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Let $A$ be a not necessarily commutative unital ring with a unique simple module (up to isomorphism). Let $\mathfrak m$ be the annihilator of this simple module, which is a two-sided ideal. We claim that $\mathfrak m$ is a maximal two-sided ideal. If $I$ is a maximal left ideal, then $A/I$ is a simple module and its annihilator is contained in $I$, since any annihilating element must kill $1+I$. If $J$ is a two-sided ideal contained in $I$, then $J$ must annihilate $A/I$, since if $x\in J, y\in A$, then $x(y+I)=xy+xI\subseteq I$, since $xy\in J\subseteq I$. Now, if $M$ is a maximal two-sided ideal (which exists by Zorn's Lemma), then there's a maximal left ideal $I$ containing $M$ (again by Zorn). Then, $R/I$ is simple and its annihilator is a two-sided ideal containing $M$ and thus equal to $M$, which also equals $\mathfrak m$ because there's a unique simple module. Hence, $\mathfrak m$ is the unique maximal two-sided ideal.

If $A$ is an Artinian ring, then $A/\mathfrak m$ is also an Artinian ring (since any infinite descending chain of left ideals in the quotient lifts to an infinite descending chain in $A$). Furthermore, $A/\mathfrak m$ is a simple ring since $\mathfrak m$ is a maximal two-sided ideal, so by Artin-Weddenburn, $A/\mathfrak m$ is isomorphic to a matrix algebra over a division ring. Is this true if we don't assume $A$ is Artinian?

Nishant
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    "This ideal is also a maximal left ideal" - False! Consider $A = M_{n \times n}(F)$, for $F$ a field. – Dustan Levenstein Dec 15 '15 at 17:54
  • Does that $A$ have a unique simple module? – Nishant Dec 15 '15 at 17:58
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    Yes, the unique simple module is $F^n$, that is, $A$ acting on column vectors by multiplication of matrices. It's semisimple, with every $A$-module a direct sum of the canonical simple module. When $A$ is considered as a left module over itself, it is a direct sum of $n$ copies of the simple module. – Dustan Levenstein Dec 15 '15 at 18:01
  • If it had been true that $\mathfrak m$ was also a maximal left ideal, then it would follow that $A/\mathfrak m$ is a ring with no nontrivial left ideals, which implies every nonzero element has a left inverse, which in turn implies every nonzero element is invertible, so it's a division ring. But this isn't true in general. That said, your question is a legitimate one, and one which I should know the answer to by now, but don't. My guess is it's not true. My comments are just to clean up your misunderstandings in the Artinian case. – Dustan Levenstein Dec 15 '15 at 18:05
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    I think I see the mistake in my proof...I assumed that if $I$ is a maximal left ideal, then the annihilator of $R/I$ contains $I$, which is false. – Nishant Dec 15 '15 at 18:08
  • Indeed, that is not the case. – Dustan Levenstein Dec 15 '15 at 18:08
  • @nishant Now what is the question? It seems to have entirely fallen apart. Do you plan to rescue it with edits somehow? That's still possible if you're quick and beat anyone writing solutions... – rschwieb Dec 15 '15 at 18:10
  • Let me try to correct my mistakes... – Nishant Dec 15 '15 at 18:11
  • I wrote up an answer I thought was in the right spirit, but if it turns out to not suit your edits I will be content to correct/remove the solution. Thanks – rschwieb Dec 15 '15 at 18:15
  • I just edited it. – Nishant Dec 15 '15 at 18:28
  • The counterexample we seek (which I don't know if it exists) amounts to finding a simple ring whose right ideals are all essential and somehow produce the same simple quotient module. – rschwieb Dec 15 '15 at 20:44

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Let $m$ be the annihilator of a simple right $A$-module called $S$.

Then $S$ becomes a simple and faithful $A/m$ module, so that $A/m$ is a right primitive ring. These may or may not be Artinian, and the Artinian ones are precisely the simple Artinian rings (square matrix rings over division rings.)

one isotype of simple module

Now additionally require $A$ to have one isotype of simple right module.

You're right that every maximal right ideal must contain one particular two sided ideal, and it is the unique maximal ideal of $A$. Furthermore, it is the Jacobson radical of $A$.

Additionally, every maximal right ideal is essential in $A$, and the unique simple module is singular and nonprojective.

rschwieb
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  • Do you know of an example of a non-artinian ring with a unique isomorphism class of simple module? Is $\operatorname{End}_F(V)$ such an example when $V$ is infinite-dimensional? – Dustan Levenstein Dec 15 '15 at 18:22
  • @DustanLevenstein The ring you mention has at least two simple right modules: the ones that are summands are projective, and the ones that are quotients by essential right ideals are not. – rschwieb Dec 15 '15 at 18:39
  • @DustanLevenstein every local ring has a unique isoclass of simple module, including the non artinian local rings. – rschwieb Dec 15 '15 at 18:44
  • Is there a local ring whose quotient by its maximal left ideal is not a matrix algebra? – Nishant Dec 15 '15 at 19:34
  • @Nishant Of course not: that quotient has to be a division ring. All of its right ideals are trivial. – rschwieb Dec 15 '15 at 20:37