The image of generator under an automorphism of a cyclic function field

Question

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

Answer 1

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$.