Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 74 (Algebraic Number Theory)

A Dedekind domain is a Noetherian integrally closed domain in which every nonzero prime ideal is maximal. The equivalent local characterization is that its localizations at nonzero prime ideals are discrete valuation rings. A field is harmless as the dimension-zero case.

Write $R$ for the domain, $F$ for its fraction field and $I$ for the nonzero fractional ideal. Its proposed inverse is an $R$-submodule of $F$. Choose $0\ne d\in R$ with $dI\subseteq R$, so $d\in I^{-1}$ and the inverse is nonzero. Choose also $0\ne a\in I$. Every $x\in I^{-1}$ satisfies $xa\in R$, hence $I^{-1}\subseteq a^{-1}R$. Writing $a=r/s$ with $r,s\in R$ gives $r{I}^{-1}\subseteq sR\subseteq R$. Thus $I^{-1}$ is a fractional ideal; the Noetherian hypothesis also makes it finitely generated.

For a nonzero maximal prime ideal $\mathfrak{p}$, the discrete valuation ring $R_{\mathfrak{p}}$ makes $I_{\mathfrak{p}}$ principal, say $I_{\mathfrak{p}}=a_{\mathfrak{p}}{R}_{\mathfrak{p}}$. We need to justify that inversion commutes with this localization. One inclusion is immediate. Conversely, if $x{I}_{\mathfrak{p}}\subseteq R_{\mathfrak{p}}$, finite generation of $I$ supplies a common $s\notin \mathfrak{p}$ such that $sxI\subseteq R$. Therefore $sx\in I^{-1}$, provingIt follows that $(I{I}^{-1}{)}_{\mathfrak{p}}=R_{\mathfrak{p}}$ at every such prime ideal. Globally $I{I}^{-1}\subseteq R$. If this were a proper ideal, it would be contained in a maximal ideal and would remain proper after localizing there, a contradiction. In the field case every nonzero fractional ideal is the whole field. Hence in all casesThis proves that every nonzero fractional ideal of a Dedekind domain is an invertible fractional ideal, rather than presupposing its inverse exists.

For pairwise comaximal ideals $I_1,\dots ,I_r$ of $R$, reduction gives the Chinese remainder theorem for ideals:Its kernel is ${\bigcap }_i{I}_i$. For two comaximal ideals, write $1=u+v$ with $u\in I$, $v\in J$. If $x\in I\cap J$, then $x=xu+xv\in IJ$, showing $I\cap J=IJ$. Induction gives the product formula: a product of ideals individually comaximal with $J$ is still comaximal with $J$, by multiplying identities modulo $J$.

Put $J_i={\prod }_{j\ne i}{I}_j$. Since $I_i+J_i=R$, choose $e_i\in J_i$ with $e_i\equiv 1(\mathrm{mod}{I}_i)$. The element ${\sum }_i{a}_i{e}_i$ has any prescribed residues $a_i(\mathrm{mod}{I}_i)$, proving surjectivity. Its residue class modulo the product is unique because we already identified the kernel. In a Dedekind domain, powers of distinct nonzero prime ideals are pairwise comaximal, giving the prime-power form. The theorem requires comaximality; it is not a statement about arbitrary lists of ideals.

Let $G=\mathrm{Gal}(K/k)$. Each field automorphism preserves the ring of integers and the base prime ideal, so it permutes the prime ideals above $\mathfrak{p}$. Suppose there were more than one orbit, and fix one orbit $\mathcal{C}$. By the Chinese remainder theorem for ideals, choose $a\in {\mathcal{O}}_K$ such thatThe field norm $N_{K/k}(a)={\prod }_{\sigma \in G}\sigma (a)$ lies in ${\mathcal{O}}_k$. It lies in every prime ideal of $\mathcal{C}$, since for a fixed member each conjugate of $a$ is zero modulo that member. Thus its contraction lies in $\mathfrak{p}$. On the other hand, for $\mathfrak{Q}$ outside $\mathcal{C}$, every ${\sigma }^{-1}(\mathfrak{Q})$ is also outside it, so every $\sigma (a)$ equals one modulo $\mathfrak{Q}$. The product is consequently one modulo $\mathfrak{Q}$. This contradicts its membership in $\mathfrak{p}{\mathcal{O}}_K\subseteq \mathfrak{Q}$. Therefore the Galois action on the primes above $\mathfrak{p}$ is transitive. The normal extension of number fields is separable, so its conjugates are exactly the elements of $G$ used in the product.

