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For an integer $n \ge 3$, consider the function $$u(n) = \frac{\sigma(n)}{n \log \log n}$$ with $\sigma$ the divisor function. Now consider the sequence (bounded below and decreasing) $$v_n = \sup_{m>n} u(m).$$

Then Robin's theorem states that the Riemann hypothesis (RH) is true if and only if the sequence $(v_n)$ reaches its limit (which means the existence of $m \in \mathbb{N}$ such that $v_m = \lim_{n \to \infty} v_n$).


To illustrate the complexity of the ring $\mathbb{Z}[\zeta_r]$ of cyclotomic integers of degree $r$ (with $\zeta_r=e^{2i\pi/r}$), here is the picture of the set $$\{\sum_{i=0}^4 a_i\zeta_5^i \ | \ a_i \in \mathbb{Z}, |a_i| \le 3 \} \subset \mathbb{Z}[\zeta_5].$$
enter image description here


Consider the quotient map $$i: \mathbb{Z}[\zeta_r] \to \mathbb{Z}[\zeta_r]/\sim$$ with $x \sim y$ if and only if there is a unit $u \in \mathbb{Z}[\zeta_r]$ such that $x=uy$. Then consider $$\mathcal{P}_r = i(\mathbb{Z}[\zeta_r] \cap \mathbb{R}),$$ where $\mathbb{Z}[\zeta_r] \cap \mathbb{R}$ is the ring of real cyclotomic integers of degree $r$.

For $x \in \mathcal{P}_r$, and $a \in i^{-1}(\{ x\})$, consider the set of divisors $$D(x)= i(\{ b \in \mathbb{Z}[\zeta_r] \cap \mathbb{R} \ | \ a/b \in \mathbb{Z}[\zeta_r] \cap \mathbb{R} \}).$$

Question 1: Is the set $D(x)$ finite (for $x$ nonzero)?

As pointed out by Wojowu in comment (before revision), the group of units can be huge, and so without quotient, the set of divisors could be infinite (for example, let $\phi$ be the golden ratio, then $\phi^k, \phi^{-k} \in \mathbb{Z}[\zeta_5]$, and $1 = \phi^k \cdot \phi^{-k}$ for all $k$). Then Question 1 asks whether the quotient is sufficient to make the set finite.

Answer to Question 1 (after the comments of François Brunault): Yes.
Proof: the ring of algebraic integers in a finite extension of $\mathbb{Q}$ is a Dedekind domain. So any nonzero proper ideal of $\mathbb{Z}[\zeta_r] \cap \mathbb{R}$ decomposes uniquely (up to permutation) into a product of (finitely many) prime ideals. Then a nonzero ideal admits only finitely many dividing ideals. But $$x \sim y \Leftrightarrow x(\mathbb{Z}[\zeta_r] \cap \mathbb{R}) = y(\mathbb{Z}[\zeta_r] \cap \mathbb{R}),$$ so the set $D(x)$ is in bijection with the set of singly generated dividing ideals of $x(\mathbb{Z}[\zeta_r] \cap \mathbb{R})$. $\square$


To define a divisor function $\sigma$, we first need to define a natural non-negative representative for each element of $\mathcal{P}_r$, by a map $$\tau: \mathcal{P}_r \to \mathbb{R}_{\ge 0}.$$ For so, first consider the surjective map: $$ s: \mathbb{Z}^{r} \to \mathbb{Z}[\zeta_r]$$ such that $s(n_0,n_1, \dots, n_{r-1}) = \sum_i n_i \zeta_r^i$.

