Let $p$ be a prime, and let $|\cdot|_p$ be the $p$-adic absolute value. Let $f(x) \in \mathbb{Z}[x]$ be a monic polynomial with $f(0) = 1$.
Question 1 What is the volume $c_n$ of the following set $$\{a \in \mathbb{Q}_p: |f(a)|_p = p^n\}$$ for any given $n$?
Question 2 For a given $p$, we can form the sequence $(c_n)_{n \in \mathbb{Z}}$. Are there infinitely many possible sequences? For a given sequence, what is the density of $p$'s with that sequence?
My thoughts For question 1: the situation where $|a| \neq 1$ are clear via ultrametric inequality: when $|a|_p < 1$, $|f(a)|_p = 1$; and when $|a|_p > 1$, $|f(a)|_p = p^{degree(f)}$.
However, I am not sure what happens when $|a|_p = 1$. A first step should be to look at $$\{a \in \mathbb{F}_p^*: f(a) \equiv 0 \mod p\}$$ and I suppose Hensel's lemma can help me lift to the situation of $\mathbb{Z}_p$. I am mostly fuzzy about a few points though:
- On how to characterize the $a$'s with $f(a) \equiv 0 \mod p$: this is probably related to reciprocity law; maybe there is no good explicit description for general $f$?
- On how to control the valuation of $f(a)$, when $f(a) \equiv 0 \mod p$; I do not know how to think about this.
- On what happens when Hensel's lemma fails, e.g. when $f \mod p$ has repeated factors.
- For question 2, as $f(a) \equiv 0 \mod p$ should be closely related to factorization of $p$ in the splitting field of $f$, it smells like Chebotarev density, but it's also different since I need to consider $f \mod p^n$, not just $f \mod p$.
Thanks!
General context The general motivation here is to compute total volume of $\mathbb{A}^1(\mathbb{Q}_p)$ with respect to the measure $|f(a)|_p da$, where $da$ is the standard Haar measure.
I'll cut to the chase, if you're wanting to integrate something of the form,
$$\int_{\mathbb{Z}_p}|f(z)|^s dz$$
You can chop it into a disjoint union $\mathbb{Z}_p=\bigsqcup_{n=0}^{p-1}n+p\mathbb{Z}_p$ to focus on the simpler $p$ pieces. We could do this more fine grained, however doing one power of $p$ at a time lets us think of this recursively like Hensel's lemma in parallel and also helps keep our feet on the floor without being needlessly abstract.
$$= \sum_{n=0}^{p-1}\int_{n+p\mathbb{Z}_p}|f(z)|^s dz$$
If we recall Hensel's lemma, if we find a $z$ such that $|f(z)|<|f'(z)|^2$ it implies that it's actually the distinct root in the open ball of radius $|f'(z)|$. This isn't too important, the main concept we really care about is that distinct roots are separable into disjoint balls. The point is, by splitting our integral up into separate integrals over disjoint balls, eventually we can ignore the other roots.
For convenience, although not necessary, let's factor $f(z)$ in $\mathbb{C}_p$, since we will have entirely linear factors, which may occur with some multiplicity. Now look at the p-adic absolute value of these factors and how they will appear to these different balls we have just dissected $\mathbb{Z}_p$ into.
I'll take a specific example so it's clear. Take $f(z)=z^4-1$ in $\mathbb{Q}_3$. If we factor it entirely in $\mathbb{C}_3$ we have $(z+1)(z-1)(z+i)(z-i)$ and now let's look at $|f(z)|$ in the three balls $3\mathbb{Z}_3$, $1+3\mathbb{Z}_3$, and $2+3\mathbb{Z}_3$. In each ball we can effectively imagine plugging in $0,1,2 \mod 3$ and seeing what gets cast out or not.
