Assuming a commutative ring, the nilradical of that ring is the intersection of all prime ideals in it. I've read somewhere that the nilradical is equal to the $(0)$ ideal.
So my question is, doesn't that mean that is always the zero ring? Or there is any example where that doesn't happen? I think it's quite silly to have a fancy name like that for the zero ring.
Thank you.
The nilradical consists of all nilpotent elements of the ring, that is, those elements $a$ such that there exists $n>0$ with $a^n=0$.
The proof can be found in every textbook on commutative algebra. One part is essentially obvious: suppose $a$ is nilpotent and let $P$ be a prime ideal. Then $a^n=0$ for some $n>0$, so $0=a^n=aa^{n-1}$, which implies that either $a\in P$ or $a^{n-1}\in P$; continuing by induction, we conclude that $a\in P$ anyway.
The other direction is a bit more complicated (and more interesting). It consists in showing that if $a$ is not nilpotent, then there is a prime ideal $P$ such that $a\notin P$.
For instance, in the ring $\mathbb{Z}/24\mathbb{Z}$ there are nilpotent elements, namely $6$, $12$ and $18$, besides $0$. The nilradical is indeed $\{0,6,12,18\}$ (a nilpotent element must have both $2$ and $3$ in its factorization, when looked at as an integer).