Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am looking for a more elementary proof of the following result:

Theorem (Hungerford, 1968): Let $R$ be a principal ideal ring. Then $R \cong \prod_{i=1}^n R_i$, where each $R_i$ is a homomorphic image of a principal ideal domain (PID).

Hungerford's article is available free online at:

http://projecteuclid.org/euclid.pjm/1102986148

What do I mean by "more elementary"? Hungerford uses the Cohen structure theory of complete local rings, which I would like to avoid (because I have notes on commutative algebra which do not discuss such things).

Note that Hungerford's theorem is a refinement of a previous result of Zariski and Samuel, which asserts that a principal ideal ring is isomorphic to a finite direct product of rings, each of which is either a PID or a "special principal ideal ring", i.e., a local Artinian principal ideal ring. The proof of this result uses primary decomposition, which is acceptable to me (in fact I put a section on primary decomposition into my notes for exactly this application).

Given the theorem of Zariski-Samuel, Hungerford's result is plainly equivalent to the fact that every Artinian local principal ideal ring is the quotient of a PID. Now doesn't that sound like you should be able to prove it without invoking the structure theory of complete local rings?

share|improve this question
1  
In answer to your question at the end: no (not to me, at least). If not equi-char, how do you expect to make any link to PID's without using the dvr coefficient ring (which is the hardest part of the proof of the structure theorem, esp. for imperfect residue field)? [There's the separate matter of why principal ideal rings are of any interest beyond the PID case, but I won't get into that here.] –  BCnrd May 23 '10 at 14:15
1  
@BCnrd -- I take your point that the answer may well be "no". (After all, I don't know how to do it myself...) As to your final question: PIRs arise naturally in many contexts (e.g. when taking the quotient of a Dedekind domain by a not necessarily prime ideal), usually in a way that makes manifest the relationship to a PID. This is certainly a useful technique for studying PIRs, so it's nice to know it's always valid. Moreover, I am 99% sure that realizing an abstractly given PIR as a quotient of a PID comes up in some paper of Mazur and/or Rubin. –  Pete L. Clark May 23 '10 at 14:52
1  
@Pete: Please confirm if the 99% is 100% (and of course that their PIR is not obviously related to PID...otherwise there's no point). –  BCnrd May 23 '10 at 15:01
1  
@Victor: It was meant in the direction of "I have no idea why anyone would care about this concept in general", though Pete suggests a context where it might come up in a non-artificial way. I am curious to see what more he has to say on that. –  BCnrd May 23 '10 at 15:31
2  
Brian may have high standards of what is interesting and almost certainly has very high standards of what is obvious, but I, at least, have been grateful that Mazur and Rubin made this remark. After all, it means that Kolyvagin systems are useful to study the ETNC for any principal artinian ring, and this is not uninteresting if one is interested in interactions between the ETNC and deformation theory. –  Olivier May 27 '10 at 8:31

1 Answer 1

up vote 2 down vote accepted

Theorem 5.2 in http://www.emis.de/journals/BAG/vol.46/no.1/b46h1her.pdf gives an answer (take the projective limit. The paper has a related one with corrections, but not for the part that is related to your question). This is for a non-commutative case, and the theorem has a non-commutative extension: a PIR is a finite direct product of prime and artinian indecomposable cases, which are matrix rings over CPU rings (Faith, Algebra II should contain all the needed references)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.