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Clearly if $A$ is a commutative domain then the torsion elements of an $A$-module form a submodule.

I'm having trouble finding an example of an $A$-module such that $A$ is a noncommutative domain and the torsion elements of the module do not form a submodule.

Is anyone aware of a good example of this?

Thanks in advance

AnotherPerson
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1 Answers1

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Let $A=\mathbb{C}\langle x,y\rangle$ be the free algebra on two (non-commuting) generators. Then $A$ is a domain.

Let $M=A/xA\oplus A/yA$ and $m=(1+xA,0)\in M$, $m'=(0,1+yA)\in M$.

Then $m$ and $m'$ are torsion elements, since $mx=0$ and $m'y=0$, but $m+m'$ is not torsion. So the torsion elements are not closed under addition, and so do not form a submodule.

The important feature here is that $A$ is a domain with two non-zero right ideals $I=xA$ and $J=yA$ with $I\cap J=0$. Basically the same construction will work for any such domain.

Jeremy Rickard
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  • I thought for sure we could use that property of nonOre domains, but didn't get there before you. Glad to see that works :) – rschwieb Sep 24 '16 at 18:41