2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.