Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 123 (Algebraic Number Theory)

Write $n=[L:K]$. The field trace gives a nondegenerate $K$-bilinear trace pairingThe inverse different, or codifferent, is the trace-dual latticeIt contains $S$, because the trace of an element integral over $R$ belongs to the integrally closed ring $R$. It is stable under multiplication by $S$: if $s,t\in S$ and $x\in {\mathfrak{D}}_{S/R}^{-1}$, then $\mathrm{Tr}(sxt)=\mathrm{Tr}(x(st))\in R$. Nondegeneracy of the trace pairing and finite generation of $S$ show that this trace dual is a finitely generated $R$-module spanning $L$. Therefore it is a fractional ideal of $S$.

Its inverseis the different ideal. Since $S\subseteq {\mathfrak{D}}_{S/R}^{-1}$, every such $x$ lies in $S$; hence ${\mathfrak{D}}_{S/R}$ is an integral ideal of $S$.

The discriminant ideal $\mathrm{disc}(S/R)$ is locally generated bywhere ${\alpha }_1,\dots ,{\alpha }_n$ is a local $R$-basis of $S$. Equivalently, it is the image of the determinant of the trace pairingThis formulation makes the definition independent of a basis, since changing a basis multiplies its discriminant by the square of the determinant of the change-of-basis matrix.

The asserted identity of ideals can be checked after localization at every nonzero prime ideal of $R$. We may therefore assume that $R$ is a discrete valuation ring and choose a basis ${\alpha }_1,\dots ,{\alpha }_n$ of $S$. Let ${\beta }_1,\dots ,{\beta }_n$ be its trace-dual basis, so $\mathrm{Tr}({\alpha }_i{\beta }_j)={\delta }_{ij}$; this is a basis of ${\mathfrak{D}}_{S/R}^{-1}$. If $T=(\mathrm{Tr}({\alpha }_i{\alpha }_j))$, thenThus the determinant measuring the inclusion $S\subseteq {\mathfrak{D}}_{S/R}^{-1}$ is $\det T$. The determinant description of the norm of a fractional ideal consequently givesLocalization then proves the equality over the original Dedekind domain.

Finally, the determinant-of-pairing map identifies the invertible $R$-module $({\det }_{R}S{)}^{\otimes 2}$ with the discriminant ideal. Hence in the ideal class groupup to the harmless inverse caused by the convention used to identify invertible modules with fractional ideals. In either convention the class is a square.

For each place of a number field $v$, let $K_v$ be the corresponding completion, and for finite $v$ let ${\mathcal{O}}_v$ be its valuation ring. The adele ring is the restricted productIts restricted product topology has basic open sets ${\prod }_v{U}_v$, where every $U_v\subseteq K_v$ is open and $U_v={\mathcal{O}}_v$ at all but finitely many finite places.

First take $K=\mathbb{Q}$. The neighborhoodof zero meets the diagonal copy of $\mathbb{Q}$ only in zero: a rational number lying in every ${\mathbb{Z}}_p$ is an integer, and the only integer in the indicated real interval is zero. Thus $\mathbb{Q}$ is discrete in ${\mathbb{A}}_{\mathbb{Q}}$.

Every rational adele is congruent modulo $\mathbb{Q}$ to an element ofIndeed, the finitely many negative $p$-adic principal parts can be removed simultaneously by subtracting a rational number, using the Chinese remainder theorem; subtracting an integer then moves the real component into $[0,1]$. This set is compact by the compactness of $[0,1]$, the compactness of every ${\mathbb{Z}}_p$, and the Tychonoff theorem. Its image covers the quotient, so ${\mathbb{A}}_{\mathbb{Q}}/\mathbb{Q}$ is compact.

Now choose a $\mathbb{Q}$-basis of the number field $K$. The given topological isomorphismidentifies the additive pair $({\mathbb{A}}_K,K)$ with $({\mathbb{A}}_{\mathbb{Q}}^n,{\mathbb{Q}}^n)$. A finite product of discrete subgroups is discrete, andis compact.

The idele group iswhere the distinguished subgroup at a finite place is ${\mathcal{O}}_v^{\times }$. It carries the corresponding restricted product topology on the idele group. The inclusion $j:J_K\to {\mathbb{A}}_K$ is continuous: the inverse image of a basic adelic open set is locally a product of open subsets of $K_v^{\times }$, and outside finitely many places every idele component already belongs to ${\mathcal{O}}_v^{\times }\subseteq {\mathcal{O}}_v$.

