Polynomials whose fractional part behaves like a logarithm

I finally found a proof that there are no more examples of logarithm-like polynomials, at least for the original ring $R = \mathbb{Z}$. That is, any logarithmic-like polynomial $P$ for some $m$ is decomposable up to equivalence as a sum $P \cong \sum_{N=0}^{N_\text{max}} a_N P_{N,m}$, where $P_{N,m}$ is the sequence of polynomials I defined in the question. The proof is almost elementary; the only non-elementary ingredients it needs are the characterization of the group of units of a cyclic ring $(\mathbb{Z}/n\mathbb{Z})^\times$, and a theorem about the prime factors of cyclotomic polynomials. Furthermore, it is a constructive proof, in the sense that it gives an explicit method to compute the expansion of any such $P$ as an integral linear combination of a specific kind of generating polynomials.

First of all, following Merosity's suggestion in the comments, we will work locally at a prime $p$ by means of the decomposition $P \cong \sum_p \{P\}_p$, where $\{ \cdot \}_p$ denotes the $p$-adic fractional part.

Note that all the denominators appearing in $\{P\}_p$ are all prime powers by definition. This implies that both the period of $\{P\}_p$ (the least number $a$ such that $\{P\}_p(x+a) \cong \{P\}_p(x)$ for all $x$) and the order of $\{P\}_p$ (the least number $b$ such that $b \{P\}_p(x) \cong 0$) are prime powers, which will be useful later on.

Since for $p|m$ we have $\{P_{N,m}\}_p = k P_{N,p}$ for some integer $k$ coprime to $p$ (and $\{P_{N,m}\}_p = 0$ for $p \not\mid m$), it will be convenient to understand the behavior of the group $\Psi_p$ generated by the polynomials $P_{N,p}$. Later, armed with this knowledge, we will show how the $p$-parts $\{P\}_p$ of any polynomial may be expressed inductively as a sum of particular elements of $\Psi_p$.

Structure of $\Psi_p$

For odd $p$, we have $\Delta_{N,p}=p^{r+1}$, where $r$ is the maximum integer such that $p^r(p-1)|N$. For $p=2$, $\Delta_{N,2}=2$ if $N$ is odd and $\Delta_{N,2}=2^{r+2}$ if $N$ is even, where $r$ is the maximum integer such that $2^r|N$. These facts are easily deduced from the expression for the exponent of the group of units of a cyclic ring (concretely, $\Delta_{N,p}$ is the biggest power of $p$ whose Carmichael function divides $N$). In any case, observe that if we put $N=u p^r(p-1)$ with $\gcd(u,p)=1$, we have

$$P_{N,p}(x) = \frac{x^N-1}{\Delta_{N,p}^2} = \frac{x^N-1}{\Delta_{p^r(p-1),p}^2} = \frac{(x^u)^{p^r(p-1)}-1}{\Delta_{p^r(p-1),p}^2} = P_{p^r(p-1),p}(x^u) \cong u P_{p^r(p-1),p}(x),$$

where the last equivalence is due to the fact that $P_{N,p}$ is logarithm-like. Hence $\Psi_p$ is actually generated by the smaller set of polynomials $P_{p^r(p-1),p}$, which we will abbreviate as $g_{r,p}$.

From the definition of the $\Delta$'s it follows that the order of $P_{N,p}$ is simply $\Delta_{N,p}$. We can thus see that $g_{r,p}$ and $p\, g_{r+1,p}$ have the same order, except when $r=0$ and $p=2$, in which case it is $g_{0,2}$ and $4 g_{1,2}$ which have order $2$. A natural question arises: are these pairs of polynomials equivalent? The answer is yes for $p$ odd, we do have $g_{r,p} \cong p\, g_{r+1,p}$. In the $p=2$ case, that equivalence never holds; we can only assert that $2 g_{r,2} \cong 4 g_{r+1,2}$ for $r>1$. Hence there is only one generator of each prime power order for $p$ odd, and two generators if $p=2$, which become the same upon multiplication by two (except if the order is $4$, for which there is only one generator $2 g_{1,2}$). See the footnote (*) for the proof of all these statements.

