Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 123 / 1

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