A topological space \(X\) is a Zariski space if it is noetherian and every nonempty closed irreducible subset has a unique generic point.
For example, let \(R\) be a discrete valuation ring and let \(T = \operatorname{sp}(\Spec R)\). Then \(T\) consists of two points: \(t_0\), the maximal ideal, and \(t_1\), the zero ideal. The open subsets are \(\varnothing\), \(\ts{t_1}\), and \(T\). This is an irreducible Zariski space with generic point \(t_1\).
Show that if \(X\) is a noetherian scheme then \(\operatorname{sp}(X)\) is a Zariski space.
Show that any minimal nonempty closed subset of a Zariski space consists of one point. We call these closed points.
Show that a Zariski space \(X\) satisfies the axiom \(T_0\): given any two distinct points of \(X\), there is an open set containing one but not the other.
If \(X\) is an irreducible Zariski space, then its generic point is contained in every nonempty open subset of \(X\).
If \(x_0 \in \cl\qty{\ts{x_1}}\), we say \(x_0\) is a specialization of \(x_1\), or that \(x_1\) is a generization of \(x_0\). Now let \(X\) be a Zariski space. Show that the minimal points for the partial ordering determined by \(x_1 > x_0\) when \(x_0\) is a specialization of \(x_1\) are the closed points, and the maximal points are the generic points of the irreducible components of \(X\). Show also that a closed subset contains every specialization of any of its points; we say closed subsets are stable under specialization. Similarly, open subsets are stable under generization.
Let \(t\) be the functor on topological spaces introduced in the proof of Hartshorne II.2.6. If \(X\) is a noetherian topological space, show that \(t(X)\) is a Zariski space. Furthermore, \(X\) itself is a Zariski space if and only if the map \(\alpha: X \to t(X)\) is a homeomorphism.
Solution
1. If \(X\) is a noetherian scheme, then its underlying topological space is a Zariski space.
By definition of a noetherian scheme, the underlying space is noetherian.
Let \[ Z\subseteq X \] be a nonempty irreducible closed subset. Hartshorne II.2.9 shows that every nonempty irreducible closed subset of a scheme has a unique generic point. Therefore \(Z\) has a unique generic point. This is exactly the definition of a Zariski space.
2. Any minimal nonempty closed subset \(Y\) of a Zariski space consists of one point.
For any \(y\in Y\), the closure \[ \overline{\{y\}} \] is a nonempty closed subset of \(Y\). By minimality, \[ \overline{\{y\}}=Y. \] Thus every point of \(Y\) is a generic point of \(Y\).
The set \(Y\) is irreducible: if \[ Y=Y_1\cup Y_2 \] with \(Y_i\) closed in \(Y\), then every nonempty \(Y_i\) is a nonempty closed subset of \(X\) contained in \(Y\), hence must equal \(Y\) by minimality. So one of \(Y_1,Y_2\) equals \(Y\).
Since \(Y\) is irreducible and the space is Zariski, \(Y\) has a unique generic point. But every point of \(Y\) is generic, so \(Y\) has exactly one point.
3. A Zariski space satisfies the \(T_0\) axiom.
Let \(x\ne y\). The closures \[ \overline{\{x\}},\qquad\overline{\{y\}} \] are irreducible closed subsets. If they were equal, then \(x\) and \(y\) would both be generic points of the same irreducible closed subset, contradicting uniqueness.
Thus the closures differ. After interchanging \(x\) and \(y\) if necessary, assume \[ y\notin\overline{\{x\}}. \] Then \[ X\setminus\overline{\{x\}} \] is an open set containing \(y\) but not \(x\). Hence \(X\) is \(T_0\).
4. If \(X\) is an irreducible Zariski space with generic point \(\eta\), then \(\eta\) lies in every nonempty open subset of \(X\).
Let \(U\subseteq X\) be nonempty. If \(\eta\notin U\), then the closed set \[ X\setminus U \] contains \(\eta\), hence contains \[ \overline{\{\eta\}}=X. \] This would force \(U=\varnothing\), contradiction.
5. Under the specialization order\[ x_1>x_0 \quad\Longleftrightarrow\quad x_0\in\overline{\{x_1\}}, \]the minimal points are exactly the closed points.
If \(x\) is closed, then \[ \overline{\{x\}}=\{x\}, \] so \(x\) has no proper specialization and is minimal.
Conversely, if \(x\) is minimal, every point of the nonempty closed set \[ \overline{\{x\}} \] is a specialization of \(x\). Minimality therefore forces \[ \overline{\{x\}}=\{x\}, \] so \(x\) is closed.
6. The maximal points for the specialization order are exactly the generic points of the irreducible components of \(X\).
Let \(x\) be maximal. The irreducible closed subset \[ \overline{\{x\}} \] is contained in some irreducible component \(C\) of \(X\). Let \(\eta_C\) be the generic point of \(C\). Then \[ x\in C=\overline{\{\eta_C\}}, \] so \(\eta_C\) is a generization of \(x\). Maximality gives \[ x=\eta_C. \]
Conversely, let \(x=\eta_C\) be the generic point of an irreducible component \(C\), and suppose \(y\) is a generization of \(x\). Then \[ x\in\overline{\{y\}}, \] so \[ C=\overline{\{x\}}\subseteq\overline{\{y\}}. \] The right side is irreducible and closed. Maximality of the irreducible component \(C\) forces \[ \overline{\{y\}}=C. \] Thus \(y\) and \(x\) are both generic points of \(C\), and uniqueness gives \(y=x\). Hence \(x\) is maximal.
