Question on Primitive Ring and $\operatorname{End}(V)$ A ring $R$ is primitive if it has a faithful irreducible module. Let's say that $R$ is primitive and it's faithful irreducible module is $V_R$. Since this module is faithful, we have that $R$ embeds naturally into $\operatorname{End}_R(V)$. Since $V_R$ is irreducible, we have that $R$ acts transitively on $V_R$.
When I first read this, it seemed that this should imply that $R \cong \operatorname{End}_R(V)$. Since the module is faithful, we can view $R$ as being embedded into $\operatorname{End}_R(V)$, and I was thinking that $R$ acting transitively on $V$ means that $R$ must be all of $\operatorname{End}_R(V)$. 
Can someone give me some examples or insight into why this isn't so? Thanks!
 A: Hm, a couple things need to be straightened out.
First of all, given a faithful left $R$ module $M$, one can always talk about the natural embedding of $R$ into $\mathrm{End}(M_\mathbb Z)$.  If it is a faithful right $R$ module, then you get an embedding of $R^{op}\to \mathrm{End}(M_\mathbb Z)$.
When $M$ is a faithful simple left $R$ module, $End(_RM)=D$ is a division ring. This can be considered as an action of $D$ on the right of $M$, and in a similar way to the above we can embed $R\to End(M_D)$. This is the normal context for considering left primitive rings.
With that out of the way, on to your questions.

it seemed that this should imply that $≅End_()$

Nope. That would imply $R$ is a division ring.  Even with the correction of replacing $_R$ with $_D$ as suggested above, there is no reason for it to be isomorphic to the full ring of transformations.  For one thing, the full ring of transformations is always primitive on both sides, but there exists a right-not-left primitive ring.

I was thinking that $$ acting transitively on $$ means that $$ must be all of $End_()$

The fact that $R$ acts transitively on $V$ just reflects that $V_R$ is simple. (You can work out that the two things are equivalent.) It doesn't say anything about how much $R$ "fills" $End_D(V)$.
