Can there be a non-trivial epimorphism (of rings) from a field? I apologize if this question is trivial, but I just cant figure it out. Let $K$ be a field and let $K\longrightarrow A$ be an epimorphism of rings. Is it necessary that $A=K$? 
 A: Assume $A$ is not just the zero ring (consisting of just one element). Via the given map $i: K \to A$, we may regard $A$ as a $K$-module or in other words a vector space over $K$. Now let us consider the two ring maps $i_1, i_2: A \to A \otimes_K A$ where $i_1(a) = a \otimes 1$ and $i_2(a) = 1 \otimes a$. Observe that for any $k \in K$, we have 
$$(i_1 \circ i)(k) = i(k) \otimes 1 = (1 \cdot k) \otimes 1 = 1 \otimes (k \cdot 1) = 1 \otimes i(k) = (i_2 \circ i)(k)$$ 
so $i_1 \circ i = i_2 \circ i$, whence $i_1 = i_2$ by epimorphicity of $i$. But now $i_1 = i_2$ forces $\dim_K A = 1$. Indeed, if $\dim_K A > 1$, we may pick an element $a$ that is linearly independent of the identity $1$. Then we can restrict $i_1, i_2$ to the subspace $V$ spanned by basis elements $e_1 = a, e_2 = 1$, and we just observe that $i_1(a) = e_1 \otimes e_2 \in V \otimes V$, $i_2(a) = e_2 \otimes e_1$, which are clearly linearly independent ($e_i \otimes e_j$ are basis elements of $V \otimes V$). Thus $i_1, i_2$ disagree at $a$, contradicting $i_1 = i_2$. 
A: This is even true for $K$ a division ring. See my question and the answers here: 
Are epimorphisms from a division ring isomorphisms ? 
Added: That epimorphisms from a field (or a division ring) are isomorphisms doesn't require the Axiom of Choice. 
For, as noted in the comment to the accepted answer in the link, $\alpha: R \to S$ is epi iff $S \otimes_R S/\alpha(R)=0$. If $R$ is a division ring, the epi $\alpha$ is injective (due to the lack of non-trivial ideals). Since every module over a division ring is flat (this doesn't need AC!), tensoring the exact sequence $0 \to R \xrightarrow[]{\alpha} S$ with $S/\alpha(R)$ yields the exact sequence $0 \to  S/\alpha(R)\to S \otimes_R S/\alpha(R)=0$. Thus $S/\alpha(R)=0$, i.e. $\alpha(R)=S$. 
