Let $K=\mathbb{F}_q(x)$ be a function field and $L = \mathbb{F}_q(X)(\sqrt[n]{f(x)})$ be an extension where $\deg(f) = d$. I want to find what the ramification index at infinity is. If we let $s = (d,n)$ then I am pretty sure the ramification index would be $n/s$. Here is some reasoning:
We can write $f(x) = (1/x)^{-d}u$, where $u$ is a unit. Then $\sqrt[n]{f(x)} = (1/x)^{-d/n}u^{1/n} = (1/x^{s/n})^{-d/s} u^{1/n}$. Hence $ord_{P_\infty}(\sqrt[n]{f(x)}) = -d/s$. So $ord_{P_\infty}(f(x)) = -nd/s = n/s* ord_{1/x}(f(x)) $. Thus $e=n/s$.
My question is, is this correct? Why would $P_\infty$, the prime lying over $1/x$, be $1/x^{s/n}$ and not $1/x^{1/n}$? Further, I think whether the prime splits or is inert is captured in the leading coefficient of $f(x)$ (i.e. whether it is a $n^th$ root or not). Is this true? What are the exact conditions?
Any solution or reference would be greatly appreciated.
If you adjoin $\sqrt[n]{x^n}$ can we even achieve a $1/x^{1/n}$ element? What about, say, $\sqrt[4]{x^2}$: can we obtain an element $1/x^{1/4}$ in this case? This is why $s={\rm gcd}(n,d)$ matters. According to the valuation, we're essentially adjoining $x^{d/n}$ (modulo things of valuation $0$, namely $u^{1/n}$), so if the value group was originally $\Bbb Z$ the new value group is $\Bbb Z+\frac{d}{n}\Bbb Z$. We can apply Bezout's lemma to see the value group is $\frac{1}{n}(n\Bbb Z+d\,\Bbb Z)=\frac{1}{n}(s\Bbb Z)=\frac{1}{n/s}\Bbb Z$ and so the ramification is in fact equal to $e=n/s$.
Not sure about $u\in K^\times$ and inertness though.