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I'm trying to prove that $\bigcap\limits_{P\in \mathcal{P}}P\subseteq Nil(R)$, where $R$ is a commutative unital ring and $\mathcal{P}$ is a collection of prime ideals of $R$.

Suppose $a\not\in Nil(R)$ and let $\mathcal{S}=\{I: I \text{ is an ideal of $R$, such that } I\ne R, a^n\not\in I \text{ for all $n>0$} \}$. Then $\mathcal{S}$ is non-empty because $0\in Nil(R)$, so $0\in I$ for some $I\in \mathcal{S}$, thus $\langle 0\rangle\in \mathcal{S}$. Also, by Zorn's Lemma, since $R$ is an upper bound for any chain in the poset $(\mathcal{S}, \subseteq)$, there exists a maximal set $M\in \mathcal{S}$ (which, I think, must be an ideal, since $\mathcal{S}$ is a collection of ideals).

Now the part I don't get. I need to prove in this part that $x,y\not\in M$ implies that $\exists n,m>0$ such that $a^n\in M\cup\{x\}$ and $a^m\in M\cup\{y\}$. We know that, since $x,y\not\in M$ and $\mathcal{S}$ contains only sets with nilpotent elements, and $M$ is maximal (and it is related to all other sets in $\mathcal{S}$ by inclusion), then $a^n$ must be equal to $x$ and $a^m$ must be equal to $y$, and that $x,y$ must be non-nilpotent. But here's the part I don't understand:

  • How can it possibly be the case that for two arbitrary $x$ and $y$, there must necessarily exist some element $a$ such that $a^n=x$ and $a^m=y$? What is the mystery here?
sequence
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  • @arctictern I'm given that if $R$ is commutative and unital then the collection of all prime ideals of $R$ contains $Nil(R)$. Now I need to prove the inclusion in the other direction. I'm also given that $\mathcal{S}$ is defined as I wrote (with $a\not\in Nil(R)$), so that is not my supposition, this is the definition I'm given. – sequence Mar 18 '17 at 23:36
  • Why is $R$ not an upper bound? It is definitely not in $\mathcal{S}$, but why can't it be an upper bound? For example, in analysis, the supremum of a set is non necessarily greater than some upper bound of the set. So I actually do need to show that $\mathcal{S}$ is not empty because it contains the zero ideal. – sequence Mar 18 '17 at 23:38
  • @arctictern We know that if $R$ is an upper bound (although not in $\mathcal{S}$), then there must exist a set $T\subset R$ such that $T\in \mathcal{S}$. How can otherwise be possible? – sequence Mar 18 '17 at 23:59
  • @arctictern Suppose a set $E$ is bounded, thus it has an upper bound $t$ and a supremum. Suppose $t\not\in E$, so that $t\ge \sup E$. There always exists an element $s\in E$ such that $s<\sup{E}$. So that $E$ is bounded above by $\sup E$. If $E$ is open then we can take $s$ arbitrarily close to $\sup E$. – sequence Mar 19 '17 at 00:04
  • I'm saying that as long as we know that an upper bound of $E$ exists, even if it is not in $E$, then we can conclude that $E$ is bounded above by $\sup{E}$. Analogously, $R$ is an upper bound of $\mathcal{S}$. – sequence Mar 19 '17 at 00:08
  • @arctictern Can you please give an analogy in analysis of what Zorn's lemma means? – sequence Mar 19 '17 at 00:11
  • I'm reading this in Wikipedia:

    "Lemma — Suppose a partially ordered set P has the property that every chain has an upper bound in P. Then the set P contains at least one maximal element."

    This must imply that every chain in $P$ is finite.

