$R$ is a prime right Goldie ring which contains a minimal right ideal. Show that $R$ must be a simple Artinian ring.

Question

$R$ (1 is not assumed to be in $R$) is a prime right Goldie ring (finite dimensional and ACC on right annihilators) which contains a minimal right ideal. Show that $R$ must be a simple Artinian ring.

This appeared in a past paper, the first 2 parts which I managed to prove were "Every essential right ideal of a semi-prime right Goldie ring contains a regular element" and "Every non-zero ideal of a prime ring must be essential as a right ideal". There is also an extra hint that I might have to use Artin-Wedderburn.

I've been trying (and failing) to find a way to use these results since I can't assume (or prove that) the minimal right ideal is an ideal (to use the previous result). I also have that the minimal ideal is of the form $eR$ where $e$ is idempotent, but I feel like i'm barking up the wrong tree. Any help is appreciated, thanks.

Answer 1

Let $E(R)$ be the right socle of $R$. $E(R)$ is an ideal of $R$, so it is essential as a right ideal and therefore contains a regular element, $c$ say.

$R$ is isomorphic to $cR$, via the isomorphism $r\rightarrow cr$. (If $cr=0$ then $r(c)=0$ implies $r=0$, injective. Surjective obvious.)

$cR$ is inside $E(R)$, so it is completely reducible, therefore $R$ is completely reducible. R is then a direct sum of irreducible submodules, this direct sum is finite as $R$ is Goldie. Therefore $R$ is right Artinian. Prime right Artinian rings are simple.

Answer 2

By primeness of $R$, the minimal right ideal of $R$ must be faithful, so the ring is actually right primitive, and its Jacobson radical is zero.

There exists, then, a collection of maximal regular right ideals $\{T_i\mid i\in I\}$ such that $J(R)=\cap T_i=\{0\}$. This allows the natural embedding of $R_R\hookrightarrow\prod_{i\in I} R/M_i$.

Going back to the assumptions, we know that $soc(R_R)$ is an essential right ideal of $R$. Since it is Goldie, the semisimple module $soc(R_R)$ is finitely generated. It's well known that if $R$ has a finitely generated essential right socle, then it is finitely cogenerated as a module. The embedding $R_R\hookrightarrow\prod_{i\in I} R/M_i$ therefore restricts to a finite subset $F$ so that $R_R\hookrightarrow\prod_{i\in F} R/M_i$.

But $\prod_{i\in F} R/M_i=\bigoplus_{i\in F}R/M_i$, so that $R$ is embedded in this finitely generated completely reducible module, and therefore is completely reducible itself, hence Artinian.

According to your notes, this prime Artinian ring has identity and is a simple Artinian ring by the Artin-Wedderburn theorem.