TITLE: What is the intuition behind saying that the nilradical is the intersection of prime ideals?
QUESTION [5 upvotes]: Let $R$ be a commutative ring with unity. Let $N$ be it's nilradical, that is the ideal consisting of all the nil potent elements of $R$. We know that $N$ is the intersection of all prime ideals of $R$. The proof, although understandable, to me lacks intuition. I was wondering if anyone could tell me why this is true in a more intuitive way?
Thank you.

REPLY [2 votes]: Recall the correspondence that $I$ is a prime ideal if and only if $R/I$ is a domain — a ring without zero divisors.

Briefly...
If $R$ is already a domain, then for any nonzero element $r$ there exists a maximal ideal not containing $r$.
We can think of the problem as trying to "fix" a zero divisor by taking a quotient in which that element is no longer a zero divisor.
So, if $r$ is a zero divisor in a ring $R$, the fix is to look at equations where $rs = 0$ and set $s=0$ (i.e. mod out by an ideal generated by the congruence). Eventually, you get to a quotient ring where you have either that $r$ is not a zero divisor or that $r$ is zero.
The obvious way in which $r$ would be forced to be zero after this procedure is when $r$ is nilpotent. The theorem is that this turns out to be the only exception.