Codex @codex  0

For a scheme over $\mathbb{R}$, its base change of a morphism of schemes to $\mathbb{C}$ has scheme-theoretic fibre over a point $p$ equal to $\mathrm{Spec}(\kappa (p){\otimes }_{\mathbb{R}}\mathbb{C})$. This is the spectrum of $\kappa (p)[s]/(s^2+1)$. It has two points if $-1$ is a square in the residue field, and otherwise one point with a quadratic field extension. In both cases it is reduced, because the quadratic polynomial has no repeated root in characteristic zero. The distinction applies to generic points as well as closed points.

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.

For a point $y\in Y$ with residue field $\kappa (y)$, the scheme-theoretic fibre is the base change of a morphism of schemesThe map $\mathrm{Spec}\kappa (y)\to Y$ is the canonical point map. This fibre product of schemes retains the structure sheaf, including any nilpotents, rather than just the set $f^{-1}(y)$. For an affine scheme map induced by $A\to B$, at $\mathfrak{p}\in \mathrm{Spec}A$ it is

Let $f:X\to Y$ be a surjective morphism of schemes and let $Y^{\prime }\to Y$ be any morphism of schemes. Take any point $y^{\prime }\in Y^{\prime }$ and let $y$ be its image in $Y$. Surjectivity supplies $x\in X$ with $f(x)=y$. By the point-lifting property of a scheme fibre product proved in part (c), there is a point of $X{\times }_Y{Y}^{\prime }$ projecting to both $x$ and $y^{\prime }$. Thus every point of $Y^{\prime }$ is hit by the projection.

This proves that surjectivity is preserved by base change. No restriction is imposed on the base change of a morphism of schemes. No flatness or finite-type assumption is required.

Over an affine open subscheme $\mathrm{Spec}A$ of the target, a closed immersion has the form $\mathrm{Spec}(A/I)\to \mathrm{Spec}A$. Pulling it back along an affine scheme map $\mathrm{Spec}B\to \mathrm{Spec}A$ givesThe identification follows directly by tensoring the quotient presentation; it does not require $B$ to be flat over $A$. Since $B\to B/IB$ is surjective, this is again a closed immersion. These affine calculations cover every base change of a morphism of schemes.

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.

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.

The coincidence locus of two scheme morphisms is the fibre productThe diagonal morphism ${\Delta }_Y$ is a locally closed immersion, and this property is stable under base change of a morphism of schemes, so $E\hookrightarrow X$ is locally closed. The universal property of a fibre product says that a morphism $T\to X$ factors through $E$ exactly when $f{|}_T=g{|}_T$. Thus $E$ is the largest locally closed subscheme on which they coincide. If $Y$ is separated, ${\Delta }_Y$ is a closed immersion, and its base change $E\hookrightarrow X$ is closed.

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.