Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 25 (Class Field Theory)

The hypotheses for Kummer theory are that $n\ge 1$ is prime to the characteristic of $K$ and that all $n$th roots of unity lie in $K$. Under these hypotheses, adjoining radicals produces abelian extensions of exponent dividing $n$, and conversely every such finite extension is obtained this way. The roots-of-unity hypothesis matters: without it, adjoining one radical need not give a Galois extension.

We first prove Hilbert's theorem 90, including its cocycle form. Distinct field automorphisms are linearly independent as functions into the field. Indeed, in a shortest nontrivial relation ${\sum }_i{c}_i{\tau }_i(y)=0$, choose $x$ with ${\tau }_1(x)\ne {\tau }_r(x)$. Subtract ${\tau }_r(x)$ times the relation at $y$ from the relation at $xy$. This eliminates the last term while leaving the first nonzero, contradicting minimality. This is linear independence of distinct field embeddings.

Let $L/K$ be finite Galois with group $G$, and suppose $a_{\sigma }\in L^{\ast }$ satisfy $a_{\sigma \tau }=a_{\sigma }\sigma (a_{\tau })$. By independence of the automorphisms, choose $c\in L$ for whichThenThus every multiplicative cocycle is a coboundary:For a cyclic group generated by $\sigma$ of order $m$, an element $a$ of norm one defines the cocycle $a_{{\sigma }^i}={\prod }_{j=0}^{i-1}{\sigma }^j(a)$; the condition at $i=m$ is exactly $N_{L/K}(a)=1$. HenceThe reverse implication follows by telescoping the product of conjugates.

Now let $\Delta$ be a finite subgroup of $K^{\ast }/K^{\ast n}$ and put $L=K(\sqrt[n]{\Delta })$. Finitely many representatives generate it; since ${\mu }_n\subset K$, the polynomials $X^n-a$ split in $L$ and are separable. Thus $L/K$ is finite Galois. For $\sigma \in G$ and $[a]\in \Delta$, choose ${\alpha }^n=a$ and define the Kummer pairingChanging the root or multiplying $a$ by an $n$th power does not change this value. It is multiplicative in both arguments. Since the radicals generate $L$, it embeds $G$ in $\mathrm{Hom}(\Delta ,{\mu }_n)$. Consequently $G$ is abelian of exponent dividing $n$.

For completeness, establish perfection rather than just this injection. Let ${\Delta }_L$ consist of all classes in $K^{\ast }/K^{\ast n}$ having an $n$th root in $L$. The same construction gives an injective map ${\Delta }_L\to \mathrm{Hom}(G,{\mu }_n)$: the character is trivial precisely when the root lies in $K$. It is surjective by Hilbert 90. Given $\chi :G\to {\mu }_n$, apply the cocycle theorem to $a_{\sigma }=\chi (\sigma {)}^{-1}$; it gives $\sigma (b)/b=\chi (\sigma )$, so $b^n\in K$ and its class has the required character. A finite abelian group of exponent dividing $n$ has as many ${\mu }_n$-characters as elements, by decomposition into cyclic factors. ThereforeIt follows that $\Delta ={\Delta }_L$ andConversely, for any finite abelian $L/K$ of exponent dividing $n$, Hilbert 90 supplies eigenvectors for every character. Their $n$th powers lie in $K$, and the intersection of their character kernels is trivial, so they generate $L$. This proves the Kummer classification and the inclusion-reversing duality between subgroups and intermediate extensions.

The simplest case is a cyclic extension of degree $m\mid n$: it has a generator $\alpha$ with ${\alpha }^m\in K$, and its automorphisms multiply $\alpha$ by $m$th roots of unity. Quadratic extensions in characteristic different from two are the case $n=2$. The cohomological summary is the exact sequence $1\to {\mu }_n\to K_s^{\ast }\stackrel{z\mapsto z^n}{\to }{K}_s^{\ast }\to 1$ together with Hilbert 90, giving $H^1(G_K,{\mu }_n)\cong K^{\ast }/K^{\ast n}$.

