Ytukyg

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.

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.