Class number of cubic subfield of $\mathbb{Q}(\zeta_p)$ is 1 mod 3?

Question

Let $K$ be the cubic subfield of $\mathbb{Q}(\zeta_p)$ for a prime $p\equiv 1\bmod 3$. Numerical data suggests that $h_K\equiv 1\bmod 3$.

Does anyone know if this is true in general and how one can prove this? I can exclude some integers as special cases by considering the Hilbert class field $H$ of $K$, but I could not find a general pattern.

For example, if we would have $h_K=2$, then $H/\mathbb{Q}$ is a Galois extension of degree $6$. But then $G=\mathrm{Gal}(H/\mathbb{Q})$ is cyclic as $K/\mathbb{Q}$ is normal, which would force $H\subset\mathbb{Q}(\zeta_p)$ as well by Kronecker-Weber. This contradicts that $H/K$ is unramified however since $p$ is totally ramified in $\mathbb{Q}(\zeta_p)$.

Any help would be appreciated!

Answer 1

It seems the hint by @Infinity needs further clarification. Below is a complete solution, but still follows closely the hint.

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Let $p,q$ be primes such that $p\equiv 1 \pmod{q}$, let $K$ be the degree $q$ subfield of $\mathbb{Q}(\zeta_p)$. Let $h$ denotes the class number of $K$, then $h \equiv 1\pmod{q}$.

Proof: A large part of the argument is group-theoretical.

Let $H$ denote the Hilbert class field of $K$, $G=\text{Gal}(H/\mathbb{Q}), N = \text{Gal}(H/K), Q=\text{Gal}(K/\mathbb{Q})$, then $$0\to N\to G\to Q\to 0$$ is exact, it splits by considering a inertia field. $N$ has a $\mathbb{Z}[Q]$-module structure induced from the sequence. Consider $N$ under the action of $Q$, orbits of size $1$ corresponds to elements in $N^Q$ (elements of $N$ invariant under $Q$), so $$h = |N| \equiv |N^Q| \pmod{q}$$

It suffices to show that $|N^Q|=1$. Writing $N$ in addition and $Q$ in multiplication, the action determines $G= N\rtimes Q$ via $$(a,x)(b,y) = (a+xb,xy)\qquad a,b\in N, x,y\in Q$$

Now the commutator $[N,Q]$ satisfies $[N,Q]\subset N$ and $G/[N,Q] \cong Q\times (N/[N,Q])$, therefore $G/[N,Q]$ is abelian. Since $K$ is ramified at only one prime, $[N,Q] = N$ (by considering an inertia field). On the other hand, $(a,1)(0,x)(a,1)^{-1}(0,x)^{-1} = ((1-x)a,1)$.

If $Q=\langle \theta \rangle$ and $I=(1-\theta)$ is an ideal inside the group ring $\mathbb{Z}[Q]$, then $IN=[N,Q]=N$. Nakayama's lemma says there exists $r\in \mathbb{Z}[Q]$ such that $r\equiv 1 \pmod{I}$ and $rN=0$. Write $r = f(\theta)(1-\theta) + 1$, where $f$ is a polynomial, for $a\in N^Q$, $a-\theta a=0$, so $$0 = ra = f(\theta)(1-\theta)a + a = a$$ hence $N^Q$ is trivial, completing the proof.

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Alternatively, as pointed out by @franz lemmermeyer, $N^Q$ is the ambiguous class group of $K/\mathbb{Q}$, so by ambiguous class number formula, it is trivial.