Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 123 (Algebraic Number Theory)

A modulus of a number field is a formal productwhere ${\mathfrak{m}}_0={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{n_{\mathfrak{p}}}$ is a nonzero integral ideal and ${\mathfrak{m}}_{\infty }$ is a product of distinct real embeddings of $K$. Only finitely many $n_{\mathfrak{p}}$ are nonzero; complex places do not occur.

Let $I_K(\mathfrak{m})$ be the group of fractional ideals prime to ${\mathfrak{m}}_0$. Let $P_K(\mathfrak{m})$ consist of principal ideals $(\alpha )$ withfor every $\mathfrak{p}\mid {\mathfrak{m}}_0$ and $\sigma (\alpha )>0$ for every real place $\sigma \mid {\mathfrak{m}}_{\infty }$. The ray class group is

LetThe ray class number formula is

The quadratic field $K=\mathbb{Q}(\sqrt{-7})$ is the unique quadratic subfield of $\mathbb{Q}({\zeta }_7)$. Its nontrivial character is the quadratic characterUnder the Artin reciprocity map, Frobenius at an unramified prime $p$ restricts trivially to $K$ exactly when ${\chi }_7(p)=1$. The quadratic residues modulo $7$ are $1,2,4$, while $5$ is a nonresidue. Thusso the Artin symbol at $5$ is the nonidentity element of $\mathrm{Gal}(K/\mathbb{Q})$. Therefore $5$ has residue degree two and is inert in $K$.

Since $5$ is inert, $\mathfrak{m}=5{\mathcal{O}}_K$ is prime andThe unit group is $\{\pm 1\}$, and only $1$ is congruent to $1$ modulo $\mathfrak{m}$, soThere are no real places and $h_K=1$. The ray class number formula givesHence the ray class field modulo $5{\mathcal{O}}_K$ has degree $\boxed{12}$ over $K$.

For an unramified prime $\mathfrak{p}$ of $K$ and a prime $\mathfrak{P}$ of $L$ above it, the Frobenius automorphism is characterized byIn an abelian extension it is independent of $\mathfrak{P}$. This element is the Artin symbol

Every ideal of $\mathbb{Z}$ prime to $8$ has a representative $n\mathbb{Z}$ with $n$ odd and positive. The Artin reciprocity map for $\mathbb{Q}({\zeta }_8)/\mathbb{Q}$ isIts kernel consists exactly of positive principal ideals generated by numbers congruent to $1$ modulo $8$. Thereforeand the four classes areComplex conjugation sends ${\zeta }_8$ to ${\zeta }_8^{-1}={\zeta }_8^7$, so it corresponds to $\boxed{7}$.

Under the preceding quotient, $H/P_{\mathbb{Q}}(8\infty )$ is the subgroupIt has order two in a group of order four. Hence $H$ contains $P_{\mathbb{Q}}(8\infty )$ and

The ideal-theoretic existence theorem of global class field theory gives an inclusion-reversing correspondence between finite abelian extensions $L/K$ and congruence subgroupsThe corresponding field satisfies

The subgroup $H/P_{\mathbb{Q}}(8\infty )=\{1,7\}$ corresponds inside $\mathbb{Q}({\zeta }_8)$ to the fixed field of $\{1,c\}$, where $c$ is complex conjugation. This is the maximal real subfieldThus the existence theorem of global class field theory associates $H$ with $\boxed{\mathbb{Q}(\sqrt{2})}$.

A Dirichlet character modulo $m$ is a homomorphismExtend it periodically to $\mathbb{Z}$ by setting $\chi (n)=0$ when $(n,m)>1$. Its Dirichlet L-function is

The Dedekind zeta function of $K$ isfor $\mathrm{Re}s>1$.

By the Kronecker–Weber theorem, choose $m$ withRestriction gives a quotientEvery character $\psi \in \hat{G}$ inflates along this quotient and has an associated primitive Dirichlet character ${\chi }_{\psi }$, whose conductor may divide $m$. Comparing Euler factors, or applying the factorization of the Artin $L$-function of the regular representation, gives

The group $(\mathbb{Z}/7\mathbb{Z}{)}^{\times }$ is cyclic of order six, generated by $3$. Put $\omega =e^{2\pi i/6}$. The six characters areand vanish on multiples of $7$. Sincetheir values are explicitlyTaking $j=0,\dots ,5$ lists all six characters.

