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Let $R$ be a ring such that $R$ is a simple $R$-module. Show that $R$ is a division ring.

I have an idea for this but I would like to make sure it is correct. My idea is that $R$-submodules of $R$ are just the same as ideals in $R$. So if we take any non-zero element $r$ in $R$, then the ideal generated by $r$ must be the whole of $R$ (by simplicity) and so $r$ must be a unit and $R$ is a division ring.

Thanks!

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    If by $R$-module you mean left $R$-module, then a submodule is the same as a _left_ ideal. If the _left_ ideal generated by $r$ is the whole of $R$, then $r$ necessarily has a _left_ inverse. It does not follow directly from this that $r$ is a unit since it may not have a _right_ inverse, so you need to do slightly more work than this.2012-05-15
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    @QiaochuYuan, are there rings with left-inverses for all elements but a missing right inverse, or equivalently with a left-inverse which does not commute with its right inverse?2012-05-15
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    @plm: no. But that requires an extra (not very long) argument.2012-05-16
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    Does your ring have an identity?2012-05-16
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    @ArturoMagidin I guess asking to prove R is a division ring implies there is an invertible element, so an identity.2012-05-16
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    @QiaochuYuan, Thank you. I think the question assumed that both left and right proper submodules of $R$ over itself are trivial. In this case Thomas' answer works: inverses exist on both sides and they are equal because $yx=xz=1\Rightarrow y=y\cdot 1=yxz=1\cdot z=z$. But it is true that only assuming simplicity on a single side to deduce the same is much more interesting.2012-05-16
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    @plm: The question is whether the ring is *assumed* to have an identity to begin with, or if that is one of the things that need to be proven *en passant*.2012-05-16
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    @ArturoMagidin, I didn't write clearly what I meant, sorry. I meant I thought the question _assumed_ the ring had an identity, for how it was phrased. Thinking about it, $2(\mathbb Z/4)$ (the "universal" ring without identity and with only 2 elements, a ring to remember :) is an example of a ring (commutative so left and right) simple over itself which does not have an identity, so the question certainly assumed $R$ has an identity.2012-05-17

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