A prime \(q\) that is unramified in \(K\) and in \(F\) is unramified in \(KF\) (Mathlib PR #36843).

Take for \(A\) an algebraic closure of \(K\), and choose a primitive \(n\)-th root of unity in \(A\) for every \(n \geq 1\). The image of \(K\) in \(A\) is an abelian extension of \(\mathbb {Q}\), so by 66 it lies in \(\mathbb {Q}(\zeta _n)\) for some \(n \geq 1\). The field \(\mathbb {Q}(\zeta _n)\) is isomorphic to the abstract cyclotomic field of level \(n\).

The group \(\mathrm{Gal}(K/F)\) 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 65, the field \(K\) is the compositum of finitely many cyclic subextensions \(C_i\) of prime power degree. By 67, each \(C_i\) lies in some \(\mathbb {Q}(\zeta _{n_i})\). By induction on the number of factors, \(K\) lies in \(\mathbb {Q}(\zeta _n)\) with \(n\) the least common multiple of the \(n_i\).

By 79, there are a cyclic extension \(F/\mathbb {Q}\) of degree \(p^k\), unramified outside \(p\), and an integer \(n \geq 1\) with \(K\mathbb {Q}(\zeta _n) = F\mathbb {Q}(\zeta _n)\). The field \(F\) lies in some \(\mathbb {Q}(\zeta _c)\): take \(c = 1\) if \(k = 0\), use 104 if \(p = 2\), and 97 if \(p\) is odd and \(k {\gt} 0\). Then

A linear endomorphism of a one-dimensional vector space is multiplication by its determinant.

Since \(\theta (\tau ) = 1\) and \(Q/Q^2\) is one-dimensional, \(\tau \) acts trivially on \(Q/Q^2\), that is, \(\tau x - x \in Q^2\) for all \(x \in Q\). From the identity

and induction on \(n\), one gets \(\tau x - x \in Q^{n+1}\) for every \(x \in Q^n\).

We now show by induction on \(n\) that \(\delta = \tau x - x\) lies in \(Q^n\) for every \(x \in S\). Assume that \(\delta \in Q^n\). By the previous step, \(\tau ^k \delta \equiv \delta \pmod{Q^{n+1}}\) for all \(k\). The sum \(\sum _{k=0}^{m-1} \tau ^k \delta \) telescopes to \(\tau ^m x - x = 0\), so \(m\delta \in Q^{n+1}\). As \(m\) is a unit modulo \(Q\), it is a unit modulo \(Q^{n+1}\), hence \(\delta \in Q^{n+1}\).

By Krull’s intersection theorem, \(\bigcap _n Q^n = 0\), so \(\tau x = x\) for every \(x \in S\). The action being faithful, \(\tau = 1\).

The order of every \(\tau \in I(Q)\) divides \(\# I(Q)\), so it is not zero in \(S/Q\). Hence \(\theta \) is injective by 70. Thus \(I(Q)\) embeds into the multiplicative group of the finite field \(S/Q\), and a finite subgroup of the multiplicative group of a field is cyclic.

The inertia group is normal in \(D(Q)\), so \(\sigma \tau \sigma ^{-1} \in I(Q)\). The action of \(\sigma \) on \(Q/Q^2\) is semilinear: \(\sigma (cv) = \sigma (c)\sigma (v)\) for \(c \in S/Q\). Using 69, for every \(v \in Q/Q^2\),

Let \(\tau \) generate the cyclic group \(I(Q)\) (71) and let \(\sigma \) be an arithmetic Frobenius at \(Q\), which exists since \(R = S^G\). As \(G\) is abelian, \(\sigma \tau \sigma ^{-1} = \tau \), so 72 gives \(\theta (\tau )^{N\mathfrak {p}- 1} = 1\). Since \(\theta \) is injective, \(\tau ^{N\mathfrak {p}- 1} = 1\) and the order \(\# I(Q)\) of \(\tau \) divides \(N\mathfrak {p}- 1\).

The map \(C \to C/H_1 \times C/H_2\) is injective, and the exponent of the target divides the least common multiple of the two orders. A cyclic group has order equal to its exponent.