Normalize each finite-place valuation to have image $\mathbb{Z}$. For a radical class $[a]$, the residue of $v(a)$ modulo $n$ is well-defined, because replacing $a$ by $a{b}^n$ adds $nv(b)$. A suitable finite set isOne can compute the last union from any finite generating set of $\Delta$. Each representative has only finitely many nonzero valuations, so this indeed defines a finite set depending only on $n$ and the classes, not on a choice of representatives. Including all infinite places safely allows possible real-to-complex ramification.

Fix $v\notin S$ and work over the completion $K_v$. A representative has $v(a)=nk$, so $a={\pi }^{nk}u$ with $u$ a unit. Adjoining an $n$th root of $a$ is the same as adjoining an $n$th root of $u$. Since $n$ is invertible in the residue field, $X^n-\bar{u}$ is separable: its derivative $n{X}^{n-1}$ has no common zero with it.

Take a finite extension of the finite residue field containing all roots of all these finitely many reduced polynomials. The corresponding finite unramified extension of $K_v$ exists by the classification proved in Question 2. Hensel's lemma lifts each simple residue root to a root of the original polynomial in that unramified extension. Thus all the required radicals, and every conjugate root, lie in an unramified local extension. Every completion of the radical field at $v$ is therefore unramified. This proves finite ramification support for radical extensions.

The local argument does not assume ${\mu }_n\subset K$: adjoining all roots proves the stronger unramified assertion for the splitting field, and hence for any field obtained by choosing the radicals. For $n=1$ the radical extension is trivial; the displayed set is still valid.

Use the following simple-root Hensel lemma. Let $R$ be a complete discrete valuation ring, with maximal ideal $\mathfrak{m}$, and let $f\in R[X]$. If $f(a_0)\in \mathfrak{m}$ and $f^{\prime }(a_0)\in R^{\ast }$, there is a unique $a\in R$ with $a\equiv a_0(\mathrm{mod}\mathfrak{m})$ and $f(a)=0$.

For the proof, define Newton's method iterates $a_{r+1}=a_r-f(a_r)/f^{\prime }(a_r)$. Each denominator stays a unit, because all iterates are congruent to $a_0$. Taylor's polynomial identity givesIf an iterate is already a root, the process stops. Otherwise these valuations tend to infinity, so the iterates form a Cauchy sequence in the complete ring. Its limit $a$ has the required residue and satisfies $f(a)=0$. If $a,b$ were two such roots, thenThe parenthesis is a unit, forcing $a=b$.

Now let $K/{\mathbb{Q}}_p$ be finite, with residue field $k={\mathbb{F}}_q$. For every $m\ge 1$ choose a monic irreducible polynomial $\bar{f}\in k[X]$ of degree $m$ and lift it to a monic $f\in {\mathcal{O}}_K[X]$. The lift is irreducible over $K$, since a factorization would reduce to a factorization of $\bar{f}$, by the monic form of Gauss's lemma over the valuation ring. If $L=K(\alpha )$ for a root, its residue field contains $k(\bar{\alpha })$ of degree $m$. The inequality $e(L/K)f(L/K)\le [L:K]=m$ forces residue degree $m$ and ramification index one. Thus it is an unramified extension of degree $m$.

Conversely, in any unramified extension of degree $m$, choose a generator of its residue field and lift its separable minimal polynomial. Hensel's lemma supplies a root with that residue. Its degree is at least $m$ and hence it generates the extension. The same unique lifting of all residue conjugates shows that the extension is Galois andOne can see uniqueness inside a fixed algebraic closure explicitly: the roots of $X^{q^m-1}-1$ lift all nonzero residue elements, and a lifted primitive residue element generates the extension. Hencewith arithmetic Frobenius acting by $\zeta \mapsto {\zeta }^q$. The case $q^m-1=1$ simply gives $K_1=K$. These are all finite unramified extensions; $K_m\subseteq K_r$ exactly when $m\mid r$. Their union is the maximal unramified extension, whose Galois group is $\hat{\mathbb{Z}}$.

