Fixed field of the linear fractional maps ($X \to \frac{aX+b}{cX+d}$) of $k(X)$ where $k$ is finite.

Question

I come up with this interesting question in Serge Lang's Algebra:

Let $k$ be a field of $q$ elements and $K=k(X)$ be the rational field of one variable. Let $G$ be the group of automorphisms obtained by the mappings $$ X \mapsto \frac{aX+b}{cX+d} $$ with $a,b,c,d \in k$ and $ad-bc \ne 0$.

(a) Show that $G$ has $q^3-q$ elements.

(b) Show that the fixed field of $K$ is $k(Y)$ with $Y=\frac{(X^{q^2}-X)^{q+1}}{(X^q-X)^{q^2+1}}$.

(a) is easy because it's basically counting the cardinality of $PGL_2(k)$.

My question is (b) however. Put $Y=Y(X)=\frac{f(X)}{g(X)} \in k(X)$, we need to show that $Y\left(\frac{aX+b}{cX+d}\right)=Y(X)$. However I can't see how to proceed other than unfeasibly computing all coefficients with even no knowledge of the degree of $f$ and $g$. Is there any properties of $K$ related to this exercise that I have missed?

Edit 1: I've noticed that $[k(X):k(Y)]=q^3+q^2>|G|=q^3-q$. It is also clear that $Y$ is fixed by $G$ (just verify it on all generators of $G$). Perhaps then the problem can be narrowed down to a simple Galois theory problem?

Answer 1

You miscalculated the extension degree $[k(X):k[Y)]$. Regardless of whether the extension is Galois or not, we always have $|G|\le|\operatorname{Aut}_{k(Y)}(k(X))|\le [k(X):k(Y)]$ due to the independence of characters. (See Theorem 13, Section II.F. of Artin's Galois Theory. In fact, the case of $q=2$ was treated in Section II.G.)

Note that the polynomial $T^q-T$ has no repeated root (since its derivative is $-1$ which itself has no root), and any root also satisfies $T^{q^2}-T=0$ (if $T^q=T$, then $T^{q^2}=(T^q)^q=T$). Therefore $T^q-T|T^{q^2}-T$ as polynomials. In particular, we know that the denominator and the numerator of $Y$ as presented are not coprime.

Indeed, after cancelling $(X^q-X)^{q+1}$, the degree of the denominator is $q^2(q+1)-q(q+1)=(q^2-q)(q+1)=q^3-q$, and the one of the numerator is $q(q^2+1)-q(q+1)=q(q^2-q)=q^3-q^2<q^3-q$. From this, we know that the minimial polynomial of $X$ in $k(Y)$ is of degree at most $q^3-q$. Hence the earlier inequalities become equalities $|G|=|\operatorname{Aut}_{k(Y)}(k(X))|=[k(X):k(Y)]$, we must have $G=\operatorname{Aut}_{k(Y)}(k(X))$ and the extension is Galois.

Answer 2

I try to put down a more completed proof using the observation made by Just a user.

Part I - Function Field and Linear Fractional Maps

Let $k$ be a field and $X$ a variable over $k$. Let $$ \varphi(X)=\frac{f(X)}{g(X)} $$ be a rational function in $k(X)$, expressed as two polynomials $f,g$ which are coprime. Define $\deg{\varphi}=\max\{\deg f,\deg g\}$. Put $U=\varphi(X)$.

(1) $\deg\varphi=[k(X):k(U)]$.

Sketch. Put $P(t)=f(t)-Yg(t) \in k(U)[t]$. Then $P(X)=0$ and it remains to show that $P$ is irreducible in $k(U)[t]$. But by Gauss's lemma, it suffices to verify irreducibility in $k[U][t] \cong k[t][U]$, but in $k[t][U]$, the polynomial $f(t)-Ug(t)$ is irreducible because it is linear.

(2) Every automorphism of $k(X)$ can be realised as a linear fractional map.

Sketch. Let $\varphi$ be an automorphism. Then $\varphi(X)=\frac{f(X)}{g(X)}$ for some coprime $f,g \in k[X]$, by definition of $k(X)$. Then $[k(X):k(\varphi(X)]=[k(X):\varphi(k(X))]=1$, and use (1).

(3) $\mathrm{Aut}(k(X))$ is generated by $\tau_a:X \mapsto aX$, $\sigma_b:X \mapsto X+b$ and $\iota:X \mapsto X^{-1}$ where $a,b \in k$.

Part II - The extension $k(X)/k(Y)$

Now consider $$ Y=\frac{(X^{q^2}-X)^{q+1}}{(X^q-X)^{q^2+1}}. $$

As the accepted answer has shown, $[k(X):k(Y)] \le q^3-q$. However, since $G$ fixes $Y$ (which can be shown by considering all generators of $G$), we have $G \subset \mathrm{Aut}(k(X)/k(Y))$, i.e. $[k(X):k(Y)] \le |G|=q^3-q \le |\mathrm{Aut}(k(X)/k(Y))|$ and therefore $k(X)/k(Y)$ is Galois with group $G$.