If $V$ is a vector space over a division ring $K$, and $A=\mathrm{End}_K(V)$, then every quotient ring of $A$ is a prime ring Let $K$ be a division ring, let $V=V_{K}$ a vector space over $K$, and let $A=\mathrm{End}_{K}(V)$. Could anyone give me an idea of ​​how to prove that every quotient ring of $A$ is a prime ring?
 A: (The question clearly is asking about quotients by proper ideals.)
The ideal structure of this sort of ring is very simple (although it will take some work for you to show that it is true.) It turns out that the proper ideals of this ring form a chain, and they are exactly of the form $I_\alpha=\{f\in A\mid \dim_K(Im(f))\leq\alpha\}$ where $\alpha$ denotes a cardinal, $\omega\leq \alpha<\dim_K(V)$ if $\dim_K(V)$ is infinite, and $0$ if $\dim_K(V)<\omega$ (is finite, naturally because in that case it's just a matrix ring over a division ring, which is simple, hence it trivially has all factor rings prime).
So you can break this task into parts:


*

*Show that each $I_\alpha$ is an ideal of $A$.

*Show that each $I_\alpha$ is a prime ideal of $A$.

*Show that the $I_\alpha$ are the only ideals.
Then you will be done: the quotient by any of these prime ideals creates a prime ring. 

FYI, there is a connection to this question which might furnish an alternative proof: Exercise about prime ideals. The ring you are talking about is an example of a ring with all ideals idempotent and forming a chain.
It is an elementary exercise to show that $A$ is von Neumann regular. Further, it is elementary to show that every ideal of a VNR ring is idempotent. You can use this to replace point #2 above, because with #1 and #3, the linked question gives you that the ideals are prime. However, it's probably just as easy to prove they are prime directly :)
