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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.

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    Yes, it is. More formally you can say $\;\operatorname{Supp}(\operatorname{Nil}(A))=\varnothing$. – 2017-01-13
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    Sorry! What do you mean? Is that if $A$ has no nilpotent element then $A_{P}$ has no nilpotent element? – 2017-01-13
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    @LêThếLong $A$ reduced means that $A$ has no nonzero nilpotents. – 2017-01-13
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    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. – 2017-01-14
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    @LêThếLong why not? If $Nil(A)=0$ then of course $(Nil(A))_P=Nil(A_P)=0$ – 2017-01-14
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    You can not conclude like that. If you take $ \frac{a}{1} \in A_{P}$ such that it is nilpotent. It means there exists $s\in R\P$ and $n\in \mathbb{N}$ such that $sa^{n}=0$. How can you deduce that A has non-zero nilpotent elements? – 2017-01-14
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    @LêThếLong if $sa^n=0$ then $(sa)^n=0$, and $sa\neq 0$ otherwise $\frac{a}{1}=0$ in $A_P$ – 2017-01-14
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    I see. That is my mistake. I remember about this one: $x$ is nilpotent iff $\frac{x}{1}$ is nilpotent. It is wrong but not relevance to this one. Thank you. – 2017-01-14

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