An algebra with no two-sided ideal has only one irreducible representation

Question

Suppose an algebra has no sided ideal. Then why does it have only one irreducible representation upto isomorphism?

Answer 1

I am interpreting "no two sided ideal" as meaning that the ring only has trivial two-sided ideals.

In fact, the statement is false for rings in general. If you take the endomorphism ring of an infinite dimensional vector space, then its quotient by its unique maximal two-sided ideal, you get a simple, nonNoetherian Von Neumann regular ring. It has maximal right ideals which are direct summands and maximal right ideals which are essential right ideals, and the two quotients give rise to nonisomorphic simple modules.

As Pedro has brought up in the comments, the problem is fine if the ring is artinian, and this will be the case of the ring has a minimal right ideal, or if the algebra is finite dimensional. In that case it is a semisimple ring.

To prove a simple artinian ring has only one type of right ideal, first show that any simple right module is isomorphic to a minimal right ideal. Then work to show that two nonisomorphic minimal right ideals annihilate each other. Draw a conclusion that this cannot occur in a simple ring, so there is really only one isotype.