Let $R={\mathcal{O}}_k$, $S={\mathcal{O}}_K$ and $n=[K:k]$. For each prime ideal $\mathfrak{P}$ of $S$ above $\mathfrak{p}$ of $R$, set $f_{\mathfrak{P}}=[S/\mathfrak{P}:R/\mathfrak{p}]$. The relative norm of a fractional ideal can be specified on the unique prime ideal factorization byHere is why this definition is the intrinsic relative ideal norm from norms of elements, and why the residue weights occur. Localize at $\mathfrak{p}$. The finite torsion-free $R_{\mathfrak{p}}$-module $S_{\mathfrak{p}}$ is free of rank $n$, because $R_{\mathfrak{p}}$ is a discrete valuation ring. Multiplication by a nonzero integral $a$ has determinant $N_{K/k}(a)$. The Smith normal form givesDecompose the quotient at the prime ideals above $\mathfrak{p}$. In $S_{\mathfrak{P}}$ each quotient ${\mathfrak{P}}^j/{\mathfrak{P}}^{j+1}$ is one-dimensional over $S/\mathfrak{P}$, so has length $f_{\mathfrak{P}}$ over $R_{\mathfrak{p}}$. ThereforeFor an integral ideal $I$, all norms of elements of $I$ have valuations at least $\sum f_{\mathfrak{P}}{v}_{\mathfrak{P}}(I)$. These minima can be attained simultaneously at the finitely many prime ideals above a fixed $\mathfrak{p}$. In fact the Chinese remainder theorem for ideals, applied inside the finitely generated module $I$, chooses an $a\in I$ with nonzero image in each $I/\mathfrak{P}I$. Equivalently $v_{\mathfrak{P}}(a)=v_{\mathfrak{P}}(I)$ at all these primes. The generated ideal of the element norms consequently has exactly the displayed valuations. Clearing a denominator extends this description to fractional ideals.

Multiplication of fractional ideals adds their prime valuations. The displayed formula therefore provesIt also gives $N_{K/k}((a))=(N_{K/k}(a))$.

Finally, reduction of the free rank-$n$ module $S_{\mathfrak{p}}$ modulo $\mathfrak{p}$ has dimension $n$ over $R/\mathfrak{p}$. The factorization $\mathfrak{p}S={\prod }_i{\mathfrak{P}}_i^{e_i}$ and the Chinese remainder theorem for ideals identify this reduction with the product of the quotients by ${\mathfrak{P}}_i^{e_i}$. Their filtrations have $e_i$ successive quotients of dimension $f_i$ over $R/\mathfrak{p}$. Adding dimensions proves the fundamental identity for prime decomposition:

Use $R={\mathcal{O}}_{k_{\mathfrak{p}}}$, the valuation ring of the completion of a number field at a prime ideal, with uniformizer $\pi$ and residue field $\kappa$. This completeness is essential for coprime-factor Hensel lifting. Normalize the two residue factors to be monic first; if the original factors have reciprocal leading units, undo the normalization by multiplying the eventual factors by reciprocal lifts of those units.

