Past exam of the mathematics course of the University of Cambridge / 2023 / 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 places. Only finitely many exponents $n_{\mathfrak{p}}$ are nonzero.

Let $L/K$ be a finite abelian extension and let $\mathfrak{m}$ be an admissible modulus, so every ramified prime divides ${\mathfrak{m}}_0$ and the Artin reciprocity mapis defined. If $\mathfrak{p}\nmid {\mathfrak{m}}_0$, then the ideal-theoretic decomposition law says that its residue degree isand, since $\mathfrak{p}$ is unramified, it hasprime factors in $L$.

Indeed, ${\theta }_{L/K}(\mathfrak{p})$ is the Artin symbol, whose restriction to the residue field is $x\mapsto x^{N\mathfrak{p}}$. The Galois group of the finite residue-field extension is generated by this Frobenius automorphism and has order $f(\mathfrak{P}/\mathfrak{p})$. The fundamental identity $efg=[L:K]$, with $e=1$, gives the formula for $g$.

The Hilbert class field $H_K$ is the maximal unramified abelian extension of $K$, including the condition that real places do not become complex when infinite ramification is included. The Artin reciprocity law induces a canonical isomorphismConsequently $[H_K:K]=h_K$.

For an unramified prime $\mathfrak{p}$, the ideal-theoretic decomposition law givesUnder the Hilbert class field isomorphism, this Artin symbol is the image of $[\mathfrak{p}]\in \mathrm{Cl}(K)$. It is trivial exactly when $[\mathfrak{p}]=1$, which means exactly that $\mathfrak{p}$ is a principal ideal.

Let $K=\mathbb{Q}(\sqrt{-6})$, whose discriminant is $-24$, and putThis is a biquadratic extension of $\mathbb{Q}$ whose three quadratic subfields have discriminants $-24$, $8$, and $-3$. The discriminant of a biquadratic field is thereforeThe relative discriminant formula givesso $H/K$ is unramified at every finite prime. Both fields are totally imaginary, so no infinite place ramifies. Since $[H:K]=2=h_K$, the extension has the full degree of the Hilbert class field, and hence

For $\mathrm{Re}s>1$, the Dedekind zeta function of $K$ isUnique factorization of ideals gives its Euler productover the nonzero prime ideals of ${\mathcal{O}}_K$.

The analytic class number formula states that ${\zeta }_K(s)$ has a simple pole at $s=1$ withwhere $(r_1,r_2)$ is the signature, $h_K$ the class number, $R_K$ the regulator, $w_K$ the number of roots of unity in $K$, and $d_K$ the field discriminant.

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

Let $K/\mathbb{Q}$ be quadratic with discriminant $d_K$, and letbe its primitive quadratic Dirichlet character. For an unramified prime $p$, the Euler factor of ${\zeta }_K$ isSince ${\chi }_{d_K}(p)$ is respectively $1$ or $-1$, this equals the product of the Euler factors of $\zeta (s)$ and $L(s,{\chi }_{d_K})$. At a ramified prime ${\chi }_{d_K}(p)=0$, and the same equality holds. Thus

Suppose $d_K<-4$. Then $K$ has signature $(0,1)$, regulator $1$, and exactly $w_K=2$ roots of unity. The analytic class number formula becomesBecause ${\zeta }_K(s)=\zeta (s)L(s,{\chi }_{d_K})$ and ${\mathrm{res}}_{s=1}\zeta (s)=1$,Putting $m=|d_K|$ into the supplied odd-character identity yields the quadratic class number formula

For $K=\mathbb{Q}(\sqrt{-11})$, the fundamental discriminant is $-11$ and ${\chi }_{-11}(2)=-1$ because $-11\equiv 5(\mathrm{mod}8)$. The quadratic residues modulo $11$ areThereforeThe quadratic class number formula gives

For a set $S$ of nonzero prime ideals of $K$, its Dirichlet density isprovided the limit exists.

