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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.

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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.

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  • $\begingroup$ I would have attempted an answer like this, but probably with even less chance of success, owing to my relative inexperience with algebraic geometry. But I'd like to help you add some ideas I had, since we are thinking along the same lines. One of my intuitions is that the elements of the nilradical are "invisible" to investigation via localizations and interpretations using the Zariski topology. That is, the spectra of $R$ and $R/nil(R)$ are homeomorphic. A similar story can be told about the Jacobson radical and MaxSpec. $\endgroup$ – rschwieb Dec 8 '17 at 14:41
  • $\begingroup$ Actually, I had the same idea about the Jacobson radical and MaxSpec, but I don't understand why the elements of the nilradical are "invisible" to investigation via localizations. However, I must say that my knowledge of algebraic geometry is close to $0$, and therefore, you're advised to write your own answer because probably it will be a better and more informative answer than mine. $\endgroup$ – stressed out Dec 8 '17 at 14:45
  • $\begingroup$ Problem is I can't say anything with confidence, really. Maybe a better analogy is that elements of the jacobson radical act trivially on simple modules, and elements of the nilradical act trivially on the modules of the form $R/P$. I've seen cases where properties of $R/P$'s affect those of $R_P$'s, but I don't understand the relationships. $\endgroup$ – rschwieb Dec 8 '17 at 15:46
  • $\begingroup$ @rschwieb: I'm having a hard time trying to imagine that. What does acting trivially imply? Picturing maximal ideals as points of a variety, the elements of Jacobson radical leave them fixed. Is that what you mean? $\endgroup$ – stressed out Dec 8 '17 at 18:01
  • $\begingroup$ By x in R acting trivially on M, I mean $xM=\{0\}$. In a sense, such elements can be thrown away, and you get an action that is “less degenerate “ Non nilpotent things act nontrivially on at least one prime ideal. $\endgroup$ – rschwieb Dec 8 '17 at 20:09
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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.

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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.

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