**Question:** Are the class numbers of $\mathbb{Q}(\cos(\frac{2\pi}m))$ $O(m^n)$ for some fixed $n$?


Evidences (e.g. a recent paper) showing that the question above is open are also OK.


**Remark:** If such $n$ exists, then $n\geq2$. By the paper [unit groups and class numbers of real cyclic octic fields](https://pdfs.semanticscholar.org/0b59/fdd291c7d9c6181fbfbd9e7dbede4767ece5.pdf) by Yuan-Yuan Shen, there exists infinitely many cyclic octic fields with conductor $32b$ and class numbers at least $c(\epsilon)b^{3-\epsilon}$ for every $\epsilon>0$. As every number field extension $E/F$ satisfies $h(F)$ divides $[E:F]h(E)$, the class numbers of such $\mathbb{Q}(\cos(\frac{2\pi}{32b}))$s are at least $c(\epsilon)b^{2-\epsilon}$.


A key feature of Yuan-Yuan Shen's octic fields is that they have regulators of size $\log^6(b)$. If such an infinite family of number fields (i.e. real abelian with poly-logarithmic regulators) exists for every fixed degree, the answer for this question will be "no".



Are the class numbers of $\mathbb{Q}(\cos(2\pi / m))$ $O(m^n)$ for some fixed $n$?
The answer is **no**.


By Proposition 2 of Gary Cornell and Michael I. Rosen 's paper [The -rank of the real class group of cyclotomic fields](http://www.numdam.org/article/CM_1984__53_2_133_0.pdf) (paraphased):



> 
> Let $L/\mathbb Q$ be an abelian -extension of the rationals with Galois group
> $G$. Assume the inertia group of $2$ is cyclic. If $G$ is the direct sum of
> $t$ cyclic groups and $s$ finite primes of $\mathbb Q$ ramify in $L$ then the -rank of
> the class group of $L / \mathbb Q \geq \frac{t(t-1)}{2}-s-e$, where $e=0$ for $$ odd.
> 
> 
> 


Let $A$ be the first $n$ primes with the property that $x^3-2$ has three roots in $\mathbb F\_a$ ($a\in A$). By Chebotarev's density theorem, the largest element of $A$ is $O(n \log n)$.


Let $L\_a$ be the cubic field with conductor $a$. Such cubic fields must exist, as the property of $a$ ensures that $1$ has three cubic roots in $\mathbb F\_a$, so $a=1 \text{ mod }3$. Notice that $2$ splits completely in $L\_a$.


Let $L$ be the compositium of all $L\_a$ s. Then $L/\mathbb Q$ is an abelian $3$-extension with Galois group $C\_3^n$, and $2$ splits completely (so the inertia group of $2$ is trivial). There are only $n$ primes that ramify in $L$, so the size of the class group is at least $\exp (c\_1n^2)$ for some $c\_1>0$.


Also the conductor is bounded above by $\exp(c\_2n\log^2 n)$ for some $c\_2$, so the extension $\mathbb Q (2 \cos (2\pi / \prod\_{a\in A} a))/L$ has degree at most $\exp(c\_2n\log^2 n)$, and thus the class number of $\mathbb Q (2 \cos (2\pi / \prod\_{a\in A} a))/L$ is at least $\exp(c\_1n^2-c\_2n\log^2 n)$. This is asymptotically larger than $\exp(c\_2n\log^2 n)^m$ for any $m$.


