# Dedekind criterion for function fields

Let $$p$$ be a prime, $$f\in \overline{\mathbb F}_p[x]$$ a polynomial of degree $$>1$$ and $$t$$ be transcendental over $$\mathbb F_p$$. Let $$i\geq 0$$ and let $$M=\overline{\mathbb F}_p(t)(\alpha)$$, where $$\alpha$$ is a root of $$f-tx^i$$.

Question 1). Suppose that $$i=0$$. I want to understand how the place corresponding to $$0\in \overline{\mathbb F}_p(t)$$ decomposes in $$M$$. Suppose that $$f=\prod_{i=1}^m(x-\alpha_i)^{e_i}$$ in $$\overline{\mathbb F}_p[x]$$, where the $$\alpha_i$$'s are pairwise distinct. Is it true that there exist exactly $$m$$ places $$P_i$$ in $$M$$ that lie above $$0$$ and the ramification index of each $$P_i$$ is $$e_i$$?

Question 2). Suppose instead that $$i>0$$. How do I find the decomposition pattern in $$M$$ of the infinite place of $$\overline{\mathbb F}_p(t)$$?

Question 3) Does anything change if I assume that $$M$$ is Galois over $$\overline{\mathbb F}_p(t)$$?

• 1) For the splitting field, all the ramification indices will be equal, and if the $e_i$s are prime to $p$ will be equal to their least common multiple. Do you really mean the splitting field and not hte field obtained by adjoining a root of $f -t x^i$ (which will be $\overline{\mathbb F_p} (t)$)? – Will Sawin Oct 9 at 15:54
• Sorry yes, of course I meant that! I'll edit, thanks! – Pirate1234 Oct 9 at 16:11
• For each root of $f$ you get a prime ideal $I=(x-\alpha_i,t)\subset \overline{\Bbb{F}}_p[t][x]$ above $(t)$, then $f(x) = (x-\alpha_i)^{e_i}g_i(x)$ with $g_i(\alpha_i) \ne 0$ thus $g_i$ is a unit in $\overline{\Bbb{F}}_p[t][x]/I$ so its $I$-adic valuation is $v_I(g_i) =0$ and $v_I(x) = \frac{v_I(t)-v_I(g_i)}{e_i} = 1/e_i$ as you claimed. – reuns Oct 9 at 19:18

First note that $$M = \overline{\mathbb F_p} (x)$$ because that field contains $$t$$ (it's $$f/x^i$$), so contains $$\overline{\mathbb F_p}(t)$$, and is generated over it by $$x$$, which is a root of $$f -t x^i=0$$.

1) This is correct:

You can observe that all the places lying over $$t=0$$ had better be places of $$M$$, which correspond to points of the projective line. If they are not the point at $$\infty$$, they lie over $$t=0$$ if and only if they are roots of $$f$$ (zero would be special here if we didn't get rid of it), and their ramification index is their multiplicity in $$f$$, as one can see by examining $$f/x^i$$, which is just $$f$$ in this case, in local coordinates.

The point $$x=\infty$$ is mapped to $$t=\infty$$ if $$\deg f>i$$, $$t=0$$ if $$\deg f, and something else if they are equal. In particular if $$i=0$$ it's mapped to $$\infty$$, so does not contribute to the ramification over $$t=0$$.

2) The only places that can possibly be sent to the infinite place are the pole $$0$$ of $$f/t^i$$, and the point $$\infty$$. As mentioned, $$\infty$$ is mapped to the infinite place if $$i < \deg f$$. The ramification index is again the order of vanishing of a local coordinate at $$t =\infty$$. We can take the local coodrinate to be $$t^{-1}$$, so we are interested in the order of vanishing of $$x^i/f(x)$$. The order of vanishing at $$x=0$$ is clearly $$i$$, unless $$x$$ happens to divide $$f$$, in which case it is lesser, so that is the ramification index at $$0$$. To get the order of vanishing at $$\infty$$, we need to choose a local coordinate at $$\infty$$, say $$y= x^{-1}$$, obtaining $$y^{-i} / f(y^{-1})$$. We can factor $$f(y^{-1})$$ as $$y^{-\deg f}$$ times a polynomial in $$y$$ nonvanishing at $$y=0$$, so this is $$y^{\deg f-i}$$ divided by a polynomial in $$y$$ nonvanishing at $$y=0$$, giving a ramification index of $$\deg f-i$$.

So there are two ramification points, one of index $$i$$ and one of index $$\deg f-i$$, with the latter point being removed if the formula for the index is not positive, and the former point having its index reduced if $$x$$ divides $$f$$.

3) The assumption can't change anything because it is a special case, but it does put very strong assumptions on $$f$$. In fact it should not be too hard to classify all $$f$$ such that this map is Galois.

• Thanks a lot for the answer! Do you think the classification you are hinting in 3) could be written down somewhere? – Pirate1234 Oct 9 at 19:30
• @Pirate1234 I'm sure someone has written it down but I wouldn't know where to look for it - it's probably easier to solve the problem than to do a literature search. – Will Sawin Oct 9 at 19:39