# Ring with Unique Simple Module

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?

• "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
• 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
• 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

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.

• 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