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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. –  Qiaochu Yuan May 15 '12 at 22:30
    
@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? –  plm May 15 '12 at 23:33
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@plm: no. But that requires an extra (not very long) argument. –  Qiaochu Yuan May 16 '12 at 0:45
    
Does your ring have an identity? –  Arturo Magidin May 16 '12 at 2:06
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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. –  plm May 16 '12 at 9:08

3 Answers 3

Well, it's been a month, so here is the argument I had in mind. By assumption $R$ is simple as a left module over itself, hence every nonzero element $r \in R$ has a left inverse $s$, thus $sr = 1$. But $s$ also has a left inverse $t$. Hence $s$ has a left and right inverse, which must agree, and hence $r$ does too.

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Your approach is correct. An ideal is in particular a sub-module. So for a non-zero element $r$, the ideal generated by $r$ is all of $R$, so $r$ has an inverse. Now you do this for the left and from the right and then you a left and a right inverse. They are equal.

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If $R$ is simple as a left $R$-module this doesn't immediately guarantee that $R$ is simple as a right $R$-module. (It does, but proving this seems to me about as hard as proving the desired proposition.) –  Qiaochu Yuan May 16 '12 at 0:44

Perhaps you can argue as follows: by Schur's lemma, $End_R(R)\,$ is a division ring as $R$ is a simple $R-$module.

Now, let us define $\,\phi:End_R(R)\to R\,$ by $\,\phi(f):=f(1)\,$ . Show now

1) $\phi\,$ is an $R-\,$module (left, right: as you wish) homomorphism

2) $\phi\,$ is bijective -- for the onto part you may want to check that $\,f_r: R\to R\,$ defined by $\,f_r(x):= xr\,$ is a (left; if you want right fix the definitions) $\,R\,$-endomorphism of $\,R\,$ . Injectivity is trivial -- .

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If $R$ is regarded as a left module over itself, then $\text{End}_R(R)$ ought to be canonically isomorphic to $R^{op}$ rather than $R$. (This is one reason to use right modules instead.) It also seems to me that $\text{End}_R(R)$ is not naturally an $R$-module. –  Qiaochu Yuan May 16 '12 at 2:50
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Qiaochu: the "if you want right fix the definitions" part should take care of that, I think: instead of $\,f(1)\,$ we could write $\,(1)f\,$, with the function's action to the right, "as algebraists that do not surrender to analists ought to do", something I heard several times from my late Prof. Shimshon Amitzur. About your second remark: $\,(rf)s:= rf(s)\,\,,\,or\,\,(s)fr:= (sf)r\,$ makes, if I'm not wrong, $\,End_R(R)\,$ into a left (right) R-module. –  DonAntonio May 16 '12 at 11:47

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