In $3\mathbb{Z}_3$ have $|f(z)|=|(z+1)(z-1)(z+i)(z-i)|=|(1)(-1)(i)(-i)|=1$
In $1+3\mathbb{Z}_3$ have $|f(z)|=|(z+1)(z-1)(z+i)(z-i)|=|(1+1)(z-1)(1+i)(1-i)|=|z-1|$
In $2+3\mathbb{Z}_3$ have $|f(z)|=|(z+1)(z-1)(z+i)(z-i)|=|(z+1)(2-1)(2+i)(2-i)|=|z+1|$
By simply zooming in on specific balls, we can safely discard other factors. This means for any sufficiently fine partition, our integral looks like and integral of $z^k$, which is a very simple integral to evaluate. (If you're wondering why I said $z^k$ and not $(z-r)^k$, recall what you know about the center of an ultrametric ball.)
Let's resume to our more general integral. Now we can do a change of variable since $z\in n+p\mathbb{Z}_p$ means we can take $z=n+pz'$ with $z' \in \mathbb{Z}_p$. The corresponding Jacobian factor is $\frac{1}{p}$, which is inherited from being a Haar measure, $\mu(n+p\mathbb{Z}_p)=\frac{1}{p}\mu(\mathbb{Z}_p)$.
$$= \frac{1}{p}\sum_{n=0}^{p-1}\int_{\mathbb{Z}_p}|f(n+pz')|^s dz'$$
At this step we have these new polynomials $f(n+pz')$ and will cast out all terms except linear terms with root congruent to $n\mod p$ like described earlier with our 3-adic example.
Now for each of these $p$ integral we rinse and repeat until we get something of the form $z^s$, which we can now integrate by these methods - and I'll walk through it here.
$$J=\int_{\mathbb{Z}_p} |z|^s dz = \frac{1}{p}\sum_{n=0}^{p-1}\int_{\mathbb{Z}_p} |n+pz|^s dz$$
$$= \frac{1}{p}\int_{\mathbb{Z}_p} |pz|^s dz + \frac{1}{p}\sum_{n=1}^{p-1}\int_{\mathbb{Z}_p} dz$$
$$= \frac{1}{p^{1+s}}\int_{\mathbb{Z}_p} |z|^s dz + \frac{p-1}{p}$$
$$J = p^{-1-s}J + 1-p^{-1}$$
Solving for $J$ with simple algebra,
$$J = \frac{1-p^{-1-s}}{1-p^{-1}}$$
Now hopefully it's a bit clearer, and makes it possible to see the answers to these questions you've asked. If you like, instead of recursively stepping you can simply imagine going directly from the integral and breaking it down to an integral over the roots, with $g(z)$ irreducible,
$$f(z)=g(z)\prod_r^n(z-r)^{m_r}$$
Now we can partition $\mathbb{Z}_p$ over the largest balls that each contain only a single distinct root, plus a little more that doesn't contian any roots,
$$\mathbb{Z}_p=A\sqcup\bigsqcup_r r+p^{\alpha_r}\mathbb{Z}_p$$
Here $\alpha_r=\max_{s\ne r} v_p(r-s)$ is the maximum valuation, which gives us the largest ball enclosing $r$ and none of the other roots $s$. $A$ is just the space leftover that doesn't fall into any of our balls.
Now if we work through it using the same ideas above, you'll find:
$$\int|f(z)|^s dz = \mu(A)+ \sum_r \frac{1}{p^{\alpha_r}}\int|(p^{\alpha_r}z)^{m_r}|^s dz$$
$$= \left(1- \sum_r \frac{1}{p^{\alpha_r}}\right) + \sum_r \frac{1}{p^{\alpha_r(1+sm_r)}}\frac{1-p^{-1}}{1-p^{-1-sm_r}} $$
$$= 1 + \sum_r \frac{1}{p^{\alpha_r}} \left( \frac{1}{p^{s\alpha_rm_r}}\frac{1-p^{-1}}{1-p^{-1-sm_r}} -1 \right)$$
This should be enough for you to now be able to answer all of your own questions, but I can see how it might not be entirely obvious and I've left out some steps. Try to work through the details as best you can and feel free to ask for clarification on anything.