Integral points of polynomials - a Furstenburg-type "topology" on $\mathbb{Z}$

Question

Given $S \subseteq \mathbb{C}$, define $\displaystyle \mathfrak{c}(S) = \bigcap_{p(x) \in \mathbb{C}[x] \wedge p(S) \subseteq \mathbb{Z}}p^{-1}(\mathbb{Z}) \supseteq S$ ("the integral points emerging from $S$"). My quesiton is, for $S \subseteq \mathbb{Z}$, how can we describe $\mathfrak{c}(S)$? Specifically, when will $\mathfrak{c}(S) = \mathbb{Z}$?

Here are some basic properties of $\mathfrak{c}(S)$ (the proofs are omitted because they are not the keypoints of my question):
(0) $A \subseteq B \implies \mathfrak{c}(A) \subseteq \mathfrak{c}(B)$; $\mathfrak{c}(\mathfrak{c}(S)) = \mathfrak{c}(S)$ since $p(S) \subseteq \mathbb{Z} \iff p(\mathfrak{c}(S)) \subseteq \mathbb{Z}$; $(\forall p(x) \in \mathbb{C}[x], p(A) \subseteq \mathbb{Z} \implies p(B) \subseteq \mathbb{Z}) \iff B \subseteq \mathfrak{c}(A)$.

(1) $\forall a,d \in \mathbb{C} \wedge d \ne 0$, one has $\mathfrak{c}(a + dS) = a + d \mathfrak{c}(S)$.

(2)$\forall F \subseteq \mathbb{C} \wedge \# F < \infty$, one has $\mathfrak{c}(F) = F$.

(3)$\mathfrak{c}(\varnothing) = \varnothing$; $\mathfrak{c}(\mathbb{Z}) = \mathbb{Z}$.

A naive candidate for $\mathfrak{c}(S)$ when $S \subseteq \mathbb{Z}$ is as follows. Endow $\mathbb{Z}$ with Furstenberg topology whose base is given by $\varnothing$ and all nontrivial residue classes (i.e. $a + d \mathbb{Z}$ where $a,d \in \mathbb{Z} \wedge d \ne 0$). One may first conjecture that $\mathfrak{c}(S) = \bar{S}$ by the reasoning below.

First case: $S$ is finite. Then $\bar{S} = S$ since Furstenberg topology is Hausdorff and $\mathfrak{c}(S) = S$ from property (2) & (3) of $\mathfrak{c}(S)$. So $\mathfrak{c}(S) = \bar{S}$ in this case.

Second case: There is some integers $a$, $d$ ($d \ne 0$) such that $S$ is dense in $a + d \mathbb{Z}$. By property (1) of $\mathfrak{c}(S)$ and the linearity of Furstenberg topology one can assume that $a = 0 \wedge d = 1$. If $S = \mathbb{Z}$ we are done. Otherwise by property (1) of $\mathfrak{c}(S)$ and the linearity of Furstenberg topology again one can assume that $0 \notin S$. I will prove that $0 \in \mathfrak{c}(S)$, then by translation $\mathfrak{c}(S) = \mathbb{Z} = \bar{S}$ in this case. According to the first case $\# S = \infty$, so $\forall p(x) \in \mathbb{C}[x]$, $p(S) \subseteq \mathbb{Z} \implies p(x) \in \mathbb{Q}[x]$. Write the aforementioned $p(x)$ as $\dfrac{q(x)}{r}$ where $q(x) \in \mathbb{Z}[x] \wedge r \in \mathbb{Z}$. Since $S$ is dense in $\mathbb{Z}$, the open set $r \mathbb{Z}$ intersects $S$, which means $\exists k \in \mathbb{Z}, p(kr) \in p(S) \in \mathbb{Z}$, i.e. $\dfrac{q(kr)}{r} \in \mathbb{Z}$. Moreover, $q(x) \in \mathbb{Z}[x] \implies (kr - 0) \mid (q(kr) - q(0)) \implies \dfrac{q(kr) - q(0)}{r} \in \mathbb{Z}$. Thus, $p(0) = \dfrac{q(0)}{r} \in \mathbb{Z}$, $0 \in \mathfrak{c}(S)$.

But things get different if $S$ is not contained in a single residue class. The following result (see the accepted answer below) shows that $\mathfrak{c}(S)$ isn't a Kuratowski closure operator: $$\forall n \in \mathbb{Z}^+, \mathfrak{c}(\mathbb{Z} \backslash n \mathbb{Z}) = \begin{cases} \mathbb{Z} \backslash n \mathbb{Z} & \text{if $n$ is a prime power} \\ \mathbb{Z} & \text{otherwise} \end{cases}$$ and a precise description of $\mathfrak{c}(S)$ seems out of my reach.

Answer 1

For $S\subset \mathbb{Z}$, we have $a\in \mathfrak{c}(S)$ if and only if the following condition $C(S,a)$ holds:

$a\in \mathbb{Z}$ and for every prime power $p^m$ there exists an element $x\in S$ congruent to $a$ modulo $p^m$.

In particular, $\mathfrak{c}(S)=\mathbb{Z}$ if and only if $S$ contains all residues modulo each prime power.

At first, if $a\notin \mathbb{Z}$, then the polynomial $p(x)=x$ shows that $a\notin \mathfrak{c}(S)$, so, let further $a\in \mathbb{Z}$.

Next, we assume that $C(S,a)$ holds and prove that $a\in \mathfrak{c}(S)$. It means that for every polynomial $p$ which maps $S$ to $\mathbb{Z}$ we must have $p(a)\in \mathbb{Z}$. The condition $C(S,a)$ yields that $S$ is infinite, thus $p$ has rational coefficients by Lagrange interpolation. So $p(t)=f(t)/d$ for some $f$ with integer coefficients and a positive integer $d$. We must prove that $d$ divides $f(a)$. For this goal, it suffices to check that every prime power $p^m$ which divides $d$ also divides $f(a)$. By $C(S,a)$, there exists $x\in S$ such that $p^m$ divides $x-a$. Then $p^m$ divides both $f(x)$ and $f(x)-f(a)$ (since $x-a$ divides $f(x)-f(a)$ for $f$ with integer coefficients). Thus, $p^m$ divides $f(a)=f(x)-(f(x)-f(a))$.

To other direction, assume that for some prime power $p^m$ there is no element in $S$ congruent to $a$ modulo $p^m$. Put $$f(t)=\frac1p \prod_{j=1}^{p^m-1}\frac{t-a-j}{j}=\frac1p{t-a-1\choose p^m-1}.$$ Obviously, $f(a)$ is not integer, and it remains to prove that $f(x)$ is integer for every integer $x$ not congruent to $a$ modulo $p^m$ (in particular, for $x\in S$). At first, $pf(x)={x-a-1\choose p^m-1}$ is an integer. It remains to prove that this integer $pf(x)$ is divisible by $p$. Note that $pf(x)\cdot \frac{x-a-p^m}{p^m}$ is also an integer (it is nothing but $x-a-1\choose p^m$). But if $pf(x)$ was not divisible by $p$, the number $pf(x)\cdot (x-a-p^m)$ would not be divisible by $p^m$, since $p^m$ does not divide $x-a$. A contradiction, showing that $pf(x)$ is divisible by $p$, thus $f(x)$ is an integer.