Locus of Nonreduced points & Associated Points.

Before, we demonstrated that reduced-ness is not an open condition (though for quasi-compact schemes, we can still check reducedness at closed points). However, for Noetherian schemes, reduced-ness is indeed an open condition.

For this post, we only use two properties of associated points of an $A$-module $M$ for $A$ Noetherian.

• The associated points of $M$ are precisely generic points of irreducible components of supports of some global sections of $\widetilde{M}$.
• $M$ has finitely many associated points.
  
  Warning: we only talk about supports of global sections.

Proposition. Let $X = \text{Spec} A$. Show the the set of nonreduced points of $X$ is the closure of the nonreduced associated points. In other words, let $S$ be the set of non-reduced points of $X$. Then
$$S = \overline{S \cap \text{Ass} X}.$$

Proof.  Let $p$ be in the closure of the set of non-reduced associated points.
Then $p \in \overline{\{q\}}$ for some non-reduced associated point $q$, i.e.$p \supset q$. Thus we have  $A_q = (A_p)_q$. If $A_p$ is reduced then so must $A_q$, therefore $A_p$ is nonreduced.

On the other hand, suppose $p$ has nonreduced stalk, i.e. there is some (not necessarily global) section $s_p \neq 0$ such that $s_p^n = 0$ for some $n$. On a sufficiently small affine open $D(f) \ni p$, we could represent $s$ as $a/f$ for some $a \in A$. Thus we have $(af)$ is nilpotent (by proof of Lemma). Now $(af)_p \neq 0$ so $p \in \text{Supp}(af)$ thus it must be in the closure of some associated point $q$. (In particular, $(af)_q \neq 0$). On the other hand, as $D(f) \ni q$, and $(af)^n = 0$ on $D(f)$. Thus  $(af)^n_q = 0$, so $q$ is a non-reduced point.



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Thus we see that in particular, for an affine scheme $X$, the ``reduced locus'' of $X$ is open.

Recall that reduced-ness in in general not an open condition. However, from the above, we see that, for Noetherian schemes $X$, the "reduced locus" of $X$ is an open subset of $X$.
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In the above proof we have used the following Lemma.

Lemma. Localization of reduced rings is reduced.

Proof. Suppose $A$ is a reduced ring. Let $B = S^{-1} A$.
Suppose $a/s$ is a nilpotent in $B$. This means $a/s \neq 0$ in $B$ and $(a/s)^m = 0$ in $B$. In other words, $x a \neq 0$ for all $x \in S$ and $ya^m = 0$ for some $y \in S$. Thus $ya \neq 0$ but $ya^m = 0$, so $ya$ is a nilpotent in $A$, contradiction.

Lemma. If $s_p \neq 0$ then $p$ is in the support of some global section.

Proof. If $s_p \neq 0$, then $s$ is not $0$ on some neighborhood of $p$. WLOG, assume $s \neq 0$ on some $D(f) \ni p$. Suppose $s$ is represented by $a/f$ on $D(f)$. Then $a_p \neq 0$.