Let \(G = \mathrm{Gal}(K/\mathbb {Q})\) and \(I = I(\mathfrak {P}) \subseteq G\). As \(p \nmid \# G\), the prime \(\mathfrak {P}\) is tame, so \(I\) is cyclic by 71, of order \(e(\mathfrak {P}\mid p)\). Write \(H_E = \mathrm{Gal}(K/E)\) and \(H_F = \mathrm{Gal}(K/F)\). The group \(I \cap H_E\) is the inertia group of \(\mathfrak {P}\) in \(K/E\), so the multiplicativity of ramification indices in \(\mathbb {Q}\subseteq E \subseteq K\) gives \([I : I\cap H_E] = e(\mathfrak {P}\cap E\mid p)\), and likewise for \(F\). Since \(EF = K\), we have \(H_E \cap H_F = 1\), and 74 applies to \(I \cap H_E\) and \(I \cap H_F\).

We first show that \(e(\mathfrak {Q}\mid q) = q - 1\). By 75, \(e(\mathfrak {Q}\mid q)\) divides the least common multiple of \(e(\mathfrak {Q}\cap K\mid q)\) and \(e(\mathfrak {Q}\cap E\mid q) = q-1\). The extension \(K/\mathbb {Q}\) is abelian and \(q \nmid p^m\), so 73 gives that \(e(\mathfrak {Q}\cap K\mid q)\) divides \(q - 1\). Hence \(e(\mathfrak {Q}\mid q)\) divides \(q - 1\). Conversely \(q - 1 = e(\mathfrak {Q}\cap E\mid q)\) divides \(e(\mathfrak {Q}\mid q)\).

It follows that \(e(\mathfrak {Q}\mid \mathfrak {Q}\cap E) = 1\). This ramification index is the order of \(I \cap \mathrm{Gal}(L/E)\), the inertia group of \(\mathfrak {Q}\) in \(L/E\), so \(I \cap \mathrm{Gal}(L/E) = 1\). Finally \(\# I \cdot \# \mathrm{Gal}(L/E) = (q-1)[L:E] = [E:\mathbb {Q}][L:E] = [L:\mathbb {Q}]\).

By the Galois correspondence, \(FE\) is the fixed field of \(I \cap \mathrm{Gal}(L/E) = 1\), that is, \(L\). The isomorphism \(G/I \cong \mathrm{Gal}(L/E)\) is induced by the inclusion of \(\mathrm{Gal}(L/E)\) in \(G\).

Put \(E = \mathbb {Q}(\zeta _q)\) and \(L = KE\). The extension \(L/\mathbb {Q}\) is abelian, and \([L:\mathbb {Q}]\) divides \([K:\mathbb {Q}][E:\mathbb {Q}] = p^m(q-1)\), so \(q \nmid [L:\mathbb {Q}]\). Fix a prime \(\mathfrak {Q}\) of \(L\) above \(q\), with inertia group \(I \subseteq \mathrm{Gal}(L/\mathbb {Q})\). By 76, the group \(I\) is a complement of \(\mathrm{Gal}(L/E)\). Let \(F = L^{I}\) be the inertia field of \(\mathfrak {Q}\). By 77, we get \(FE = L = KE\) and

Restriction to \(K\) is injective on \(\mathrm{Gal}(L/E)\), since \(KE = L\). Hence \(\mathrm{Gal}(F/\mathbb {Q})\) is isomorphic to a subgroup of the cyclic group \(\mathrm{Gal}(K/\mathbb {Q})\), so it is cyclic of order dividing \(p^m\).

The prime \(q\) is unramified in the inertia field \(F\) (Mathlib PR #36733). Let \(q'\) be a prime unramified in \(K\). Then \(q' \neq q\), so \(q'\) is also unramified in \(E = \mathbb {Q}(\zeta _q)\), hence in the compositum \(L = KE\) (Mathlib PR #36843), and finally in the subfield \(F\) of \(L\).

Let \(S\) be the finite set of prime divisors of the discriminant of \(K\). Every prime \(q \neq p\) outside \(S\) is unramified in \(K\). We prove the statement by induction on \(S\), for all cyclic extensions \(K\) of \(p\)-power degree in which every prime \(q \neq p\) outside \(S\) is unramified.

