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 $\overline{\mathbb F}_p(t)$?

  • $\begingroup$ 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 $\overline{\mathbb F_p} (t)$)? $\endgroup$ – Will Sawin Oct 9 at 15:54
  • $\begingroup$ Sorry yes, of course I meant that! I'll edit, thanks! $\endgroup$ – Pirate1234 Oct 9 at 16:11
  • $\begingroup$ 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. $\endgroup$ – 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 $\deg f<i$, 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.

  • $\begingroup$ Thanks a lot for the answer! Do you think the classification you are hinting in 3) could be written down somewhere? $\endgroup$ – Pirate1234 Oct 9 at 19:30
  • $\begingroup$ @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. $\endgroup$ – Will Sawin Oct 9 at 19:39

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