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.
I think there's an intuition for it coming from algebraic geometry, which I know nothing about rigorously, but it goes like this:
Varities in $\mathbb{C}$ or an algebraically closed field can be put into correspondence with radical ideals. This correspondence also reverses the subset order, meaning that the bigger an ideal is, the smaller its variety is. This is the content of Hilbert's Nullstellensatz theorem, I guess. Therefore, a dictionary between the language of geometry and commutative algebra has been given to us by Hilbert.
In this language, a prime ideal is a variety (because prime ideals are radical) that is indecomposable. A maximal ideal is like a point. Intersection of ideals is translated into union of their varieties. Therefore, I think your theorem tells you, in a sense, about the decomposition of a variety into a union of indecomposable sub-varieties.
A related question is this question on MSE which is about the case when $R$ is Noetherian and therefore, we have a restriction on the ascending chains of ideals in the ring, which in our language is translated as a descending restriction on the corresponding varieties. It says that every radical ideal of a Noetherian ring is the intersection of a finite number of prime ideals which is intuitively close to what we expect from this algebro-geometric language.
If $x$ is nilpotent, then clearly $x$ belongs to every prime ideal, because $x^n=0$ for some $n$. This result is just claiming the converse, which should at least seem plausible intuitively.
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.
It is immediate from the definitions that if $A$ is a commutative ring, and $f\in A$ is nilpotent, then $f$ belongs to every prime ideal of $A$. The converse is more difficult. We shall prove this by contrapositive. If $f$ is not nilpotent, then the set $S=\{1,f,f^2,\dots\}$ does not contain $0$, and hence the localisation $S^{-1}A$ is a nonzero ring. Since every nonzero ring has a prime (even maximal) ideal, there exists a prime ideal $\mathfrak a$ in $S^{-1}A$. The inverse image of $\mathfrak a$ under the map $A\to S^{-1}A$ is a prime ideal not containing $f$.
