I have a homework exercise: Let M be a simple right module over the Weyl Algebra A_1(C); Show that the dimension of M as a C-vector space is infinite. C means complex numbers` My solution: char C=0 so that Weyl algebra is simple domain. Assume M is finite dimensional vector space over C. Its base over C is among set {1,d/dx,d^2/dx,d^3/dx....}. Assume n is the greatest n such that d^n/dx belongs to base of M over C. Take r from R which is none-constant and take m from M - non zero and having non-zero coefficient by d^n/dx. Then mr belongs to M (cause M is right R module) however mr has nonzero coefficient by some d^m/dx for some m strictly greater than n, cause R is domain so that coefficient does not goes to zero after multiplying. Hence, we have contradiction.
I assume it is wrong cause I did not use that M is simple. If You are more eloquent than me, I will be happy if You will verify my solution and give a hint if it is wrong.
Is it right or is it wrong?
Firstly: the title is wrong. A domain with a simple submodule is a division ring, and this is not a division ring.
But it certainly has simple right modules, as you wrote in the body.
You seem to have continued the mistake in the title by assuming elements of the well algebra are contained in $M$, but this is not the case. So the step supposing what the basis elements look like is also no good.
I think the characterization of right primitive rings might be a promising direction, since the dimension is linked to the size of the ring you are embedding into.