Here we view $R$ as an $R$-module of itself.
I'm trying to show the nilpotent radical of $R$ is the intersection of all the associated primes of $R$, i.e., all the primes of Ass($R$). We know that for $R$ Noetherian, any every minimal prime of Ann($M$) is in Ass($M$).
I'm trying to show this by showing the nilpotent radical of $R$ is the intersection of all minimal primes of Ann($R$). We know that the nilpotent radical is the intersection of all primes of $R$. If every prime of $R$ contains a minimal prime of Ann($R$), then this would give the result.
This is a homework problem from Aluffi's Algebra: Chapter 0 (Problem VI.4.10) for a course I'm sitting in on, but not registered for.