Exercise 6.1.4

Proof.

Let $\phi :\alpha \to {\alpha }^{\prime }$. By hypothesis, for all $\alpha ,\beta \in \mathrm{\Omega },$

(a)

By hypothesis, there exists $\alpha \in \mathrm{\Omega }$ such that ${\alpha }^{\prime }=\phi (\alpha )\ne 0$. Then $\phi (\alpha )=\phi (\alpha \cdot 1)=\phi (\alpha )\phi (1)$, and since $\phi (\alpha )\ne 0$, ${\alpha }^{\prime }=\phi (\alpha )$ has an inverse in $\mathrm{\Omega }$, thus $$\phi (1)=1.$$

$\phi$ is so a ring homomorphism between two fields.

(b)

We show that $\phi$ is injective:

If $a\ne 0$, there exists an inverse $b$ of $a$: $\mathit{ab}=1$, thus $\phi (a)\phi (b)=\phi (\mathit{ab})=\phi (1)=1$, therefore $\phi (a)\ne 0$. The kernel of $\phi$ is null, thus $\phi$ is injective.

As ${\mathrm{\Omega }}^{\prime }=\{\phi (\alpha ),\alpha \in \mathrm{\Omega }\}$, $\phi$ is surjective. So $\phi :\mathrm{\Omega }\to {\mathrm{\Omega }}^{\prime }$ is a field isomorphism.