Is the p-adic density of the image of a polynomial always rational?

Question

Let $P(x)$ be a polynomial with integer coefficients, and let $p$ be a prime number. For $n\in\mathbb N$, let $I_n$ be the number of integers $i\in\{1,\dots,p^n\}$ such that there is an integer $x$ for which $P(x)\equiv i\mod p^n$. Now define $$\delta:=\lim_{n\rightarrow\infty}\frac{I_n}{p^n}.$$ Remark that this limit exists since $\frac{I_n}{p^n}\geq \frac{I_{n+1}}{p^{n+1}}$ for all $n$. One could say that $\delta$ is `the p-adic density of the image of $P$'.

Now I have the following question: is $\delta$ a rational number for all polynomials $P$ and primes $p$?

Answer 1

Using the strategies suggested by @Merosity on MSE and @Gro-Tsen and @RP_ on MO, I have found a proof that the density is indeed always rational.

Let $P$ be a polynomial with integer coefficients, and let $p$ be a prime number. If $P$ is a constant polynomial, then we obtain $\delta=0$ which is rational. So assume that $P$ is nonconstant. We shall prove that $\delta$ is rational by means of a chain of lemmas. Each lemma only uses the previous lemma.

Lemma

Let $g:\mathbb Z_p\rightarrow\mathbb Z_p$ be a power series $g(x)=\sum_{i=0}^{\infty}g_ix^i$ with $g_i\in\mathbb Z_p$ for all $i$ and $g_i\rightarrow0$ as $i\rightarrow\infty$. Suppose that $g'(0)=1$. Then the restriction of $g$ to $p\mathbb Z_p$ has image $g(0)+p\mathbb Z_p$.

Proof

The proof is analoguous to that of Hensel's lemma.$\tag*{$\blacksquare$}$

Now let $v\in\mathbb Z_p$ for which $P'(v)=0$, and let $n_v\geq2$ be the largest integer such that $(x-v)^{n_v}$ divides $P(x)-P(v)$. Let $Q_v(x)\in\mathbb Z_p[x]$ be the polynomial such that $P(x)=P(v)+(x-v)^{n_v}Q_v(x-v)$. Now $Q_v(0)\neq0$.

Lemma For all $v\in\mathbb Z_p$ such that $P'(v)=0$, there is an integer $N_v\in\mathbb N$ such that $$P(v+p^{N_v}\mathbb Z_p)=P(v)+p^{n_vN_v}Q_v(0)f_v(\mathbb Z_p),$$ where $f_v:\mathbb Z_p\rightarrow\mathbb Z_p$ is defined by $f_v(z):=z^{n_v}$.

Proof Define the function $R:\mathbb Z_p\rightarrow\mathbb Z_p$ by $$R(x)=\sum_{i=0}^{\infty}r_ix^i:=\sum_{i=0}^{\infty}p^{2i}n_v^{i}\binom{\frac{1}{n_v}}{i}x^i.$$ For all $i$, we have $v_p(r_i)=v_p(p^{2i}n_v^{i}\binom{\frac{1}{n_v}}{i})\geq v_p(p^{2i})-v_p(i!)>i$, so all $r_i$ are $p$-adic numbers and $\lim_{i\rightarrow\infty}r_i=0$. Therefore, $R(x)$ is well-defined.\ It follows from the definition of $R$ that $R(x)^{n_v}=1+p^2n_vx$ for all $x\in\mathbb Z_p$. Now define the function $K:\mathbb Z_p\rightarrow \mathbb Z_p$ by $$K(x):=R\left(\frac{Q_v(p^2n_vQ_v(0)x)-Q_v(0)}{p^2n_vQ_v(0)}\right).$$ Then $K(x)=\sum_{i=0}^{\infty}k_ix^i$ for coefficients $k_i\in\mathbb Z_p$ with $\lim_{i\rightarrow\infty}k_i=0$. It follows for all $x\in\mathbb Z_p$ that $$Q_v(0)K(x)^{n_v}=Q_v(0)\left(1+p^2n_v\frac{Q_v(p^2n_vQ_v(0)x)-Q_v(0)}{p^2n_vQ_v(0)}\right)=Q_v(p^2n_vQ_v(0)x).$$ Therefore, we see that \begin{align*}P(v+p^2n_vQ_v(0)x)&=P(v)+(p^2n_vQ_v(0)x)^{n_v}Q_v(p^2n_vQ_v(0)x)\\ &=P(v)+(p^2n_vQ_v(0)x)^{n_v}Q_v(0)K(x)^{n_v}\\ &=P(v)+Q_v(0)(p^2n_vQ_v(0)xK(x))^{n_v}\end{align*} for all $x\in\mathbb Z_p$. Since $\frac{d(xK(x))}{dx}\big|_{x=0}=K(0)=R(0)=r_0=1$, we can use the previous lemma to see that the image of $xK(x)$ restricted to $p\mathbb Z_p$ is $p\mathbb Z_p$. Therefore, the image of $P$ restricted to $v+p^3n_vQ_v(0)\mathbb Z_p$ is $P(v)+Q_v(0)(p^3n_vQ_v(0))^{n_v}f_v(\mathbb Z_p)$. So the lemma holds for $N_v:=v_p(p^3n_vQ_v(0))$. $\tag*{$\blacksquare$}$

