Exercise 7.5.14

Proof. Consider the automorphisms of $L=F(t)$ defined by

and $G=⟨\sigma ,\tau ⟩$.

(a)

Note that ${\sigma }^2={\tau }^2=e$, $\sigma$ and $\tau$ have order 2. Let $\rho =\sigma \circ \tau =\mathit{\sigma \tau }$. For all $\alpha (t)\in F(t)$, $$\begin{array}{llll}\hfill \rho (\alpha (t))& =\sigma (\alpha (1-t))& \hfill & \\ \hfill & =\alpha \left(1-\frac{1}{t}\right),& \hfill & \\ \hfill {\rho }^2(\alpha (t))& =\alpha \left(1-\frac{1}{1-\frac{1}{t}}\right)& \hfill & \\ \hfill & =\alpha \left(\frac{1}{1-t}\right),& \hfill & \\ \hfill {\rho }^3(\alpha (t))& =\alpha \left(\frac{1}{1-\left(1-\frac{1}{t}\right)}\right)& \hfill & \\ \hfill & =\alpha (t).& \hfill & \end{array}$$

Thus $\rho$ is of order 3, and as $\tau =\mathit{\sigma \rho }$, $G=⟨\sigma ,\rho ⟩.$

Moreover, for all $\alpha (t)\in F(t)$,

$$\begin{array}{llll}\hfill (\mathit{\rho \sigma })(\alpha (t))& =\rho \left(\alpha \left(\frac{1}{t}\right)\right)& \hfill & \\ \hfill & =\alpha \left(\frac{t}{t-1}\right),& \hfill & \\ \hfill (\sigma {\rho }^{-1})(\alpha (t))& =(\sigma {\rho }^2)(\alpha (t))& \hfill & \\ \hfill & =\sigma \left(\alpha \left(\frac{1}{1-t}\right)\right)& \hfill & \\ \hfill & =\alpha \left(\frac{1}{1-\frac{1}{t}}\right)& \hfill & \\ \hfill & =\alpha \left(\frac{t}{t-1}\right).& \hfill & \\ \hfill & & \hfill \end{array}$$

Thus $\mathit{\rho \sigma }=\sigma {\rho }^{-1}$.

To summarise, $G=⟨\sigma ,\rho ⟩$, with ${\sigma }^2={\rho }^3=e(\sigma \ne e,\rho \ne e),\mathit{\rho \sigma }=\sigma {\rho }^{-1}$, therefore

$$G\simeq D_6\simeq S_3.$$

(b)

By the isomorphism $\mathrm{PGL}(2,F)\simeq \mathrm{Gal}(F(t)∕F)$ described in Section 7.5.C, $\sigma$ corresponds to $[\gamma ]\in \mathrm{PGL}(2,F)$, where ${\gamma }^{-1}=\gamma =\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$ and $\tau$ corresponds to $[\delta ]$, where ${\delta }^{-1}=\delta =\left(\begin{array}{cc}\hfill -1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$, and $\rho =\mathit{\sigma \tau }$ to $[𝜀],𝜀={(\mathit{\gamma \delta })}^{-1}=\left(\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right),$

so $[𝜀]\in \mathrm{PGL}(2,F)$ is of order 3 (but not $𝜀\in \mathrm{GL}(2,F):{𝜀}^3=-I_2$).

By $[\gamma ],[\delta ]$ acting on $\hat{F}$,

$$[\gamma ]\cdot 0=\infty ,[\gamma ]\cdot 1=1,[\gamma ]\cdot \infty =0.$$

$$[\delta ]\cdot 0=1,[\delta ]\cdot 1=0,[\delta ]\cdot \infty =\infty .$$

Thus $Ĝ=⟨[\gamma ],[\delta ]⟩\simeq G$ maps $\{0,1,\infty \}$ on itself.

Conversely, let $[\xi ]=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]\in \mathrm{PGL}(2,F)$ mapping $A=\{0,1,\infty \}$ on itself. We show that $[\xi ]$ lies in $Ĝ$.

We know by Exercise 7, which proves uniqueness in Theorem 7.5.8, that there exists at most an element in $\mathrm{PGL}(2,F)$ sending $0,1,\infty$ on three fixed points in $\hat{F}$. As the elements of $Ĝ$ map $\{0,1,\infty \}$ on itself, the group homomorphism sending $G$ on the group $S_A$ of permutions of $A$ is injective by this uniqueness. Moreover $|Ĝ|=6=|S_A|$, so this is a group isomorphism, so the elements of $Ĝ$ give the 6 possible permutations of $\{0,1,\infty \}$. As $[\xi ]$ has the same images for the elements of $A$ that an element $[\zeta ]$ of $Ĝ$ and as $[\xi ]$ is uniquely determined by these images, $[\xi ]=[\zeta ]$, so $[\xi ]\in Ĝ$.

Conclusion: $G$ corresponds to the subgroup of $\mathrm{PGL}(2,F)$ consisting of all elements that map the subset $\{0,1,\infty \}\subset \hat{F}$ to itself.

(c)

We verify that $\alpha (t)=\frac{{(t^2-t+1)}^3}{t^2{(t-1)}^2}\in L_G$: $$\begin{array}{llll}\hfill \sigma (\alpha (t))& =\frac{{({(\frac{1}{t})}^2-(\frac{1}{t})+1)}^3}{{(\frac{1}{t})}^2{((\frac{1}{t})-1)}^2}& \hfill & \\ \hfill & =\frac{{(1-t+t^2)}^3}{t^2{(1-t)}^2}& \hfill & \\ \hfill & =\alpha (t),& \hfill & \\ \hfill \tau (\alpha (t)& =\frac{{({(1-t)}^2-(1-t)+1)}^3}{{(1-t)}^2{((1-t)-1)}^2}& \hfill & \\ \hfill & =\frac{{(t^2-t+1)}^3}{{(t-1)}^2{t}^2}& \hfill & \\ \hfill & =\alpha (t).& \hfill & \end{array}$$

As $G=⟨\sigma ,\tau ⟩,\alpha (t)\in L_G$, thus

$$F\left(\frac{{(t^2-t+1)}^3}{t^2{(t-1)}^2}\right)\subset L_G\subset F(t).$$

By Theorem 7.5.3, $L_G\subset F(t)$ is a Galois extension and $[F(t):L_G]=|G|=6$, and by Theorem 7.5.5,

$$\left[F(t):F\left(\frac{{(t^2-t+1)}^3}{t^2{(t-1)}^2}\right)\right]=\max <!--\; FUNCTION\; APPLICATION\; -->(\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->{(t^2-t+1)}^3),\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(t^2{(t-1)}^2))=6,$$

so $\left[L_G:F\left(\frac{{(t^2-t+1)}^3}{t^2{(t-1)}^2}\right)\right]=1$ , therefore

$$L_G=F\left(\frac{{(t^2-t+1)}^3}{t^2{(t-1)}^2}\right).$$

Conclusion: $F\left(\frac{{(t^2-t+1)}^3}{t^2{(t-1)}^2}\right)\subset F(t)$ is a Galois extension of Galois group $G\simeq S_3$.