Exercise 6.4.5

Proof.

If $f=p∕q\in K$, then the fraction $\varphi (p)∕\varphi (q)$ doesn’t depends of the choice of the representative $(p,q)$ of the fraction: if $f=p∕q=r∕s$, then $\mathit{ps}=\mathit{qr}$, thus $\varphi (p)\varphi (s)=\varphi (\mathit{ps})=\varphi (\mathit{qr})=\varphi (q)\varphi (r)$, and so $\varphi (p)∕\varphi (q)=\varphi (r)∕\varphi (s)$. Therefore there exists a map $\tilde{\varphi }:K\to K$ defined for all $p∕q\in K$ by

In particular, if $p\in R$, $\tilde{\varphi }(p)=\tilde{\varphi }(p∕1)=\varphi (p)∕\varphi (1)=\varphi (p)$ : $\tilde{\varphi }$ extends $\varphi$.

$\tilde{\varphi }$ is a ring homomorphism: $\tilde{\varphi }(1)=1$ since $1\in R$ and $\varphi (1)=1$.

If $\tilde{\varphi }(p∕q)=0$, then $\varphi (p)∕\varphi (q)=0$, thus $\varphi (p)=0$, $p=0$, $p∕q=0$ : $\tilde{\varphi }$ is injective.

If $g=u∕v\in K$, as $\varphi$ is surjective, $u=\varphi (p),v=\varphi (q),p,q\in R$. Then $g=\varphi (p)∕\varphi (q)=\tilde{\varphi }(p∕q),p∕q\in K$ : $\tilde{\varphi }$ is surjective.

$\tilde{\varphi }:K\to K$ is a field automorphism.

If $\psi :K\to K$ is any field automorphism which extends $\varphi$, then for any fraction $p∕q\in K$,

so $\psi =\tilde{\varphi }$:

every ring isomorphism $\varphi :R\to R$ extends uniquely to an automorphism of $K$. □