If \(S\) is empty, take \(F = K\) and \(n = 1\). Otherwise write \(S = \{ r\} \cup S'\). If \(r = p\), or if \(r\) is unramified in \(K\), the induction hypothesis applies to \(K\) and \(S'\). Otherwise 78 gives a cyclic extension \(E\) of \(p\)-power degree, unramified at \(r\) and at every prime unramified in \(K\), with \(K\mathbb {Q}(\zeta _r) = E\mathbb {Q}(\zeta _r)\). The induction hypothesis applied to \(E\) and \(S'\) gives \(F\) and \(n\) with \(E\mathbb {Q}(\zeta _n) = F\mathbb {Q}(\zeta _n)\). Then

Let \(G = \mathrm{Gal}(KK'/\mathbb {Q})\). By the Galois correspondence, it suffices to show that \(\mathrm{Gal}(KK'/K') \subseteq \mathrm{Gal}(KK'/K)\). In the cyclic group \(G\), a subgroup is contained in another if and only if its order divides the order of the other. The orders are \([KK':K'] = [KK':\mathbb {Q}]/[K':\mathbb {Q}]\) and \([KK':K] = [KK':\mathbb {Q}]/[K:\mathbb {Q}]\), and the first divides the second since \([K:\mathbb {Q}]\) divides \([K':\mathbb {Q}]\).

By the Galois correspondence, \([C^{H} : \mathbb {Q}] = [\mathrm{Gal}(C/\mathbb {Q}) : H] = p^m\) and \(\mathrm{Gal}(C^{H}/\mathbb {Q}) \cong \mathrm{Gal}(C/\mathbb {Q})/H\), which is cyclic. A subfield of \(C\) is unramified outside \(p\) because \(C\) is.

The group \(\mathrm{Gal}(KK'/\mathbb {Q})\) is an abelian \(p\)-group, since \([KK':\mathbb {Q}]\) divides \(p^{2m}\). If it were not cyclic, it would have two distinct subgroups of index \(p\). Their fixed fields would be two distinct cyclic subfields of \(KK'\) of degree \(p\), unramified outside \(p\) by 63, a contradiction. So \(KK'/\mathbb {Q}\) is cyclic, and 80 gives \(K \subseteq K'\). The degrees being equal, \(K = K'\).

Since \([F:\mathbb {Q}] = p - 1\), the tower formula gives \([L:F] = p\). The extension \(L/F\) is Galois as \(L/\mathbb {Q}\) is, and a group of prime order is cyclic.

The field \(F\) contains the \(p\)-th roots of unity, so by Kummer theory \(L = F(\alpha )\) with \(\alpha ^p = x \in F\). Choose a nonzero integer \(d\) with \(dx \in \mathcal{O}_F\). Then \(\mu = d^p x\) lies in \(\mathcal{O}_F\), the element \(d\alpha \) is a root of \(X^p - \mu \), and \(L = F(d\alpha )\) is the splitting field of \(X^p - \mu \). Finally \(\mu \neq 0\), since otherwise \(L = F\).

Take a root \(\alpha \in L\) of \(X^p - \mu \). It is integral since \(\alpha ^p = \mu \) is, and nonzero since \(L \neq F\).

Let \(\alpha \in L\) be a root of \(X^p - \mu \), let \(\tau \) generate \(\mathrm{Gal}(L/F)\) and write \(\tau (\alpha ) = \zeta \alpha \), with \(\zeta \) a primitive \(p\)-th root of unity. Extend \(\sigma _a\) to \(\tilde\sigma \in \mathrm{Gal}(L/\mathbb {Q})\). As \(\tilde\sigma (\alpha )^p = \sigma _a(\mu ) \in F\), Kummer theory gives \(\tilde\sigma (\alpha ) = c\alpha ^j\) with \(c \in F\) and \(0 \leq j {\lt} p\). Since \(L/\mathbb {Q}\) is abelian, \(\tilde\sigma \tau = \tau \tilde\sigma \). Applying both sides to \(\alpha \) gives \(\zeta ^a c \alpha ^j = c\zeta ^j\alpha ^j\), so \(j = a\). Therefore \(\sigma _a(\mu ) = \tilde\sigma (\alpha )^p = c^p\mu ^a\).

