Is every non-commutative ring a subring of $\operatorname{End}(V)$ for some vector space $V$?

Question

Let $R$ be any non-commutative ring without any non-zero zero divisor. Can we say that there exists a vector space $V=V(R)$ over some suitable field $K$ such that $R$ is isomorphic to some subring of $\operatorname{End}(V)$. (The ring R may or may not have an identity element and $\operatorname{End}(V)$ is the set of endomorphisms of $V$.)

Answer 1

For a ring with identity, the answer is affirmative.

In a prime ring with identity, the center is an integral domain, and the prime subring is a field $F$. So $R$ itself is and $F$ vector space, and $R\cong End(R_R)\subseteq End(R_F)$.

I will have to think longer about the possibility of not having an identity. I can't see anything that precludes it or confirms it, at the moment.

Also, it works for rngs without nonzero zero divisors, since every such ring embeds in a ring with identity and no nonzero zero divisors, and those are prime rings, so the above applies again.

I am not sure if there is an analogue for prime rings but I plan to look into that.