     – sequence Mar 19 '17 at 00:15
  • At least if every element in $P$ has countably many elements in it and every chain has an upper bound in $P$, then, even if a chain is infinite, it must approach this upper bound set. – sequence Mar 19 '17 at 00:21
  • Well, because if you consider an inclusion chain of sets in the collection of sets $\mathcal{S}$ then you will see that if this chain has a maximum set, let's call it $J$, then this set contains all the sets in the chain. Thus $J$ is the last set in the chain. Hence, this chain must have an end, even if between the start and the end of the chain there are infinitely many sets. Also, $\mathbb{N}$ is countable, and $(\mathbb{N}, \subseteq)$ is a poset. – sequence Mar 19 '17 at 01:06
  • Why are you considering $M\cup{x}$ and $M\cup{y}$ instead of $M+\langle x\rangle$ and $M+\langle y\rangle$? – Xam Mar 19 '17 at 01:22
  • @Xam Because this is what I'm explicitly given in my problem - to prove such and such given this and this. – sequence Mar 19 '17 at 01:25
  • That makes little sense because you've defined $S$ as a set of certain proper ideals of $R$, but since $x,y\notin M$ neither $M\cup {x}$ nor $M\cup {y}$ are ideals of that ring. – Xam Mar 19 '17 at 01:43
  • @Xam Correct, I don't think they are ideals, but do they have to be?

    After egreg's pointing out that $a$ is just a single element, not "every element", I could prove the theorem almost easily, except that I couldn't prove that $\exists n,m >0$ such that $x=a^n$, $y=a^m$ if $x,y\not\in M$. In fact, I argued in that part that if $x\not\in M$ then $x\in R\backslash M$, but consider $p\in R\backslash M$ such that $p\ne a$ is prime or irreducible, and let $x:=ap$, then $x\ne a^n$. Maybe I missed some detail, but that's the way I saw it then.

     – sequence Mar 20 '17 at 03:43
  • @arctictern Yes, I did prove that yesterday. But I don't think that the question I was asking you had anything to do with tautology. I was trying to clarify a concept, but you weren't quite in the mood to explain. That's fine with me, I can always seek help elsewhere. Thanks anyway. – sequence Mar 20 '17 at 03:47
  • @sequence they have to because the idea is to show that neither $M+\langle x\rangle$ nor $M+\langle y\rangle$ belong to $S$ so you can get your contradiction following the idea given in egreg's answer. – Xam Mar 20 '17 at 03:48
  • @sequence how do you know such $p$ prime or irreducible exists? – Xam Mar 20 '17 at 03:50
  • @Xam We don't know, but what if it does exist? For the coset $M+\langle x \rangle$, I'm not sure, because the proof in my text asks to prove that if $x\not\in M$ then $\exists n,m>0$ such that $x\in M\cup {x}$. So I was concentrating on that, not on the cosets. – sequence Mar 20 '17 at 05:19

1 Answers1

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You can prove $\bigcap_{P\in\mathcal{P}}P\subseteq\operatorname{Nil}(R)$ if $\mathcal{P}$ is the collection of all prime ideals of $R$.

The union of a chain of ideals is an ideal: given two elements in the union, they belong to one and the same element of the chain, so their sum too; the other condition is easy. Thus Zorn’s lemma applies to the set $\mathcal{S}$ consisting of the ideals that contain no power of $a$, where $a\notin\operatorname{Nil}(R)$. This set is not empty, because $(0)\in\mathcal{S}$.

It's clear that the union of a chain in $\mathcal{S}$ is an element of $\mathcal{S}$, so the family contains a maximal element $M$.

We want to show that $M$ is prime, that is,

for all $x,y\in R$, if $x\notin M$ and $y\notin M$, then $xy\notin M$.

Suppose $x\notin M$ and $y\notin M$.

Note: we have to consider $M+(x)$ and $M+(y)$, not $M\cup\{x\}$ and $M\cup\{y\}$ (which are not ideals).

Since $x\notin M$, we have that $M+(x)$ properly contains $M$. By maximality of $M$ in $\mathcal{S}$, we conclude $M+(x)\notin\mathcal{S}$, so $M+(x)$ must contain $a^m$, for some $m$. Similarly, $a^n\in M+(y)$ for some $n$. Thus $a^m=r+xs$ and $a^n=t+yu$, for $r,t\in M$ and $s,u\in R$. Consequently $$ a^{m+n}=rt+xst+yru+xysu $$ If $xy\in M$, then we conclude that $a^{m+n}\in M$, because $rt,xst,yru\in M$. Contradiction. Therefore $xy\notin M$.

A simpler approach. We want to show that if $a$ is not nilpotent, then there exists a prime ideal that contains no power of $a$. Consider the multiplicative set $S=\{a^n:n\in\mathbb{N}\}$. Then the ring $S^{-1}R$ is nonzero, so it contains a maximal ideal $M$. The inverse image of $M$ under the canonical ring homomorphism $R\to S^{-1}R$ is the required prime ideal.

egreg
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  • Thanks. Can you please clarify?

