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When $\Bbb{Q}(e^{2\pi i/n})$ = $Kn$ is the $n$th cyclotomic field, and $\Phi(n)$ is the $n$th cyclotomic polynomial, the primes $p$ such that $\Phi(n)$ is factored into $\phi(n)$ linear factors when factored over the finite field $\Bbb{F}_p$ of order $p$ are those primes $p$ congruent to $1$ $\pmod n$. The only primes $n$ with class number $h=1$ are $2, 3, 5, 7, 11, 13, 17, 19$. The class numbers for $Kn$ are here.

Now let $Sn =$ $\Bbb{Q}(e^{\pi i/n})$ be a subfield of $Kn$ the $n$th cyclotomic field where the polynomial $P(n)$ defines $Sn$ (the minimal polynomial of $e^{2\pi i/n}$ + $e^{{2\pi i/n}^{n-1}}$). The primes $p$ such that $P(n)$ is factored into $\phi(n)$ linear factors when factored over the finite field $\Bbb{F}_p$ of order $p$ are those primes $p$ congruent to $±1$ $\pmod n$. Is it true that for all primes n, the class number $h$ of $\Bbb{Q}(e^{\pi i/n})$ is $1$? This is proven for primes $n < 179$, and belived to be true for all prime $n$. There is no OEIS sequence for the class numbers of the fields $\Bbb{Q}(e^{\pi i/n})$, subfields of the $n$th cyclotomic fields.

J. Linne
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  • Could you add a reference for the proof for $n < 179$ ? – lhf Nov 23 '17 at 18:24
  • What's the point of bringing up the factorizations of $\Phi(n)$? Anyway, the real subfield of conductor 163 has class number divisible by 4 since the cubic field with this conductor has class number 4. You can play the same game with quadratic subfields. The latest edition of Washington's Cyclotomic fields has tables of class number factors of real subfields computed by Schoof. –  Nov 23 '17 at 18:44
  • $\Bbb Q(e^{i\pi/n})$ is not a subfield of $\Bbb Q(e^{2i\pi/n})$, it is the other way around. – mercio Nov 29 '17 at 14:04

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