this is my first question here.
I have this problem regarding the residue class fields of points being the images by finite, separable morphisms of integral normal proper curves.
However, I will formulate the problem in the language of the corresponding function fields:
Let $F_1/F, F_2/F$ be finite Galois function field extensions having the same prime degree $p$ of transcendental degree 1 over the finite field $\mathbb{F}_q$ with $p$ and $q$ coprime.
Furthermore let $F_1/F, F_2/F$ be linearly disjoint, so the composition field $F' = F_1 F_2$ has degree $p$ over $F_1$ and $F_2$.
And now let $P', P_1, P_2$ and $P$ be places of $F', F_1, F_2$ and $F$, respectively, lying above each other, such that relative degrees $f(P_1|P)$ and $f(P_2|P)$ are 1, so the residue class fields of this places are all the same.
I claim or have that feeling or better need, that the relative degree $f(P'|P)$ is 1, too, so the residue class fields of all places are the same.
This would be enough. Actually I have that feeling (but do not need) that even without the assumption at the relative degrees the composition field of the residue class fields of $P_1$ and $P_2$ is the residue class field of $P'$, hence the title. Here I believe the only important assumption would be that the characteristic and one of the ramification indices of $P_1$ or $P_2$ over $P$ are coprime, so the Abhyankar's lemma is applicable.
However, I could prove my claim for the case that the ramification indices of $P_1$ or $P_2$ over $P$ are coprime. I just used some basic facts about the Galois group of the composition field and showed that the Decomposition groups of $P'$ over $P_2$ and $P_1$ over $P$ have the same order. By Abhyankar's lemma this yielded the equality of the relative degrees (even without the "lower relative degrees are 1" assumption).
But without the coprime assumption the ramification indices of $P'$ over $P_2$ and $P_1$ over $P$ are different and one does not know, if the order of the decomposition groups are the same or, and this is what I hope for, the order first order is smaller than the second one.
My approach to prove this is to show that the first decomposition group is exactly the inherit group (under the "lower relative degrees are 1" assumption), if this holds for the second decomposition group.
Yeah, that's it, I think. I would even be thankful for rough ideas. Thank you :).