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February 5th, 2018, 12:52 PM   #1
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Artinian module

I want to prove the following result:

Let $M$ be an artinian $ \mathbb{Z}-$module. Show for every $x\in M$ there is some $n\ge 1$ such that $nx=0$

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February 5th, 2018, 01:26 PM   #2
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Suppose for a contradiction that there is an $x \in M$ such that $nx \neq 0$ for all $n \geq 1$, and consider the following descending chain of subgroups of $M$:

$\langle x \rangle \supseteq \langle 2x \rangle \supseteq \langle 4x \rangle \supseteq \langle 8x \rangle \supseteq \dots$

I'll leave it to you to prove that this chain doesn't terminate. To do this, you might like to show that every inclusion is strict.

Last edited by cjem; February 5th, 2018 at 01:35 PM.
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February 5th, 2018, 01:36 PM   #3
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That is exactly what I am blocked in. I started as you did, but I didn't get a contradiction.

Last edited by skipjack; February 5th, 2018 at 07:48 PM.
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February 5th, 2018, 01:46 PM   #4
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Well, suppose $\langle 2^m x \rangle = \langle 2^{m+1} x \rangle$ for some $m$. Then $2^m x \in \langle 2^{m+1} x \rangle$ so there is an integer $k$ such that $2^m x = k 2^{m+1} x$. Then we have $0 = (2^m - k 2^{m+1})x = 2^m(1 - 2k)x$. But since $nx \neq 0$ whenever $n$ is a non-zero integer, this implies $2^m (1-2k) = 0$ and so $2k = 1$. But this is impossible as $k$ is an integer.

This shows that the inclusions of the chain are strict, so it's non-terminating. This contradicts $M$ being Artinian.
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February 5th, 2018, 01:57 PM   #5
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thanks a lot cjem for your explanation!
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February 6th, 2018, 07:27 AM   #6
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You're welcome.
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