Let the two monic residue degrees be $d_1,d_2$. Start with monic lifts $g_1,h_1$ satisfying $f\equiv g_1{h}_1(\mathrm{mod}\pi )$. Suppose monic lifts $g_r,h_r$ of these degrees satisfy $f-g_r{h}_r\in {\pi }^{r}R[X]$. Put $E_r=(f-g_r{h}_r)/{\pi }^r$, of degree less than $d_1+d_2$. Over $\kappa$, the linear mapis injective: a zero image makes ${\varphi }_1$ divide $u$, by coprimality, forcing $u=0$ by its degree bound, and then $v=0$. The dimensions agree, so this map is a bijection. Choose lifts $u_r,v_r$ whose image is ${\overline{E}}_r$, and setTheir product agrees with $f$ modulo ${\pi }^{r+1}$; the new cross term has factor ${\pi }^{2r}$ and $2r\ge r+1$. The degree bounds preserve monicity. Each coefficient is a Cauchy sequence in the complete discrete valuation ring $R$, so the limits $f_1,f_2$ satisfyIf ${\mathfrak{o}}_{\mathfrak{p}}$ were interpreted only as an ordinary localization rather than the completed valuation ring, the assertion would be false. For example $X^2-2$ splits modulo seven but is irreducible over $\mathbb{Q}$. The completion notation in the next part provides the intended local-field interpretation.

Set $L=K_{\mathfrak{P}}$ and $F=k_{\mathfrak{p}}$, with valuation rings $S,R$. Let $e$ be the ramification index, $f$ the residue-field degree, and $p$ the residue characteristic. A base uniformizer $\pi$ generates ${\mathfrak{P}}^e$ in $S$, up to a unit. The free $R$-module $S$ has rank $ef$. For an integral $a\in S$, its field trace is the matrix trace of multiplication by $a$ on this module. Reduction modulo $\pi$ preserves the matrix trace.

The quotient $S/\pi S$ has a filtrationEach of the $e$ successive quotients is a one-dimensional vector space over ${\kappa }_L=S/\mathfrak{P}$. On each, multiplication by $a$ acts as multiplication by its residue $\bar{a}$, whose trace over ${\kappa }_F=R/\pi R$ is ${\mathrm{Tr}}_{{\kappa }_L/{\kappa }_F}(\bar{a})$. A basis adapted to the filtration makes the multiplication matrix block triangular, so its trace is the sum of these quotient traces. This proves trace reduction in a ramified local extension:Since the hypothesis says $p\mid e$, the right side is zero in ${\kappa }_F$. ThusNo normality hypothesis is needed. The original PDF only requires divisibility of $e$, not divisibility exactly once; the converted TeX has corrupted that condition and several local subscripts.

For $R={\mathcal{O}}_k$ and $S={\mathcal{O}}_K$, define the inverse different, or codifferent, by the trace-dual latticeThe trace pairing is nondegenerate in this separable field extension, so this is a full fractional ideal of $S$. The relative different ideal is its inverse:Every integral $x$ has integral traces against $S$, so $S\subseteq {\mathfrak{D}}^{-1}$ and $\mathfrak{D}\subseteq S$.

For $k=\mathbb{Q}$, take an integral basis $b_1,\dots ,b_n$ and its trace pairing matrix $T=(\mathrm{Tr}(b_i{b}_j))$. In these coordinates the codifferent is $T^{-1}{\mathbb{Z}}^n$, andThe ideal norm of $\mathfrak{D}$ equals this index, by inversion of fractional ideal norms. Therefore the norm of the different ideal relation isHere $r_2$ is the number of conjugate pairs of complex embeddings. The different is an ideal of the upper field, whereas the relative discriminant ideal is its ideal norm in the base field; over $\mathbb{Q}$ the ordinary signed field discriminant also records the signature.

Use $R={\mathcal{O}}_k$, $S={\mathcal{O}}_L$, $T={\mathcal{O}}_K$, and denote their relative codifferents by $C_{L/k}$, $C_{K/L}$ and $C_{K/k}$. Transitivity of the field trace givesFor the forward direction, testing an element $t\in T$ against every $s\in S$ is legitimate because $st\in T$. For the reverse direction, take $s=1$ and use trace transitivity.

Localize at a nonzero base prime ideal. The intermediate ring is semilocal Dedekind domain, so its invertible fractional ideal $C_{L/k}$ is principal there, say $C_{L/k}=d^{-1}S$ locally. This principality follows by choosing, with the Chinese remainder theorem for ideals, a generator at each of its finitely many maximal ideals. The displayed criterion now becomesInvert these fractional ideals using Solution 1. Equality at every base localization gives the global different in a tower formulaThe second factor is extended to the upper ring of integers; that extension is implicit in the printed formula. At a local prime the exponents consequently satisfy $d(K/k)=d(K/L)+e(K/L)d(L/k)$.

