I have a question about Exercise 5.5.E in Vakil's algebraic geometry notes. Here's the statement:

Show that the locus on $\text{Spec } A$ of points $[p]$ where $\mathcal{O}_{\text{Spec } A;[p]} = A_p$ is nonreduced is the closure of those associated points of $\text{Spec } A$ whose stalks are non reduced.

The statement of the problem further indicates that we are supposed to do this using the following two properties of associated points:

A. The associated points of $\text{Spec }A$ are precisely the generic points of irreducible components of the support of some element of $A$.

B. $\text{Spec }A$ has finitely many associated points.

My problem is that I seem to have solved it using A but not B!

Question: Where did I go wrong?

Here's my purported solution. I will use A repeatedly (without comment), but I will never use B:

Assume first that $[p]$ is an associated point of $\text{Spec } A$ where $A_p$ is non-reduced and $[q]$ is a point in the closure of $[p]$. We then have $p \subset q$, and thus $A_p$ is a localization of $A_q$. If $A_q$ were reduced, then this would imply that $A_p$ is reduced; we therefore deduce that $A_q$ is non-reduced.

Now assume that $[r]$ is a point of $\text{Spec } A$ where $A_r$ is non-reduced. Our goal is to prove that $[r]$ is in the closure of some associated point $[s]$ where $A_s$ is non-reduced. Since $A_r$ is non-reduced, there exists some $f \in A$ and $n \geq 2$ and $x \in A \setminus r$ such that $x f^n = 0$, but where $y f \neq 0$ for all $y \in A \setminus r$. Set $g = x f$. We then have $g^n = 0$, but $y g \neq 0$ for all $y \in A \setminus r$. This implies that $[r] \in \text{Supp } g$. Moreover, since $g$ itself is nilpotent it witnesses the fact that $A_{r'}$ is non-reduced for all points $[r']$ in $\text{Supp } g$. In particular, if $[s]$ is the generic point of an irreducible component of $\text{Supp } g$ containing $[r]$, then $[s]$ has the desired property.

  • 2
    $\begingroup$ I guess one issue is basic topology: if you have an infinite family of sets then it may not be true that closure commutes with union. $\endgroup$ – Hoot Sep 7 '16 at 6:21

To make an answer out of the above comment, the first paragraph of your solution shows that the union of $\overline{\mathfrak{p}}$ over $\mathfrak{p}\in \text{Ass } \text{Spec }A$ such that $A_\mathfrak{p}$ is nonreduced is a subset of the locus on $\text{Spec } A$ of points $[p]$ where $A_{\mathfrak{p}}$ is nonreduced. The second paragraph shows the converse, but the implicit assumption in both directions is that the closure of the set of associated points of Spec A whose stalks are non reduced is the same as the union of the closures of those associated points. $\bf{(B)}$ allows you to make this assumption, as the closure of a finite union of sets is the union of the closures of each.

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