Counterexample to non-invariance of division rings for primitive rings I'm studying "Basic Algebra 2" by N. Jacobson. At page 200, after stating and proving the (Jacobson) density theorem, he warned the reader saying that the division ring of Proposition 4.6 (that is, $B/I \simeq \text{End}_R(R/I)$, where $B$ is the idealizer of the left ideal $I$ in $R$), is $\textit{not}$ an invariant of the primitive ring. 
Moreover he states, without proving, that "A given primitive ring may have non-isomorphic irreducible modules giving faithful representations and even some irreducible modules whose endomorphism division rings are not isomorphic", and then he focuses on left artinian rings, in  which the division ring $\textit{is}$ an invariant of the primitive ring.
I tried to find an example of Jacobson's warning, but I definitively not capable of: obviously it has to be a non-artinian rings, but a part from this I cannot construct a specific counterexample.
Any hint, solution or reference would be much appreciate. Thanks in advance to everybody. 
 A: Regretfully this is not complete, but I think it may be the best starting point.
One obvious place to start $R=End(V_k)$ where $V$ is a right $k$-vector space with dimension $\aleph_0$. This is actually a right and left primitive ring but we'll focus on left primitivity since I want you to view $V$ as a left $R$ module in the natural way.
The module $_RV$ is indeed a faithful simple left module for $R$. Now, there are clearly maximal left ideals that are summands: for example, you can select a basis for $V$ and choose the projection which zeros out all coordinates except the first coordinate. That projection generates a direct summand of $R$ which obviously has codimension $1$ (since its kernel has dimension $1$) and so the kernel is a maximal left ideal $L_1$.
On the other hand, it must have maximal left ideals that are not summands, because otherwise $R$ would be semisimple Artinian. For example, $R$ is known to have one proper ideal made up of the transformations with finite dimensional image, and this ideal is essential as a left ideal. It is necessarily contained in a maximal left ideal $L_2$, which must be essential as well, and an essential left ideal cannot be a summand.
It can be proven that $R/L_2$ is not projective and $R/L_1$ is, so they are definitely not isomorphic. This establishes the claim "A given primitive ring may have non-isomorphic irreducible modules giving faithful representations."
Now, I presently don't know a slick way to determine whether or not $End(_RR/L_1)$ and $End(_RR/L_2)$ are isomorphic as division rings, but it would be a good thing to check first.
