I am reading a book that suggests to find what $N(R)$ is - the nilradical of $R$ where $R$ is Artinian you just need to find a nilpotent ideal $I $ of $R$ and then show that $R/I $ has no nonzero nilpotents. Is this enough to show that $I=N(R)$?
I’m not sure why if so.
I know that in an Artinian ring is the largest nilpotent ideal of $R$ and I know that $N(R)$ is a nilpotent ideal of $R$ and $R/N(R)$ has no nil ideals. But could we have another nilpotent of ideal of $R$, say $J$ with $R/J$ having no nil ideals with $J \neq N(R)$?
To exclude the possibility of a $J$ such as you ask about, show two things:
a) Every nilpotent ideal $J$ of $R$ is contained in $N(R)$.
[So that leaves only the possibility $J \subsetneq N(R)$ if we assume $J \neq N(R)$.]
b) If $J \subsetneq N(R)$, then $R/J$ does contain nilpotent elements / a nil ideal.