Find the invariant of a subgroup of automorphisms on $\mathbb Q(X)$ Let $\sigma,\tau\in\operatorname{Aut}_{\mathbb Q}(\mathbb Q(X))$, where $\sigma(X)=1/X$ and $\tau(X)=1-X$. Let $\rho=\sigma\tau$. I have to determine $\mathbb Q(X)^{\langle\rho\rangle}$.
So I just learned the fundamental theorem of Galois theory.  I know that a field extension $K\subset L$ is called finite Galois, if there is a finite subgroup $G\subset\operatorname{Aut}(L)$ of automorphisms of $L$ with invariant field $L^G=K$. So it seems then that I have to find $K\subset\mathbb Q(X)$, such that $\mathbb Q(X)$ is a finite Galois over $K$. From the fundamental theorem of Galois theory I know that the order of $\mathbb Q(X)$ over $K$ would then have to be the number of elements in $\operatorname{Aut}_K(\mathbb Q(X))$.
It's easily shown that $\langle\rho\rangle$ has order 3, so it seems to me then that we would have $[\mathbb Q(X):K]=3$.
However, I'm not sure how to proceed. Any help?
 A: I think this question is important enough that it deserves an answer.
What I’ve noticed from hand computation is that in almost every case I’ve looked at, the following happens. If a field is $K=k(\theta)$ and a finite group $G$ of $k$-automorphisms acts on $K$, then the fixed field can be described as either $k(\text{Norm}_G(\theta))$ or $k(\text{Trace}_G(\theta))$.
By $\text{Norm}_G(\theta)$ I mean $\prod_{g\in G}g(\theta)$ and by $\text{Trace}_G(\theta)$ I mean $\sum_{g\in G}g(\theta)$.
In this case, you’re taking $k=\Bbb Q$, $\theta=X$, and $K=\Bbb Q(X)$. For $G$, you’re taking the finite group generated by $\rho$, which sends $X$ to $\frac1{1-X}$, while $\rho^2$ sends $X$ to $\frac{X-1}X$. As you see, the Norm here is $-1$, so that’s not going to help, but the trace is
$$
X+\frac1{1-X}+\frac{X-1}X\,,
$$
which I’ll call $\xi$. Necessarily $\xi$ is in the fixed field, which I’ll call $F$. The only question is whether $\Bbb Q(\xi)$ is smaller than $F$. You show that they’re equal by constructing a cubic $\Bbb Q(\xi)$-polynomial of which $X$ is a root. Well, set the above display equal to $\xi$, clear of fractions, and get $X^3-\xi X^2+(\xi-3)X+1=0$, which means that $$T^3-\xi T^2+(\xi-3)T+1$$
is a $\Bbb Q(\xi)$-polynomial having $X$ for a root.
