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Suppose that $(A,m)$ is a Noetherian local ring, $M$ is an $A$-finite module. Assume that $x_1, ..., x_n$ are elements in $m$. Is the following equality true:

$$ \mbox{ann}(M/(x_1, ..., x_n)M) = (x_1, ..., x_n) + \mbox{ann}(M). $$

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There are easy counterexamples. –  Martin Brandenburg Jun 26 '11 at 8:59

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up vote 4 down vote accepted

By modding out ${ann} (M)$ one can assume that $ann(M)=0$. Then the following is true:

$$I \subseteq ann(M/IM) \subseteq \bar I $$

Here $\bar I$ denotes the integral closure of $I$. You can prove it using the determinantal trick (the one used in the proof of Nakayama's Lemma). In particular equality happens if $I$ is integrally closed.

Now, a simple counter-example for the equality you wrote is $R=k[[x,y]]/(x^2-y^3)$ and $M$ be the ideal $(x,y)$. Then you can check that $y^2M \subseteq xM$, thus $y^2 \in ann(M/xM)$. Note that $y^4-yx^2=0$, so $y^2 \in \overline{(x)}$.

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Thanks guys! I've figured it out myself too. I guess one can make it a little bit sharper, replacing the integral closure by the radical as follows. $$ (x_1, ..., x_n) + \mbox{ann}(M) \subset \mbox{ann}(M/(x_1, ..., x_n)M) \subset \sqrt{(x_1, ..., x_n) + \mbox{ann}(M)}. $$ –  mr.bigproblem Jun 27 '11 at 4:37
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The radical of $I$ is generally substantially larger than the integral closure of $I$. In a normal ring for example, every principal ideal is integrally closed, but not every principal ideal is radical certainly. –  Karl Schwede Jun 27 '11 at 6:51
    
That's right, my bad! I messed up with the definition of the integral closure. Thanks! –  mr.bigproblem Jun 27 '11 at 14:58

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