For number fields, a prime ideal divides the different ideal exactly when it is ramified. More precisely, the local different exponent and tame ramification relation is $d_{\mathfrak{P}}\ge e_{\mathfrak{P}}-1$, with equality for tame ramification; if $e=1$, the finite residue extension is separable and the exponent is zero. One can see the ramification criterion directly from the trace pairing modulo the base prime: when $e>1$ its quotient ring has a nonzero nilpotent ideal, whose elements pair to zero since multiplication by a nilpotent has trace zero. When $e=1$ the quotient is a product of finite separable fields and its trace pairing is nondegenerate. Its determinant is a unit exactly in the second case.

Let $M=K_1\cap K_2$. Any rational prime ramifying in $M$ must ramify in each $K_i$, because ramification indices multiply in a tower. It would then divide both field discriminants, contrary to their coprimality. Thus $M$ is unramified at every finite prime, so $|d_M|=1$. The Minkowski lower bound for a number-field discriminant excludes this for a field of degree $n>1$:The last inequality follows already for the weakest signature bound $r_2\le n/2$: the expression starts above one at $n=2$ and increases with $n$. Consequently $M=\mathbb{Q}$.

For completeness, normality converts this intersection statement into the desired degree statement. Put $E=K_1{K}_2$. Restriction injects $\mathrm{Gal}(E/K_2)$ into $\mathrm{Gal}(K_1/\mathbb{Q})$. The fixed field of its image in $K_1$ is exactly $K_1\cap K_2$, because $E/K_2$ is Galois. The Galois correspondence therefore gives $[E:K_2]=[K_1:M]=[K_1:\mathbb{Q}]$. This proves coprime discriminants imply linear disjointness:

For a number field $k$, its ordinary Hilbert class field $H_k$ is the maximal finite abelian extension unramified at every finite prime and with every real place splitting completely. The Artin reciprocity map inducesThe image of an unramified prime ideal is its Frobenius automorphism. Thus a prime splits completely in $H_k$ precisely when its ideal class is trivial. This field packages ideal-class obstructions into splitting behavior. Ideals become principal in $H_k$ by the principal ideal theorem. Specifying the real-place convention matters for real fields: omitting it leads instead to the narrow class-field problem. Both illustrative fields below are imaginary quadratic, so there are no real places to discuss.

First consider $k=\mathbb{Q}(\sqrt{-23})$, whose quadratic discriminant is $-23$. The correspondence between ideal classes and reduced primitive positive definite binary quadratic forms lets us compute its class number without guessing a class field. A reduced form $(a,b,c)$ obeys $|b|\le a\le c$, $b^2-4ac=-23$, and $a\le \sqrt{23/3}<3$; a boundary case uses $b\ge 0$. For $a=1$ the only class is $(1,1,6)$; for $a=2$ the two possibilities are $(2,1,3)$ and $(2,-1,3)$. Hence $h_k=3$ and its ideal class group is cyclic of order three.

Take $g(X)=X^3-X-1$. Its only possible rational roots are $\pm 1$, neither a root, so it is irreducible. Its polynomial discriminant is $4-27=-23$, which is nonsquare. Thus its splitting field $H$ has Galois group $S_3$ and the quadratic fixed field of $A_3$ is $k$, generated by the square root of its discriminant. Therefore $H/k$ is cyclic of degree three and $H=k(\theta )$ for a root $\theta$ of $g$.

At a prime other than $23$, the separable residue polynomial splits over a finite residue-field extension; Hensel's lemma puts all its roots in an unramified local extension. At $23$ there is the exact factorizationThe simple factor lifts by Hensel's lemma. The remaining quadratic factor has discriminant valuation one, because the cubic discriminant is $-23$ and the resultant with the simple factor is a unit. Its local splitting field is ramified quadratic, with inertia group of order two. Its intersection with $A_3$ is trivial, so the relative extension $H/k$ is unramified at $23$ as well. There are no real places. An unramified abelian extension of degree $h_k=3$ must be all of $H_k$, givingThis is the Hilbert class field of Q of square root minus twenty-three.

