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Let $A$ be a commutative ring. Suppose $P \subset A$ is a minimal prime ideal. Then it is a theorem that $P$ consists of zero-divisors.

This can be proved using localization, when $A$ is noetherian: $A_P$ is local artinian, so every element of $PA_P$ is nilpotent. Hence every element of $P$ is a zero-divisor. (As Matt E has observed, when $A$ is nonnoetherian, one can still use a similar argument: $PA_P$ is the only prime in $A_P$, hence is the radical of $A_P$ by elementary commutative algebra.)

Can this be proved without using localization?

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Why bother? It's not like you're going to get very far in commutative algebra without using localization... –  user126 Jul 22 '10 at 22:53
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I don't agree Harry, if you had said Alg Geom, then I'd agree, but comm alg is pretty big... But this is purely opinion of course, you likely weigh the type of ring theory using localization much higher due to your inclinations :D –  BBischof Jul 22 '10 at 22:59
    
Harry: A friend had asked me about this from his algebra course, where they hadn't yet covered localization. –  Akhil Mathew Jul 23 '10 at 0:06
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Because it's good to have more than one perspective on a problem! –  Noah Snyder Jul 23 '10 at 3:40

4 Answers 4

up vote 10 down vote accepted

Denote set complements in $\rm A $ by $\rm\,\bar T = A - T.\, $ Consider the monoid $\rm\,S\,$ generated by $\rm\,\bar P\,$ and $\rm\,\bar Z,\ $ for $\rm\,Z = $ all zero-divisors in $\rm A $ (including $0).\,$ Note $\rm\,0\not\in S\ $ (else $\rm\, 0 = a\,b,$ $\rm\ a\in \bar P,$ $\rm\ b\in \bar Z\ $ $\rm \Rightarrow b\in Z)$ so we can enlarge $\,0\,$ to an ideal $\rm\,Q\,$ maximally disjoint from $\rm\,S.\, $ Since $\rm\,S\,$ is a monoid, $\rm\,Q\,$ is prime. $\rm\, S\,\supset\, \bar P \cup \bar Z\ \Rightarrow\ Q \subset \bar S \subset P\cap Z,\, $ so by minimality of $\rm\,P\,$ we infer $\rm\, P = Q \subset Z.\quad$ QED

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Dear Bill, Very nice, thanks! –  Akhil Mathew Mar 18 '11 at 4:10
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I only discovered your question now via a search (I was not a member when it was asked). Almost surely this is in Kaplansky's Commutative Rings (probably in a more general form). –  Bill Dubuque Mar 18 '11 at 5:02
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In some sense, this is "secretly" a localization argument since you are manipulating multiplicatively closed subsets of $A$. –  Amitesh Datta Jun 1 '11 at 11:52
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@Amitesh True, but one could probably say essentially the same of almost any "elementary" argument. –  Bill Dubuque Jun 1 '11 at 12:48

This is a comment on the proof sketch in the question (I don't have sufficient rep. to actually leave comments): the localization $A_P$ is local of dimension zero (its unique maximal ideal is also a minimal prime ideal, and hence is the unique prime ideal of $A_P$), but need not be Artinian, as far as I can tell. E.g. if $A =\mathbb C[[x^{1/2},x^{1/3},\dots]]/(x)$, then $A$ is local with a unique prime ideal (namely the ideal generated by all the $x^{1/n}$), but is not Artinian (equivalently, not Noetherian), since if $(a_i)$ is any strictly descending sequence of rational numbers in the interval $(0,1)$, then the ideals $x^{a_i}A$ form a strictly descending sequence of ideals in $A$.

(Hopefully I'm not blundering here; if I am, someone please let me know!)

(Also, I should add that it is still the case that since $PA_P$ is the unique minimal prime of $A_P$, every element of $PA_P$ is nilpotent, and hence every element of $P$ is a zero divisor, so my comment is very nitpicky: it is just about the use of the terminology Artinian.)

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Right, what I said only works when $A$ is noetherian. (+1) –  Akhil Mathew Jul 28 '10 at 3:01
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Dear Akhil, No, I think you are okay: A_P has a unique maximal ideal, which is also a minimal prime, and hence has a unique minimal prime too (PA_P is simultaneously uniquely maximal and minimal), hence PA_P is the nilradical of A_P, and so what you wrote is okay (unless I am blundering too). –  Matt E Jul 28 '10 at 3:25
    
Yes, this makes sense. I'll add a reference to your comment. –  Akhil Mathew Jul 30 '10 at 21:42

In the early days of commutative algebra people did not use the language of localization, so probably such proof exists. In particular, I have been told Kaplansky's early book (around 1950?) might be such reference, but I don't have it right now.

In any case, Exercise 2.3 of Eisenbud tells you how to cheat and localize without admitting it. I might be wrong here, but what you want to prove is about some multiplicative properties of elements in $R$, so perhaps any proof you can find is just localization in disguise.

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If A has an identity then what about this argument:

Take p in P and look at pP. This is an ideal of A and a subset of P so we must have that pP = P by minimality. Thus pa = p for some a in P. a cannot be one for otherwise P = A is not a prime ideal. Thus p(a-1) = 0 so that p is a zero divisor.

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pP isn't necessary prime though, right? –  Akhil Mathew Jul 22 '10 at 22:04
    
@Akhil: pP=P as it is an ideal and a subset of P and P is a minimal ideal. Since P is prime, pP is too –  Casebash Jul 22 '10 at 22:15
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P is a minimal prime, not a minimal ideal... –  user126 Jul 22 '10 at 22:51
    
Ah right...I had in my mind that P was a minimal and prime ideal, not a minimal prime ideal. –  Eric O. Korman Jul 23 '10 at 0:06

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