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For morphisms of schemes $f:X\to Y$ and $g:Y\to Z$, the graph is the displayed map. It is the pullback of the diagonal morphism $Y\to Y{\times }_{Z}Y$ along $(f\circ p_X,p_Y)$. Thus a separated morphism $g$ makes ${\Gamma }_f$ a closed immersion. If also $g\circ f$ is a closed immersion, the projection $X{\times }_{Z}Y\to Y$ is a closed immersion by base change of a morphism of schemes, and its composite with ${\Gamma }_f$ proves that $f$ is a closed immersion.

A finite morphism is a finite type morphism. It is also a separated morphism: on an affine scheme chart $\mathrm{Spec}B\to \mathrm{Spec}A$, its diagonal morphism is represented by the surjective multiplication ring homomorphism $B{\otimes }_{A}B\to B$, and is therefore a closed immersion by Question 1.

We prove the lifting condition in the valuative criterion for properness. Choose an affine open subset of $Y$ containing the image of the closed point of $\mathrm{Spec}V$. Every point of $\mathrm{Spec}V$ is a generalization of its closed point, so the whole map factors through this affine open subset. Its inverse image under the finite morphism is $\mathrm{Spec}B$, with $B$ a finitely generated $A$-module. Every $b\in B$ is integral over $A$: apply the determinant trick to multiplication by $b$ on a finite set of module generators to obtain a monic polynomial which annihilates all generators and hence annihilates $1$.

In the given valuative square, we have compatible ring homomorphisms $A\to V$ and $B\to K$. The image in $K$ of each $b\in B$ satisfies a monic polynomial over $V$. To see directly why it belongs to $V$, suppose $x\in K\setminus V$. The defining property of a valuation ring gives $x^{-1}\in V$, and it lies in the maximal ideal since it cannot be a unit. A monic equation for $x$, multiplied by $x^{-n}$, would then express $1$ as a sum of elements in that maximal ideal, a contradiction. This proves that valuation rings are integrally closed and that $B\to K$ factors through $B\to V$. The factorization is unique because $V\hookrightarrow K$ is injective. Thus the lift exists uniquely, proving

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 $f:X\to Y$ is a separated morphism if its diagonal morphismis a closed immersion. The diagonal morphism is induced by the pair of identity maps on $X$. Thus separatedness is a condition relative to the given target $Y$, not a requirement that $X$ be absolutely separated.

Since $g$ is a separated morphism, the diagonal morphism ${\Delta }_g:Y\to Y{\times }_{Z}Y$ is a closed immersion. The squareis cartesian. Indeed, on test schemes its pullback condition is $(f(x),y)=(t,t)$, forcing $y=t=f(x)$, and leaving exactly the choice of $x$. The graph morphism of schemes ${\Gamma }_f$ is therefore a base change of a morphism of schemes of ${\Delta }_g$. Part (c) shows that ${\Gamma }_f$ is a closed immersion.

Now assume additionally that $h=g\circ f:X\to Z$ is a closed immersion. The projection $p_Y:X{\times }_{Z}Y\to Y$ is its base change of a morphism of schemes along $g$, so is a closed immersion by part (c). Sincepart (b) shows that $f$ is a closed immersion. This covers the further assumption in the final unheaded paragraph.

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.

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.

A morphism $f:X\to S$ is a separated morphism when its diagonal morphismis a closed immersion.

For the requested example, take the affine plane with doubled origin: glue two copies $U,V\cong {\mathbb{A}}_k^2$ by the identity onThe opens $U$ and $V$ are affine, while their intersection is the punctured affine plane, which is not affine. Indeed, its regular functions are still $k[x,y]$; if it were affine, the canonical map to $\mathrm{Spec}k[x,y]={\mathbb{A}}_k^2$ would be an isomorphism, contrary to the missing origin. The resulting scheme is not separated: in a separated scheme, the intersection of two affine opens is the inverse image of the closed diagonal inside their affine product and is therefore affine.

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.

A scheme over a base $S$ is separated when its structure morphism to $S$ is a separated morphism. An absolute scheme is usually called separated when it is separated over $\mathrm{Spec}\mathbb{Z}$.