# Wild ramification in composite fields

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

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 $$E$$.

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.

• Right, thanks. This was very helpful. Maybe, one follow up question: I guess, in your last sentence, you are refering to a theorem which states the existence of an extension of function fields with such an inertia group. What happens if we assume $F_1 / F$ and $F_2 / F$ to be linearly disjoint? Let's say, the degree of $E / F$ is $3^2$ and the characteristic is $3$. Can we find places $Q$ and $Q'$ which lie over $P$ and $P_i$ in $E$ such that $e( P_i | P ) = 3$ and $e( Q | P ) = 6$ and $e( Q' | P ) = 3$.... Feb 13 at 10:03
• Since the extension can not be Galois, your method needs to be applied on the Galois closure $E^*$ of $E / F$. So, for instance, find an inertia group $H$ of order 6 and a conjugated subgroup $H'$ in $E^* / F$ such that $\operatorname{ord}( H \cap G_1 \cap G_2 ) = 1$ and $\operatorname{ord}( H' \cap G_1 \cap G_2 ) = 2$ where $G_i$ is the Galois group of $E^* / F_i$. If I'm not mistaken Feb 13 at 10:18
• @diddy You can make the local picture whatever you want while keeping the extensions linearly disjoint. Adjoin a root of $x^3 -t x = 1$ and then a root of $x^3 - tx = 1 + t^{1000}$ or something. These will have the same local Galois groups, so the prime $P$ lifts to primes $P_1,P_2$, each of which lift two two primes, one of degree $6$ and one of degree $3$, and the degree $6$ one will give you a counterexample. Feb 13 at 15:05
• @diddy To work with non-Galois extensions, you usually want to think of them as sets with an action of the Galois group. Then there's not a big difference between Galois vs. not Galois. And the Galois group can be pretty much anything with a fixed inertia group, as long as it contains the inertia group. Feb 13 at 15:20
• Thanks again. The Example and the provided intuition work for me. Feb 13 at 21:05