Finally prove norm-surjectivity on units using the Herbrand quotient. For a cyclic group $G$ of order $m$, writeThe quotient is multiplicative on short exact sequences, equals one for finite modules, equals $m$ on the trivial module $\mathbb{Z}$, and equals one on induced regular modules. For a finite module, the equality follows directly from $|M^G||(\sigma -1)M|=|M|=|\ker N||NM|$.

Let $U_L={\mathcal{O}}_L^{\ast }$. For sufficiently large $r$, the equivariant p-adic logarithm identifies $U_L^r=1+{\mathfrak{p}}_L^r$ with ${\mathfrak{p}}_L^r$ additively. In the unramified case this is an induced regular lattice. Indeed choose a normal basis of the finite residue extension and lift its generating element. Its $m$ Galois conjugates reduce to a basis and hence form an ${\mathcal{O}}_K$-basis of ${\mathcal{O}}_L$; their coordinate determinant is a unit. Since a uniformizer of $K$ is also one of $L$, multiplication by its $r$th power gives the same regular-module structure on ${\mathfrak{p}}_L^r$. For a regular module, invariants are norm images and a vector of coefficient sum zero is a cyclic difference, so both indicated Tate cohomology groups vanish. Consequently $h_G(U_L^r)=1$. The quotient $U_L/U_L^r$ is finite, so $h_G(U_L)=1$.

Apply multiplicativity toIt gives $h_G(L^{\ast })=m$. Hilbert 90 makes ${\hat{H}}^{-1}(G,L^{\ast })$ trivial, so $[K^{\ast }:N_{L/K}{L}^{\ast }]=m$. On the other handso the quotient maps onto $\mathbb{Z}/m\mathbb{Z}$. As its order is already $m$, its unit kernel is trivial. A norm which is a unit comes from a unit, by the same valuation formula. ThereforeThe standard normal-basis theorem for finite fields and the convergent logarithm/exponential on sufficiently deep principal units are the ingredients used in evaluating the Herbrand quotient; the norm conclusion itself has been deduced here.

In the absolute-value description, a place of a number field is an equivalence class of nontrivial absolute values, where equivalent absolute values give the same topology, or differ by a positive power. More generally this defines the rank-one absolute-value places used in weak approximation for a field. For a number field, finite places correspond to nonzero prime ideals of its ring of integers, while infinite places correspond to real embeddings and pairs of conjugate complex embeddings. The completion at $v$ is denoted $K_v$.

A valuation place can also be described by a valuation ring ${\mathcal{O}}_v\subset K$ and its residue map into $k_v$, extended by sending elements outside that ring to infinity. The finite places above are the discrete-valuation instances of this description.

The weak approximation theorem states that for finitely many pairwise inequivalent nontrivial absolute values $v_1,\dots ,v_r$, the diagonal image of $K$ is dense in ${\prod }_i{K}_{v_i}$. Equivalently, for arbitrary $a_i\in K_{v_i}$ and ${ϵ}_i>0$, there is $a\in K$ such thatThe places must be distinct: two requirements at the same completion need not be compatible. In the number-field setting this permits simultaneous prescribed finite congruences and archimedean approximations.

The idele group is the restricted productwith respect to ${\mathcal{O}}_v^{\ast }$ at the finite places. Thus $a=(a_v)$ is an idele if all components are nonzero and $a_v$ is a unit at all but finitely many finite places. Its topology has basic open sets which equal the full unit subgroup outside finitely many places; it is not just the subspace topology of the unrestricted product. The group is locally compact, and the diagonal $K^{\ast }$ consists of the principal ideles. The idèle class group is $C_K=J_K/K^{\ast }$.

The finite valuations give the idele-to-ideal homomorphismIts kernel is the product of the finite unit groups and the infinite multiplicative groups; principal ideles map to principal fractional ideals. Thus further quotienting gives the ideal class group. The product of normalized local moduli defines $|a{|}_J$; at complex places the normalized modulus is the square of the ordinary absolute value. The product formula says that principal ideles have modulus one. These constructions package ideals, units and archimedean conditions into one group.

