If \(X\) is a topological space and \(Z\) an irreducible closed subset of \(X\), a generic point for \(Z\) is a point \(\zeta\) such that \(Z = \cl\qty{\ts{\zeta}}\). If \(X\) is a scheme, show that every nonempty irreducible closed subset has a unique generic point.
Solution
Let \[ Z\subseteq X \] be a nonempty irreducible closed subset.
1. Choose an affine open subset\[ U=\operatorname{Spec}A\subseteq X \]such that\[ Z\cap U\ne\varnothing. \]Then \(Z\cap U\) is a nonempty irreducible closed subset of \(U\).
An affine open cover of \(X\) covers the nonempty set \(Z\), so some affine open \(U\) meets it.
Since \(Z\) is closed in \(X\), the intersection \(Z\cap U\) is closed in the subspace \(U\). It is irreducible because every nonempty open subset of an irreducible space is irreducible: if \[ Z\cap U=C_1\cup C_2 \] with \(C_1,C_2\) closed in \(Z\cap U\), then taking closures in \(Z\) would write the dense open subset \(Z\cap U\) as contained in the union of two proper closed subsets unless one \(C_i\) were the whole intersection. Equivalently, the standard topological lemma says a nonempty open subspace of an irreducible space is irreducible.
2. There is a unique prime ideal\[ \mathfrak p\subseteq A \]such that\[ \boxed{Z\cap U=V(\mathfrak p).} \]
Any closed subset of \(\operatorname{Spec}A\) has the form \[ V(I) \] for some ideal \(I\subseteq A\). The closed subset \(V(I)\) is irreducible exactly when its radical \[ \sqrt I \] is prime.
Indeed, if \(\sqrt I\) is prime then \[ V(I)=V(\sqrt I) \] is the closure of the point \(\sqrt I\), hence irreducible. Conversely, if \(V(I)\) is irreducible and \[ ab\in\sqrt I, \] then \[ V(I)\subseteq V(ab)=V(a)\cup V(b). \] Irreducibility forces \[ V(I)\subseteq V(a) \quad\text{or}\quad V(I)\subseteq V(b), \] so \[ a\in\sqrt I \quad\text{or}\quad b\in\sqrt I. \] Thus \(\sqrt I\) is prime.
Take \[ \mathfrak p=\sqrt I. \] Uniqueness follows because for prime ideals \[ V(\mathfrak p)=V(\mathfrak q) \] implies \[ \mathfrak p =\sqrt{\mathfrak p} =\sqrt{\mathfrak q} =\mathfrak q. \]
3. Let\[ \zeta\in U \]be the point corresponding to \(\mathfrak p\). Then\[ \overline{\{\zeta\}}^{\,U} =Z\cap U. \]
In an affine spectrum, the closure of the point corresponding to a prime \(\mathfrak p\) is \[ V(\mathfrak p). \] Apply 2.
4. The closure of \(\zeta\) in \(X\) is exactly \(Z\):\[ \boxed{ \overline{\{\zeta\}}^{\,X}=Z. } \]
Since \(Z\) is closed in \(X\) and contains \(\zeta\), one has \[ \overline{\{\zeta\}}^{\,X} \subseteq Z. \]
Intersect this closure with the open set \(U\). Closure in an open subspace satisfies \[ U\cap\overline{\{\zeta\}}^{\,X} =\overline{\{\zeta\}}^{\,U} =Z\cap U \] by 3.
The subset \(Z\cap U\) is a nonempty open subset of the irreducible space \(Z\), hence is dense in \(Z\). The closed subset \[ \overline{\{\zeta\}}^{\,X} \subseteq Z \] contains this dense subset, so it contains all of \(Z\). Together with the first inclusion, equality follows.
5. Thus every nonempty irreducible closed subset of a scheme has a generic point.
The point \(\zeta\) constructed in 3 satisfies the defining condition by 4.
6. If \(\zeta\) is a generic point of \(Z\), then \(\zeta\) belongs to every nonempty open subset of \(Z\).
Let \[ W\subseteq Z \] be nonempty and open. Suppose \[ \zeta\notin W. \] Then \[ Z\setminus W \] is a closed subset of \(Z\) containing \(\zeta\). Hence it contains the closure of \(\zeta\) in \(Z\), which is all of \(Z\). This would force \(W=\varnothing\), contradiction.
7. The generic point of \(Z\) is unique.
Suppose \[ \zeta,\eta\in Z \] are both generic points. Choose an affine open neighborhood \[ \zeta\in U=\operatorname{Spec}A. \] Then \[ U\cap Z \] is a nonempty open subset of \(Z\), so 6 applied to the generic point \(\eta\) gives \[ \eta\in U. \]
The closures of \(\zeta\) and \(\eta\) inside \(U\) are both \[ Z\cap U. \] If they correspond to prime ideals \(\mathfrak p\) and \(\mathfrak q\) of \(A\), then \[ V(\mathfrak p)=V(\mathfrak q). \] As in 2, equality of these closed sets for prime ideals implies \[ \mathfrak p=\mathfrak q. \] Therefore \[ \zeta=\eta. \]
8. Hence every nonempty irreducible closed subset \(Z\) of a scheme has a unique generic point:\[ \boxed{ Z=\overline{\{\zeta\}} \quad\text{for a unique }\zeta\in Z. } \]
Existence is 5 and uniqueness is 7.
9. Q.E.D.
Step 8 is the assertion of the exercise.