# Surjective homomorphism $\varphi$ from field to ring implies $\varphi$ isomorphism

Let $\varphi: F\to R$ be a surjective homomorphism from a field to a non-trivial ring. Suppose $\exists a\in F, a\ne 0_F$ such that $\varphi(a)=0_R$, then, since $F$ is a field $\varphi(a^{-1})=\varphi(a)^{-1}=0^{-1}$, which is impossible, since $0^{-1}$ does not exist. Thus $\ker \varphi =\{0_F\}$. Hence, $\varphi$ is an isomorphism.

Do you think this is correct?

• Why do you assume it is surjective ? – Max Apr 12 '17 at 9:29
• @Max: That's part of the task. – Henning Makholm Apr 12 '17 at 9:30
• Oh I'm sorry, you wanted the isomorphism, my bad (I thought you only wanted the injectivity) – Max Apr 12 '17 at 9:32

That looks good except for a tiny nitpick: $0^{-1}$ exists in the zero ring, leaving the possibility that $\varphi:F \rightarrow R$ is the trivial homomorphism $x \mapsto 0$ for all $x \in F$. You have to use the given surjectivity to preclude this.
An alternate route would be to take advantage of the fact that the only ideals of a field are $\{0\}$ and the field itself. Why? If we have $a \in I$ and $a \neq 0$, then $1 \in I$ since $a^{-1}$ exists and $aa^{-1} \in I$. One can prove that kernels of a ring homomorphisms are an ideals, which means that homomorphisms out of a field are either trivial or injective.