Number fields with finite maximal unramified $p$-extensions

Question

As discussed in the question number fields with no unramified extensions, it is an open question whether there are an infinite number of number fields that have no unramified extensions. Inspired by this, I would like to ask the weaker question as follows:

Question:

1. Let $p$ be a fixed prime. Are there infinitely many number fields such that its maximal unramified $p$-extension is finite?
2. Is there a prime $p$ such that $\mathbb{Q}(\zeta_{p^m})$ admits a finite maximal unramified $p$-extension for all positive integer $m$?

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Update: the answer for question $1$ is Yes. In Theorem I of Construction of maximal unramified p-extensions with prescribed Galois groups by Manabu Ozaki, it's proved that

for any prime $p$ and any given finite $p$-group $G$, there exists a number field $F$ such that the Galois group of the maximal unramified $p$-extension over $F$ is isomorphic to $G$.

Since there are infinitely many finite $p$-groups, we see that there are infinitely many number fields such that its maximal unramified $p$-extension is finite.

Answer 1

One can do this without any recent theorems. Let $\ell$ be a prime congruent to $1$ mod $p$. Then there is a Galois extension $F$ of $\mathbb Q$ with Galois group $\mathbb Z/p$ ramified only at $\ell$, arising as a subfield of $\mathbb Q(\mu_\ell)$. Consider the maximal unramified $p$-extension $K$ of $F$.

$K$ is necessarily Galois over $\mathbb Q$ because the condition of being a $p$-extension is stable under automorphisms of $\mathbb Q(\mu_\ell)$, it is a $p$-extension, and it is ramified only at $\ell$.

The abelianization of $\operatorname{Gal}(K/F)$ is associated to some abelian $p$-extension of $\mathbb Q$ ramified only at $\ell$. By Kronecker-Weber, such an extension is contained in the $\ell$-power roots of unity and has Galois group a quotient of $\mathbb Z_\ell^\times$. In particular its Galois group is cyclic.

But since $p$-groups are nilpotent, any $p$-group with cyclic abelianization is itself cyclic, and since its abelianization is finite, must be finite. (In fact, this shows $\operatorname{Gal}(K/\mathbb Q)$ is $\mathbb Z/p$ so $\operatorname{Gal}(K/\mathbb Q)$ is trivial.)

By Dirichlet's theorem there are infinitely many primes congruent to $1$ mod $p$ and we are done.

Answer 2

The answer to question 2 is also yes. To see this, we need the following two facts.

Fact 1: Let $\zeta_{p^{m}}$ be a primitive $p^m$-th root of unity. Then the class number of $\mathbb{Q}(\zeta_{p^{m}})$ is divisible by $p$ iff the class number of $\mathbb{Q}(\zeta_p)$ is. See Example 12.4 on p.136 of [Koch, H., 2002. Galois theory of p-extensions. Springer Science & Business Media.]

Fact 2: A number field $K$ can be imbeded into another number field with class number prime to $p$ iff the degree of the maximal unramified $p$-extension (or $p$-class field tower) of $K$ is finite. See Proposition 2 on p.232 of [Cassels, J.W.S. and Fröhlich, A., Algebraic number theory, London, 1986.]

By Fact 1 and 2, those primes $p$ satisfying question 2 are exactly those primes such that the class number of $\mathbb{Q}(\zeta_p)$ is prime to $p$, i.e. they are exactly regular primes.