    1) Does $\mathcal{S}$ contain the same elements as $(\mathcal{S}, \subseteq)$? I don't see why they should be equal.

    2) How is a union of a chain of ideals defined? Why can't we use $R$ as an upper bound? In analysis, an upper bound of a set is not necessarily in a set. Why should this be different in algebra? Both branches of mathematics use the same set theory. 3) I'm actually explicitly asked to prove that $\exists n,m$ such that $a^n\in M\cup {x}$ and $a^m\in M\cup {y}$.

     – sequence Mar 18 '17 at 23:51
  • 4) You say that by maximality of $M$, the ideal $M+(x)$ must contain $a^m$. I don't see why this is so.

    5) Can we just say that $xy=a^{n+m}$, but since $a^s\not \in I$ for any $s$, where $I\in \mathcal{S}$, $a^{n+m}\not\in M$.

    Thanks again.

     – sequence Mar 18 '17 at 23:52
  • @sequence $M$ is maximal in the set of ideals that contain no power of $a$; since $M+(x)$ properly contains $M$, it cannot be in that set of ideals, so it must contain a power of $a$. No, we cannot say $xy=a^{n+m}$. – egreg Mar 19 '17 at 00:14
  • If $x=a^n$, $y=a^m$, $R$ is commutative, then $xy=a^n a^m = a^{n+m}$, correct? @egreg – sequence Mar 19 '17 at 00:25
  • @sequence Yes, of course, but from the data in the argument you *cannot* assume $x$ and $y$ to be powers of $a$. – egreg Mar 19 '17 at 00:28
  • I see, your approach is to first prove that $xy\not \in M$ and then to conclude that neither $x$ nor $y$ are in $M$. But for this we also need to know that $M$ is prime. In my problem, however, the sequence of steps is different. – sequence Mar 19 '17 at 00:30
  • @sequence Not at all. I assume $x,y\notin M$ and conclude that $xy\notin M$. Therefore $M$ is prime. What you say about “$\mathcal{S}$ only contains sets with nilpotent elements” is wrong. – egreg Mar 19 '17 at 00:32
  • But the definition of $\mathcal{S}$ says that if $a\not\in Nil(R)$ then $a^n\not\in I\in \mathcal{S} $ for any natural number $n$. Doesn't this obviously mean that only nilpotent elements are in $I$? @egreg – sequence Mar 19 '17 at 00:49
  • @sequence No, why? The element $a$ is a particular nonnilpotent element. – egreg Mar 19 '17 at 00:50
  • Do you mean that $a$ is just some single element and not all elements that are not in $Nil(R)$? Then this would be really confusing for a definition, very ambiguous. @egreg – sequence Mar 19 '17 at 00:52
  • @sequence That's exactly what the definition says. There is no "for all $a\notin\operatorname{Nil}(R)$" – egreg Mar 19 '17 at 09:21
  • I'd have hoped for the statement to be defined a little bit less ambiguously. After your pointing out that it was just for a single $a$, I was able to almost easily prove the theorem. Except that I couldn't prove that $\exists n,m >0$ such that $x=a^n$, $y=a^m$ if $x,y\not\in M$. In fact, I argued in that part that if $x\not\in M$ then $x\in R\backslash M$, but consider $p\in R\backslash M$ such that $p\ne a$ is prime or irreducible, and let $x:=ap$, then $x\ne a^n$. Maybe I missed some detail, but that's the way I saw it then. @egreg – sequence Mar 20 '17 at 03:39
  • @sequence You were not able to prove it because it is generally false. If $a=1$, and $x\notin M$, would you try and prove $x=1^m$? – egreg Mar 20 '17 at 07:34
  • If $x=1^m$, then this satisfies that $\exists m$ such that $x=1^m$. @egreg – sequence Mar 23 '17 at 07:36
  • @sequence “if unicorns exist, then unicorn exist” is a true statement, but it doesn't prove the existence of unicorns. – egreg Mar 23 '17 at 07:52
  • What I mean is that we know that $1\not\in M$, so if $x=1$, then the case would be trivial. @egreg – sequence Mar 24 '17 at 00:39