Let $P(\mathbb{Z}^{r})$ be the set of subsets of $\mathbb{Z}^{r}$. Consider the minimizers map $$m: P(\mathbb{Z}^{r}) \to P(\mathbb{Z}^{r})$$ such that $$m(S) = \{ v \in S \ | \ \Vert v \Vert_2 \le \Vert w \Vert_2, \ \forall w \in S \},$$ next consider the lexicographical order on $\mathbb{Z}^{r}$ defined by $$(n_0, n_1, \dots, n_{r-1}) \le (n'_0, n'_1, \dots, n'_{r-1}) \Leftrightarrow $$ $$n_{r-1} < n'_{r-1} \text{ or } (n_{r-1} = n'_{r-1} \text{ and } n_{r-2} < n'_{r-2}) \text{ or } \dots$$ Finally consider the lexicographical first map $l: P(\mathbb{Z}^{r}) \to \mathbb{Z}^{r}$ such that $l(S) = min(S,\le)$.

Now we can defined the teased map $\tau: \mathcal{P}_r \to \mathbb{R}_{\ge 0}$ as follows: $$ \tau(x) = s \circ l \circ m \circ s^{-1} (i^{-1} (\{x \})\cap \mathbb{R}_{\ge 0}).$$

Note that $\tau$ is unital, i.e. for every unit $u \in \mathbb{Z}[\zeta_r] \cap \mathbb{R}$ then $\tau(i(u))=1$.

For $x \in \mathcal{P}_r$, consider the divisor function $$\sigma(x) = \sum_{y \in D(x)} \tau(y),$$ next, for $\tau(x)>e$, consider the function $$u(x) = \frac{\sigma(x)}{\tau(x) \log \log \tau(x)},$$ and finally consider the sequence $$v_n = \sup_{\tau(x)>n} u(x).$$ Question 2: Is $v_n$ finite?


Assume that Question 2 admits a positive answer (otherwise the definition of $u$ should be modified using a more appropriate asymptotic). Then a degree $r$ cyclotomic Riemann hypothesis (CRH$_r$) can be stated as follows:

The decreasing sequence $v_n$ (defined above with $\mathcal{P}_r$) reaches its limit.

Note that for $r=1,2,3,4,6$, (CRH$_r$) is obviously (RH) because then $\mathbb{Z}[\zeta_r] \cap \mathbb{R} = \mathbb{Z} $.

Question 3.1: Is (CRH$_r$) equivalent to (RH) for all $r$?
Question 3.2: Is there $r$ such that (CRH$_r$) is false?


Note that we can get alternative versions of the post replacing $\mathbb{Z}[\zeta_r] $ by any Noetherian subring of the algebraic integers.

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    $\begingroup$ I'm fairly certain the cyclotomic units are dense in $\mathbb R$ for all $r$ a prime larger than $5$, and hence $D(x)$ is going to be very much infinite. This should follow from Dirichlet Unit Theorem and the classical fact that all units are a root of unity times a real number. $\endgroup$
    – Wojowu
    Commented Jun 25, 2020 at 15:27
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    $\begingroup$ @Wojowu: I just did some modifications according to your comment. Now Q1 asks whether the quotient by the unit group makes the divisor set finite. Next to define the divisor function, and I introduced a natural representative for each quotient class. $\endgroup$ Commented Jun 26, 2020 at 13:33
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    $\begingroup$ You are (essentially) looking at the ring of integers of the maximal real subfield $\mathbb{Q}(\zeta_r)^+$ of the cyclotomic field. There are only finitely many ideals dividing a given nonzero ideal. $\endgroup$ Commented Jun 27, 2020 at 9:01
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    $\begingroup$ @SebastienPalcoux This is true because of uniqueness of factorisation into prime ideals. More generally, this holds in Dedekind rings. $\endgroup$ Commented Jun 27, 2020 at 10:52
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    $\begingroup$ Yes, I misunderstood what you meant by representative. Your map is well-defined, but I believe the expected asymptotic will be different as soon as the degree is bigger than $1$ (so $n$ not $1,2,3,4,6$.) This is because your quantity is essentially a small power of the norm. You'll get something like $e^{ log n^{ 1-1/d} / \log \log n}$ if I did the math right, where $d$ is the degree. It may not be as sensitive to the Riemann hypothesis in this case. $\endgroup$
    – Will Sawin
    Commented Jun 27, 2020 at 13:49

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