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." 



Are epimorphisms from a division ring isomorphisms ? 
I think the question can be answered affirmatively. 


Since $0,R$ are the only ideals of $R$, $\alpha$ is injective (I assume $\alpha(1\_R)=1\_S$) and $R$ can be considered as subring and as left $R$-submodule of $S$. Let $\mu:S \otimes\_R S \to S$ be the multiplication. A splitting of $\mu$ is given by $j: S \to S \otimes\_R S,\; s \mapsto s \otimes 1$. Hence $S \otimes\_R S = \ker(\mu) \oplus \text{im}(j)$ and as the image of $j$ equals the image of $S \otimes\_R R \to S \otimes\_R S$, we have  

$$\ker(\mu) = \frac{S \otimes\_R S}{\text{im}(j)}=\frac{S \otimes\_R S}{S \otimes\_R R}=S \otimes\_R (S/R)$$ 
Since $R$ is a division ring, $S$ is a free right $R$-module, say, $S=\bigoplus\_i R$. Thus 
$\ker(\mu) = \bigoplus\_i (S/R)$ and in particular $\ker(\mu)=0 \Leftrightarrow R=S$. 


But, by Prop. 1.1 (cited in the OP's question), $\ker(\mu)=0$ since $\alpha$ is epi and the assertion follows. 


