Localization at a prime ideal is a reduced ring

Here is the question that I came up with, which I am having trouble proving or disproving:

Let $A$ be a ring (commutative). Let $p \in Spec(A)$ such that $A_p$ is reduced. Then there exists an open neighborhood of $U \subset Spec(A)$ containing $p$ such that $\forall q \in U$, $A_q$ is reduced.

Here is some background to my question:

I am basically trying to prove that if the stalks at all closed points of a quasicompact scheme are reduced rings, then the scheme is reduced.

Since the closure of every point of a quasicompact scheme contains a closed point of that scheme, proving the above commutative algebra statement (if it is true) will yield a proof of this statement about reducedness of quasicompact schemes.

If the statement in bold is true, then I guess the neighborhood $Spec(A)-V(A-p)$ should suffice (this is just a guess), but I am running into some problems trying to use this neighborhood to show that the localization at every point of $Spec(A)-V(A-p)$ gives me a reduced ring. So there might be some other neighborhood of $p$ that I am missing, or the statement in bold is not true. Either way, some help would be appreciated (if the statement in bold is true, then I would appreciate hints and not complete answers).

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Okay, I managed to show that localization at the generalizations of $p$ are reduced, and to prove the statement about quasicompact schemes, it seems this is all you need. But, my original question still stands. – Rankeya Feb 21 '12 at 6:34
But, now my guess is that the statement in bold may not be true. – Rankeya Feb 21 '12 at 6:46

Your boldface statement is false in general. Consider $$A=F[X_1, \dots, X_n, ...., Y_1, \dots, Y_n, ....]/(X_n^2, X_nY_n)_n$$ over a field $F$. Denote by $x_n, y_n$ the images of $X_n, Y_n$ in $A$. Let $\mathfrak p$ be the ideal generated by the $x_n$'s. As $A/\mathfrak p$ is $F[y_1, \dots, y_n, ....]$ which is integral, $\mathfrak p$ is a prime ideal, generated by nilpotent elements, thus is the minimal prime ideal of $A$. As every $x_ny_n=0$, $\mathfrak pA_{\mathfrak p}=0$, hence $A_{\mathfrak p}$ is reduced.

If there were a reduced open neighborhood $U$ of $\mathfrak p$, then $U$ contains a non-empty reduced principal open subset $D(f)$. So $A_f$ is reduced and $\mathfrak p A_f=0$. Therefore for all $n\ge 1$, there exists $r_n\ge 1$ such that $f^{r_n}x_n=0$. Now it is easy to check that $f\in y_nA+\mathfrak p$ and $\cap_n (y_nA +\mathfrak p)=\mathfrak p$. Thus $f$ is nilpotent and $D(f)=\emptyset$. Contradiction.

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This looks good to me. Thanks. – Rankeya Feb 22 '12 at 1:29

The basic equation is the not completely trivial equality (do you want a proof?) $$(NilA)_{\mathfrak p}=Nil(A_{\mathfrak p})$$
If Nil(A) is a finitely generated ideal of $A$ (automatic for noetherian $A$), you deduce that $$Supp (NilA)=V(Ann (Nil A))$$
is closed, as required.

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Ignore my earlier comment. What I mean is that if the nilradical is not finitely generated, then does my result still hold? – Rankeya Feb 21 '12 at 13:49
Dear @Rankeya, the answer to your question whether the result holds for non finitely generated $Nil A$ is simple and clear: I don't know! – Georges Elencwajg Feb 21 '12 at 17:28

If the localizations of $A$ at all of its maximal ideals are reduced, then the annihilator of any nilpotent element of $A$ is not contained in any maximal ideal, and so contains $1$.

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I know this. But, how does this answer my question? This just shows that to prove reducedness of Spec(A) it suffices to look at closed points. In fact, just proving this does not (I think it doesn't) yoeld a proof for quasicompact schemes in general. – Rankeya Feb 21 '12 at 13:43
Of course it answers your question for quasicompact schemes --- cover your scheme with affines. Concerning your more general question, on each affine the set you are looking for the complement of the supports of the annihilators of all nilpotents in the ring. If the ring is noetherian, you only need to take a finite set of generators of the nilradical, so it's open, but otherwise probably not. – anon Feb 21 '12 at 19:17
Wait, but all I know is that the stalks at the closed points of my scheme are reduced. When I cover my scheme with affines how does this imply that for every closed point of the affine piece, the localization at that closed point (the closed point being a maximal ideal) is reduced? The affine piece may have closed points that are not closed in my original scheme. Am I missing something obvious? – Rankeya Feb 21 '12 at 22:01
But here is the thing. If a point is closed in an affine open, it need not be closed in the scheme. This is true and there are schemes with no closed points. It's simple topology: If a point is closed in an open subset of a topological space, then it need not be a closed point of the whole space. So, an affine open could have closed points that are not closed in my scheme. – Rankeya Feb 23 '12 at 2:05
@anon Rankeya is saying that a point of a scheme $X$ which is closed in some open $U$ need not be closed in $X$. You seem to be saying that a point of $x$ which is closed in every open in an open cover is necessarily closed in $X$. Both statements are true. For an example of what Rankeya is saying, look at the spectrum of a DVR. It has two points, the generic point and the closed point. The generic point $\{\eta\}$ is open, and certainly it is closed in itself, but not in the ambient scheme. However, in a finite type scheme over a field, closed points of affine opens are closed in – Keenan Kidwell Feb 24 '12 at 2:23