\subseteq K$ be any valuation ring containing $A^{\prime }$, with fraction field $K$. The canonical morphism of schemes $\mathrm{Spec}K\to \mathrm{Spec}B$ and the ring homomorphism $A\to A^{\prime }\hookrightarrow D$ form a valuative problem for $f$. Since $f$ is a proper morphism, there is an extension. On coordinate rings it is a ring homomorphism $\beta :B\to D$ such thatis the original inclusion of $B$ into $K$. Therefore every element of $B$ lies in $D$. Since $D$ was arbitrary, the stated description of integral closure as an intersection of valuation rings givesThus every element of $B$ is an integral element over $A^{\prime }$, and hence over $A$: lift the coefficients of its monic polynomial along the surjection $A\to A^{\prime }$. The ring homomorphism $\alpha$ is therefore an integral extension in the sense of the question.

A proper morphism is of finite type, so in this affine situation $B=A[b_1,\dots ,b_s]$. If $b_i$ satisfies a monic polynomial of degree $n_i$, repeatedly reducing its powers shows that the finitely many monomialsgenerate $B$ as an $A$-module. Equivalently, apply the allowed equivalence between finite and integral plus finite type. Hence

The dimension of an algebraic set is the supremum of lengths of strict chains of nonempty irreducible closed subsets. It is the maximum of the dimensions of the irreducible components. On an affine variety $X=\mathrm{Spec}A$, it equals the Krull dimension of $A$; if $X$ is irreducible, Noether normalization identifies this with the transcendence degree of its function field over $k$. Nonempty Zariski-open subsets of an irreducible algebraic variety have the same dimension, so the definition is compatible with affine charts and with birational maps. In particular, $\dim {\mathbb{A}}^n=\dim {\mathbb{P}}^n=n$ and $\dim (X\times Y)=\dim X+\dim Y$ for irreducible algebraic varieties.

The fiber dimension theorem says that a dominant morphism of algebraic varieties $f:X\to Y$ between irreducible algebraic varieties has generic fibre dimensionEvery irreducible component of every nonempty fibre over a closed point has dimension at least $r$. There is a nonempty Zariski-open subset of $Y$ on which the fibres are nonempty and are equidimensional algebraic varieties of dimension $r$. Dominance is required; for a general morphism of algebraic varieties, replace $Y$ by the closure of its image. This gives the dimension of image of a morphism formula.

For the generic fibre, the formula is the additivity of transcendence degree in the tower $k\subset k(Y)\subset k(X)$. The lower bound on closed fibres comes from cutting by at most $\dim Y$ local parameters after a finite Noether normalization of an affine target chart: the principal hypersurface dimension lemma drops dimension by at most one per equation, and the finite normalization fibre separates the finitely many target points. For the generic upper bound, take finitely many affine source charts and apply Noether normalization to each coordinate ring over $k(Y)$. Clearing denominators in $k[Y]$ makes each chart finite over ${\mathbb{A}}^r$ relative to a suitable open target chart. Its fibres have dimension at most $r$. Intersecting these finitely many target opens and using the lower bound gives generic equidimensionality of fibres.

The upper semicontinuity of local fibre dimension is a statement on the source: for a finite-type morphism of schemes, the points $x$ satisfying ${\dim }_x{X}_{f(x)}\le d$ form an open set. For a proper morphism, the maximum fibre dimension is upper semicontinuous on the target as well. The properness qualification matters; one should not assert the latter for every finite-type morphism of schemes.

A first application is dimension counting for plane sections and intersections. On an irreducible affine variety, imposing $q$ polynomial equations gives each nonempty component algebraic codimension at most $q$, by repeated principal ideal theorem. On a smooth variety this underlies the intersection dimension bound on a smooth variety. A second application is to images: if a morphism of algebraic varieties has finite nonempty fibres, its image closure has the same dimension as the source. If it is also proper, that image is closed. Thus an isolated fibre point forces dominance in equal dimensions when the irreducible source and target have equal dimension.

For the cubic-surface application, projective lines in ${\mathbb{P}}^3$ form the Grassmannian $G=\mathrm{Gr}(2,4)$ of dimension four. Cubic forms form ${\mathbb{P}}^{19}$, since there are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ degree-three monomials in four variables. The cubic surface line incidence varietyhas fibre ${\mathbb{P}}^{15}$ over every $L$: restricting a cubic to a projective line gives four independently prescribable coefficients. It is therefore an irreducible projective-space bundle of dimension $4+15=19$. The other projection $\pi :I\to {\mathbb{P}}^{19}$ is a proper morphism, so its image $Z$ is closed and irreducible.

To show $Z={\mathbb{P}}^{19}$, dimension counting alone is not enough; exhibit a fibre with an isolated point. TakeA nearby projective line is the graph $x_2=a{x}_0+b{x}_1$, $x_3=c{x}_0+d{x}_1$. Restriction givesIt vanishes only when $a=b=c=d=0$. Thus $L_0$ is an isolated reduced point of the fibre. Applying the lower bound in the fiber dimension theorem to $I\to Z$ gives $0\ge 19-\dim Z$. Since $Z\subseteq {\mathbb{P}}^{19}$, its dimension is exactly 19, and closedness forces $Z={\mathbb{P}}^{19}$. This is the incidence proof that a cubic surface contains a line. The auxiliary cubic $F_0$ need not be smooth: the argument proves existence for every cubic form, and hence in particular for every smooth cubic surface.