The following tables show the first few generator polynomials and their properties for $p$ odd and $p=2$ respectively:

The $p$ odd case

How does all this help us? Let's deal with odd $p$ first. Suppose that $\{P\}_p$ has order $p^e$ and period $p^f$. We can think of it as a function from $(\mathbb{Z}/p^f\mathbb{Z})^\times$ to $\mathbb{Q}/\mathbb{Z}$. Since $(\mathbb{Z}/p^f\mathbb{Z})^\times$ is known to be a cyclic group, let's say generated by $\gamma$, all the values of $\{P\}_p$ are determined by the value of $\{P\}_p$ in (any integral lift of) $\gamma$, since by the logarithm-like property $\{P\}_p(x) \cong \{P\}_p(\gamma^t) \cong t \{P\}_p(\gamma)$.

Now choose an integer $\alpha_{e-1}$ such that $p^{e-1}\{P\}_p(\gamma) \cong \alpha_{e-1}\, g_{0,p}(\gamma)$ (this can be done because both $p^{e-1}\{P\}_p$ and $g_{0,p}$ have order $p$). Since $g_{0,p} \cong p^{e-1} g_{e-1,p}$, one can rewrite this as $p^{e-1}(\{P\}_p-\alpha_{e-1}\, g_{e-1,p}) = 0 \pmod 1$, that is, the polynomial $\{P\}_p-\alpha_{e-1}\, g_{e-1,p}$ has order $p^{e-1}$. Repeating this procedure with this new polynomial we obtain a new one of order $p^{e-2}$, and so on, until in the end we get $\{P\}_p-\alpha_{e-1}\, g_{e-1,p}-\alpha_{e-2}\, g_{e-2,p} \ldots \cong 0$, i.e., we found a decomposition

$$\{P\}_p \cong \alpha_{e-1}\, g_{e-1,p} +\alpha_{e-2}\, g_{e-2,p} +\ldots \alpha_0\, g_{0,p},$$

which shows that $\{P\}_p \in \Psi_p$ as we wanted.

The $p=2$ case

Now set $p=2$, and assume $\{P\}_2$ has order $2^e$ and period $2^f$. As before, $\{P\}_2$ can be thought of as a function from $(\mathbb{Z}/2^f\mathbb{Z})^\times$ to $\mathbb{Q}/\mathbb{Z}$. The group $(\mathbb{Z}/2^f\mathbb{Z})^\times$ is known to be trivial for $f=1$, cyclic for $f=2$ (generated by $-1$), and the product of two cyclic groups for $f>2$ (generated respectively by $-1$ and $5$). Note that modulo $1$ we have $g_{0,2}(-1) \cong 4g_{1,2}(5) \cong 1/2$ and $g_{0,2}(5) \cong 4g_{1,2}(-1) \cong 0$.

Now, since $2^{e-1}\{P\}_p$ has order $2$, modulo $1$ we must have $2^{e-1}\{P\}_p(-1)\cong\beta_{e-1}/2$ and $2^{e-1}\{P\}_p(5)\cong\alpha_{e-1}/2$ for some integers $0 \le \alpha_{e-1}, \beta_{e-1} \le 1$. By the generation property we then have $2^{e-1}\{P\} - \beta_{e-1}\, g_{0,2} - 4\alpha_{e-1}\, g_{1,2} \cong 0$. If $e=1$, we are done; otherwise note that $\beta_{e-1}$ has to be zero, as for any logarithm-like polynomial $2P(-1) \cong P((-1)^2) = P(1) \cong 0$, so since $4 g_{1,2} \cong 2^{e-1} g_{e-2,2}$, one can rewrite this as $2^{e-1}(\{P\}_2- \alpha_{e-1} g_{e-2,2}) = 0 \pmod 1$, that is, the polynomial $\{P\}_2- \alpha_{e-1} g_{e-2,2}$ has order $2^{e-1}$. Repeating this procedure in the same way as before, we find the decomposition

$$\{P\}_2 \cong \alpha_{e-1}\, g_{e-2,2} + \alpha_{e-2}\, g_{e-3,2} + \ldots + (\alpha_2 + 2\alpha_1 + 4 \alpha_0) g_{1,2} + \beta_0\, g_{0,2},$$

which shows that $\{P\}_2 \in \Psi_2$.