7. Closed subsets are stable under specialization, and open subsets are stable under generization.
Let \(F\subseteq X\) be closed and let \(x\in F\). If \(x_0\) is a specialization of \(x\), then \[ x_0\in\overline{\{x\}}\subseteq F \] because \(F\) is closed. Thus \(F\) is stable under specialization.
Now let \(U\) be open, let \(x\in U\), and let \(y\) be a generization of \(x\). If \(y\notin U\), then the closed complement \(X\setminus U\) contains \(y\) and therefore all its specializations, including \(x\), contradiction. Hence \(y\in U\).
8. Let \(X\) be a noetherian topological space. Recall that \(t(X)\) is the set of nonempty irreducible closed subsets of \(X\), with closed subsets\[ t(Y)=\{Z\in t(X):Z\subseteq Y\} \]for closed \(Y\subseteq X\). Then \(t(X)\) is noetherian.
The assignment \[ Y\longmapsto t(Y) \] preserves inclusions and arbitrary intersections. It also preserves finite unions: if an irreducible closed subset \(Z\) is contained in \[ Y_1\cup Y_2, \] then \[ Z=(Z\cap Y_1)\cup(Z\cap Y_2), \] so irreducibility forces \(Z\subseteq Y_1\) or \(Z\subseteq Y_2\). Hence \[ t(Y_1\cup Y_2)=t(Y_1)\cup t(Y_2). \] Thus these \(t(Y)\) are exactly the closed sets of the topology on \(t(X)\).
If \[ t(Y_1)\supseteq t(Y_2)\supseteq\cdots \] is a descending chain of closed subsets of \(t(X)\), then \[ Y_1\supseteq Y_2\supseteq\cdots. \] Indeed, if \(x\in Y_{n+1}\) then the irreducible closed set \(\overline{\{x\}}\) lies in \(t(Y_{n+1})\subseteq t(Y_n)\), so \(x\in Y_n\).
Since \(X\) is noetherian, the chain \(Y_n\) stabilizes, hence so does the chain \(t(Y_n)\). Therefore \(t(X)\) is noetherian.
9. Every nonempty irreducible closed subset of \(t(X)\) has a unique generic point.
Let \[ t(Y)\subseteq t(X) \] be nonempty and irreducible. If \(Y\) were reducible, \[ Y=Y_1\cup Y_2 \] with proper closed subsets \(Y_i\), then \[ t(Y)=t(Y_1)\cup t(Y_2) \] would be reducible. Hence \(Y\) is irreducible.
Thus \(Y\) itself is a point of \(t(X)\). Its closure in \(t(X)\) is \[ \overline{\{Y\}}=t(Y), \] because a closed set \(t(F)\) contains the point \(Y\) exactly when \(Y\subseteq F\), and then it contains every irreducible closed subset of \(Y\). So \(Y\) is a generic point of \(t(Y)\).
If another point \(Z\in t(X)\) had the same closure, then \[ t(Z)=t(Y). \] Since \(Z\in t(Z)=t(Y)\), one has \(Z\subseteq Y\), and similarly \(Y\subseteq Z\). Thus \(Z=Y\). The generic point is unique.
10. Hence \(t(X)\) is a Zariski space.
Step 8 gives noetherianity and 9 gives unique generic points for nonempty irreducible closed subsets.
11. Define\[ \alpha:X\longrightarrow t(X), \qquad x\longmapsto\overline{\{x\}}. \]Then \(\alpha\) is continuous and\[ \alpha^{-1}(t(Y))=Y \]for every closed subset \(Y\subseteq X\).
For \(x\in X\), \[ \alpha(x)\in t(Y) \iff \overline{\{x\}}\subseteq Y. \] Because \(Y\) is closed, this is equivalent to \(x\in Y\). Thus the inverse image of the closed set \(t(Y)\) is the closed set \(Y\), proving continuity and the formula.
12. If \(X\) is a Zariski space, then \(\alpha:X\to t(X)\) is a bijection.
Every point of \(t(X)\) is an irreducible closed subset \(Y\subseteq X\). Since \(X\) is a Zariski space, \(Y\) has a generic point \(y\), and \[ \alpha(y)=\overline{\{y\}}=Y. \] Thus \(\alpha\) is surjective.
If \(\alpha(x)=\alpha(y)\), then \(x\) and \(y\) are both generic points of the same irreducible closed subset. Uniqueness gives \(x=y\). Thus \(\alpha\) is injective.
13. If \(X\) is a Zariski space, then \(\alpha\) is a homeomorphism.
By 12, \(\alpha\) is bijective, and by 11 it is continuous.
Let \(Y\subseteq X\) be closed. For every irreducible closed subset \(Z\subseteq Y\), the Zariski-space property gives a generic point \(z\in Z\subseteq Y\), and \[ \alpha(z)=Z. \] Hence \[ \alpha(Y)=t(Y), \] which is closed in \(t(X)\). Therefore \(\alpha\) is a closed continuous bijection and hence a homeomorphism.
14. Conversely, if \(X\) is noetherian and \(\alpha:X\to t(X)\) is a homeomorphism, then \(X\) is a Zariski space.
By 10, \(t(X)\) is a Zariski space. The property of being noetherian with a unique generic point for every nonempty irreducible closed subset is topological and is preserved by homeomorphism. Hence \(X\) is a Zariski space.
15. Therefore, for a noetherian space \(X\),\[ \boxed{ X\text{ is a Zariski space} \iff \alpha:X\xrightarrow{\sim}t(X)\text{ is a homeomorphism}.} \]
Steps 13 and 14 prove the two implications.
16. Q.E.D.
Steps <1>1–<1>7 prove parts (a)–(e), and steps <1>8–<1>15 prove part (f).