I'd like to check that my solution to the following exercise of Vakil's FOAG is correct. I am bothered that we have to use (B) in it, and I want to make sure that I used it in the end for the correct reason.
5.5.E. EXERCISE (ASSUMING (A) AND (B)). Show that the locus on Spec A of points [p] where $\mathcal O_{\operatorname{Spec} A,[p]} = A_p$ is nonreduced is the closure of those associated points of Spec A whose stalks are nonreduced. (Hint: why do points in the closure of these associated points all have nonreduced stalks? Why can’t any other point have a nonreduced stalk?)
(A) and (B) here are:
(A) The associated primes/points of M are precisely the generic points of irreducible components of the support of some element of M (on Spec A).
(B) M has a finite number of associated primes/points.
My solution:
Let $\{\mathfrak n_i\}$ be the points with non-reduced stalks and $\{\mathfrak g_i\}$ the associated points with non-reduced stalks. We want to show the two containments of the equality
$$\overline{\bigcup \mathfrak g_i} = \bigcup \mathfrak n_i$$
By (B), the union on the left is finite, so we can rewrite this as:
$$\bigcup \overline{\mathfrak g_i} = \bigcup \mathfrak n_i$$
The containment "$\supset$" follows from: if $f \in A_{\mathfrak n}$ is a nilpotent, it is still a nilpotent on the generic point of its support ($f$ is nonzero on $\mathfrak g:= \operatorname{Supp}f$ by definition, and is contained in all prime ideals contained in $\mathfrak g$ since it is contained in all prime ideals contained in $\mathfrak n \supset \mathfrak g$).
For the opposite containment "$\subset$" note that if $\mathfrak p \in \mathfrak g$ and $0\neq \tilde f = \frac f b \in A_{\mathfrak g}$ is such that $a\tilde f^n =0$ for some $a\not \in \mathfrak g$, then $abf$ is also a nonzero nilpotent, and it lives in $A_\mathfrak p$.
