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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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    In your question, $K$ has to be strictly larger than $E$.2011-11-24
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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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    I thought about this before I posted my question initially but I couldn't figure out how to use it. Firstly, how would I prove $\mathrm{PGL}_2(E)$ has elements of infinite order? $x \longmapsto x+1$ works for characteristic 0 fields, but there are infinite fields with finite characteristic, so I don't know what to do there. More importantly though, how does the order of an element relate to the degree of the extension?2011-11-24
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    I have been trying to figure out your hint for the past hour and have made no progress. Could you please explain a bit more?2011-11-24
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    You've shown that for any intermediate field $K$, $[E(x):K]$ is finite. If the fixed field is an intermediate field (that is, not equal to $E$), then the Galois group of $E(x)$ over $K$ is finite (because it is a finite extension), but it is $\mathrm{Gal}(E(x),E)$ (because that's how you got $K$: it's the fixed field of this group). So you would necessarily need $\mathrm{Gal}(E(x),E)$ to be finite. So if you can prove that the group is *infinite* (e.g., if it has elements of infinite order), then the fixed field cannot be an intermediate field, so it must be $E$.2011-11-24
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    Ahh yes! This is very helpful. Now to the question of constructing (or proving the existence of) an element of infinite order. There must be a trick for fields with nonzero characteristic.2011-11-24
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    @Kb100: For characteristic $p$, if there is an element $\alpha$ that is transcendental over the prime field, then $x\mapsto \alpha x$ will do. If not, then there must be elements of arbitrarily large degree over the prime field, and this will translate into being able to find elements of arbitrarily large order in the group you have; this will suffice.2011-11-24
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    @Pete: To be honest, I wasn't sure if I could find something of infinite order in $\overline{\mathbb{F}_p}$.... but I realized I could find things of arbitrarily larger order. (-:2011-11-24
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    Okay last bit... How are you constructing elements of arbitrarily large order in a field of characteristic $p$ which has no transcendental element? I tried proving it myself couldn't come up with a proof.2011-11-24
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    If it has no transcendental elements, then it is the limit (union) of its subfields of finite degree over the base field $\mathbb{F}_p$ - which are all finite. Now can there be finitely many such subfields if $E$ is infinite? Show that if there are infinitely many such subfields then there are elements of arbitrary large order in $E$.2011-11-24
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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