The Galois group \(\mathrm{Gal}(K/\mathbb {Q})\) is a finite abelian group, hence a direct product of cyclic groups of prime power order. Each factor corresponds to a cyclic subextension of prime power degree, and \(K\) is their compositum.

By 56, any finite abelian extension of \(\mathbb {Q}\) is a compositum of cyclic extensions of prime power degree. By 57, each such extension can be replaced by a cyclotomic extension times a cyclic extension of prime power degree unramified outside \(p\). By 64 and 63, the latter is cyclotomic.

If \(K \not\subseteq K'\) and \(K' \not\subseteq K\), then \(KK'\) contains a subgroup isomorphic to \((\mathbb {Z}/p\mathbb {Z})^2\), contradicting the assumption that \(\mathrm{Gal}(KK'/\mathbb {Q})\) is cyclic.

A quadratic extension of \(\mathbb {Q}\) unramified outside \(2\) has discriminant a power of \(2\), so it is of the form \(\mathbb {Q}(\sqrt{d})\) with \(d \in \{ -1, \pm 2\} \). This gives exactly the three fields listed. Among these, only \(\mathbb {Q}(\sqrt{2})\) is real.

If \(m = 1\) we are done by 62. If \(m \geq 2\), assume first that \(K\) is nonreal. Then \(K(i)/K\) is a quadratic extension, and its maximal real subfield \(M\) is cyclic of degree \(2^m\) over \(\mathbb {Q}\). Since \(K/\mathbb {Q}\) is cyclotomic if and only if \(M\) is, we may assume that \(K\) is totally real.

Let \(K'\) be the maximal real subfield of \(\mathbb {Q}(\zeta _{2^{m+2}})\). If \(K'K\) is not cyclic, then it contains three real quadratic subfields unramified outside \(2\), contradicting 62. Thus \(K'K\) is cyclic, and 59 implies \(K = K'\).

Let \(K'\) be the subfield of degree \(p^m\) in \(\mathbb {Q}(\zeta _{p^{m+1}})\). If \(K'K\) is not cyclic, then it contains a subfield of type \((p, p)\) unramified outside \(p\), which contradicts 65. Thus \(K'K\) is cyclic, and 59 implies \(K = K'\).

Let \(K/\mathbb {Q}\) be a cyclic extension of prime degree \(p\) unramified outside \(p\). By the sequence of lemmas below, \(L = KF = \mathbb {Q}(\zeta _{p^2})\) (see 73). Therefore \(K \subseteq \mathbb {Q}(\zeta _{p^2})\). Since \([K:\mathbb {Q}] = p = [\mathbb {Q}(\zeta _{p^2}):F]\) and \(K \cap F = \mathbb {Q}\), \(K\) is the unique subfield of degree \(p\) of \(\mathbb {Q}(\zeta _{p^2})\) linearly disjoint from \(F\), which is the subfield of degree \(p\) of \(\mathbb {Q}(\zeta _{p^2})\) over \(\mathbb {Q}\). Since any two cyclic extensions of degree \(p\) unramified outside \(p\) are contained in \(\mathbb {Q}(\zeta _{p^2})\), the maximal such extension is exactly this subfield.

Since \([K:\mathbb {Q}] = p\) and \([F:\mathbb {Q}] = p-1\) are coprime, \(K \cap F = \mathbb {Q}\), so \([L:F] = [KF:F] = [K:\mathbb {Q}] = p\). The extension \(L/F\) is Galois with cyclic Galois group of order \(p\), and \(F\) contains the \(p\)-th roots of unity \(\mu _p\). By Kummer theory (IsCyclic + Kummer), \(L = F(\sqrt[p]{\mu })\) for some \(\mu \in F^\times \). Multiplying by a \(p\)-th power if necessary, we may assume \(\mu \in \mathcal{O}_F\).

For the ramification: let \(q \neq p\) be a prime. Since \(K/\mathbb {Q}\) is unramified at \(q\) and \(F/\mathbb {Q}\) is unramified at \(q\) (as \(F = \mathbb {Q}(\zeta _p)\) is only ramified at \(p\)), the compositum \(L = KF\) is also unramified at \(q\).

Let \(\sigma _a\) be an element of the decomposition group of \(\mathfrak {q}\) in \(\mathrm{Gal}(F/\mathbb {Q})\), so \(\sigma _a(\mathfrak {q}) = \mathfrak {q}\). Since \(L/\mathbb {Q}\) is abelian, by 67 there exists \(\xi \in F^\times \) such that \(\sigma _a(\mu ) = \xi ^p \mu ^a\). Since \(\sigma _a(\mathfrak {q}) = \mathfrak {q}\), applying \(\sigma _a\) to the factorization \((\mu ) = \mathfrak {q}^r \mathfrak {a}\) gives \((\xi ^p \mu ^a) = \mathfrak {q}^r \sigma _a(\mathfrak {a})\), hence \(\mathfrak {q}^r \| \xi ^p \mu ^a\). But also \((\mu ^a) = \mathfrak {q}^{ar}\mathfrak {a}^a\), so \(\mathfrak {q}^r \| \xi ^p \mu ^a\) implies \(r \equiv ar \pmod{p}\). Since \(p \nmid r\), this forces \(a \equiv 1 \pmod{p}\), so \(\sigma _a = 1\) and \(\mathfrak {q}\) splits completely in \(F/\mathbb {Q}\).

