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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."

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So what about $S=0$. – Martin Brandenburg Feb 6 '13 at 0:21
You can assume $0 \neq 1 \in S$. – tj_ Feb 6 '13 at 0:49
up vote 5 down vote accepted

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.

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In fact, $\alpha : R \to S$ is epi iff $S \otimes_R S/\alpha(R) = 0$. – Martin Brandenburg Feb 6 '13 at 18:31

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 $S = \alpha(R)$.

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Thanks for the additional background and the alternative proof. – tj_ Feb 7 '13 at 11:38
You're welcome. Btw, using the theory in Stenström, loc. cit., here is a slick proof: Over the skew field R, every module is flat, so by Popescu-Spircu, S is a localisation of R for some Gabriel topology. As one might hope and guess, the only Gabriel topologies on a skew field are the trivial and the discrete one, and their corresponding localisations are 0 and R, respectively. – Torsten Schoeneberg Feb 10 '13 at 11:00
... R (trivial) and 0 (discrete), that should be. (Note to myself: If you have double-checked it because you knew you mix it up anyway, triple-check it.) – Torsten Schoeneberg Mar 17 '13 at 14:17

Lemma (Silver, Bergman). If $R \to S$ is an epimorphism in $\mathsf{Ring}$, then $s \otimes 1 = 1 \otimes s$ in $S \otimes_R S$ for all $s \in S$.

Proof. We make the $S$-bimodule $S \otimes_R S$ into a ring $S \oplus (S \otimes_R S)$ with the obvious multiplication so that $S \otimes_R S$ has square zero and the unit is $(1,0)$. The two maps $f,g : S \to S \oplus (S \otimes_R S)$ defined by $f(s)=(s,0)$ and $g(s)=(s,s \otimes 1- 1 \otimes s)$ are ring homomorphisms, which agree on $R$, and therefore also on $S$. $\square$

Claim. When $R$ is a division ring, and $\alpha : R \to S$ is an epimorphism in $\mathsf{Ring}$, then $\alpha$ is surjective, i.e. $S=0$ or $\alpha$ is an isomorphism.

Proof: If $s \in S$ doesn't lie in the image of $\alpha$, we can extend $1,s$ to an $R$-basis of $S$. But then $s \otimes 1 = 1 \otimes s$ is a contradiction. $\square$

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I don't understand which contradiction you obtain. Could you please give some more details. Thanks. – tj_ Feb 6 '13 at 0:48
I use the following elementary fact: If $M$ is a free $R$-module and $m,n$ are part of a basis such that $m \otimes n = n \otimes m$ in $M \otimes_R N$, then $m=n$. – Martin Brandenburg Feb 6 '13 at 1:41
Martin, pushouts in Ring are not given by tensor product as in CommRing. – Tom Goodwillie Feb 6 '13 at 1:52
One surely has to be careful as a priori the maps $i_j$ are not homomorphisms, but under the given criterion they are; maybe one also has to extend $1, s$ once to a left and once to a right basis, and then check Bourbaki, Algebra ch. II §3 no. 7 cor. 1 and following remarks to see that what Martin writes in his first comment is true after making a left-/right-distinction. – Torsten Schoeneberg Feb 6 '13 at 13:56
@Tom: Sure, but the condition follows from the criterion mentioned by TJ. I've added the proof. – Martin Brandenburg Feb 6 '13 at 13:57

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