The global case

Finally, to deal with the case of composite $m$ we simply note that $g_{r,p}=P_{p^r(p-1),p}$ is an integer multiple of $P_{p^r(p-1),m}$ if $p$ is among the factors of $m$ (since in general if $p|m$ then $P_{N,p}$ is a multiple of $P_{N,m}$, as follows from the definition of the associated $\Delta$'s), so as a consequence any $P \cong \sum_p \{P\}_p$ will be expressible as an integral linear combination of $P_{N,m}$.

Final comments

A possible drawback of this proof is that I'm not sure it can be generalized to the case of orders in arbitrary global fields; for example, in $\mathbb{F}_2[t]$ the denominators of the $p$-adic parts of $P^{(\mathbb{F}_2[t])}_N$ depend on the factorization properties of $\Delta^{(\mathbb{F}_2[t])}_N = 2^N-1$, i.e. on the theory of Mersenne numbers, which (to my limited knowledge) do not yet seem to have an easy characterization.

(*) First, it is easy to show that for any prime (even or odd) and $r>0$ we have $p\, g_{r,p} \cong p^2 g_{r+1,p}$; e.g. for $p$ odd we have:

$$p^2 g_{r+1,p}(x) = p^2 \frac{x^{p^{r+1}(p-1)}-1}{p^{2(r+1)+2}} = \frac{(x^p)^{p^r(p-1)}-1}{p^{2r+2}} = g_{r,p}(x^p) \cong p\, g_{r,p}(x),$$

whereas for $p=2$ the denominators contains an extra factor of $4$, but the proof goes through identically. To prove that in the $p$ odd case we can refine this relation to $g_{r,p} \cong p\, g_{r+1,p}$, consider the factorization

$$p\, g_{r+1,p}(x) = \frac{x^{p^{r+1}(p-1)}-1}{p^{2r+3}} = \frac{x^{p^r(p-1)}-1}{p^{2r+2}} \cdot \frac{1+x^{p^r(p-1)}+x^{2p^r(p-1)}+\cdots+x^{(p-1)p^r(p-1)}}{p} = g_{r,p}(x) \frac{A(x)}{p}.$$

The second factor $A(x)/p$, which is an integer, must clearly be congruent to $1$ modulo $p^{r+1}$ for the equivalence to hold, since the order of $g_{r,p}$ in $\mathbb{Q}/\mathbb{Z}$ is $p^{r+1}$. That is, we must have

$$A(x)=1+x^{p^r(p-1)}+x^{2p^r(p-1)}+\cdots+x^{(p-1)p^r(p-1)} = p \pmod {p^{r+2}}.$$

This congruence certainly holds modulo $p^{r+1}$, since by Euler's strengthening of Fermat's little theorem $x^{p^r(p-1)}=1 \pmod {p^{r+1}}$ (recall that $x$ is always coprime to $p$), and there are $p$ summands. Hence $A(x) = p + k p^{r+1} \pmod {p^{r+2}}$ for some $k$ depending on $x$. But note now that $A(x) = \Phi_{p^{r+1}}(x^{p-1})$, where $\Phi_{p^{r+1}}$ is a cyclotomic polynomial. It is known that any prime factor of $\Phi_{n}(a)$ for any integers $n, a$ is either a factor of $n$ or congruent to $1$ modulo $n$. Hence $A(x)=p^s (1+l p^{r+1})$ for some $s$, $l$ depending on $x$. Comparing with the previous expression, we have $s=1$, so $A(x)=p (1+l p^{r+1}) = p + l p^{r+2} = p \pmod {p^{r+2}}$, as we needed.

In the case $p=2$, the obstruction that prevents this proof from working is precisely the extra factor of $4$ in the denominators, which means that the order of $g_{r,2}$ is twice the "expected" one. There seems to be numerical evidence that an equivalence similar to the odd case, namely $2g_{r+1,2} \cong g_{r,2} + 4 g_{1,2}$, might hold for $r>1$ (recall that $4 g_{1,2}$ has order $2$), but this is not needed for the proof above.