Write \(v_{\mathfrak {q}}(\mu ) = r\). If \(\mathfrak {q} \nmid p\), then \(\mathfrak {q}\) does not split completely in \(F/\mathbb {Q}\) (since \(F/\mathbb {Q}\) is totally ramified at \(p\) and unramified elsewhere, so primes above \(q \neq p\) have nontrivial decomposition group). By 68, \(p \mid r\). If \(\mathfrak {q} = \mathfrak {p}_p = (1-\zeta _p)\), the same argument applies: \(\mathfrak {p}_p\) does not split completely in \(F/\mathbb {Q}\), so \(p \mid r\).

By 69, \(p \mid v_{\mathfrak {q}}(\mu )\) for every prime \(\mathfrak {q}\). Therefore \((\mu ) = \mathfrak {a}^p\) where \(\mathfrak {a} = \prod _{\mathfrak {q}} \mathfrak {q}^{v_{\mathfrak {q}}(\mu )/p}\). In particular, \(v_{\mathfrak {p}_p}(\mu ) \equiv 0 \pmod{p}\), so either \(v_{\mathfrak {p}_p}(\mu ) = 0\) (i.e. \((1-\zeta _p) \nmid \mu \)) or \(v_{\mathfrak {p}_p}(\mu ) \geq p\). The case \(v_{\mathfrak {p}_p}(\mu ) \geq p\) would mean \((1-\zeta _p)^p \mid \mu \), but since \(L/F\) is unramified outside \(p\) and \(\mu \) generates the Kummer extension, replacing \(\mu \) by \(\mu /(1-\zeta _p)^p\) (which generates the same extension) we may assume \(v_{\mathfrak {p}_p}(\mu ) = 0\), i.e. \((1-\zeta _p) \nmid \mu \).

From \((\mu ) = \mathfrak {a}^p\) and \(\sigma _a(\mu ) = \xi ^p \mu ^a\) (by 67), we get \(\sigma _a(\mathfrak {a})^p = \sigma _a(\mathfrak {a}^p) = \sigma _a((\mu )) = (\xi ^p \mu ^a) = \mathfrak {a}^{pa}\). Since the ideal monoid is torsion-free, \(\sigma _a(\mathfrak {a}) = \mathfrak {a}^a\), hence \(\sigma _a([\mathfrak {a}]) = [\mathfrak {a}]^a\) for every \(1 \leq a {\lt} p\).

Now, \(\mathfrak {a}\) is coprime to \(p\) by 70. Write \(\mathfrak {a} = \prod _i \mathfrak {p}_i\) with each \(\mathfrak {p}_i \nmid p\). By 54, each \(\mathfrak {p}_i^{\theta }\) is principal in \(\mathbb {Q}(\zeta _p)\), hence \(\mathfrak {a}^{\theta }\) is principal. Therefore \(\mathfrak {a}^{p\theta }\) is principal. Since \(p\theta = \sum _{a=1}^{p-1} a\sigma _a^{-1} \in \mathbb {Z}[G]\), we compute \([\mathfrak {a}]^{p\theta }\):

where we used \(\sigma _a^{-1}([\mathfrak {a}]) = [\mathfrak {a}]^{a^{-1}}\) (since \(\sigma _{a^{-1}}([\mathfrak {a}]) = [\mathfrak {a}]^{a^{-1}}\)). Therefore \([\mathfrak {a}]^{-1} = 1\), i.e. \([\mathfrak {a}] = 1\).

By 71, \(\mathfrak {a} = (\alpha )\) for some \(\alpha \in \mathcal{O}_F\). Then \((\mu ) = \mathfrak {a}^p = (\alpha ^p)\), so \(\mu /\alpha ^p \in \mathcal{O}_F^\times \). Setting \(\eta = \mu /\alpha ^p\) gives \(\mu = \alpha ^p \eta \).

Write \(\eta = \zeta _p^t \varepsilon \) where \(\varepsilon \) is a unit in the maximal real subfield of \(F\), so \(\sigma _{-1}(\varepsilon ) = \varepsilon \). Since \(L/\mathbb {Q}\) is abelian, by 67 applied to \(\sigma _{-1}\), there exists \(\xi \in F^\times \) such that \(\sigma _{-1}(\mu ) = \xi ^p \mu ^{-1}\). Since \(\mu = \alpha ^p \zeta _p^t \varepsilon \),

Equating, \(\varepsilon ^2 = (\xi /\alpha )^{-p}\), so \(\varepsilon \) is a \(p\)-th power in \(F^\times \). Therefore \(\mu = \alpha ^p \zeta _p^t \varepsilon = (\alpha ')^p \zeta _p^t\) for some \(\alpha ' \in \mathcal{O}_F\), and \(L = F(\sqrt[p]{\mu }) = F(\sqrt[p]{\zeta _p^t}) \subseteq \mathbb {Q}(\zeta _{p^2})\). Since \([L:F] = p = [\mathbb {Q}(\zeta _{p^2}):F]\), we conclude \(L = \mathbb {Q}(\zeta _{p^2})\).