About Nilradical and Localization

Question

I've been trying to answer the following question:

If the nilradical of $A_{P}$ is zero for all prime ideals $P\subset A$, then the nilradical of $A$ is also zero.

I tried to prove that it is true, but I couldn't came up with any proof. In the other hand, I coulnd find any counterexample in books or in the internet.

Can anyone give me a hint? Thanks in advance.

Answer 1

Let $x \in A$ be nilpotent.

Then $x$ is nilpotent in every $A_p$, so is zero in every $A_p$.

Let $I=\{y \in A,\,xy=0\}$, $I$ is a nonzero ideal of $A$.

Let $p$ be any prime ideal of $A$: since $x=0$ in $A_p$, by definition $I$ is not a subset of $p$.

In other words, $I$ is not contained in any maximal ideal of $A$. So $I=A$ thus $x=0$.

Answer 2

The nilradical $N$ of $A$ is the intersection of all the prime ideals of $A$ if $n\neq 0\in N$, and $f_P:A\rightarrow A_P$ the localisation morphism, $f_P(n)=0$ implies that there exists $s_P\in A-P$ such that $sn=0$ for every $P$ and $n=0$, since the sheaf of regular functions on $Spec(A)$ is well defined, and $f_P(s)$ is the value of $n\in O_{Spec(A)}(Spec(A))$ at $P$.

We can interpret the classical proof of the fact mentioned above here; $P$ is not an element of $V(s_P)$ thus $\cap_PV(s_P)$ is empty this implies $A$ is generated by $s_P,P\in Spec(A)$ and there exists $P_1,...,P_n, u_1,..,u_n$ such that $u_1s_{P_1}+..u_ns_{P_n}=1$, this implies that $1.n=0$.

Answer 3

Hint:

The nilradical of $A_\mathfrak{p}$ is the localisation of the nilradical $N_\mathfrak p$. If it is $0$ for all prime ideals, $\;\operatorname{Supp}(N)=\varnothing$.

Answer 4

There are many ways to approach this, as the other answers show. Normally, I wouldn't want to muddy the waters, but I feel like there is a nice approach that no one has yet mentioned (though Tsemo Aristide's answer is somewhat close).

First, note that for any commutative ring $A$, the natural map $\varphi \colon A \to \prod_{P \in \mathrm{Spec}(A)} A_{P}$ is an injection of rings. There are many ways to show this: if, for instance, you are familiar with the construction of the usual structure sheaf of rings on $\mathrm{Spec}(A)$, then this is an immediate consequence of the general result that for any separated presheaf of sets (or groups, rings, etc.) $\mathcal{F}$ on a topological space $X$ and any open subset $U$ of $X$, the canonical map $\mathcal{F}(U) \to \prod_{p \in U} \mathcal{F}_{p}$ is an injection. One can avoid the machinery of sheaves in name and reason directly, however.

Suppose $x \in A$ such that $\varphi(x) = 0$. Then $x/1$ is zero in $A_{P}$ for every $P \in \mathrm{Spec}(A)$, so there exists $s_{P} \in A \setminus P$ such that $s_{P} \cdot x = 0$. Since $P \in D(s_{P})$ for each $P \in \mathrm{Spec}(A)$, it follows that $\{D(s_{P})\}_{P \in \mathrm{Spec}(A)}$ is an open cover of $\mathrm{Spec}(A)$. Since $\mathrm{Spec}(A)$ is (quasi)compact, there exist finitely many $s_{1}, \ldots, s_{n}$ such that $\mathrm{Spec}(A) = \bigcup_{i=1}^{n} D(s_{i})$, and so $s_{1}, \ldots, s_{n}$ generate the unit ideal of $A$. Letting $a_{1}, \ldots, a_{n} \in A$ be such that $a_{1}s_{1} + \cdots + a_{n}s_{n} = 1$, we then see that $x = 1 \cdot x = (a_{1}s_{1} + \cdots + a_{n}s_{n}) \cdot x = 0$, since $s_{i} \cdot x = 0$ by hypothesis.

Armed with the fact above, the proof is very simple. Note that any product of reduced rings is reduced, and any subring of a reduced ring is reduced. In particular, since $\varphi$ embeds $A$ as a subring of $\prod_{P \in \mathrm{Spec}(A)} A_{P}$, we are done, since $A_{P}$ is reduced for each prime $P$ of $A$ by hypothesis.