Here is the precise reciprocity equivalence. For a finite abelian extension $L/K$, the global law is a continuous Artin reciprocity mapwhich kills $K^{\ast }$ and sends a uniformizer idele at an unramified place to arithmetic Frobenius. Its full class-field-theoretic norm statement is $C_K/N_{L/K}{C}_L\cong \mathrm{Gal}(L/K)$. In the local formulation, the mapsare continuous, kill local units when unramified, send a local uniformizer to the same Frobenius, are compatible with restriction in towers, and obeyHere $D_v$ is the decomposition group of any prime above $v$, independent of that choice because the extension is abelian. The usual local norm theorem identifies $K_v^{\ast }/N_{L_w/K_v}{L}_w^{\ast }$ with $D_v$. The principal-idele condition is the essential global compatibility; existence of unrelated local norm isomorphisms alone would not establish global reciprocity.

Given the compatible local maps, defineOnly finitely many factors are nontrivial: there are finitely many ramified places, and an idele is a unit almost everywhere. The product is a homomorphism. Continuity follows by choosing local kernel neighbourhoods at those exceptional places and the full unit groups elsewhere. The boxed compatibility kills $K^{\ast }$, so the map descends to $C_K$. At every unramified place it has the required Frobenius value. The local norm maps also show that idele norms are killed, since the local component of an idele norm is ${\prod }_{w\mid v}{N}_{L_w/K_v}(b_w)$. Together with the class-field norm-index statement this gives the global norm-kernel formulation.

Conversely, embed $K_v^{\ast }$ in $J_K$ with every other component one, and set $r_v=\theta \circ {\iota }_v$. Each restriction is continuous and has the required unramified normalization. An open kernel of the finite-valued global map contains a basic neighbourhood whose unrestricted finite-place factors are full unit groups outside a finite set. Hence those local restrictions kill units. For an arbitrary idele, remove the finitely many exceptional and nonunit components; the remainder lies in that kernel neighbourhood. It follows directly that $\theta$ is exactly the product of its local restrictions, not merely a map agreeing on single-component ideles. Its triviality on $K^{\ast }$ is exactly the boxed product relation. With the decomposition and norm assertions of the reciprocity theorem, these restrictions are the local reciprocity maps into $D_v$ with kernel $N_{L_w/K_v}{L}_w^{\ast }$.

One can explicitly connect this idelic formulation with the classical ideal law without assuming an idelic law in advance. A modulus of a number field records finitely many nonnegative exponents at finite places and a selected subset of real places. For a modulus $\mathfrak{m}$, let $U_{\mathfrak{m}}$ have factors $1+{\mathfrak{p}}_v^{m_v}$ at finite places dividing the modulus, full units at other finite places, positive real numbers at specified real places, and full infinite multiplicative groups otherwise. Weak approximation for number fields supplies a global $a\in K^{\ast }$ making $a^{-1}j$ lie in the specified local factors at the finitely many modulus places. Its valuations outside the modulus define an ideal prime to $\mathfrak{m}$. A different such $a$ changes that ideal by a ray-principal ideal, and every ideal prime to $\mathfrak{m}$ is obtained. HenceAn ideal Artin map killing $P_K(\mathfrak{m})$ therefore extends to $J_K$ via this isomorphism; restricting it gives the local maps just constructed. Conversely, continuous local maps satisfying the product condition have an open product kernel containing some $U_{\mathfrak{m}}$ and hence make the ideal Artin map kill $P_K(\mathfrak{m})$. This proves the idelic local-global reciprocity equivalence at the reciprocity-law level as well as the formal conversion of maps. The arithmetic Frobenius normalization and product relation agree with Milne's local-global reciprocity formulation.

