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Suppose that $R$ is a ring, and suppose that $\lambda_R(_RR)<\infty$ and $\lambda_R(R_R)<\infty$ (where $\lambda_R$ is the length of an $R$-module). Is it true that then $\lambda_R(_RR)=\lambda_R(R_R)$? If $R$ is semisimple I know that it is true, but in general? Do you know a proof or a counterexample?

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There is a counterexample.

Take a field endomorphism $\sigma:\mathbb{F}\rightarrow \mathbb{F}$ such that $[\mathbb{F}:\sigma(\mathbb{F})]=n>1$. Form the skew polynomial ring $\mathbb{F}[x;\sigma]$. This just means you are using a noncommutative polynomial arithmetic, where $xa=\sigma(a)x$ for all $a\in \mathbb{F}$. Let $R=\mathbb{F}[x;\sigma]/(x^2)$. The image of $x$ in the quotient will be denoted by $\overline{x}$.

I invite you to verify that $Rx=\mathbb{F}\overline{x}$ is a minimal and maximal left ideal of $R$, and that it is also is a maximal right ideal and as a right $R$ module it is semisimple with composition length $n$. (Corrections made to this paragraph.)

After you have shown this, you will have seen that the composition length of $_R R$ is 2, but the composition length of $R_R$ is $n+1>2$.

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I'm making some mistakes somewhere. Define $I=\{a\bar{x}:a\in\mathbb{F}\}$ and $I^\prime=\{\bar{x}a:a\in\mathbb{F}\}$. Then $I$ and $I^\prime$ are maximal because $R/I\cong R/I^\prime\cong\mathbb{F}$, right? Now, take $0\neq J\subset I$, a left ideal, then $0\neq a\bar{x}\in J$, and so $a^{-1}a\bar{x}=\bar{x}\in J$ and so $J=I$ and $I$ is minimal, the problem is that it seems to me that for the same reason $I^\prime$ is minimal, where is my mistake? –  Alex M May 17 '12 at 19:19
    
No mistake yet, but we need to look a little more closely at what happens on the right. If $a\overline{x}\in J$, then it could be that $a^{-1}\notin Im(\sigma)$, so you cannot pull the same trick to show $J$ is a simple right module. That means it might not be possible to "pass $a^{-1}$ across $x$" to invert $a$. To get you going: let $\{\sigma(a_i)\mid i=1\dots n\}$ be a $\sigma(F)$ basis of $F$. Show $\sigma(a_i)\overline{x}F$ are simple right $R$ modules for every $i$. –  rschwieb May 17 '12 at 21:26
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I'm sorry but I'm not understanding yet, let $I^\prime$ be as above, and take $0\neq J^\prime\subset I^\prime$ right ideal, then there is $0\neq \bar{x}a\in J^\prime$, and so $\bar{x}=\bar{x}aa^{-1}\in J^\prime$ and so $J^\prime=I^\prime$, where is the mistake here? –  Alex M May 17 '12 at 21:48
    
My apologies, I accidentally overcomplicated my example. The ideal $F\overline{x}$ (and never $\overline{x}F$) is all we need. You have already shown it is simple as a left module, but this is the one that is semisimple length $n$ as a right $R$ module. You are of course correct that $\overline{x}F$ is simple :) My hint two comments ago is still the right hint to follow. –  rschwieb May 18 '12 at 11:05
    
Let $\{a_1,\ldots,a_n\}$ ($a_i\in\mathbb{F}$) be an $\sigma(\mathbb{F})$-basis of $\mathbb{F}$. I was trying to prove that $\mathbb{F}a_i\bar{x}$ is simple for every $i$, i.e. take an element $fa_i\bar{x}$ with $f\in\mathbb{F}$ I want an element $r\in R$ such that $fa_i\bar{x}r=a_i\bar{x}$, but I'm having problem in finding this $r$, do you have any suggestions? –  Alex M May 23 '12 at 20:01
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