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I am trying to prove that if $E$ is an infinite field, then the fixed field of $Gal(E(x)/E)$ is $E$.

The first part of the question was to find all automorphisms $x\longmapsto \frac{P(x)}{Q(x)}:E(x)\to E(x),$ which I did. They are of the form $ x \longmapsto \frac{ax+b}{cx+d} $ with $ad-bc \neq 0$.

The second part said to prove that if $E \subset K \subseteq E(x)$ for an intermediate field $K$, then $[E(x):K]$ is finite. I did this.

The third part (which I am stuck on) says to prove that if $E$ is infinite that the fixed field of $Gal(E(x)/E)$ is $E$.

I don't really see how this relates to the previous parts, nor do I know how to finish the proof. Since the fixed field of $Gal(E(x)/E)$ is an intermediate field between $E(x)$ and $E$ I could try to prove that the degree of $E(x)$ over the fixed field of $Gal(E(x)/E)$ is infinite and then apply the second part, but this seems very complicated and I don't know how I would go about it.

Does anyone have any suggestions for me? They would be very much appreciated.

If anyone could elaborate on Pete's hint, I have been trying to figure it out for quite a while now and haven't made any progress. Thank you!

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    @KCd: I think that's implied in the use of $\subset$ vs. $\subseteq$; note he write $E\subset K\subseteq E(x)$, suggesting first inclusion is proper and second does not have to be.2011-11-24

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Hint: The group $\operatorname{PGL}_2(E)$ of linear fractional transformations $x \mapsto \frac{ax+b}{cx+d}$ with $ad-bc \neq 0$ has elements of infinite order if and only if $E$ is infinite. For instance, if $E$ has characteristic $0$, then $x \mapsto x+1$ has infinite order.

Added: Oops! As Arturo Magidin points out, the above statement should be:

Proposition: For a field $E$, the following are equivalent:
(i) The supremum of the orders of elements of $\operatorname{PGL}_2(E)$ is infinite.
(ii) $E$ is infinite.

(My previous assertion is false iff $E$ is an infinite algebraic extension of a finite field.)

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    @Kb100: For any given prime $p$ and for each $n\gt 0$, there is one and only one extension of $\mathbb{F}_p$ of degree $n$, and it contains all elements of degree $d$ over $\mathbb{F}_p$, where $d$ ranges over the divisors of $n$ (since the field of order $p^d$ is contained in the field of order $p^n$ if and only if $d|n$). If the field is infinite, it must contain elements of arbitrarily large degree over the prime field, or else there is a bound $m$ for the degree, and all elements will be contained in the extension of degree $m!$, and so the field will be finite.2011-11-24