A prime ring whose socle is nonzero and of finite length is simple Artinian.

Question

This is a part of an exercise (Sect. 14 Exercise 11) in Anderson & Fuller's book "Rings and Categories of Modules", and I'm a graduate level student in Turkey.

I want to prove that if $R$ is a prime ring with $0\not=Soc(_RR)$ which is of finite length, then $R$ is a simple Artinian ring.

I proved, as the preceeding part of the exercise, that a prime ring $R$ with nonzero socle is primitive (since the socle is the sum of minimal left ideals and it is nonzero, then there exists a simple left ideal, say $_RI$, which is faithful since R is prime). I also see that $Soc(_RR)\le$ $_RR$ is essential and generated by $_RI$.

I also proved that a left Artinian prime ring is simple, so it suffices to prove that if socle is nonzero and of finite length, then $_RR$ is Artinian. But I can't prove this on my own, in fact, I can't see how the length of the socle is related to $R$. I need this information because the next part asks me to prove that if socle is simple, then $R$ is a division ring.

A hint would be better than a complete proof. Thanks.

Answer 1

If you look back a little ways (Prop 10.7), since $_RR$ has a finitely generated essential socle, it is finitely cogenerated. Since $R$ is left primitive, the intersection of maximal left ideal is zero,

and you get the natural injection of left $R$ modules $R\to \prod U_i$. Since $_RR$ is finitely cogenerated, $R\to \prod_{i\in F} U_i$ for a finite index set $F$. But this product is now the same as a direct sum, so $R$ is a submodule of a semisimple module, and is semisimple itself.