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Is there any characterization for the left primitive rings which are left duo ? Clearly division rings satisfy this property.

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up vote 2 down vote accepted

The class of rings in the question is exactly the class of Division rings. Here is another proof based on the characterization of primitive rings.

Proof. Since you have asked about primitive rings I can assume that you are familiar with the Jacobson's Density Theorem. Therefore $R$ can be regarded as a dense subring of linear transformations of a vector space $V$ over a division ring $D$. It is enough to show that $\mbox{dim}_D V = 1$ (since then $R$ would be equal to $D$) . Suppose not and consider two linearly independent vectors $u,v \in V$. By density there are $f,h \in R$ such that $f(u) = u , f(v) = 0$ and $h(u) = u , h(v) = u$. Now Take $I$ to be the left ideal generated by $f$ in $R$. We show that $fh \not \in I$ (which is a contradiction since by assumptions any left ideal of $R$ is two-sided):

If $fh \in I$ then there exists $g \in R$ such that $fh = gf$. Hence $u = fh(v) = gf(v) = g(0) = 0$ a contradiction !

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@Chatish: Thanks. –  user30818 Feb 18 '13 at 18:25
    
I accept this solution since the argument works even if $R$ assumed to be a right duo ring. –  user30818 Feb 19 '13 at 8:47
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Unfortunately, I'm afraid these rings aren't very interesting: a left primitive ring $R$ is left duo if and only if it is a division ring.

For the nontrivial direction, suppose $R$ is left duo. Fix a faithful simple left $R$-module, which up to isomorphism is of the form $R/M$ for a maximal left ideal $M$. As $R$ is left duo, $M$ is in fact a two-sided ideal. But for $R/M$ to be simple, this means that $M = 0$. Thus $M = 0$ is a maximal (left) ideal, and $R$ is a division ring.

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@Manny Reyes: Thanks for your celever solution. But I accepted chatish's solution since his argument works even if we assume that $R$ is a right duo ring (instead of left duo ring). –  user30818 Feb 19 '13 at 8:50
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