Lemma For all $v\in\mathbb Z_p$ such that $P'(v)=0$, there is a finite set $S_v\subset\mathbb Z$ such that $$P(v+p^{N_v}\mathbb Z_p)=P(v)+p^{n_vN_v}Q_v(0)\left(\{0\}\cup\left(\bigcup_{s\in S_v}\bigcup_{i=0}^{\infty}p^{in_v}(s+p^{2v_p(n_v)+1}\mathbb Z_p)\right)\right).$$

Proof Define $S_v:=\{a^{n_v}\mid 0<a<p^{v_p(n_v)+1},p\nmid a\}$. Then it follows from arguments similar to the proof of Hensel's lemma that for all $i\geq0$, the image of $f_v$ restricted to $p^i\mathbb Z_p\backslash p^{i+1}\mathbb Z_p$ is equal to $p^{in_v}\bigcup_{s\in S_v}(s+p^{2v_p(n_v)+1}\mathbb Z_p)$. Taking the union over all $i$, we see that the image of $f_v$ equals $\{0\}\cup\left(\bigcup_{s\in S_v}\bigcup_{i=0}^{\infty}p^{in_v}(s+p^{2v_p(n_v)+1}\mathbb Z_p)\right)$. Therefore, this lemma follows from the previous lemma. $\tag*{$\blacksquare$}$

Lemma For all $\sigma\in P(\mathbb Z_p)$, there is an integer $M_{\sigma}\geq0$ such that $$P(\mathbb Z_p)\cap(\sigma+p^{M_{\sigma}}\mathbb Z_p)$$ has rational $p$-adic density.

Proof First, suppose that there is an $x\in\mathbb Z_p$ such that $P(x)=\sigma$ and $P'(x)\neq0$. Then it follows from arguments similar to Hensel's lemma that $P(\mathbb Z_p)$ contains a neighbourhood of $\sigma$. This immediately proves the lemma.\ Now suppose that for all $x\in\mathbb Z_p$ such that $P(x)=\sigma$, we have $P'(x)=0$. Let $V_{\sigma}:=P^{-1}(\sigma)$, then $V_{\sigma}$ is a finite set since $P$ is nonconstant. Since $P$ is continuous, we can choose $M_{\sigma}\in\mathbb Z_{\geq0}$ such that $P^{-1}(\sigma+p^{M_{\sigma}}\mathbb Z_p)$ is contained in $\bigcup_{v\in V_{\sigma}}(v+p^{N_v}\mathbb Z_p)$. Now it follows that $$P(\mathbb Z_p)\cap(\sigma+p^{M_{\sigma}}\mathbb Z_p)=\bigcup_{v\in V_{\sigma}}P(v+p^{N_v}\mathbb Z_p)\cap(\sigma+p^{M_{\sigma}}\mathbb Z_p).$$ Using the previous lemma, we see that this set equals $$\bigcup_{v\in V_{\sigma}}\left(\sigma+p^{n_vN_v}Q_v(0)\left(\{0\}\cup\left(\bigcup_{s\in S_v}\bigcup_{i=0}^{\infty}p^{in_v}(s+p^{2v_p(n_v)+1}\mathbb Z_p)\right)\right)\right)\cap(\sigma+p^{M_{\sigma}}\mathbb Z_p).$$ Let $n:=\mathrm{lcm}_{v\in V_{\sigma}}(n_v)$. Then there exists an integer $C>0$ and a finite collection of numbers $a_l\in \mathbb Z_p\backslash p^{C}\mathbb Z_p$, $1\leq l\leq L$, such that our set can be written as $$P(\mathbb Z_p)\cap (\sigma+p^{M_{\sigma}}\mathbb Z_p)= \{\sigma\}\cup\bigcup_{l=1}^{L}\left(\sigma+\bigcup_{i=0}^{\infty}p^{in}(a_l+p^{C}\mathbb Z_p)\right).$$ The $p$-adic density of this set is $$\lvert\{a_l\bmod p^{C}\mid 1\leq l\leq L\}\rvert\cdot \frac{p^n}{p^n-1}\cdot p^{-C}$$ which is a rational number. $\tag*{$\blacksquare$}$

Theorem The $p$-adic density of $P(\mathbb Z_p)$ is rational.

