Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 81 / 3

For a number field $K$ and a modulus of a number field $\mathfrak{m}={\mathfrak{m}}_0{\mathfrak{m}}_{\infty }$, let $I_K(\mathfrak{m})$ be the group of fractional ideals prime to ${\mathfrak{m}}_0$, and let $P_{K,1}(\mathfrak{m})$ consist of principal fractional ideals $(a)$ with $a\equiv 1$ at the finite modulus and $a>0$ at the real places in ${\mathfrak{m}}_{\infty }$. The ray class group is ${\mathrm{Cl}}_{\mathfrak{m}}(K)=I_K(\mathfrak{m})/P_{K,1}(\mathfrak{m})$.

The Main theorem of class field theory states that there is a unique finite abelian ray class field $K_{\mathfrak{m}}/K$ for which the Artin reciprocity map inducesIt is unramified outside the finite modulus and has no ramification at real places omitted from ${\mathfrak{m}}_{\infty }$. More generally, subgroups $H$ with $P_{K,1}(\mathfrak{m})\subseteq H\subseteq I_K(\mathfrak{m})$ correspond, with inclusions reversed, to intermediate abelian fields, with Galois group $I_K(\mathfrak{m})/H$. Every finite abelian extension occurs for some modulus. Equivalently, $K_{\mathfrak{m}}$ is the maximal abelian extension with conductor of an abelian extension dividing $\mathfrak{m}$. This states both reciprocity and the existence theorem of global class field theory.

There is no infinite component in the modulus $(5)$. The ordinary ideal class group of $\mathbb{Q}$ is trivial, and its unit group is $\{\pm 1\}$. The ray class exact sequence therefore givesConcretely, a fractional ideal prime to $5$ has a rational generator $r=a/b$ with $5\nmid ab$. Reducing $r$ modulo $5$ is well-defined up to multiplication by the units $\pm 1$. Its ray class is trivial precisely when one of the two generators $r,-r$ is congruent to one. Thus the quotient has the two classes $\{1,4\}$ and $\{2,3\}$. No positivity restriction is imposed on a generator.

To identify the corresponding field, put $E=\mathbb{Q}({\zeta }_5+{\zeta }_5^{-1})=\mathbb{Q}(\sqrt{5})$. The maximal real subfield of the fifth cyclotomic field is fixed by complex conjugation, the element $-1$ of $(\mathbb{Z}/5\mathbb{Z}{)}^{\times }$. For a prime $\ell \ne 5$, the Frobenius automorphism of $\mathbb{Q}({\zeta }_5)/\mathbb{Q}$ is ${\zeta }_5\mapsto {\zeta }_5^{\ell }$. Its restriction to $E$ thus corresponds exactly to the residue class of $\ell$ modulo $\{\pm 1\}$. Multiplicativity extends this to the Artin reciprocity map on all fractional ideals prime to $5$, whose kernel is exactly $P_{\mathbb{Q},1}((5))$. Its field discriminant $5$ also confirms that $E$ is unramified outside $5$; moreover $E$ is a totally real number field. The uniqueness in the Main theorem of class field theory now gives

Including the real place changes the ray class group: a principal generator must then be positive as well as congruent to one. The residue and signature group of a modulus givesHere the isomorphism sends $(u,\epsilon )$ to $\epsilon u$. Consequently ${\mathbb{Q}}_{(5)\infty }=\mathbb{Q}({\zeta }_5)$, of degree four. The finite modulus gives the real quadratic field; the modulus with the real place gives the full cyclotomic field. The printed finite-modulus assertion needs no correction.

Set $K=\mathbb{Q}(\sqrt{pq})$ and $E=\mathbb{Q}(\sqrt{p},\sqrt{q})$. The distinct primes give independent square classes, so $[E:\mathbb{Q}]=4$ and $[E:K]=2$. In particular $K(\sqrt{p})=K(\sqrt{q})=E$: these are the same quadratic extension of $K$. We prove it is unramified by an explicit field discriminant calculation.

Since $p,q\equiv 1(\mathrm{mod}4)$, put $a=(1+\sqrt{p})/2$, $b=(1+\sqrt{q})/2$. Both are algebraic integers. The ring $R=\mathbb{Z}[a,b]\subseteq {\mathcal{O}}_E$ has basis $1,a,b,ab$. Up to the ordering of the basis, its trace pairing matrix is the tensor product of the quadratic trace matrices, because the four embeddings of $E$ choose the two radical signs independently. The two quadratic matrices areThusOn the other hand, $pq$ is squarefree and congruent to one modulo four, so $d_K=pq$. Let $j=[{\mathcal{O}}_E:R]$. The discriminant-index formula for an integral lattice and the relative discriminant tower formula giveThe ideal norm on the right is a positive integer, while $j$ is a positive integer. Hence $N({\mathfrak{d}}_{E/K})=1/j^2$ forces $j=1$ and ${\mathfrak{d}}_{E/K}={\mathcal{O}}_K$. Its prime divisors are exactly the ramified finite primes, so $E/K$ is unramified at every finite prime, including primes above $2,p,q$. This also proves $d_E=p^2{q}^2$, agreeing with the discriminant of a biquadratic field formula.

Every embedding of $E$ is real, so the real places of $K$ split as well. The extension is quadratic, hence abelian. The ordinary Hilbert class field characterization therefore impliesFinally, the Artin reciprocity map identifies $\mathrm{Cl}(K)$ with $\mathrm{Gal}(H_K/K)$. Restriction onto $\mathrm{Gal}(E/K)\simeq C_2$ is surjective, so the ideal class group has a quotient of order two. Therefore $\boxed{|\mathrm{Cl}(K)|\text{ is even}}$.