# Proof check: being reduced is a local property (Atiyah-Macdonald 3.5)

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.

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

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 $$(3)\implies (1)$$.

• $\def\nil{\operatorname{Nil}}$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. Mar 21 at 10:27