A central simple algebra over $K$ is a finite-dimensional associative unital $K$-algebra $A$ with no nonzero proper two-sided ideals and center exactly $K$. Its dimension is a square, ${\dim }_{K}A=n^2$. The Artin–Wedderburn theorem writes it as $M_r(D)$ for a uniquely determined central division algebra $D$ up to isomorphism. One obtains this structure by taking a minimal left ideal and its commuting division endomorphism ring; the simple algebra acts as all endomorphisms over that ring. Over a suitable finite splitting field it becomes a full matrix algebra.

The Brauer group identifies two central simple algebras if their division-algebra parts agree, equivalently if tensoring each with suitable matrix algebras makes them isomorphic. The product is tensor product, the identity is the class of $K$, and the inverse of $[A]$ is $[A^{\mathrm{op}}]$. Indeedis an isomorphism for a central simple algebra, so the product class is split. This also explains why matrix size is irrelevant to the Brauer class. The order of the class is its period, while the degree of its division-algebra representative is its index; these are different notions in general.

For the cyclic algebra construction, take a cyclic extension $L/K$ of degree $m$, a generator $\sigma$ and $a\in K^{\ast }$. SetThe multiplication is obtained by moving coefficients past $u$ and replacing every $u^m$ by $a$. It is associative: reducing a product of three monomials gives the same total exponent, the same power of $a$ and the same iterated automorphisms in either order. The scalar $a$ is fixed by $\sigma$ and ${\sigma }^m=1$, so reduction is consistent.

An element commuting with every $\ell \in L$ has only its $u^0$ coefficient nonzero, because ${\sigma }^i\ne 1$ for $0<i<m$. Commuting with $u$ then forces that coefficient into $K$, proving that the center is $K$. To prove simplicity, choose a nonzero element of a nonzero ideal with the fewest nonzero monomial coefficients. Since $u$ is invertible and each nonzero coefficient in $L$ is invertible, normalize one term to be the constant one. If another term remains, its automorphism differs from the identity; commuting with a suitable element of $L$ eliminates the constant term but not that other term. This gives a shorter nonzero element, a contradiction. The minimal element is therefore a unit, and the ideal is all of $A$. Thus $A$ is central simple of dimension $m^2$.

It splits over $L$. For example, over $L$ represent $\ell$ by the diagonal matrix with entries $\ell ,\sigma (\ell ),\dots ,{\sigma }^{m-1}(\ell )$, and let $u$ shift the standard basis down one place, sending the first vector to $a$ times the last. These matrices satisfy the defining relations. A primitive element of the separable extension $L/K$ has distinct conjugates; the Vandermonde matrix of their powers is invertible. Thus the diagonal images span all diagonal matrices after scalar extension to $L$. Multiplying the diagonal matrix units by powers of the shift gives every matrix unit, proving $A{\otimes }_{K}L\cong M_m(L)$.

The particularly useful splitting criterion isIf $a=N(b)$, let the algebra act on $L$ by multiplication for $\ell$ and by the semilinear operator $b\sigma$ for $u$. Its $m$th power is multiplication by $N(b)=a$, so this gives an isomorphism with ${\mathrm{End}}_K(L)$, by simplicity and dimension. Conversely, a split algebra acts on an $m$-dimensional $K$-space $V$. Restriction to the subfield $L$ makes $V$ one-dimensional over $L$, and $u$ is a nonzero $\sigma$-semilinear operator. In an $L$-basis it is $b\sigma$, so $u^m=a$ forces $a=N(b)$. More generally the relative Brauer group split by a cyclic $L/K$ is $K^{\ast }/N_{L/K}{L}^{\ast }$, with multiplication of parameters corresponding to addition of classes.

For a finite field, a finite division algebra is commutative by Wedderburn's little theorem. Therefore every central simple algebra is a matrix algebra and $\boxed{\mathrm{Br}({\mathbb{F}}_q)=0}$. The cyclic-algebra viewpoint gives the same conclusion for cyclic classes because the norm between finite fields is surjective on their cyclic multiplicative groups.

