The first Goldie theorem says that if $R$ is an order in a simple Artinian ring, then $R$ is a prime Goldie ring.
This should be an immediate consequence of the second Goldie theorem which asserts:
A ring R is semiprime goldie iff R is an ore ring and an order in a semisimple ring.
I can see that from this follows that $R$ is a semiprime Goldie ring but I don't see why $R$ is prime.
I tried to lifts ideals of $R$ to the left localisation $S^{-1}R$ but this didn't seem to work.
Any help is appreciated!
It seems clear to me that if $e$ is a nontrivial central idempotent of $R$, then $(1, e)$ is automatically a nontrivial central idempotent of the left classical ring of quotients.
Therefore $R$ has no such idempotents, and is simple.