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In Atiyah-Macdonald, in the proof of Theorem 10.17 (Krull's intersection theorem), the authors go through a 4-line chain of arguments to show that that the kernel $E=\bigcap_{n=1}^{\infty}\mathfrak a^nM$ of $M\to\hat M$ satisfies $\mathfrak aE=E$. When checking some other sources (for example Prof. May's notes), I see that the approach is similar: use some form of Artin-Rees lemma to conclude $\mathfrak aE=E$.

Question: Doesn't it follow trivially from the equality $E=\bigcap_{n\ge1}\mathfrak a^nM$ that $\mathfrak aE=\mathfrak a\left(\bigcap_{n\ge1}\mathfrak a^nM\right)=\bigcap_{n\ge2}\mathfrak a^nM=E$? (basically the first term $\mathfrak aM$ of the intersection can be ignored, as it contains every $\mathfrak a^nM$ anyway). Am I missing something trivial here?

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    $\begingroup$ yes. you are right. but how do you get the second equality without artin ress lemma? $\endgroup$
    – user119882
    Apr 20, 2014 at 2:35
  • $\begingroup$ thanks @user119882. You are absolutely right. It doesn't follow directly. $\endgroup$
    – aytio
    Apr 20, 2014 at 3:02

1 Answer 1

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The equality $\mathfrak a\left(\bigcap_{n\ge1}\mathfrak a^nM\right)=\bigcap_{n\ge2}\mathfrak a^nM$ assumes that multiplication by ideals and intersection of submodules commute. This is unfortunately not the case.

In general we only have $\mathfrak{a} \left(\bigcap_{n\ge1} M_n\right) \subset \left(\bigcap_{n\ge1}\mathfrak{a} M_n\right)$, but not equality.

The Artin-Rees lemma tells you that you have equality in the special case where $M_n = \mathfrak a^nM$.

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  • $\begingroup$ Thanks, you are very right. In fact, when I read the proof of Artin-Rees lemma from the book, I wondered why the statement $\mathfrak a(M'\cap M_n)\subseteq \mathfrak aM'\cap\mathfrak aM_n$ is not an equality and I even came up with a counterexample then, but by the time I got to the proof of Krull's Intersection Theorem, it completely slipped my mind until I saw user119882's comment and your answer. :) $\endgroup$
    – aytio
    Apr 20, 2014 at 3:00

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