The unique order-two character is ${\chi }_3$, the Legendre symbolIt is $1$ on $1,2,4$, is $-1$ on $3,5,6$, and is zero on multiples of $7$. Since the unique quadratic subfield of $\mathbb{Q}({\zeta }_7)$ is $\mathbb{Q}(\sqrt{-7})$, this is the character corresponding to $K$.

For $K=\mathbb{Q}(\sqrt{-7})$,The character is odd, ${\chi }_3(2)=1$, andThe supplied odd-character formula gives

The analytic class number formula for this imaginary quadratic field isbecause $w_K=2$ and $d_K=-7$. Since ${\mathrm{Res}}_{s=1}\zeta (s)=1$, comparison with $L(1,{\chi }_3)$ yields $\boxed{h_K=1}$.

The idele group is the restricted productover all places, with respect to ${\mathcal{O}}_v^{\times }$ at finite places. Thus an idele has a nonzero component in every completion and belongs to ${\mathcal{O}}_v^{\times }$ at all but finitely many finite places.

For $\alpha =({\alpha }_v)\in {\mathbb{I}}_K$, the congruence $\alpha \equiv 1(\mathrm{mod}\mathfrak{m})$ meansAt finite primes outside the modulus require ${\alpha }_{\mathfrak{p}}\in {\mathcal{O}}_{\mathfrak{p}}^{\times }$. These ideles form ${\mathbb{I}}_K(\mathfrak{m})$. Their imageinside the idèle class group is the idelic congruence subgroup.

Write $m={\prod }_p{p}^{n_p}$. For the modulus $m\infty$,This is understood as the corresponding restricted-product subgroup of ${\mathbb{I}}_{\mathbb{Q}}$.

Multiplying by a positive rational number normalizes the real component and all valuations, leaving the finite unit residue modulo $m$. This identifiesLocally,when $n_p>0$. Multiplying gives

The assumed inclusion and the class-field norm-index theorem giveThe outer subgroup has index $\phi (m)$ by part (d), whileTwo nested subgroups of the same finite index are equal. Thus

The Kronecker–Weber theorem says that every finite abelian extension of $\mathbb{Q}$ lies in some $\mathbb{Q}({\zeta }_m)$.

Indeed, global class field theory assigns to a finite abelian $L/\mathbb{Q}$ the open norm group of an abelian extension $N_{L/\mathbb{Q}}{C}_L$. It contains a congruence subgroup $C_{\mathbb{Q}}(m\infty )$. By part (e), this is the norm group of $\mathbb{Q}({\zeta }_m)$. The inclusion-reversing class-field correspondence therefore gives

For a set $S$ of prime ideals, its Dirichlet density iswhen this limit exists.

Let $G=\mathrm{Cl}(K)$ and let $\hat{G}$ be its character group. Character orthogonality givesFor $\mathrm{Re}s>1$, the prime term of $\log L(s,\chi )$ isup to a function bounded as $s\to 1^{+}$, since higher prime powers converge there. The trivial character contributeswhereas every nontrivial character contributes $O(1)$ because its $L$-function is nonzero at $1$. Thereforeand the density is $\boxed{1/h_K}$.

The Chebotarev density theorem states that for a finite Galois extension $L/K$ and a conjugacy class $C\subseteq \mathrm{Gal}(L/K)$, the unramified primes whose Frobenius conjugacy class is $C$ have Dirichlet density

PutIt contains $K=\mathbb{Q}(\sqrt{26})$ and has degree two over $K$. The three quadratic subfields have discriminants $8$, $13$, and $104$, so the biquadratic discriminant formula givesThe relative discriminant formulatherefore gives ${\mathfrak{d}}_{H/K}={\mathcal{O}}_K$: no finite prime ramifies. The extension is totally real, so no infinite prime ramifies either. Since $h_K=2$, the Hilbert class field has degree two over $K$. Consequently

A prime ideal of $K$ is principal exactly when its Artin symbol inis the identity. The group has order two. Applying the Chebotarev density theorem to the identity conjugacy class gives density