Write \(\sigma = \sigma _a\) and \(r = v_{\mathfrak {q}}(\mu )\). By 86, \(\sigma (\mu ) = \xi ^p\mu ^a\). Choose a nonzero integer \(d\) such that \(\beta = d\xi \) lies in \(\mathcal{O}_F\); then \(\sigma (d^p\mu ) = \beta ^p\mu ^a\). Since \(\sigma (\mathfrak {q}) = \mathfrak {q}\), the valuations at \(\mathfrak {q}\) of \(\sigma (x)\) and \(x\) agree, so

Reducing modulo \(p\) gives \(r \equiv ar \pmod p\). As \(p \nmid r\), this forces \(a \equiv 1 \pmod p\), that is, \(\sigma = 1\).

Let \(q\) be the prime number below \(\mathfrak {q}\).

Assume first that \(q = p\). The prime \(p\) is totally ramified in \(F\), so the stabilizer of \(\mathfrak {q}\) in \(\mathrm{Gal}(F/\mathbb {Q})\) has order \(e(\mathfrak {q}\mid p) f(\mathfrak {q}\mid p) = p - 1 {\gt} 1\), as \(p\) is odd. It contains some \(\sigma \neq 1\), and 87 gives \(p \mid v_{\mathfrak {q}}(\mu )\).

Assume now that \(q \neq p\), and let \(\mathfrak {Q}\) be a prime of \(L\) above \(\mathfrak {q}\). The prime \(q\) is unramified in \(L\) and in \(F\), so \(e(\mathfrak {Q}\mid \mathfrak {q}) = 1\). Let \(\alpha \in \mathcal{O}_L\) with \(\alpha ^p = \mu \) (85). Then

By 88, write \(v_{\mathfrak {q}}(\mu ) = p\, n_{\mathfrak {q}}\) for every prime \(\mathfrak {q}\), and take \(\mathfrak {a} = \prod _{\mathfrak {q}} \mathfrak {q}^{n_{\mathfrak {q}}}\).

Let \(c\) be the class of \(\mathfrak {a}\) in \(\mathrm{Cl}(F)\). Since \(\mathfrak {a}^p = (\mu )\) is principal, \(c^p = 1\).

Let \(a \in (\mathbb {Z}/p\mathbb {Z})^\times \). By 86, \(\sigma _a(\mu ) = \xi ^p\mu ^a\) with \(\xi \in F^\times \). Hence, as fractional ideals,

and since the group of fractional ideals is torsion-free, \(\sigma _a(\mathfrak {a}) = (\xi )\mathfrak {a}^a\). Thus \(\sigma _a(c) = c^a\).

By 61, applied to \(F \cong \mathbb {Q}(\zeta _p)\),

where \(\sigma _a^{-1} = \sigma _{a^{-1}}\) and \(c^{a^{-1}a} = c\) because \(c^p = 1\). Together with \(c^p = 1\), this gives \(c = 1\).

Write \(\mathfrak {a} = (\alpha )\). Then \((\alpha ^p) = (\mu )\), so \(\mu = \alpha ^p\eta \) with \(\eta \) a unit, and \(\alpha \neq 0\) since \(\mu \neq 0\).

By 86 with \(a = p - 1\), there is \(\xi \in F^\times \) with \(\sigma _{-1}(\mu ) = \xi ^p\mu ^{p-1} = (\xi \mu )^p\mu ^{-1}\). Substituting \(\mu = \alpha ^p\eta \),

whence \(\eta \, \sigma _{-1}(\eta ) = \big(\xi \mu /(\alpha \, \sigma _{-1}(\alpha ))\big)^p\).

The field \(F\) is a CM field, and the automorphism \(\sigma _{-1}\) is its complex conjugation. Every unit of a CM field has a square of the form \(\zeta _0\varepsilon _0\), with \(\zeta _0\) a root of unity and \(\varepsilon _0\) a real unit (NumberField.IsCMField.exists_torsion_realunits_pow_two_eq_mul). Squaring, \(\eta ^4 = \zeta \varepsilon \) with \(\zeta = \zeta _0^2\) and \(\varepsilon = \varepsilon _0^2\). The roots of unity of \(F\) form a group of order \(2p\), so \(\zeta ^p = \zeta _0^{2p} = 1\).

