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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 $N=I^0N$?

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Are you sure that you want $I^0$? It seems that $I^0=I$ and the statement is true... –  Bugs Bunny Oct 25 '10 at 10:30
@Tmobius: to clarify, your ring is not necessarily Noetherian, right? –  Hailong Dao Oct 25 '10 at 10:39
@TmobiusX: I asked the above because, if $R$ is local and Noetherian and $I$ is proper, then $N=0$, so it would be a little strange to state Krull's theorem this way. –  Hailong Dao Oct 25 '10 at 11:09
(cont) However, if $R$ is not Noetherian, then I am not sure this version of Krull is true. Do you have some reference for your second sentence? –  Hailong Dao Oct 25 '10 at 11:14
@Hailong Dao: Sorry, the ring I mentioned is commutative noetherian local ring. This version of Krull intersection theorem can be found in Theorem 2.5.7 of Enochs and Jenda's book "Relative Homological algebra". –  TmobiusX Oct 25 '10 at 12:25

1 Answer 1

up vote 5 down vote accepted

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.

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