Let \(X\) be a Zariski topological space. A constructible subset of \(X\) is a subset which belongs to the smallest family \(\mathcal{F}\) of subsets such that every open subset is in \(\mathcal{F}\), a finite intersection of elements of \(\mathcal{F}\) is in \(\mathcal{F}\), and the complement of an element of \(\mathcal{F}\) is in \(\mathcal{F}\).
A subset of \(X\) is locally closed if it is the intersection of an open subset with a closed subset. Show that a subset of \(X\) is constructible if and only if it can be written as a finite disjoint union of locally closed subsets.
Show that a constructible subset of an irreducible Zariski space \(X\) is dense if and only if it contains the generic point. Furthermore, in that case it contains a nonempty open subset.
A subset \(S\) of \(X\) is closed if and only if it is constructible and stable under specialization. Similarly, a subset \(T\) of \(X\) is open if and only if it is constructible and stable under generization.
If \(f: X \to Y\) is a continuous map of Zariski spaces, then the inverse image of any constructible subset of \(Y\) is a constructible subset of \(X\).
Solution
1. Every locally closed subset of \(X\) is constructible.
A locally closed subset has the form \[ U\cap F \] with \(U\) open and \(F\) closed. The open set \(U\) belongs to the defining family of constructible sets, and \[ F=X\setminus(X\setminus F) \] is constructible because \(X\setminus F\) is open and constructible sets are closed under complements. Hence \(U\cap F\) is constructible.
2. Every finite union of locally closed subsets is constructible.
Constructible sets are closed under finite unions because \[ A\cup B=X\setminus\bigl((X\setminus A)\cap(X\setminus B)\bigr). \] Apply 1 to the finitely many locally closed subsets.
3. Every constructible subset is a finite disjoint union of locally closed subsets.
Any given constructible set is obtained from finitely many open subsets \[ U_1,\ldots,U_n \] by finitely many intersections and complements. Consider the \(2^n\) Boolean atoms \[ A_\epsilon = \bigcap_{i=1}^n E_i, \] where each \(E_i\) is either \(U_i\) or \(X\setminus U_i\).
The atoms are pairwise disjoint and partition \(X\). Each atom can be written as \[ \left(\bigcap_{\epsilon_i=1}U_i\right) \cap \left(\bigcap_{\epsilon_i=0}(X\setminus U_i)\right), \] an intersection of an open subset with a closed subset, hence is locally closed.
Every Boolean expression in the \(U_i\) is a union of those atoms on which the expression is true. Since there are only finitely many atoms, the constructible set is a finite disjoint union of locally closed subsets.
4. Thus a subset of \(X\) is constructible if and only if it is a finite disjoint union of locally closed subsets.
Step 2 proves one direction and 3 proves the other.
5. Let \(X\) be irreducible with generic point \(\eta\), and let \(C\subseteq X\) be constructible and dense. Then\[ \boxed{\eta\in C.} \]
By 4, write \[ C=C_1\amalg\cdots\amalg C_r, \qquad C_i=U_i\cap F_i \] with \(U_i\) open and \(F_i\) closed.
Since \(C\) is dense, \[ X=\overline C = \overline{C_1}\cup\cdots\cup\overline{C_r}. \] The space \(X\) is irreducible, so one of these finitely many closed subsets equals \(X\). Say \[ \overline{C_i}=X. \] But \(C_i\subseteq F_i\) and \(F_i\) is closed, so \[ X=\overline{C_i}\subseteq F_i, \] hence \(F_i=X\). Therefore \[ C_i=U_i \] is a nonempty open subset of \(X\).
By Hartshorne II.3.17(d), the generic point \(\eta\) belongs to every nonempty open subset of an irreducible Zariski space. Thus \[ \eta\in U_i=C_i\subseteq C. \]
6. Conversely, if a subset \(C\subseteq X\) contains the generic point \(\eta\), then it is dense.
Since \[ \overline{\{\eta\}}=X \] and \(\eta\in C\), \[ X=\overline{\{\eta\}}\subseteq\overline C. \] Hence \(\overline C=X\).
7. A constructible subset of an irreducible Zariski space is dense if and only if it contains the generic point; when dense, it contains a nonempty open subset.
The equivalence is <1>5–<1>6. In the proof of 5, density produced a locally closed piece \(C_i\) for which the closed factor was all of \(X\), so \[ C_i=U_i \] was a nonempty open subset contained in \(C\).
8. Every closed subset of \(X\) is constructible and stable under specialization.
A closed subset is the complement of an open set, hence constructible. Hartshorne II.3.17(e) proves that closed subsets are stable under specialization.
9. Let \(S\subseteq X\) be constructible and stable under specialization, and put\[ Z=\overline S. \]Then \(S\) contains the generic point of every irreducible component of \(Z\).
Because \(Z\) is a closed subset of the noetherian space \(X\), it has finitely many irreducible components \[ Z=Z_1\cup\cdots\cup Z_r. \] Let \(\eta_i\) be the generic point of \(Z_i\).
Remove the other components: \[ O_i = Z_i\setminus\bigcup_{j\ne i}Z_j. \] This is a nonempty open subset of the irreducible space \(Z_i\), and it contains \(\eta_i\).
Since \(S\) is dense in \(Z\), the intersection \[ S\cap O_i \] is dense in \(O_i\). It is constructible in the open subspace \(O_i\). The space \(O_i\) is an irreducible Zariski space: it is an open subspace of a noetherian space, and its irreducible closed subsets acquire generic points by taking closures in \(Z_i\).
Applying 7 to \(S\cap O_i\) gives \[ \eta_i\in S\cap O_i\subseteq S. \]
10. The subset \(S\) from 9 equals its closure \(Z\) and hence is closed.
By 9, \(S\) contains every generic point \(\eta_i\) of an irreducible component \(Z_i\). Since \(S\) is stable under specialization, it contains every specialization of \(\eta_i\), i.e. all of \[ \overline{\{\eta_i\}}=Z_i. \] Therefore \[ Z=\bigcup_iZ_i\subseteq S. \] The reverse inclusion \(S\subseteq Z\) is automatic, so \(S=Z\) is closed.
11. Hence\[ \boxed{ S\text{ is closed} \iff S\text{ is constructible and stable under specialization}.} \]
Step 8 proves one direction and <1>9–<1>10 prove the converse.
12. A subset \(T\subseteq X\) is open if and only if it is constructible and stable under generization.
The complement of a constructible set is constructible. Moreover \(T\) is stable under generization exactly when its complement \[ X\setminus T \] is stable under specialization.
Thus 11 applied to \(X\setminus T\) gives \[ T\text{ open} \iff X\setminus T\text{ closed} \iff T\text{ constructible and generization-stable}. \]
13. If \(f:X\to Y\) is continuous, the inverse image of every constructible subset of \(Y\) is constructible in \(X\).
Let \[ \mathcal G = \{C\subseteq Y:f^{-1}(C)\text{ is constructible in }X\}. \] For every open \(U\subseteq Y\), continuity makes \(f^{-1}(U)\) open, hence constructible, so \(U\in\mathcal G\).
The family \(\mathcal G\) is closed under complements because \[ f^{-1}(Y\setminus C)=X\setminus f^{-1}(C), \] and under finite intersections because inverse images commute with intersections. Therefore \(\mathcal G\) contains the smallest family generated from open subsets under complements and finite intersections, namely all constructible subsets of \(Y\).
14. Q.E.D.
Step 4 proves part (a), 7 proves part (b), <1>11–<1>12 prove part (c), and 13 proves part (d).