Let $A$ be a commutative ring with unity. Let the radical $\operatorname{Rad}(A)$ of $A$ be the ideal consisting of all nilpotent elements of $A$.
Is $\operatorname{Rad}(A)$ of $A$ the same as the intersection of all maximal ideals of $A$?
I think this should be a standard result so I'd grateful if someone could point me to a reference or a hint on how to prove it will be even better!
Thank you!
No. The intersection of maximal ideals is the Jacobson radical (denoted by $J(R)$), and this can differ from the nilradical $N(R)$. For instance, if $R$ is a local integral domain (e.g. $R=K[[X]]$) with maximal ideal $m$, then $N(R)=0$ and $J(R)=m$.
The ideal of all nilpotent elements is the nilradical of A. It is the intersection of all prime ideals of $A$.
The (Jacobson) radical of $A$ is the intersection of all maximal ideals. It is also the set of elements $x$ such that $1-\lambda x$ is a unit for any $\lambda\in A$.
