February 5th, 2018, 12:52 PM  #1 
Senior Member Joined: Jan 2015 From: usa Posts: 101 Thanks: 0  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 $nx=0$ Thanks 
February 5th, 2018, 01:26 PM  #2 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 226 Thanks: 75 Math Focus: Algebraic Number Theory, Arithmetic Geometry 
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. 
February 5th, 2018, 01:36 PM  #3 
Senior Member Joined: Jan 2015 From: usa Posts: 101 Thanks: 0 
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. 
February 5th, 2018, 01:46 PM  #4 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 226 Thanks: 75 Math Focus: Algebraic Number Theory, Arithmetic Geometry 
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 nonzero integer, this implies $2^m (12k) = 0$ 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 nonterminating. This contradicts $M$ being Artinian. 
February 5th, 2018, 01:57 PM  #5 
Senior Member Joined: Jan 2015 From: usa Posts: 101 Thanks: 0 
thanks a lot cjem for your explanation!

February 6th, 2018, 07:27 AM  #6 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 226 Thanks: 75 Math Focus: Algebraic Number Theory, Arithmetic Geometry 
You're welcome.


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