Suppose we have a commutative local ring $R$ with unit. I'm curious about whether the following statements are correct:

1- every proper finite ideal is nilpotent.

2-every proper finitely generated ideal is nilpotent.

3- If maximal ideal be finite then it's nilpotetnt.

4- If maximal ideal be finitely generated then it's nilpotetnt.

  • 1
    $\begingroup$ Finite as in has only finitely many elements? Or finitely generated? If you can provide context where you found these questions, it would also be good to provide it for the best possible help. $\endgroup$ – rschwieb Mar 5 '14 at 20:50
  • $\begingroup$ So by adding these two new points, you're now asking for both versions? $\endgroup$ – rschwieb Mar 5 '14 at 20:59
  • $\begingroup$ @rschwieb , Finite means finitely many elements. 1 and 3 are my questions. $\endgroup$ – Stella Mar 5 '14 at 21:05
  • $\begingroup$ Any partial progress on this fountain of questions? :) $\endgroup$ – rschwieb Mar 5 '14 at 21:06
  • $\begingroup$ Dear Stella : Please don't jump to conclusions: I meant no disrespect. Do you want to eliminate the entire ring from being an ideal? Otherwise any finite ring is a counterexample to the first two. $\endgroup$ – rschwieb Mar 5 '14 at 21:14

For $1$:

Suppose $I$ is a proper finite ideal and $|I|=n$. Take $x\in I$, $x\neq 0$ and think about $\{x,x^2,\ldots,x^n\}\subseteq I$. Suppose none of these powers of $x$ in this set are zero: then $x^k=x^m\neq 0$ for some $m> k$, but then $(x^{m-k}-1)x^k=0$, but the left hand factor must be a unit. See the end?

For $2$ look at $(x)$ in $\Bbb R[[x]]$.

For $3$ look at $1$.

For $4$ look at $2$.

Update: Just so there is no mistake, "the end" is not "right away." Where I left off, one should be able to conclude that $x^n=0 $ for all $x\in I$, and then conclude that $I$ is nilpotent by a pigeonhole-principle argument.

  • 1
    $\begingroup$ Excellent: +1 . $\endgroup$ – Georges Elencwajg Mar 5 '14 at 21:30
  • $\begingroup$ @Stella Yes: that's how I concluded $1-x^{m-k}$ is a unit. $\endgroup$ – rschwieb Mar 5 '14 at 22:26
  • $\begingroup$ @rschwieb , I think NAKAYAMA lemma is useful. $\endgroup$ – Stella Mar 5 '14 at 22:32
  • $\begingroup$ @Stella Yes, the Nakayama lemma is extremely useful (and would trivialize this exercise, if we weren't trying to prove it by elements :) ) $\endgroup$ – rschwieb Mar 5 '14 at 23:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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