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

This question was previously posted here on MSE.

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 $$i\in\{1,\dotsc,p^n\}$$ such that there is an integer $$x$$ for which $$P(x)\equiv i\bmod p^n$$. 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 $$p$$?

This question is connected to another question Cardinality of the image of a polynomial modulo $p^n$ on MathOverflow, which asks for general information on the behavior of $$I_n$$ as $$n\rightarrow\infty$$.

• Note: barring finitely many critical values, the fact that $y$ is in the image of $P$ can, by Hensel's lemma, be decided from a finite approximation of $y$. So it suffices to answer the problem in a ball around a critical value. Oct 1, 2021 at 16:54
• It could be connected to the limit of the Igusa zeta function as $s>0$ approaches $0$. Oct 2, 2021 at 11:08
• Around critical points, the density locally equals that of $f(z)=z^n$ around $z=0$, for some $n>1$ (as can be seen by letting $z$ be an appropriate power series in $x$). It should not be so hard to prove that this is rational.
– R.P.
Oct 2, 2021 at 11:08
• The strategy worked! I hope to soon submit the proof to the site. Oct 3, 2021 at 20:35
• For power series, doesn't it immediately follow from the case of polynomials via Weierstraß preparation? Or can there be convergence issues? Oct 7, 2021 at 16:00

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 $$S_v:=\{a^{n_v}\mid 0. Then it follows from arguments similar to the proof of Hensel's lemma that for all $$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.$$\tag*{\blacksquare}$$