Fixed field of C(t) subfield

Question

$$ \sigma:t \rightarrow \frac{t+i}{t-i} $$

$$ \tau :t \rightarrow \frac{it-i}{t+1} $$

We assume that group $G$ is generated by $\sigma$, $\tau$.

Prove that: $G$ is isomorphic with the alternating group $A_4$. Moreover, compute the fixed field of $G$.

$\sigma^3=1$ and $\tau^3=1$ and $\sigma\tau\sigma\tau=1$ by Todd-Coxeter Algorithm I can prove it $A_4$

By find the track of t $\{t,-t, \frac{1}{t}, -\frac{1}{t}, \frac{t+i}{t-i},\frac{t-i}{t+i}, -\frac{t+i}{t-i},-\frac{t-i}{t+i}, i\frac{t+1}{t-1},i\frac{t-1}{t+1},-i\frac{t+1}{t-1},-i\frac{t-1}{t+1}\}$

After very long computation and I am not sure whether it right. $[x^4-(t^2+\frac{1}{t^2})x^2+1][x^4+2(\frac{t^4+6t^2+1}{(t^2-1)^2})x^2+1][x^4-2(\frac{t^4-6t^2+1}{(t^2+1)^2})x^2+1]$ I still cannot find out the fixed field.

Answer 1

Promoting the comments to an alternative argument for the fact that the group $G=\langle \sigma,\tau\rangle\simeq A_4$:

• $z_1=(1+i)(1-\sqrt3)/2$ is a fixed point of $\sigma$. If we declare $z_2=\tau(z_1)$, $z_3=\sigma(z_2)=-z_1$ and $z_4=\sigma(z_3)=-z_2$, then $z_4$ is a fixed point of $\tau$, and both $\sigma$ and $\tau$ act on the set $X=\{z_1,z_2,z_3,z_4\}$ as 3-cycles. More precisely as $(234)$ and $(123)$ (on the subscripts).
• From this action we get a homomorphism $\phi:G\to Sym(X)$. As the generators act via 3-cycles the image only consists of even permutations of $X$. It is well known and easy to verify that any two 3-cycles with distinct fixed points of $A_4$ generate all of $A_4$. So $\phi:G\to A_4$ is surjective.
• Any Möbius transformation with more than two fixed points is necessarily the identity. But the elements in the kernel of $\phi$ have at least four fixed points. Thus we can conclude that $\phi$ is also injective.

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On with the task of identifying the fixed field. Here I confess to using Mathematica to speed up the processing of these rational functions as well as to diminish the chance of errors. The list of the twelve elements of $G$ was already figured out by the OP, so I simply expanded the polynomial $$ m(x)=\prod_{\alpha\in G}(x-\alpha(t)) $$ that we expect to be the minimal polynomial of $t$ over the fixed field $K$.

Upon inspection we see that the coefficients of $m(x)$ can be written in terms of the rational function $$ \begin{aligned} w&=\frac{(t^2-2 t-1) (t^2+2 t-1) (t^4+1) (t^4+6 t^2+1)}{t^2(t^4-1)^2}\\ &=\frac{t^{12}-33t^8-33t^4+1}{t^{10}-2t^6+t^2}, \end{aligned} $$ and reads $$ m(x)=x^{12}-wx^{10}-33x^8+2wx^6-33x^4-wx^2+1. $$ Using the factorizations of the numerator and the denominator of $w$ should make it easier to verify that $w$ is, indeed, in the fixed field of $G$. The details are a bit taxing though, so as an alternative we can use the observation that essentially $w$ is the sum of squares of the orbit of $t$: $$2w=\sum_{\alpha\in G}\alpha(t)^2.$$

Anyway, the rest is easy. The coefficients of $m(x)$ are in the subfield $L=\Bbb{C}(w)\subseteq K$. Furthermore, $t$ is a zero of a degree $12$ polynomial with coefficients in $L$, so $[\Bbb{C}(t):L]=[L(t):L]\le 12$. On the other hand, Galois theory together with the tower law tells us that $$ [\Bbb{C}(t):K]=|G|=12=[\Bbb{C}(t):L]/[K:L]\le 12/[K:L]. $$ The only possibility is thus that we have equalities $K=L$ and $[\Bbb{C}(t):L]=12$. A consequence of this is that $m(x)$ must be the minimal polynomial of $t$ over $K=L$.

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We knew in advance that we must have $K=\Bbb{C}(f)$ for some element $f\in\Bbb{C}(t)$. By Lüroth's theorem all the intermediate fields between $\Bbb{C}$ and $\Bbb{C}(t)$ are of that form.