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.

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 $$x=0$$.

Hint:

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

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 $$1.n=0$$.

• sorry, but, once $A$ is not necessarily an integral domain, how can I assure that $s$ must be zero? – ArkPDEnational Sep 17 at 0:08

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 a sheaf injects into the product of its stalks. 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.