TITLE: Are epimorphisms from a division ring isomorphisms ? 
QUESTION [6 upvotes]: According to Corollary 1.2(3) of the paper  Silver: Noncommutative Localizations and Applications. J. of Alg. 7(1964), 44-67:  
If $R$ is a (commutative) field and $\alpha: R \to S$ an epimorphism in the category of rings, then $\alpha$ is an isomorphism.
Question: Is $\alpha$ also an isomorphism if $R$ is not a field but just a division ring ? 
For convenience, I repeat the short proof from Silver (which doesn't seem to work for division rings): First it is noted (Prop. 1.1) that a homomorphism $\alpha: R \to  S$ of (not necessarily commutative) rings with identity is epi iff multiplication $S \otimes_R S \to S$ is an isomorphism (of abelian groups). Then: 
"For $x\in S$, consider the subring $R[x]$ of $S$ generated by $R$ and $x$. Since $R$ is a field, one can easily see that $R \to R[x]$ is an epimorphism using 1.1. If $x$ is transcendental over $R$, then $\beta: R[x] \to R[x]$ defined by $\beta(f)=f(0)$ agrees on $R$ with the identity map of $R[x]$. So $x$ cannot be transcendental over $R$ by definition of an epimorphism. Finally, if $[R[x]:R] < \infty$, then by 1.1, $[R[x]:R]^2=[R[x]:R]$, so $[R[x]:R]=1$ and $x \in R$. Hence $\alpha$ is an isomorphism as desired."

REPLY [7 votes]: Yes, and there are more general results available. E.g. in the paper of H. H. Storrer, Epimorphic Extensions of Non-Commutative Rings, at the bottom of p. 74 there are references given for the following fact: Among those rings $R$ such that any epic monomorphism of (unital, associative) rings $\phi: R \rightarrow S$ is an isomorphism, there are the von Neumann regular rings and the self-injective rings. Compare also the first paragraphs of chapter XI in Bo Stenström's book Rings of Quotients.
And another proof in the vein of Martin's and Ralph's: If $0 \neq \alpha: R \rightarrow S$ is an epimorphism, every $R$-linear endomorphism of $S$ (as left $R$-module, say) has to be $S$-linear. (In fact, $\alpha_*: S-Mod \rightarrow R-Mod$ being full is another equivalent criterion for $\alpha$ being an epi). Now as $R$ is a skew field, $\alpha(R)$ is a direct summand in $S$. If it had a non-trivial complement, we could certainly define non-identical $R$-linear endomorphisms of $S$ whose restrictions to $\alpha(R)$ are identical; but any $S$-linear endomorphism of $S$ is determined by what it does on $1_S = \alpha(1_R)$. So $S = \alpha(R)$.