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Proposition: If $F$ is a field, let $F[x]$ be the ring of all polynomials whose coefficients are in $F$. The fraction field of $F[x]$, denoted $F(x)$, is defined to be the ratios $r(x) = f(x)/g(x)$ for $f(x),g(x)\in F[x]$ with $g(x) \not= 0$. The $F$-automorphisms$\dagger$ of $F(x)$ are linear-fractional changes of variables $r(x) \mapsto r((ax+b)/(cx+d))$ where $a,b,c,d\in F$ with $ad-bc \not= 0$.

${\dagger}$An $F$-automorphism of $F(x)$ is an automorphism of this field that keeps each element of $F$ fixed.

Question 1: Since the motions of hyperbolic surface correspond to the general Mobius transformation $\frac{ax+b}{cx+d}$ where $a,b,c,d \in \mathbb C$ with $ad-bc \not= 0$ (Poincare,1883), I want to know if the above statement has any geometric importance. Or is there any work related is done?

Question 2: What is the role of this result in the theory of conformal mappings? Or is there any related work (and references)?

Unrelated Curiosity If, further, $F$ is a commutative ring with identity, is the Proposition still true?

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up vote 2 down vote accepted

Here is a relationship between your question and conformal mappings. For $X$ and $Y$ arbitrary Riemann surfaces, any $\mathbb{C}$-homomorphism between their fields of meromorphic functions comes from a unique holomorphic map $f:X\to Y$. In particular, $\mathbb{C}$-isomorphisms come from biholomorphisms, i.e. Riemann surfaces are determined by their fields of meromorphic functions. See Theorem 3.1 in http://www.ams.org/journals/tran/1968-132-01/S0002-9947-1968-0226043-5/S0002-9947-1968-0226043-5.pdf.

The field of meromorphic functions on the Riemann sphere is the field of rational functions. The above theorem says that every $\mathbb{C}$-automorphism of that field is induced by an automorphism of the Riemann sphere, that is a Moebius transformation.

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I rethink the question and found this is related to the global conformality. The Mobius trans. over the field $\mathbb C$ is the only globally conformal mapping on the Riemann sphere. Many works can be traced from there. And thanks you for your time, thanks @KConrad 's reformatting! –  Henr.L May 25 '13 at 10:33

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