For a nonarchimedean local field, the local Brauer invariant givesAn unramified cyclic extension of degree $m$, with arithmetic Frobenius generator, gives invariant $v_K(a)/m$ to its cyclic algebra. Question 2 shows that all units are norms; norm valuations are multiples of $m$. Thus the relative classes are $\mathbb{Z}/m\mathbb{Z}$, represented by powers of a uniformizer. Allowing all $m$ realizes all invariants. At the infinite local fields, $\mathrm{Br}(\mathbb{C})=0$, while $\mathrm{Br}(\mathbb{R})\cong \mathbb{Z}/2\mathbb{Z}$, with the nonzero class represented by Hamilton's quaternions and invariant $1/2$.

For a number field, the Albert-Brauer-Hasse-Noether theorem supplies the exact sequenceHence a global algebra is determined by its finitely many nonzero local invariants, whose sum is zero; it is split precisely when all its local classes vanish. Every division algebra over a number field is cyclic, so cyclic algebras provide representatives for the global classes as well. Their norm parameters translate splitting into local norm conditions and the reciprocity constraint on the sum of invariants. The local invariant theorem, Wedderburn's little theorem and the global invariant theorem are the standard structural results used here; their role in the requested computation is explicit.

The existence theorem of global class field theory, together with reciprocity, gives an inclusion-reversing bijection between finite abelian extensions $L/K$ inside a fixed algebraic closure and open finite-index subgroups of the idèle class group $C_K$. The subgroup for $L$ is $N_{L/K}{C}_L$, andEquivalently, for each modulus $\mathfrak{m}$, subgroups between the ray-principal group $P_K(\mathfrak{m})$ and the ideal group $I_K(\mathfrak{m})$ classify abelian extensions whose conductor divides $\mathfrak{m}$. Their quotients are identified with the corresponding Galois groups by the Artin map. Changing the modulus gives the same extension when the induced congruence subgroup is the same.

The conductor criterion says that a finite place has conductor exponent zero exactly when the extension is unramified there; a real place is absent from the conductor exactly when it stays real. Thus conductor one means unramified at every finite place and no real-to-complex ramification.

Apply the existence theorem to the open subgroup of $C_K$ which is the image ofThe idele-to-ideal map identifies the quotient by this subgroup with the ordinary ideal class group, which is finite. Therefore its class field $H$ exists and satisfiesIts conductor is one, so it is everywhere unramified. Conversely any everywhere-unramified abelian extension has conductor one; its Artin map kills $U$, hence factors through this quotient and corresponds to an extension contained in $H$. This proves that $H$ is exactly the Hilbert class field, the maximal abelian everywhere-unramified extension. Here the ordinary ideal class group is appropriate because the infinite factors include all $K_v^{\ast }$, rather than only the positive real factors used in a narrow ray class group.

Put $\omega =(1+\sqrt{229})/2$. Since $229\equiv 1(\mathrm{mod}4)$ is squarefree,We first compute the ideal class group, rather than assume the proposed extension has the full Hilbert-class-field degree.

The real quadratic Minkowski bound for ideal classes is $\sqrt{229}/2<8$. Thus every ideal class has an integral representative of norm at most seven. The prime two is inert because $229\equiv 5(\mathrm{mod}8)$, while seven is inert because $229\equiv 5(\mathrm{mod}7)$ is a nonsquare. The inert ideal $(2)$ has norm four and is principal; a prime above seven has norm forty-nine. The only possible nonprincipal prime factors of these small representatives are therefore primes above three and five.

Reduction of $X^2-X-57$ gives roots $0,1$ modulo three and $2,4$ modulo five. WriteTheir conjugate pairs multiply to $(3)$ and $(5)$. Direct norm calculations giveThe element $6+\omega$ vanishes at the residue roots $0$ modulo three and $4$ modulo five, and not at the conjugate roots. Its absolute norm forces each exponent to be one. Likewise $15+2\omega$ vanishes only at the root zero modulo three, and its norm forces exponent three. ThusThese also give the factorizations suggested by the hint: $(13+\sqrt{229})=(2)PQ$ and $(16+\sqrt{229})=P^3$. The classes of $P,Q$ generate the ideal class group; the first relation makes $[Q]=[P{]}^{-1}$ and the second makes $[P{]}^3=1$. Hence the class number divides three.

