Let \(f: X \to Y\) be a morphism of separated schemes of finite type over a noetherian scheme \(S\). Let \(Z\) be a closed subscheme of \(X\) which is proper over \(S\). Show that \(f(Z)\) is closed in \(Y\), and that \(f(Z)\) with its image subscheme structure is proper over \(S\).
Hint: factor \(f\) into the graph morphism \(\Gamma_f: X \to \fiberprod{X}{S}{Y}\) followed by the second projection \(p_2\), and show that \(\Gamma_f\) is a closed immersion.
Solution
Let \[ h=f|_Z:Z\longrightarrow Y. \]
1. The graph morphism\[ \Gamma_h:Z\longrightarrow Z\times_SY, \qquad z\longmapsto(z,h(z)), \]is a closed immersion.
The graph is the base change of the diagonal of \(Y/S\). Indeed, in the Cartesian square \[ \begin{array}{ccc} Z&\xrightarrow{\Gamma_h}&Z\times_SY\\ \downarrow{\scriptstyle h}&&\downarrow{\scriptstyle h\times\id_Y}\\ Y&\xrightarrow{\Delta_{Y/S}}&Y\times_SY, \end{array} \] the top map is precisely the pullback of the diagonal.
Because \(Y\) is separated over \(S\), \[ \Delta_{Y/S}:Y\longrightarrow Y\times_SY \] is a closed immersion. Closed immersions are stable under base change, hence \(\Gamma_h\) is a closed immersion.
2. The second projection\[ p_2:Z\times_SY\longrightarrow Y \]is proper.
The projection \(p_2\) is the base change of the proper structure morphism \[ Z\longrightarrow S \] along the morphism \(Y\to S\). Proper morphisms are stable under base change, so \(p_2\) is proper.
3. The closed immersion \(\Gamma_h\) is proper.
A closed immersion is finite: affine-locally it is induced by a quotient ring map. Hartshorne II.4.1 proves that every finite morphism is proper. Therefore \(\Gamma_h\) is proper.
4. The morphism\[ \boxed{h:Z\longrightarrow Y} \]is proper.
By construction, \[ h=p_2\circ\Gamma_h. \] Both morphisms on the right are proper by <1>2–<1>3. Proper morphisms are stable under composition, so \(h\) is proper.
5. The set-theoretic image\[ \boxed{h(Z)=f(Z)} \]is closed in \(Y\).
A proper morphism is universally closed, hence in particular closed. Since \(Z\) is closed in itself, its image under the closed map \(h\) is a closed subset of \(Y\).
6. Let\[ i:Y'\hookrightarrow Y \]be the scheme-theoretic image of \(h\). Then\[ |Y'|=h(Z)=f(Z). \]
The morphism \(h\) is proper, hence quasi-compact and quasi-separated. Therefore its scheme-theoretic image is defined by \[ \mathcal I = \ker\bigl(\mathcal O_Y\longrightarrow h_*\mathcal O_Z\bigr). \]
We show directly that the underlying closed set is the closure of \(h(Z)\). Work on an affine open \[ V=\Spec A\subseteq Y \] and put \[ W=h^{-1}(V). \] Because \(h\) is quasi-compact, \(W\) is quasi-compact. On \(V\), the ideal of the scheme-theoretic image is \[ I=\ker\bigl(A\longrightarrow\Gamma(W,\mathcal O_W)\bigr). \] Every element of \(I\) vanishes at every point of \(h(W)\), so \[ \overline{h(W)}\subseteq V(I). \]
Conversely, let \(\mathfrak p\in V\) lie outside \(\overline{h(W)}\). Choose \[ a\in A\setminus\mathfrak p \] such that \[ D(a)\cap h(W)=\varnothing. \] Then \[ h^{-1}(D(a))=\varnothing. \] Thus the pullback of \(a\) to \(W\) belongs to every prime ideal at every point of \(W\), hence is locally nilpotent. A finite affine cover of the quasi-compact scheme \(W\) supplies one exponent \(N\) such that the pullback of \(a^N\) is zero globally. Therefore \[ a^N\in I. \] Since \(a\notin\mathfrak p\), also \(a^N\notin\mathfrak p\), so \[ \mathfrak p\notin V(I). \] Hence \[ V(I)=\overline{h(W)}. \] These affine equalities glue, proving \[ |Y'|=\overline{h(Z)}. \]
Here 5 shows that the image is already closed, so \[ |Y'|=\overline{h(Z)}=h(Z). \]
7. The induced morphism\[ q:Z\longrightarrow Y' \]is surjective on underlying topological spaces.
By definition \(h=i\circ q\). Hence \[ i(q(Z))=h(Z). \] The closed immersion \(i\) is a homeomorphism of \(|Y'|\) onto its image in \(Y\). By 6 that image is exactly \(h(Z)\). Therefore \[ q(Z)=|Y'|. \]
8. The structure morphism\[ Y'\longrightarrow S \]is separated and of finite type.
The closed immersion \[ Y'\hookrightarrow Y \] is finite, hence of finite type, and \(Y\to S\) is of finite type. Thus their composite \(Y'\to S\) is of finite type.
Likewise a closed subscheme of a separated \(S\)-scheme is separated over \(S\): its diagonal is obtained by composing a closed immersion with the pullback of the closed diagonal of \(Y/S\).
9. For every base change\[ S'\longrightarrow S, \]the induced map\[ q_{S'}:Z_{S'}\longrightarrow Y'_{S'} \]is surjective.
Surjectivity of morphisms of schemes is stable under base change. Concretely, let \(y'\in Y'_{S'}\) map to \(y\in Y'\). Since \(q\) is surjective, choose \(z\in Z\) over \(y\). The fibre of \[ Z_{S'}\longrightarrow Y'_{S'} \] over \(y'\) is the spectrum of a tensor product of nonzero residue fields over \(\kappa(y)\), hence is nonempty. Thus some point of \(Z_{S'}\) maps to \(y'\).
10. The morphism\[ Y'\longrightarrow S \]is universally closed.
Let \(S'\to S\) be arbitrary and let \[ C\subseteq Y'_{S'} \] be closed. Its inverse image \[ q_{S'}^{-1}(C)\subseteq Z_{S'} \] is closed.
Since \(Z\to S\) is proper, its base change \[ Z_{S'}\longrightarrow S' \] is closed. Hence the image \[ \operatorname{im}\bigl(q_{S'}^{-1}(C)\to S'\bigr) \] is closed in \(S'\).
By surjectivity of \(q_{S'}\) from 9, every point of \(C\) has a preimage in \(q_{S'}^{-1}(C)\). Therefore \[ \operatorname{im}(C\to S') = \operatorname{im}\bigl(q_{S'}^{-1}(C)\to S'\bigr), \] which is closed. Thus every base change of \(Y'\to S\) is a closed map.
11. The image subscheme \(Y'\) is proper over \(S\).
By 8, \(Y'\to S\) is separated and of finite type. By 10, it is universally closed. These are exactly the three defining conditions for properness.
12. Therefore\[ \boxed{ f(Z)\text{ is closed in }Y, \qquad f(Z)\text{ with its scheme-theoretic image structure is proper over }S. } \]
5. The second is 6 and
11.
13. Q.E.D.
Step 12 is the required conclusion.