The other alternative, $k=\mathbb{Q}(\sqrt{-30})$, has quadratic discriminant $-120$. Reduced forms have $a\le \sqrt{40}<7$ and even $b$. Testing $1\le a\le 6$, $|b|\le a$, and $c=(b^2+120)/(4a)\ge a$ gives exactlyThus $h_k=4$. Each class is its own inverse, because replacing $b$ by $-b$ leaves these representatives unchanged; consequently the ideal class group is $C_2\times C_2$.

Set $H=\mathbb{Q}(\sqrt{2},\sqrt{-3},\sqrt{5})$. The square classes $2,-3,5$ are independent, so $[H:\mathbb{Q}]=8$. Their product gives $\sqrt{-30}$, and $H/k$ is abelian of degree four. Its seven quadratic subfields have discriminantsFor a multiquadratic abelian extension, the conductor-discriminant formula takes the product of these quadratic character conductors, namely the absolute quadratic discriminants. Hence $|d_H|=120^4$. The relative discriminant tower formula givesThe integral relative discriminant ideal therefore has norm one and is the unit ideal. Thus $H/k$ is everywhere unramified, and its degree equals the class number. This proves the other explicit answer:It agrees with the Hilbert class field of Q of square root minus thirty. Both alternatives have been computed independently from their class numbers and ramification.

The original PDF has cube roots throughout this question; the ninth roots in the converted TeX are transcription errors. Put $\alpha =\sqrt[3]{m_1{m}_2^2}$ and $\beta ={\alpha }^2/m_2$. The given condition excludes the trivial cube case. Both generators are integral, becauseMoreoverso $A=\mathbb{Z}+\mathbb{Z}\alpha +\mathbb{Z}\beta$ is an order in the cubic field. The field traces of $\alpha ,\beta ,{\alpha }^2,{\beta }^2$ are zero, while $\mathrm{Tr}(\alpha \beta )=3m_1{m}_2$. Thus the trace pairing matrix of this basis isWe must prove that this is the maximal order, not just compute the discriminant of a suborder. The discriminant-index formula for an integral lattice restricts any index prime to those dividing $3m_1{m}_2$. If $p\mid m_1$, the cubic for $\alpha$ is Eisenstein at $p$, while $m_2$ is a unit there and $A\otimes {\mathbb{Z}}_p={\mathbb{Z}}_p[\alpha ]$. If $p\mid m_2$, use instead the Eisenstein polynomial for $\beta$, and $\alpha ={\beta }^2/m_1$. In either case an Eisenstein generator is a uniformizer of the totally ramified local field and generates its full valuation ring. Indeed expansion in powers of that uniformizer using base residue representatives, and grouping powers modulo three, expresses every integral local element in its power basis.

These arguments also cover three when $3\mid m_1{m}_2$. If $3\nmid m_1{m}_2$, every unit modulo nine has cube congruent to $1$ or $-1$. Therefore the condition $m_1\not\equiv \pm m_2(\mathrm{mod}9)$ implies $m_1{m}_2^2\not\equiv \pm 1(\mathrm{mod}9)$: multiply a contrary congruence by the inverse of $m_2^2$ and use $m_2^3\equiv \pm 1$. Choose $r\in \{1,-1\}$ with $r^3\equiv m_1{m}_2^2(\mathrm{mod}3)$. The polynomialis Eisenstein at three, since its constant term is divisible by three but not nine. Its root $\alpha -r$ again generates the full local valuation ring. There is no index prime left. We have proved the integral basis of a cubefree pure cubic field and the requested discriminant:For $K=\mathbb{Q}(\sqrt[3]{6})$ write ${\alpha }^3=6$. Here ${\mathcal{O}}_K=\mathbb{Z}[\alpha ]$ and $d_K=-972$. The norm of $x+y\alpha +z{\alpha }^2$ is the determinant of its multiplication matrix, givingIn particularAn integral element of norm one has an integral inverse, by its monic characteristic polynomial. Here direct multiplication gives $u^{-1}=109+60\alpha +33{\alpha }^2$. Nontriviality also follows from linear independence of $1,\alpha ,{\alpha }^2$.

