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.
 A: 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 nonempty 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 $u \in \text{Ann}(A)$. 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:


*

*$A$ is reduced

*$A_\mathfrak{p}$ is reduced for every prime $\mathfrak{p} \subset A$

*$A_\mathfrak{m}$ 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 $(3)\implies (1)$.