It is not one. The algebraic integerso it is a unit. If $P$ were principal, its generator $a=x+y\omega$ would have norm $\pm 3$. Multiply $a$ by a suitable integer power of $ϵ$ so thatIts conjugate then also has absolute value at most $\sqrt{3ϵ}$ because $|a{a}^{\prime }|=3$. ThereforeThe final inequality is $12ϵ<229$, immediate from the displayed value of $ϵ$. Thus the integer $y$ is zero, which makes $|N(a)|=x^2=3$ impossible. ConsequentlyThis proves the ideal class group of Q of square root 229. No claim that $ϵ$ is a fundamental unit is needed. Its norm minus one also realizes all real sign patterns after multiplication by $-1$, so the ordinary and narrow class numbers agree.

Now let $f(X)=X^3-4X-1$ and let $H$ be its splitting field. The possible rational roots $\pm 1$ are not roots, so $f$ is irreducible. Its discriminant isThus its Galois group is $S_3$, since an irreducible cubic has group $A_3$ precisely when its discriminant is a square. Its quadratic subfield is $\mathbb{Q}(\sqrt{229})=K$, and $H/K$ is cyclic of degree three. If $\alpha$ is one root, then $H=K(\alpha )$.

For every rational prime $\ell \ne 229$, the reduction of $f$ is separable. Its roots lift in an unramified extension containing its residue splitting field, by Hensel's lemma. Hence its splitting field is unramified over $\mathbb{Q}$ at all these primes, and consequently over $K$ there.

At 229 there is exactly one double residue root and one simple root:The simple root lifts to $r\in {\mathbb{Q}}_{229}$, and division gives $f=(X-r)q$ with $q$ monic quadratic. The discriminant-product identity isHere $q(r)=f^{\prime }(r)$ is a unit, since $r$ has the simple residue root. Therefore the quadratic discriminant has valuation one, and its root field is the ramified quadratic field ${\mathbb{Q}}_{229}(\sqrt{229})$. This is already the completion of $K$. The local splitting field has no further extension over it: the simple root lies in ${\mathbb{Q}}_{229}$ and the other two lie in that quadratic field. Thus $H/K$ is unramified even at 229. Equivalently its rational inertia is a transposition and has trivial intersection with $A_3$.

The positive cubic discriminant gives three distinct real roots, so $H$ is totally real and there is no infinite ramification over $K$. We have exhibited an everywhere-unramified cyclic extension of degree $h_K=3$. The Hilbert-class-field characterization in part (i) now proves

Finally let $p\ne 229$ be rational prime. A prime ideal of norm $p$ is principal if and only if its Artin symbol in $H_K/K$ is trivial. If it is principal, a generator has norm $\pm p$; multiplying by the norm-minus-one unit if necessary gives norm $+p$. Conversely an element of norm $p$ generates a prime ideal of norm $p$. Hence representation by $x^2+xy-57y^2$ is equivalent to $p$ splitting in $K$ with trivial Artin symbol in $H_K/K$, or to complete splitting of $p$ in $H_K/\mathbb{Q}$.

Let ${\tau }_p\in S_3$ be its rational Frobenius, defined up to conjugacy. Its restriction to the discriminant quadratic field is its sign. The splitting criterion for that quadratic field isFor odd $p$ this follows by quadratic reciprocity from the character of discriminant 229; for $p=2$ both criteria are negative since $229\equiv 5(\mathrm{mod}8)$. The polynomial is separable modulo $p$, and the degrees of its factors are the cycle lengths of ${\tau }_p$. Thus a root modulo $p$ means that ${\tau }_p$ fixes a root. An even permutation of three letters with a fixed point is the identity: a nonidentity three-cycle has none. Therefore the two stated conditions together are exactly complete splitting in $H_K$, andThis is prime representation by the norm form of Q of square root 229, with the norm sign and the prime two explicitly covered.