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.



In a noetherian commutative ring with only one associated prime, is the nilradical locally free?
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 $N=0$.


