I'm reading the proof of Lemma 4.1 [1] which says:


"Let $F = K(x,y), y^q = f(x)$, where $q$ is a prime different from characteristic of $K$.
Let $Z := Gal(F/K(x))$ and we have $Z < G < Aut(F/K)$ Then:


$Z < Z(G) \iff$ $Z$ is normal in $G$ and for all $\sigma \in G$ there exists $0 \neq B\_{\sigma} \in K(x)$ with $\sigma(y) = y.B\_{\sigma}$."


My problem is just on the second line, so I'm writing everything up to that point:


"Proof:
"$\Rightarrow$"
Let $\phi: y \mapsto \xi y, x \mapsto x$ with $\xi$ primitive $q$-th root of unity. Then $Z = <\phi>$. Now let $\sigma \in G$, then $\sigma(y) = y^k.B\_{\sigma}$, with $k \in \mathbb{Z}, (k,q) =1$ and $0 \neq B\_{\sigma} \in K(x)$..." (the rest is routine to show $k=1$)


My problem is that why only one power of $y$ appears in the image of $\sigma(y)$ while all powers of $y$ makes a basis for $F/K(X)$ why not $\sigma(y) = \sum\_{i = 0}^{q-1}y^i.B\_{i}$.


Thank you very much indeed for helping me.


[1] Rolf Brandt, *Über die Automorphismengruppen von algebraischen Funktionenkörpern*,Universität-Gesamthochschule Essen, 1988



The image of generator under an automorphism of a cyclic function field
OK, finally, I think I got it, but it is not that trivial to simply be omitted from the proof (If I complicated it and there's is a straight forward way to see it please tell me):


We have $\sigma(y)^q \in K(X)$. Expanding $\sigma(y)^q$, we see that every term in the expansion has the form $y^{\sum s\_i i}B\_i^{s\_i}$ such that $\sum s\_i = q$ but also we needs that $\sum s\_i i \equiv 0 \mod q$ for all terms. These two equations, generate a non-singular homogeneous system in $\mathbb{Z}/q\mathbb{Z}$ and hence the only solution to the system is $s\_i \equiv 0$, for all $i$'s. Which means $s\_i = k\_iq$ but they are non-negative and sum-up to $q$, so the only legitimate term in the expansion is when for one $i = k, s\_i = q$ and for $ i \neq k, s\_i = 0$.