Use the graph morphism of schemesIt is a closed immersion: the square with ${\Delta }_g:Y\to Y{\times }_{Z}Y$ and the map $X{\times }_{Z}Y\to Y{\times }_{Z}Y$, $(x,y)\mapsto (f(x),y)$, is a fibre product of schemes square. The separatedness of $g$ makes ${\Delta }_g$ a closed immersion, and base change preserves a closed immersion.

The projection $\pi :X{\times }_{Z}Y\to Y$ is the base change of the proper morphism $gf:X\to Z$, so it is a proper morphism. The closed immersion ${\Gamma }_f$ is a finite morphism, hence a proper morphism by part (i). Since $f=\pi {\Gamma }_f$, part (ii) provesThis argument supplies finite type for $f$ as well; it need not be separately assumed.

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.

An elliptic curve works over every algebraically closed field, including one algebraic over a finite field. Give it explicitly in ${\mathbb{P}}^2$ byIn the first case the only point at infinity is $O=[0:1:0]$, where the derivative with respect to $z$ is nonzero. In the affine chart $z=1$, a singular point of an algebraic variety would have $y=0$, $x^3-x=0$, and $3x^2-1=0$. The possible roots $x=0,1,-1$ cannot satisfy the last equation in characteristic different from $2$: at $0$ it is $-1$, and at $\pm 1$ it is $2$. In the second case the three homogeneous partial derivatives are $x^2,z^2,y^2$, up to harmless signs, and cannot vanish simultaneously at a projective point. Thus both cubics are smooth varieties. A reducible plane cubic would have intersecting positive-degree components by Bézout theorem, and be singular at an intersection; therefore these smooth cubics are irreducible varieties. The genus-degree formula gives $g=(3-1)(3-2)/2=1$. With the marked point $O$ they are elliptic curves.

For a smooth algebraic curve, its divisor class group is its Picard group. Its degree-zero subgroup isThe map $P\mapsto [P-O]$ gives an isomorphism of abelian groupsTo recall why it is a bijection, Riemann-Roch theorem in geometric genus one says every degree-one divisor class has exactly one independent section: a canonical divisor has degree zero, so the dual degree-$-1$ term has no section. Thus the class contains one effective divisor on an algebraic curve of degree one, namely a point $P$, and that point is unique. Translating by $-O$ gives the claimed bijection; this is precisely the chord-and-tangent group law on the elliptic curve.

Choose a prime integer $\ell$ different from $\mathrm{char}k$. The multiplication-by-n morphism $[\ell ]:E\to E$ has derivative $\ell$ times the identity at $O$, so it is nonconstant. A nonconstant morphism of algebraic varieties between smooth projective curves is surjective: its image is closed by proper morphism and cannot have algebraic dimension zero. Since $k$ is an algebraically closed field, this makes $[\ell ]$ surjective on $k$-points. ThereforeYet $E(k)$ is infinite. For each $x\in k$, the affine equation has a solution for $y$ in the algebraically closed field; since $k$ is infinite, this gives infinitely many points.

By the Fundamental theorem of finitely generated abelian groups, a finitely generated abelian group $A$ satisfying $A=\ell A$ has no free part and is a finite torsion group of order prime to $\ell$. The infinite group $E(k)$ cannot have this property while being finitely generated. Thus ${\mathrm{Pic}}^0(E)$ is not finitely generated. A subgroup of a finitely generated abelian group is finitely generated, soThis proof uses divisibility and infinitude, rather than uncountability, so it remains valid over countable algebraically closed fields.

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.

Because ${\mathcal{A}}_{\bullet }$ is coherent and locally generated by ${\mathcal{A}}_1$, there is locally a finite-rank coherent sheaf $\mathcal{E}$ and a surjection of graded algebrasThe Relative Proj construction turns this into a closed immersionConsequently $Y\to X$ is a projective morphism and therefore a proper morphism. Since $X\to \mathrm{Spec}k$ is proper and proper morphisms are closed under composition, $Y\to \mathrm{Spec}k$ is 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.

Write $f:U={\mathbb{A}}_{\mathbb{C}}^1\setminus \{1\}\to {\mathbb{A}}_{\mathbb{C}}^1$ for $f(t)=t^2$. For any closed point $a\in \mathbb{C}$, the fiber is the zero-dimensional closed subscheme of $U$ cut out by $t^2-a$. It is finite over $\mathbb{C}$, and every finite morphism is proper. This includes $a=1$, whose fiber in $U$ consists only of $t=-1$.

The morphism itself fails the valuative criterion for properness. Take the discrete valuation ring $R=\mathbb{C}[[s]]$ with fraction field $K=\mathbb{C}((s))$. The $K$-point $t=1+s$ lies in $U(K)$ and lies over the $R$-point of the target defined byAny extension $\mathrm{Spec}R\to {\mathbb{A}}^1$ of that $K$-point must still be given by $t=1+s$, whose closed point maps to the deleted point $t=1$. It therefore cannot factor through $U$. The required lift does not exist, so $f$ is not proper. This is an instance of the fact that proper closed-point fibers do not imply properness.

Let $R$ be a valuation ring with fraction field $K$. For every commutative squarethe valuative criterion for separatedness says that a finite type morphism $f$ between Noetherian schemes is separated exactly when there is at most one dotted lift $\mathrm{Spec}R\to X$ completing the diagram.

Under the same finiteness hypotheses, the valuative criterion for properness says that $f$ is proper exactly when every such square has a unique lift. Thus separatedness supplies uniqueness, while properness supplies existence as well.