It is not a homeomorphism onto its image. Let $p_i$ be the $i$th rational prime and define the idele $x^{(i)}$ to equal $p_i$ at the place over $p_i$ and $1$ everywhere else. In the adele topology, $x^{(i)}\to 1$: the difference is zero at every fixed place once $i$ is large, while $p_i-1\in {\mathbb{Z}}_{p_i}$ at the single moving place. In the idele topology the sequence does not converge to $1$, because the open neighborhoodcontains no $x^{(i)}$: its $p_i$-component has positive valuation and is not a unit. Hence the inverse of $j$ on its image is not continuous.

The Schwartz-Bruhat space $\mathcal{S}(F)$ of a non-Archimedean local field $F$ is the vector space of locally constant, compactly supported complex-valued functions on $F$. Fix a nontrivial continuous additive character $\psi :F\to {\mathbb{C}}^{\times }$ and a Haar measure $dx$. With the sign convention required in the question, the Fourier transform over a local field is

Letbe the annihilator of the valuation ring. Translation invariance givesIndeed, the integral is the volume when the character is trivial; otherwise translation by an element on which the character is nontrivial multiplies the integral by a scalar different from one, forcing it to vanish. With the usual character of conductor ${\mathcal{O}}_F$ and the normalization $\mathrm{vol}({\mathcal{O}}_F)=1$, this becomesFor the canonical character induced from ${\mathbb{Q}}_p$, the annihilator is instead the inverse different and the displayed general formula applies.

If $g(x)=f(ax+b)$ with $a\ne 0$, the substitution $u=ax+b$ and the scaling rule $dx=|a{|}^{-1}du$ give

Every locally constant compactly supported function is a finite linear combination of characteristic functions of cosets $b+a{\mathcal{O}}_F$: compactness extracts finitely many cosets on which the function is constant. The formula just proved, together with the transform of $1_{{\mathcal{O}}_F}$, shows that the transform of each such characteristic function is again locally constant and compactly supported. Therefore the Fourier transform over a local field maps $\mathcal{S}(F)$ to itself.

Suppose that $L/\mathbb{Q}$ were a Finite Galois extension with Galois group $S_3$ and unramified away from $7$. Its quadratic subfield is the fixed field of the alternating subgroup $A_3$. A quadratic field ramified only at $7$ must be $\mathbb{Q}(\sqrt{-7})$: the classification by fundamental discriminants shows that $-7$ is the only nontrivial quadratic discriminant supported at $7$. In particular, $7$ ramifies in this quadratic subfield.

Because $7$ does not divide $|S_3|=6$, all ramification at $7$ is tame ramification. Its inertia group is therefore cyclic. Its image in $S_3/A_3\simeq C_2$ is nontrivial because the quadratic subfield is ramified, so the inertia group is generated by a transposition and has order two; it cannot have order six because it is cyclic.

The discriminant exponent of a tame Galois extension is thereforeThere is no other finite ramification, so $|d_L|=7^3=343$.

On the other hand, the Minkowski bound for ideal classes implies the Minkowski lower bound for a number-field discriminantSince $0\le r_2\le 3$, the right side is smallest at $r_2=3$ and is greater than $980$. This contradicts $|d_L|=343$, so no such extension exists.

Put $s=\sqrt{-6}$, so ${\mathcal{O}}_K=\mathbb{Z}[s]$ and ${\mathcal{O}}_K^{\times }=\{\pm 1\}$. The primes $2$ and $3$ are ramified:Thus the finite modulus in the question is

The ray class exact sequence, with no real component in the modulus, givesThe Chinese remainder theorem for unit groups and the ramification relations yieldThe image of $-1$ kills the $C_2$ coming from ${\mathbb{F}}_3^{\times }$. Consequently the kernel of the map from the ray class group to the ordinary class group isSince the given class number is two, $|{\mathrm{Cl}}_{\mathfrak{m}}(K)|=12$.

It remains to distinguish $C_{12}$ from $C_6\times C_2$. LetThe prime $5$ splits in $K$, and $\mathfrak{q}$ represents the nontrivial ordinary ideal class because no element of $\mathbb{Z}[s]$ has norm $5$. Direct multiplication, or comparison of norms and valuations at the two primes over $5$, givesModulo $2$, the element $1+2s$ is $1$. Modulo $3$, its class $1+2ϵ$ is nontrivial and has order three because ${ϵ}^2=0$. Hence the ray class of $\mathfrak{q}$ has order six: its square is a nontrivial element of order three in the congruence kernel. The remaining order-two factor of that kernel, coming from $({\mathcal{O}}_K/(2){)}^{\times }$, is independent of $⟨[\mathfrak{q}]⟩$. Therefore