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).
 A: 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.
A: 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.
A: 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 $1$.