The signature is $(1,1)$, so the Minkowski bound for ideal classes isEvery class therefore has an integral representative of norm at most eight. Its prime factors have norms at most eight, and lie above $2,3,5,7$. Since the index of $\mathbb{Z}[\alpha ]$ is one, reduction of $X^3-6$ gives their factorizations by the Kummer-Dedekind theorem. At two and three the polynomial is $X^3$, so there is one prime of norm $2$ or $3$, respectively. They are principal, becauseAt five the factorization is $(X-1)(X^2+X+1)$ with the quadratic irreducible. The norm-five prime is $(\alpha -1)$ because $N(\alpha -1)=5$; the norm-twenty-five prime is outside the bound.

At seven the roots are $3,5,6$, giving three primes ${\mathfrak{q}}_r=(7,\alpha -r)$ of norm seven. The elements $\alpha +1$ and ${\alpha }^2+\alpha -5$ have norm seven and vanish at roots six and three, respectively; hence they generate ${\mathfrak{q}}_6$ and ${\mathfrak{q}}_3$. Since ${\mathfrak{q}}_3{\mathfrak{q}}_5{\mathfrak{q}}_6=(7)$, the remaining prime is principal too. Every prime which can divide a Minkowski representative is now principal, soThis proves class number one for Q of cube root six with all primes within the bound accounted for.

The hypotheses make $K$ a CM-field, with nontrivial automorphism $c$ over the totally real number field $k$. In every complex embedding of $K$, this automorphism acts as complex conjugation. Write $E_K={\mathcal{O}}_K^{\times }$, $E_k={\mathcal{O}}_k^{\times }$ and ${\mu }_K$ for the finite cyclic group of roots of unity in $K$.

For any $u\in E_K$, the element $u/c(u)$ is an algebraic integer, indeed a unit, and every conjugate has absolute value one. The Kronecker theorem on algebraic integers in the unit disk makes it a root of unity. For clarity, the essential proof is finite: the conjugates of all positive powers have modulus one, so the integer coefficients of their monic minimal polynomials are bounded by fixed binomial coefficients. There are only finitely many such polynomials and roots; two powers coincide, forcing a root of unity.

Thusis a homomorphism. Its kernel is exactly $E_k$, since a conjugation-fixed element lies in $k$ and both it and its inverse are integral there. For $\zeta \in {\mu }_K$, $c(\zeta )={\zeta }^{-1}$ and $\varphi (\zeta )={\zeta }^2$. Consequently $\varphi$ induces an injectionTo check injectivity explicitly, if $\varphi (u)={\zeta }^2$, then $\varphi (u/\zeta )=1$, so $u/\zeta \in E_k$. The cyclic group ${\mu }_K$ has even order because it contains $-1$, and its quotient by squares has order two. Therefore the unit index of a CM-field satisfies

Let $H_k$ be the ordinary Hilbert class field of $k$. Its real places split completely, so $H_k$ is totally real. Since $K/k$ is quadratic and $K$ is totally imaginary, $H_k\cap K$ cannot equal $K$; hence it equals $k$. Therefore the compositum $K{H}_k$ hasand is an abelian extension of $K$, by restriction of the Galois group to $H_k$. At finite primes it is unramified because unramified extensions remain unramified under base change. The field $K$ has no real places, so there is no infinite-place ramification condition left. Thus $K{H}_k\subseteq H_K$, where $H_K$ is the Hilbert class field of $K$. Since $[H_K:K]=h_K$, the tower degree formula proves class number divisibility for a CM extension:This argument uses the ordinary class field with split real places; replacing it silently by the narrow class field would invalidate the claim that $H_k$ is totally real.