Let $V$ be an infinite dimensional vector space, $A$ its algebra of linear endomorphisms (also known as a full matrix ring). The algebra $A$ is primitive because $V$ is a faithful simple (left) $A$-module.

Question 1: does $A$ admit a non-isomorphic to $V$ faithful simple module?

I understand that $A$ has many simple modules. For instance, take the ideal $I$ of linear maps of finite rank and any maximal left ideal $L$ containing $I$. Then $A/L$ is a simple, but not faithful $A$-module. What are the others?

Questions 2: Is there a classification of simple $A$-modules or/and a classification of simple faithful $A$-modules?


Question 1:

The subset of transformations with finite dimensional images forms the (left and right) socle of this ring, a nontrivial two-sided ideal. The socle is the intersection of all essential right ideals, so the essential maximal right ideals cannot provide simple modules which are faithful.

You're left with the non-essential maximal ideals, where are necessarily direct summands of $A_A$. Then $A=S\oplus M$ where $M$ is a maximal right ideal and $S$ is a minimal right ideal, so faithful simple right modules necessarily embed into $A$. (Another confirmation that $M$ doesn't contain the socle is here: $M$ doesn't contain $S$, which is in the socle.)

So now the question is whether or no you can have two minimal right ideals $S_1\ncong S_2$ in $A$. The answer is "no", because when you sum the minimal ideals of a single isoclass, you get an ideal, and if you had two such ideals $I_1$ and $I_2$, their product is necessarily $\{0\}$ (these are basic ideas you encounter with the Artin-Wedderburn theorem.) But this is a prime ring (in fact a semiprimitive one) so that you can't have a product of nonzero ideals being zero.

So, there is only one isoclass of faithful simple right module for this ring.

Question 2+:

Well, we classified the faithful simple ones.

As for classifying the nonfaithful simple modules, I don't know offhand if it's been done comprehensively. The two-sided ideal structure of these rings is well-known (the two-sided ideals are linearly ordered, corresponding to cardinals between $\aleph_0$ and $\dim(V)$) so by using the same argument you used, you can find simple right modules annihilated by an ideal of your choice.

But also, since the ring has Jacobson radical zero, given any nontrivial ideal $I$, you can find a maximal ideal $M$ such that $I\nsubseteq M$, so that $I$ doesn't annihlate $R/M$, and you can find simple modules that are not annihilated by an ideal of your choice.

Maybe this can be stitched together into a complete description.

  • $\begingroup$ Thanks! It looks like a useful and complete answer to Q1. I will check your math later. $\endgroup$
    – Bugs Bunny
    Dec 6 '17 at 12:06

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