7
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Welcome the Mathematics Stack Exchange! A quick tour of the site (math.stackexchange.com/tour) will help you get the most of your time here. $\endgroup$ – dantopa Dec 27 '18 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.