Closed immersions are stable under base extension: if \(f: Y \to X\) is a closed immersion and \(X' \to X\) is any morphism, then \(f': \fiberprod{Y}{X}{X'} \to X'\) is also a closed immersion.
If \(Y\) is a closed subscheme of an affine scheme \(X = \Spec A\), then \(Y\) is also affine, and in fact \(Y\) is the closed subscheme determined by a suitable ideal \(\mfa \subseteq A\) as the image of the closed immersion \(\Spec A/\mfa \to \Spec A\).
Let \(Y\) be a closed subset of a scheme \(X\) and give \(Y\) the reduced induced subscheme structure. If \(Y'\) is any other closed subscheme of \(X\) with the same underlying topological space, show that the closed immersion \(Y \to X\) factors through \(Y'\). We express this by saying that the reduced induced structure is the smallest subscheme structure on a closed subset.
Let \(f: Z \to X\) be a morphism. Then there is a unique closed subscheme \(Y\) of \(X\) with the following property: the morphism \(f\) factors through \(Y\), and if \(Y'\) is any other closed subscheme of \(X\) through which \(f\) factors, then \(Y \to X\) factors through \(Y'\) also. We call \(Y\) the scheme-theoretic image of \(f\). If \(Z\) is a reduced scheme, then \(Y\) is just the reduced induced structure on the closure of the image \(f(Z)\).
Hint
For part (b): first show that \(Y\) can be covered by a finite number of open affine subsets of the form \(D(f_i) \intersect Y\) with \(f_i \in A\). By adding some more \(f_i\) with \(D(f_i) \intersect Y = \varnothing\) if necessary, arrange that the \(D(f_i)\) cover \(X\). Next show that \(f_1, \ldots, f_r\) generate the unit ideal of \(A\). Then use the affineness criterion of [@Har10a, Exercise II.2.17(b)] to show that \(Y\) is affine, and [@Har10a, Exercise II.2.18(d)] to show that \(Y\) comes from an ideal \(\mfa \subseteq A\). A second proof using sheaves of ideals is given in [@Har10a, Corollary II.5.10].
Solution
1. Closed immersions are stable under arbitrary base change.
Let \[ i:Y\hookrightarrow X \] be a closed immersion and let \(X'\to X\) be any morphism. The assertion is local on \(X\) and \(X'\). Thus take \[ X=\Spec A, \qquad Y=\Spec(A/I), \qquad X'=\Spec B. \] Then \[ Y\times_XX' \cong \Spec\bigl((A/I)\otimes_AB\bigr) \cong \Spec(B/IB). \] The projection to \(X'=\Spec B\) is induced by the quotient map \[ B\twoheadrightarrow B/IB, \] so it is a closed immersion.
2. Let \(i:Y\hookrightarrow X=\Spec A\) be a closed subscheme. The scheme \(Y\) has a finite affine cover of the form\[ Y\cap D(f_1),\ldots,Y\cap D(f_r), \qquad f_i\in A. \]
The underlying space of \(Y\) is closed in the quasi-compact space \(X\), hence is quasi-compact. Cover \(Y\) by affine open subsets of \(Y\).
For each \(y\in Y\), choose an affine open \(W\subseteq Y\) containing \(y\). Write \(W=Y\cap O\) for an open \(O\subseteq X\). Choose a distinguished open \[ y\in D(f)\subseteq O. \] Then \[ Y\cap D(f)=W\cap D(f) \] is distinguished inside the affine scheme \(W\), hence is affine. These opens cover \(Y\), and quasi-compactness gives a finite subcover.
3. We may enlarge the list in 2 so that the distinguished opens \(D(f_i)\) cover all of \(X\), while every \(Y\cap D(f_i)\) remains affine.
The opens from 2 cover \(Y\). The complement \(X\setminus Y\) is covered by distinguished opens \(D(g)\) contained in \(X\setminus Y\).
Together these form an open cover of the affine scheme \(X\), hence a finite subcover. Add the finitely many selected \(g\)’s to the list. For every added one, \[ Y\cap D(g)=\varnothing, \] which is affine.
4. The functions \(f_1,\ldots,f_r\) from 3 generate the unit ideal of \(A\).
Since the \(D(f_i)\) cover \(\Spec A\), \[ V(f_1,\ldots,f_r)=\varnothing. \] Therefore \[ (f_1,\ldots,f_r)=A. \]
5. The closed subscheme \(Y\) is affine.
Let \[ B=\Gamma(Y,\mathcal O_Y). \] The restrictions of the \(f_i\) to \(Y\) are elements of \(B\) and generate the unit ideal there, because the relation \[ \sum_i a_if_i=1 \] from 4 restricts to \(Y\).
3. Hartshorne II.2.17(b) therefore implies that \(Y\) is affine.
6. There is an ideal \(I\subseteq A\) such that\[ Y\cong\Spec(A/I) \]over \(X=\Spec A\).
By 5, write \[ Y=\Spec B. \] The closed immersion induces a ring homomorphism \[ \phi:A\longrightarrow B. \] By definition, a closed immersion is a homeomorphism onto a closed subset and its map of structure sheaves is surjective. Hartshorne II.2.18(d) therefore shows that \(\phi\) is surjective.
Let \[ I=\ker\phi. \] Then \(B\cong A/I\), and the given closed immersion is the standard map \[ \Spec(A/I)\hookrightarrow\Spec A. \]
7. Let \(Y\subseteq X\) be closed and give it the reduced induced structure. If \(Y'\subseteq X\) is any closed subscheme with the same underlying topological space, then\[ Y\longrightarrow X \]factors uniquely through \(Y'\).
