Let f be an automorphism of some integral domain R. Show that f admits a unique extension to an automorphism of the quotient field. Quotient Field (field of fractions) - Let R be an integral domain. Then the quotient field of R is the set (field) of all elements a/b for all a,b in R.
Could anyone help me please prove this?
Thank you,
Daniele
 A: If $f$ can be extended to an automorphism $\tilde{f}$ of the quotient field of $R$, then we must have
$$ \tilde{f}(r)\tilde{f}\Big(\frac{1}{r}\Big)=f(1)=1$$
for all $0\neq r\in R$, hence $\tilde{f}(\frac{1}{r})=\frac{1}{\tilde{f}(r)}=\frac{1}{f(r)}$. This implies that we must have
$$ \tilde{f}\Big(\frac{r}{s}\Big)=\tilde{f}(r)\tilde{f}\Big(\frac{1}{s}\Big)=\frac{f(r)}{f(s)}$$
This shows that the extension, if it is well-defined, is unique. So all that's left to show is that $\tilde{f}$ as defined above is in fact an automorphism of the field of fractions.
A: It suffices to proove the following: 

Universal property. If $A$ is an integral ring, $K$ is a field and $f:A\hookrightarrow K$ is an injective ring morphism, then there exists a unique field morphism $\overline{f}:\textrm{Frac}(A)\hookrightarrow K$ such that $f=\overline{f}\circ i$, where $i : A\hookrightarrow\textrm{Frac}(A)$ is the inclusion.

Proof. $f=\overline{f}\circ i$ imposes that $\overline{f}(a/b)=f(a)/f(b)$. The only thing to do is to show that $\overline{f}(a/b)$ does not depends on the representant chosen for $a/b$. $\Box$
This result boils down to the wanted property when taking $K=\textrm{Frac}(A)$.
