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The title says it all.

I suspect that the answer in general is no, although my intuition tells me that a jump in the dimension of the fibre of the nilradical at some point of Spec(A) can occur only when one meets an associated prime cycle.

Note the easiest case: when the ring is generically reduced then it has to be reduced hence its nilradical is 0 hence free. This follows easily from the fact that Ass={pt} implies that the ring is the union of its nilpotents and its nonzerodivisors.

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    $\begingroup$ Isn't $k[x]/(x^2)$ (or more generally, any artinian local ring) a counterexample? Actually, I'm having more trouble coming up with examples (where the nilradical is nonzero) than counterexamples. $\endgroup$
    – Eric Wofsey
    Commented Oct 7, 2015 at 20:37
  • $\begingroup$ you are absolutely right that my question is nonsense. I guess that what I have in mind is something along the following lines : if X is a scheme of finite type over a field, and N is the sheaf of nilpotent regular functions, then the dimension d(x)=dim_k(N_x)=_dim_k(Nil(O_{X,x})) is finite, and in case X has only one associated point, the function d should be constant. Does that make sense to you ? $\endgroup$
    – Matthieu Romagny
    Commented Oct 7, 2015 at 21:03
  • $\begingroup$ Well, as you're defining it, $d(x)$ is usually not finite (if $A=B\otimes C$ where $B$ is a domain and $C$ is artinian local, then $N_x$ is finite rank over a localization of $B$, not over $k$). I have a sense of the intuition you're going for with $d(x)$, but I don't know the right way to define it in general. Maybe you want to say $d(x)$ is the length of a maximal chain of ideals in $\mathcal{O}_{X,x}$ which are all annihilators of submodules of the nilradical, or something similar? $\endgroup$
    – Eric Wofsey
    Commented Oct 7, 2015 at 21:21
  • $\begingroup$ All right, I see I'm a bit messed up. I have to think about it a bit more. $\endgroup$
    – Matthieu Romagny
    Commented Oct 8, 2015 at 9:39

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Actually, this never happens unless the nilradical is $0$ (i.e., unless the ring is a domain). Let $A$ be a Noetherian ring with only one associated prime whose nilradical $N$ is locally free. Then $A$ has only one minimal prime, so $N$ is prime. Now localize at $N$. We now have a $0$-dimensional Noetherian local ring $A_N$, so it is artinian. In particular, $A_N$ has finite length over itself, and the length of the maximal ideal $NA_N$ is strictly smaller than the length of $A_N$. But we are assuming $N$ is locally free, so $NA_N$ is free over $A_N$; the only way this can happen is if $NA_N=0$. But, as you observe, every element of $A\setminus N$ is a nonzerodivisor, so the canonical map $A\to A_N$ is injective. Thus $NA_N=0$ implies $N=0$.

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