Let $A$ be a commutative ring with unit. Show that $A$ is reduced iff for every prime ideal $\mathfrak{p}\subseteq A$, $A_{\mathfrak{p}}$ is reduced.

This corresponds more or less to exercise 5, chapter 3 of Atiyah-Macdonald. It is useful to remember that $\text{Nil}(A_{\mathfrak{p}})=(\text{Nil}(A))_{\mathfrak{p}}$ (*).

($\Rightarrow$) Obvious by (*).

($\Leftarrow$) By the fact that "being $0$" is a local property which is satisfied by $\text{Nil}(A)$, again because of (*).

I wonder if the argument for the $\Leftarrow$ is sufficient.

  • $\begingroup$ Yes, it is. More formally you can say $\;\operatorname{Supp}(\operatorname{Nil}(A))=\varnothing$. $\endgroup$
    – Bernard
    Jan 13, 2017 at 11:14
  • $\begingroup$ Sorry! What do you mean? Is that if $A$ has no nilpotent element then $A_{P}$ has no nilpotent element? $\endgroup$
    – Soulostar
    Jan 13, 2017 at 12:36
  • $\begingroup$ @LêThếLong $A$ reduced means that $A$ has no nonzero nilpotents. $\endgroup$
    – Xam
    Jan 13, 2017 at 15:23
  • $\begingroup$ So, the problem means that $A$ has no nonzero nilpotent elements iff $A_{P}$ is too with $P\in Spec{A}$? If it is that then the "$\Rightarrow"$ is not true. $\endgroup$
    – Soulostar
    Jan 14, 2017 at 3:18
  • $\begingroup$ @LêThếLong why not? If $Nil(A)=0$ then of course $(Nil(A))_P=Nil(A_P)=0$ $\endgroup$ Jan 14, 2017 at 8:02

1 Answer 1


This question was posted a while ago, but I'll add a more explicit answer just for my own reference in the future.

Let $A$ be a nonzero commutative unital ring, so that $A_\mathfrak{p}$ is reduced for every prime $\mathfrak{p}$. Suppose (towards a contradiction) that $x \in A$ is nilpotent and nonzero. Then its annihilator $\text{Ann}(x) \subsetneq A$ is a non-zero ideal of $A$ (and is proper since it does not contain 1). By Zorn's lemma, it is contained in a maximal ideal $\mathfrak{m}$. Then the image of $x$ in the localization $A \to A_{\mathfrak{m}}$ is zero. This means that $\frac{x}{1} \sim \frac{0}{1}$, so there exists a $u \in A-\mathfrak{m}$ so that $0 = (1\cdot 0 - x\cdot 1)u = xu$, hence $u \in \text{Ann}(x)$. But this is a contradiction since this says $u \in \mathfrak{m}$, hence we conclude that $x=0$, and so $A$ is reduced. In fact we have an even stronger statement:

Theorem: The following are equivalent:

  1. $A$ is reduced
  2. $A_\mathfrak{p}$ is reduced for every prime $\mathfrak{p} \subset A$
  3. $A_\mathfrak{m}$ is reduced for every maximal ideal $\mathfrak{m} \subset A$

The proofs $(1)\implies (2)\implies (3)$ are trivial, and the content of the discussion above shows $(3)\implies (1)$.

  • $\begingroup$ $\def\nil{\operatorname{Nil}}$Another way of proving your theorem is by applying 3.8 of Atiyah, MacDonald, combined with the observation that $S^{-1}\nil A=\nil S^{-1}A$, where $\nil A=\sqrt{0}$ is the nilradical, and $S\subset A$ is a multiplicatively closed subset. More generally, $S^{-1}\sqrt{I}=\sqrt{S^{-1}I}$, for $I\subset A$ an ideal. $\endgroup$ Mar 21 at 10:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .