most recent 30 from http://mathoverflow.net 2013-05-21T22:03:22Z http://mathoverflow.net/feeds/question/43499 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43499/a-question-arising-from-the-krull-intersection-theorem TmobiusX 2010-10-25T09:11:41Z 2010-10-25T11:37:41Z

<p>Let R be a local ring, I an ideal, M a finitely generated module and $N=\cap _nI^nM$. Then the Krull intersection theorem states that $N=IN$. Now if R is a local ring of characteristic $p>0$, for each $e\geq 0$ let $I^e$ denote the ideal generated by the $p^e$-th power of the elements of $I$. let $N=\cap _eI^eM$. Is it ture that $N=I^0N$?</p>

http://mathoverflow.net/questions/43499/a-question-arising-from-the-krull-intersection-theorem/43503#43503 Qing Liu 2010-10-25T10:34:46Z 2010-10-25T10:34:46Z

<p>First Krull's theorem is for Noetherian (not necessarily local) rings. Let $n\ge 1$. If $I$ is generated by $r$ elements $x_1, \dots, x_r$, then the usual $n$-power $I^n$ of $I$ is contained in your $I^e$ if $n/r \ge p^e$ (any element of $I^n$ is a combination of $x_1^{a_1}...x_r^{a_r}$ with $a_1+\dots + a_r=n$, so $\max_i{a_r} \ge n/r$). Therefore your $N$ is equal to the usual $\cap_n I^nM$ and the answer to your question is yes. </p>