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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 $n+1>2$.