I'm stuck at the very beginning of an exercise I have to do for my algebra class.
We're looking at the field of $\mathbb{C}(X)$ and it's automorphisms.
Let $a \in \mathbb{C}^*$, $ b \in \mathbb{C} $ and $ \sigma : \mathbb{C}(X) \to \mathbb{C}(X) $ given by $ X \mapsto aX + b$.
a) Explain why $\sigma$ is a field-automorphism.
I wrote that $\sigma$ is a coordinate transform in $\mathbb{C}(X)$ and therefore must be a field-automorphism.
Is this explanation correct?
b) Let $ H = \langle \sigma \rangle \subseteq Aut(\mathbb{C}(X)) $. Find the field fixed by $H$, i.e. $\mathbb{C}(X)^H$.
I know that $ \mathbb{C}(X)^H = \{ f \in \mathbb{C}(X) \; | \; \tau(f) = f \;\; \forall \tau \in H \} $ but I did not really get around how to decode this automorphism into something I can use to find the field fixed.
Any hints how to find it would be greatly appreciated.
Edit:
Since I am still stuck I wanted to add some thoughts of me. I'm working on b).
I tried some cases for $a$, $a=1$ is obvious.
$a = -1$:
When we only take a look at monoms $f(X) = X^i$ we have to conclude that $\sigma^i(f) = f$ for $i \in \mathbb{N}$ holds only for even exponents of $X^i$ and $\mathbb{C}(X)^H = \mathbb{C}(X^2)$. (Where we write $\sigma^i$ for $\sigma(\sigma(...))$.)
$a = i$:
With the same argumentation we see that $\mathbb{C}(X)^H = \mathbb{C}(X^4)$.
I now assumed that: $|a| = 1$ (hence $a$ is a root of unity) or else the field fixed would only consist of $\mathbb{C}$. I came up with the explanation, that when $|a| \not = 1$ you may take a look at any monom $X^j$. When you evaluate $\sigma$ $i$-times on $X^j$ you get: $\sigma^i(X^j) = a^{i+j} X$.
If $|a| > 1 $ or $|a| < 1$ you are either making the coefficient very large or close to zero by evaluation $\sigma$ many times. It can't hold that $\sigma^i(f) = f$ therefore.
What I couldn't find out if there are other fixed fields (besides $a = -i$).
As Jyrki Lahtonen pointed out, there are solutions where $ b \not = 0$. For example $ X \mapsto -X - 1$ and $ -X(X+1) $ is invariant under $\sigma$. I generalized that a bit and came up with $$ \sigma : X \mapsto -X - b \\ \sigma(-X(b+X)) = -(-X-b)(b - X + b) = -X(b+X) \\ \text{And clearly: } ord \; \sigma = 2 $$ For $b=1$ The field fixed for this is $ \mathbb{C}(X^2+X) $.
What I'm now trying to do is proof that: $$ \mathbb{C} \subsetneq \mathbb{C}(X)^H \implies ord \; \sigma \lt \infty $$ The argument for that lies within the answer of Michael Zieve though I couldn't figure it out until now: Intersection of two subfields of the Rational Function Field in characteristic $0$
About a), you should probably say better that $\sigma$ is an invertible coordinate transformation, but why not just checking the axiomatic definition of field automorphism?
About b), let $f\in\Bbb C(X)$ such that $f(aX+b)=f(X)$. Recall that $f$ is a quotient of polynomials. What can we say about the set of zeroes and poles of $f$?