Proof When we define $B_{\sigma}:=\sigma+p^{M_{\sigma}}\mathbb Z_p$ for all $\sigma\in P(\mathbb Z_p)$, then $\{B_{\sigma}\mid \sigma\in P(\mathbb Z_p)\}$ is an open cover of $P(\mathbb Z_p)$. Since $\mathbb Z_p$ is a compact set and $P$ is continuous, the image $P(\mathbb Z_p)$ is also a compact set. Therefore, the open cover has a finite subcover $\{B_{\sigma_1},\dots,B_{\sigma_q}\}$ which is minimal. The sets in this subcover must be pairwise disjoint, so it follows that $P(\mathbb Z_p)$ is the disjoint union of the sets $B_{\sigma_i}\cap P(\mathbb Z_p)$ for $1\leq i\leq q$. Therefore, the $p$-adic density of $P(\mathbb Z_p)$ is the sum of the densities of these sets. By the last lemma, all those densities are rational. Therefore their sum is also rational.$\tag*{$\blacksquare$}$

Answer 2

EDIT: The answer I gave was a bit too terse and left out some critical details which left a hole in the proof. Fortunately it can be salvaged with some caveats based on where the zeroes of the derivative lie which are described after the proof of the following lemma.

First let's prove a lemma that establishes an equality of sets,

$$P(\mathbb{Z}_p) = \bigcup_{x \in \mathbb{Z}_p} B_{\le |pP'(x)^2|}(P(x))$$

Here the notation $B_{\le r}(c) :=\{z \in \mathbb{Z}_p : |z-c|\le r\}$. Note that $r=0$ is allowed, except it's no longer an open ball, but the singleton, $B_{\le 0}(c) = \{c\}$.

We clearly see that because $P(x)$ is the center of every ball or singleton in the union, the image is contained within it:

$$P(\mathbb{Z}_p) \subseteq \bigcup_{x \in \mathbb{Z}_p} B_{\le |pP'(x)^2|}(P(x))$$

To show the other containment, take an arbitrary point of a ball, $y \in B_{\le |pP'(x)^2|}(P(x))$. This satisfies the inequality (when $P'(x) \ne 0$),

$$|P(x)-y|\le |pP'(x)^2| < |P'(x)|^2$$

If we pick $f(x)=P(x)-y$ we have that the general criteria for Hensel's lemma is satisfied for $|f(x)|<|f'(x)|^2$, and so $x_{n+1}=x_n-\frac{P(x_n)-y}{P'(x_n)}$ converges to some $z$ that makes $P(z)-y=0$, which proves $y=P(z) \in P(\mathbb{Z}_p)$. Just as a quick remark, the singleton case when $P'(x)=0$ forces the inequality $|P(x)-y|\le 0$ and so we have simply $P(x)=y$ and no Hensel lemma required.

Now that we have established this lemma, here are where the caveats will begin to apply. The singleton sets are precisely the points $\{P(x)\}$ where $P'(x)=0$. There are only finitely many of these problem points for any given polynomial, so some possibilities for these to not cause problems are when the only $x$ which gives $P'(x)=0$ lies outside $\mathbb{Z}_p$, possibly in $\mathbb{Q}_p$ or a finite extension (for instance $px^2+x$ or $x^3-x$), or the set $\{P(x)\}$ is contained in one of the open balls (I don't have an example of this last case off hand). Checking the Newton polygon of $P'$ is a simple way to confirming/creating any "nice" polynomial that meets the first criteria.

Supposing now that we have no problem singleton sets, the lemma shows $P(\mathbb{Z}_p)$ is a union of infinitely many open balls. Since $P$ is continuous and $\mathbb{Z}_p$ is compact, we know $P(\mathbb{Z}_p)$ is compact. These open balls form a cover that must contain some finite subcover by balls, of which there is some minimum radius. In other words, we can represent the image for some constant $p^{-n}$ and some centers $\{c_t\}_{t=1}^N$ as,

$$P(\mathbb{Z}_p) = \bigcup_{t=1}^N B_{\le p^{-n}}(c_t)$$

Specifically we can choose $c_t \in \{0,\dots,p^n-1\}$ and use this to compute $I_n=N$ because $P(x)=c_t \mod p^n$ for some p-adic integer $x$ which, by density of the integers in $\mathbb{Z}_p$ lets us pick a regular integer to do so. So our ratio is $\frac{I_n}{p^n} = \frac{N}{p^n}$.

We can further subdivide these balls evenly into the disjoint balls,

$$B_{\le p^{-n}}(c_t) = \bigcup_{a=0}^{p-1} B_{\le p^{-n-1}}(c_t+ap^n)$$

Each of these balls contains $p$ more integer points which gives us solutions to all of $P(x)=c_t+ap^n \mod p^{n+1}$ for each $a \in \{0,\dots,p-1\}$. Each of these $N$ balls when subdivided this way gets us $p$ times as many points, so we have $\frac{I_{n+1}}{p^{n+1}} = \frac{Np}{p^{n+1}} =\frac{N}{p^n}$. We can repeat the subdivision process and the numerator and denominator will continue to multiply by the equal powers of $p$, leaving us with,

$$\delta = \frac{N}{p^n}$$

Unfortunately the singleton sets where the derivative is 0 is a problem still, but maybe with a bit more work these can be resolved too.