Let $ F $, $ F\_i $ for $ i = 1, 2 $ and $ E $ be function fields over a finite field with characteristic $ p $ such that $ F \subseteq F\_i \subseteq E $ and $ E = F\_1 \cdot F\_2 $ is the composite field of $ F\_1 $ and $ F\_2 $. Let $ Q $ be a place in $ E $, $ P\_i = Q \cap F\_i $ and $ P = Q \cap F $. Assume that $ P\_i / P $ is wildly ramified for $ i = 1, 2 $. Is the following claim true? The ramification index $ e( Q | P\_1 ) $ a divisor of $ e( P\_2 | P ) $.


Why the question? First, by Abhyankar's Lemma, the claim is true if the extension $ P\_i / P $ is tamely ramified for some $ i = 1, 2 $. Second, by using completions, the estimate $ e( Q | P\_1 ) \le e( P\_2 | P ) $ can at least be proven. Third, this estimate implies the claim for linearly disjoint Galois extensions $ F\_1 / F $ and $ F\_2 / F $ such that their degrees are powers of $ p $.


Anyway, I have the feeling that the claim is false. I looked in the proof of Abhyankar's Lemma in the book of Stichtenoth "Algebraic Function Fields and Codes" and I don't see how it could be modified to prove the claim. Also, a google search only yield that the claim is true for very specific Galois extensions.


Thanks



Wild ramification in composite fields
Your intuition is good - it is not true.


We can take $F = \mathbb F\_q(t)$ and define $F\_1$ and $F\_2$ to be adjoining distinct roots of a cubic $x^3 - t x =1$. Then if $E = F\_1 ( \sqrt{t})$ since $x, x+ \sqrt{t}, x- \sqrt{t}$ are the three roots.


The prime $t=0$ has ramification degree $3$ in $F\_1$ and $F\_2$, and then ramification degree $2$ when we pass from $F\_1$ to $E$ or $F\_2$ to $E$.




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The best way I know to solve problems like this is to express everything in terms of the Galois actions on the roots - in particular, the ramification degrees are sizes of orbits of the inertia subgroup of the Galois group acting on the roots.


Using $e(Q |P\_1) e( P\_1|P) = e(Q|P)$, we're trying to find a situation where $e(Q|P)$ doesn't divide $e (P\_1|P) e(P\_2|P)$ - i.e. when the size of an orbit of the inertia group on the product of two sets doesn't divide the product of the sizes of its orbits on the two sets.


This divides the problem into finding a group with orbits that behave like this - $S\_3$ acting on the product of two sets with $3$ elements does the trick, with an orbit of size $6$ - and then finding an extension of fields with that inertia group - which we can do because $S\_3$ has a normal $p$-subgroup with a cyclic, prime-to-$p$ quotient.