The automorphism \(\sigma _{-1}\) fixes \(\varepsilon \) and inverts \(\zeta \), so \(\sigma _{-1}(\eta )^4 = \zeta ^{-1}\varepsilon \). Multiplying, \(\varepsilon ^2 = (\eta \, \sigma _{-1}(\eta ))^4\). By 92, \(\eta \, \sigma _{-1}(\eta ) = w^p\), so \(\varepsilon ^2 = (w^4)^p\). As \(\gcd (2,p) = 1\), it follows that \(\varepsilon = \delta ^p\) for some nonzero \(\delta \in F\).

By 90 and 91, \(\mu = \alpha ^p\eta \) with \(\alpha \neq 0\) and \(\eta \) a unit, and by 93, \(\eta ^4 = \zeta \delta ^p\). As \(\gcd (4,p) = 1\), the polynomial \(X^p - \mu ^4\) is irreducible over \(F\) and \(L\) is its splitting field. Since \(\mu ^4 = \zeta (\alpha ^4\delta )^p\), the field \(L\) is also the splitting field of \(X^p - \zeta \). In particular \(\zeta \neq 1\), since otherwise \(L = F\), so \(\zeta \) is a primitive \(p\)-th root of unity. A root \(\xi \in L\) of \(X^p - \zeta \) is then a primitive \(p^2\)-th root of unity. Finally \(\mathbb {Q}(\xi ) \subseteq L\) and both fields have degree \(p(p-1)\) over \(\mathbb {Q}\), so \(L = \mathbb {Q}(\xi )\).

Put \(F = \mathbb {Q}(\zeta _{p^2}^p)\), so that \(F \cong \mathbb {Q}(\zeta _p)\), and \(M = KF\). The field \(F\) is unramified outside \(p\), hence so is \(M\) by 63, and \(M/\mathbb {Q}\) is abelian. Since \([K:\mathbb {Q}] = p\) and \([F:\mathbb {Q}] = p-1\) are coprime, \(K \cap F = \mathbb {Q}\) and \([M:\mathbb {Q}] = p(p-1)\). By 83, the extension \(M/F\) is cyclic of degree \(p\), and by 84, \(M\) is the splitting field of \(X^p - \mu \) for some nonzero \(\mu \in \mathcal{O}_F\). The polynomial \(X^p - \mu \) is irreducible: otherwise \(\mu \) would be a \(p\)-th power in \(F\), and \(X^p - \mu \) would split over \(F\), giving \(M = F\).

By 89, \((\mu ) = \mathfrak {a}^p\) with \(\mathfrak {a} \neq 0\). By 94, the field \(M\) is the cyclotomic field of level \(p^2\), so \(M = \mathbb {Q}(\zeta _{p^2})\). Hence \(K \subseteq M = \mathbb {Q}(\zeta _{p^2})\).

By 95, both fields lie in \(\mathbb {Q}(\zeta _{p^2})\), which has degree \(p(p-1)\). So \([K_1K_2 : K_1] \leq p - 1\). As \(K_2/\mathbb {Q}\) is Galois, \([K_1K_2 : K_1]\) divides \([K_2:\mathbb {Q}] = p\), hence \([K_1K_2 : K_1] = 1\) and \(K_2 \subseteq K_1\). The degrees being equal, \(K_1 = K_2\).

Put \(C = \mathbb {Q}(\zeta _{p^{m+1}})\). It is unramified outside \(p\), and since \(p\) is odd, \(\mathrm{Gal}(C/\mathbb {Q}) \cong (\mathbb {Z}/p^{m+1}\mathbb {Z})^\times \) is cyclic of order \(p^{m}(p-1)\). Let \(\sigma \) be a generator and \(H = \langle \sigma ^{p^m}\rangle \), a subgroup of index \(p^m\) with cyclic quotient. By 81, the field \(K' = C^{H}\) is a cyclic extension of \(\mathbb {Q}\) of degree \(p^m\), unramified outside \(p\). By 96, the hypothesis of 82 holds, so \(K = K' \subseteq C\).

Let \(d\) be the discriminant. A prime number divides \(d\) if and only if it ramifies, so \(2\) is the only possible prime divisor of \(d\). The discriminant of a quadratic field is a fundamental discriminant (Mathlib PR #43490), so \(d \equiv 0\) or \(1 \pmod4\).

