# Show that if $\phi(F) \neq \{0\}$ then $F \cong R$.

Let $F$ be a field. Let $R$ be a ring and suppose $\phi : F \rightarrow R$ is an onto ring homomorphism. Show that if $\phi(F) \neq \{0\}$ then $F \cong R$. (Prove $F$ isomorphic to $F/\{0\}$ first)

How do I prove $F$ isomorphic to $F/\{0\}$ ? Is it because $\{0\}$ is the trivial ideal of F and there are only possible ideal of $F$ are $\{0\}$ and $I$ ? For the question, since $\phi$ is onto, then the kernel of $\phi$ is an ideal by the first ring homomorphism theorem. Because $\phi(F)\neq \{0\}$, then $\phi(F)=R$. How could I apply $F$ isomorphic to $F/\{0\}$ to show $F\cong R$.

Can anyone give me a hit to connect all information so that I can write a proof?

Thanks

$F$ is isomorphic to $F/\{0\}$ since the kernel of the identity homomorphism $\mathrm{id}: F \rightarrow F$ is trivial, whence by the first isomorphism theorem we have $F/\{0\} \cong F$.
For the second part, you're on the right track. Consider the given surjective homomorphism $\phi: F \rightarrow R$. The kernel of $\phi$ is an ideal of $F$, so as you observed it's either trivial or all of $F$. Since $\phi(F) \neq \{0\}$, it must be trivial. Now we apply the first isomorphism theorem to conclude...
Note that $\ker\phi$ is an ideal of $F$. Also, for any field the only ideals are $F$ and $\{0\}$. Since $\phi(F)\neq \{0\}$ we have that $\ker\phi=\{0\}$ so $\phi$ is injective. Also, $\phi$ is surjective by definition.