The Chebotarev density theorem says that if $L/K$ is a finite Galois extension with group $G$ and $C\subseteq G$ is a conjugacy class, then the unramified primes $\mathfrak{p}$ whose Frobenius conjugacy class equals $C$ have Dirichlet density

Let ${\mathfrak{f}}_{L/K}$ be the conductor of an abelian extension and let $\mathfrak{m}$ be divisible by it. For any $\sigma \in \mathrm{Gal}(L/K)$, the Chebotarev density theorem supplies infinitely many unramified prime ideals $\mathfrak{p}$ withOnly finitely many primes divide ${\mathfrak{m}}_0$, so $\mathfrak{p}$ may be chosen prime to ${\mathfrak{m}}_0$. It then belongs to $I_K(\mathfrak{m})$ and its image under the Artin reciprocity map is $\sigma$. Hence the Artin map is surjective.

A prime splits completely in $L$ exactly when its Frobenius conjugacy class is $\{1\}$. Applying the Chebotarev density theorem to the identity class gives

By Question 1, a prime ideal of $K$ is principal exactly when it splits completely in the Hilbert class field $H_K$. Question 3(4) therefore gives

The adele ring is the restricted productAddition and multiplication are componentwise. If $x_v,y_v\in {\mathcal{O}}_v$ outside finite sets, then $x_v+y_v,x_v{y}_v\in {\mathcal{O}}_v$ outside their union, so these operations are closed on ${\mathbb{A}}_K$. The componentwise ring axioms make it a ring. Its restricted product topology has basic open sets ${\prod }_v{U}_v$ with each $U_v\subseteq K_v$ open and $U_v={\mathcal{O}}_v$ at all but finitely many finite places.

The idele group isthe restricted product with respect to the local unit groups.

Underthe preimage of a basic product neighbourhood $U\times V$ isAt almost all finite places, $U_v=V_v={\mathcal{O}}_v$, and the condition $x_v,x_v^{-1}\in {\mathcal{O}}_v$ is exactly $x_v\in {\mathcal{O}}_v^{\times }$. Thus these preimages form the usual restricted-product basis for ${\mathbb{I}}_K$. Conversely, every basic idele neighbourhood is obtained by choosing suitable $U_v$ and $V_v$ locally. Hence the restricted product topology on the idele group equals the subspace topology induced by $\varphi$.

The diagonal mapis an injective homomorphism because every completion map $K\to K_v$ is injective. The diagonal copy of $K$ is discrete in the adele ring. Choose an adelic neighbourhood $W$ of $1$ with $W\cap K=\{1\}$. By the subspace description of the idele topology, $W\cap {\mathbb{I}}_K$ is an idele neighbourhood of $1$ meeting diagonal $K^{\times }$ only at $1$. Translation proves that $K^{\times }$ is a discrete subgroup of ${\mathbb{I}}_K$.

For each place $v$ of $K$ and each $w\mid v$ of $L$, the inclusion $K_v\hookrightarrow L_w$ definesAt all but finitely many finite $v$, the component $x_v$ is a unit, and its image is a unit at every $w\mid v$, so the image is an idele. The map is a homomorphism and is injective because every local inclusion is injective.

The idele norm has $v$-componentFor the principal idele associated with $\alpha \in L^{\times }$, compatibility of local and global field norms givesThe right side is the principal idele of the element $N_{L/K}(\alpha )\in K^{\times }$.

Let $G=\mathrm{Gal}(L/K)$ and letbe the fixed field of the commutator subgroup. Then $M/K$ is the maximal abelian subextension of $L/K$, withThe nonabelian norm-residue kernel theorem identifies the kernel of the global reciprocity mapwith $N_{L/K}{C}_L$. Applied to the abelian extension $M/K$, the Artin reciprocity law identifies the kernel of the same map with $N_{M/K}{C}_M$. Thereforeso the norm group of the Galois extension $L/K$ is the norm group of the abelian extension $M/K$.