If \(d \equiv 1 \pmod4\), then \(d\) is odd. By Minkowski’s bound \(|d| {\gt} 2\), so \(d\) has a prime divisor, which is odd: a contradiction. Hence \(d = 4m\). Since \(4m\) is a fundamental discriminant, the integer \(m\) is squarefree and \(m \not\equiv 1 \pmod4\). As the only prime dividing \(m\) is \(2\), this leaves \(m \in \{ -1, 2, -2\} \), that is, \(d \in \{ -4, 8, -8\} \).

By 98, each discriminant is \(-8\), \(-4\) or \(8\). The discriminant of a totally real quadratic field is positive (Mathlib PR #43493), so both discriminants equal \(8\). Two quadratic fields with the same discriminant are isomorphic (Mathlib PR #43490), and a quadratic field is normal over \(\mathbb {Q}\), so \(K_1 = K_2\) as subfields of \(A\).

An element of \(L\) lies in \(L^+\) if and only if it is fixed by \(c\). Since \(c\) has order \(2\), being fixed by \(c\) is the same as being fixed by \(\langle c\rangle \).

Complex conjugation sends every root of unity to its inverse, so it sends \(\zeta _n\) to \(\zeta _n^{-1} = \zeta _n^{n-1}\).

The subgroups generated by \(5\) and by \(-1\) are complementary in \((\mathbb {Z}/2^{n+2}\mathbb {Z})^\times \), so the quotient is isomorphic to the cyclic group generated by \(5\).

Put \(C = \mathbb {Q}(\zeta _{2^{m+2}})\), a CM field unramified outside \(2\), and let \(c\) be its complex conjugation. The subgroup \(\langle c\rangle \) has order \(2\), hence index \(2^m\) in \(\mathrm{Gal}(C/\mathbb {Q})\). By 101, the quotient \(\mathrm{Gal}(C/\mathbb {Q})/\langle c\rangle \) is isomorphic to \((\mathbb {Z}/2^{m+2}\mathbb {Z})^\times /\langle -1\rangle \), which is cyclic by 102. By 81 and 100, the maximal real subfield \(K' = C^+\) is a cyclic extension of \(\mathbb {Q}\) of degree \(2^m\), unramified outside \(2\), and it is totally real.

The compositum \(KK'\) is totally real, so every subfield of \(KK'\) is totally real. By 99, any two quadratic subfields of \(KK'\) unramified outside \(2\) are equal. Hence 82 gives \(K = K' \subseteq C\).

If \(K\) is totally real, this is 103. Otherwise let \(L = K\mathbb {Q}(i)\), with \(i = \zeta _4\). It is a totally complex abelian extension of \(\mathbb {Q}\), unramified outside \(2\) by 63, of degree \(2^l\) with \(l \leq m+1\); it is a CM field. Let \(M = L^+\) be its maximal real subfield, of degree \(2^k\) with \(k \leq l\). As \(L\) is totally complex, \(M \neq L\), so \(k \leq m\). The field \(M\) is abelian and unramified outside \(2\).

The field \(M\) is cyclic over \(\mathbb {Q}\). Indeed, \(\mathrm{Gal}(M/\mathbb {Q}) \cong \mathrm{Gal}(L/\mathbb {Q})/\langle c\rangle \), where \(c\) is the complex conjugation of \(L\). The restriction map \(\mathrm{Gal}(L/\mathbb {Q}) \to \mathrm{Gal}(K/\mathbb {Q}) \times \mathrm{Gal}(\mathbb {Q}(i)/\mathbb {Q})\) is injective, so \(\mathrm{Gal}(M/\mathbb {Q})\) embeds into the quotient of \(\mathrm{Gal}(K/\mathbb {Q}) \times \mathrm{Gal}(\mathbb {Q}(i)/\mathbb {Q})\) by the image of \(\langle c\rangle \). The restriction of \(c\) to \(\mathbb {Q}(i)\) generates \(\mathrm{Gal}(\mathbb {Q}(i)/\mathbb {Q})\), so this quotient is generated by the image of the cyclic group \(\mathrm{Gal}(K/\mathbb {Q}) \times 1\). Hence it is cyclic, and so is \(\mathrm{Gal}(M/\mathbb {Q})\).

By 103, \(M \subseteq \mathbb {Q}(\zeta _{2^{k+2}})\). Since \([L:M] = 2\) and \(i \notin M\), we get \(L = M\mathbb {Q}(i)\). Therefore