$R$ (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.

  • $\begingroup$ Are you sure identity is not assumed? It is very unusual in this context to talk about rngs. If you have identity, then your answer is here. $\endgroup$ – rschwieb Apr 9 '17 at 4:30
  • $\begingroup$ At times like this I wish I could remember if Jacobson proved his density theorem for rngs. Using the linked question it's not hard to prove $R$ is a submodule of a direct sum of simple modules, but again I don't recall if the characterization of semisimple rings holds up without explicitly saying there is an identity. $\endgroup$ – rschwieb Apr 9 '17 at 4:32
  • $\begingroup$ You can't prove the minimal right ideal is an ideal because that is obviously not necessary. Just look at square matrix rings over fields. A better idea would be to look at the sum of all right ideals isomorphic to it, which is an ideal. Immediately you know the ring has an essential right socle, which must be finitely generated by your assumption it is right Goldie... $\endgroup$ – rschwieb Apr 9 '17 at 4:35
  • $\begingroup$ The question about the density theorem for rngs has come up before $\endgroup$ – rschwieb Apr 9 '17 at 4:39
  • $\begingroup$ @rschwieb In my course, Artinian rings are assumed to have 1, but Noetherian rings are not, but here we have ACC not to mention only on right annihilators. I also double checked the defs for Goldie ring, but that does not mention 1 either. Also in the exam it always specifies when 1 is included i.imgur.com/njeqoL4.png for example. $\endgroup$ – Henry Apr 9 '17 at 10:46

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.

  • $\begingroup$ At least, I think everything here works in rings without identity. It is so rare to have to check... $\endgroup$ – rschwieb Apr 18 '17 at 14:50

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.

  • $\begingroup$ There are a lot of strange things about what you've written that I don't understand. "$cR$ is inside $E(R)$, so it is completely reducible, therefore $R$ is completely reducible." Well if that logic seems valid to you... $R$ is inside $E(R)$ from the beginning, so what is the point of $cR$? $\endgroup$ – rschwieb Apr 18 '17 at 12:56
  • $\begingroup$ And I'm not sure why being contained in $E(R)$ would imply $R$ is completely reducible. (It might be very simple and I just don't know the argument.) But certainly you can take whatever random non-completely-reducible module $M$ that you like and $E(M)$ contains it and it doesn't prove the complete reducibility of $M$. $\endgroup$ – rschwieb Apr 18 '17 at 12:58
  • $\begingroup$ I thought you were not satisfied by the result of proving $R$ is completely reducible. I will write out the strategy that I mentioned above as a solution, then. $\endgroup$ – rschwieb Apr 18 '17 at 13:03
  • $\begingroup$ @rschwieb $c\in E(R)$ implies $cR\subset E(R)$ as $E(R)$ is an ideal. Otherwise i'm not sure why $R$ would be inside $E(R)$? We define a module to be completely reducible if it is a sum of (not necessarily direct but equivalent) irreducible submodules and $E(R)$ is a sum of irreducible submodules and we have proven $R=E(R)$. $\endgroup$ – Henry Apr 18 '17 at 13:34
  • $\begingroup$ oh, there has been some misunderstanding... $E(R)$ is standard notation for the injective envelope of $R$. What ideal do you mean? Oh must be the right socle? (Very strange choice of notation) that certainly is an interesting way to get a copy of $R$ in the socle :) $\endgroup$ – rschwieb Apr 18 '17 at 14:03

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