This is local on \(X\). Take \(X=\Spec A\). By 6, write \[ Y'=\Spec(A/I). \] Its underlying closed subset is \[ V(I). \] The reduced induced structure on the same set is \[ Y=\Spec(A/\sqrt I). \] Since \[ I\subseteq\sqrt I, \] there is a quotient homomorphism \[ A/I\longrightarrow A/\sqrt I, \] and therefore a factorization \[ Y\longrightarrow Y'\longrightarrow X. \] It is unique because \(Y'\to X\) is a monomorphism, as every immersion is.
8. Let \(f:Z\to X\) be any morphism. Consider all quasi-coherent ideal sheaves\[ \mathcal I_\lambda\subseteq\mathcal O_X \]such that \(f\) factors through the corresponding closed subscheme \(Y_\lambda\subseteq X\), and put\[ \mathcal I=\sum_\lambda\mathcal I_\lambda. \]Then \(\mathcal I\) is a quasi-coherent ideal sheaf.
The family is nonempty because \(X\) itself is a closed subscheme of \(X\), corresponding to the zero ideal sheaf.
On an affine open \(U=\Spec A\subseteq X\), each restriction \(\mathcal I_\lambda|_U\) corresponds to an ideal \(I_\lambda\subseteq A\). Their sheaf-theoretic sum restricts to the sheaf associated to \[ \sum_\lambda I_\lambda\subseteq A, \] because localization commutes with arbitrary sums of submodules. Hence \(\mathcal I\) is quasi-coherent.
9. Let \(Y\subseteq X\) be the closed subscheme defined by \(\mathcal I\) from
8. Then \(f\) factors through \(Y\).
Factoring through \(Y_\lambda\) means that the pullback of \(\mathcal I_\lambda\) to \(Z\) maps to zero in \(\mathcal O_Z\). Therefore the pullback of their sum \[ \mathcal I=\sum_\lambda\mathcal I_\lambda \] also maps to zero.
Affine-locally, if \(Y\cap U=\Spec(A/I)\), this says that the ring map defining \(f\) kills \(I\), hence factors through \(A/I\). These local factorizations glue, giving a factorization \[ Z\longrightarrow Y\longrightarrow X. \]
10. The closed subscheme \(Y\) from 9 is contained in every closed subscheme \(Y'\subseteq X\) through which \(f\) factors.
Let \(\mathcal I_{Y'}\) be the ideal sheaf of such a \(Y'\). It is one of the summands used to define \(\mathcal I\), so \[ \mathcal I_{Y'}\subseteq\mathcal I. \] The inclusion of ideals reverses under \(\Spec\) and gives a unique factorization \[ Y\longrightarrow Y'\longrightarrow X. \]
11. Hence \(Y\) is the unique scheme-theoretic image of \(f\).
Steps 9 and 10 show that \(Y\) has the defining minimality property. If two closed subschemes had that property, each would factor through the other over \(X\), and the resulting maps would be inverse because closed immersions are monomorphisms. Thus the scheme-theoretic image is unique.
12. Assume now that \(Z\) is reduced, and put\[ T=\overline{f(Z)}\subseteq X. \]Then \(f\) factors through the reduced induced subscheme \(T_{\mathrm{red}}\).
This is local on \(X\). Let \(U=\Spec A\subseteq X\) and write \[ T\cap U=V(J) \] with \(J\) radical, so \[ T_{\mathrm{red}}\cap U=\Spec(A/J). \]
Take \(a\in J\). For every point \[ z\in f^{-1}(U), \] the point \(f(z)\) lies in \(T\cap U\), so the germ of \(f^*a\) at \(z\) maps to zero in the residue field \(\kappa(z)\).
On an affine open \(W=\Spec C\subseteq f^{-1}(U)\), the section \(f^*a\) corresponds to an element \(c\in C\) lying in every prime ideal of \(C\). Hence \(c\) is nilpotent. Since \(Z\) is reduced, \(C\) is reduced, so \(c=0\).
Thus the ring map \(A\to\Gamma(W,\mathcal O_Z)\) kills \(J\) on every affine \(W\), and \(f|_{f^{-1}(U)}\) factors through \(\Spec(A/J)\). These local factorizations glue.
13. The reduced induced subscheme \(T_{\mathrm{red}}\) is contained in every closed subscheme \(Y'\subseteq X\) through which \(f\) factors.
If \(f\) factors through \(Y'\), then \[ f(Z)\subseteq|Y'|. \] Since \(|Y'|\) is closed, \[ T=\overline{f(Z)}\subseteq|Y'|. \]
Affine-locally, write \[ Y'\cap U=\Spec(A/I), \qquad T\cap U=V(J) \] with \(J\) radical. The inclusion \(T\subseteq V(I)\) means \[ I\subseteq J. \] Hence \[ A/I\longrightarrow A/J \] induces a factorization \[ T_{\mathrm{red}}\cap U\longrightarrow Y'\cap U. \] These factorizations glue over \(X\).
14. If \(Z\) is reduced, the scheme-theoretic image of \(f\) is exactly\[ \boxed{\overline{f(Z)}_{\mathrm{red}}.} \]
Step 12 shows that \(f\) factors through \(T_{\mathrm{red}}\), and 13 shows that this closed subscheme is contained in every other closed subscheme through which \(f\) factors. By uniqueness in 11, it is the scheme-theoretic image.
15. Q.E.D.
Step 1 proves part (a); steps <1>2–<1>6 prove part (b); 7 proves part (c); and steps <1>8–<1>14 prove part (d).