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For an unramified prime $\mathfrak{p}$ in a finite abelian extension $L/K$, the residue degree equals the order of its Artin symbol. Consequently the number of primes above $\mathfrak{p}$ is $[L:K]$ divided by that order.

Let ${\zeta }_n$ be a primitive nth root of unity. The cyclotomic polynomial isInduction on $n$ shows that ${\Phi }_n$ is monic with integer coefficients. Here is the essential irreducibility of cyclotomic polynomials argument. Let $f$ be the minimal polynomial of ${\zeta }_n$ and let $\ell \nmid n$ be a prime. If ${\zeta }_n^{\ell }$ had a different monic irreducible polynomial $g$, then $f\mid g(X^{\ell })$. Reducing modulo $\ell$ gives $\bar{f}\mid {\bar{g}}^{\ell }$, so $\bar{f}$ and $\bar{g}$ have a common factor. As $fg\mid {\Phi }_n$, this would give a repeated factor of $X^n-1$ modulo $\ell$, impossible because its derivative is $n{X}^{n-1}$. Thus ${\zeta }_n^{\ell }$ is conjugate to ${\zeta }_n$. Every positive integer prime to $n$ is a product of such primes, so all primitive nth roots of unity are conjugate. ConsequentlyIn particular the cyclotomic field is an abelian extension. Complex conjugation corresponds to $-1$; for $n>2$ its fixed field is $\mathbb{Q}({\zeta }_n+{\zeta }_n^{-1})$, the maximal real subfield, of degree $\phi (n)/2$.

The ring of integers is $\mathbb{Z}[{\zeta }_n]$. One way to see that there is no missing index is to work at each prime $p$. Write $n=p^{a}m$ with $p\nmid m$. The prime-to-$p$ roots of unity have separable reduction; their local factors generate unramified extensions with full integer rings. Over each such factor, adjoining ${\zeta }_{p^a}$ is generated by the Eisenstein polynomial ${\Phi }_{p^a}(1+T)$, with uniformizer ${\zeta }_{p^a}-1$, and hence gives the full integer ring of that totally ramified factor. The coprimality of $p^a$ and $m$ means that ${\zeta }_n$ generates both sets of roots of unity. Thus ${\mathbb{Z}}_p[{\zeta }_n]$ is maximal at every $p$, proving the integral assertion.

The same local description computes ramification and splitting. For $p\nmid n$, the ramification index is one, the residue degree is the multiplicative order of $p$ modulo $n$, and there are $\phi (n)/f$ primes above $p$. If $p\mid n$, put $m=n/p^a$; thenwhere ${\mathrm{ord}}_1(p)=1$. Indeed, the prime-to-$p$ part is unramified, and the $p$-power part is totally ramified by the preceding Eisenstein polynomial. Notice the exception $p=2,a=1$: its ramification index is one, consistently with $\mathbb{Q}({\zeta }_{2m})=\mathbb{Q}({\zeta }_m)$ for odd $m$. Thus finite ramification occurs only at the indicated conductor primes, with that exception.

We now prove cyclotomic Artin reciprocity over the rational numbers, using arithmetic Frobenius automorphisms. If $\ell \nmid n$ and $\mathfrak{L}$ lies over $\ell$, then Frobenius sends the reduction of ${\zeta }_n$ to its $\ell$th power. Distinct nth roots of unity remain distinct modulo $\mathfrak{L}$, since $X^n-1$ has separable reduction. Hence the Artin symbol is exactlyExtend multiplicatively to the group $I(n)$ of rational fractional ideals prime to $n$. Such an ideal has a unique positive rational generator $q$ whose numerator and denominator are prime to $n$. The formula isEvery unit residue modulo $n$ has a positive integer representative prime to $n$, so this Artin map is surjective. Its kernel is precisely the subgroup $P_{n\infty }$ generated by positive $q\equiv 1(\mathrm{mod}n)$. ThereforeThis proves the Artin reciprocity law for this extension, including its principal-ray kernel. The infinite component of the modulus records positivity and therefore the possible action of complex conjugation.

A modulus of a number field is $\mathfrak{m}={\mathfrak{m}}_0{\mathfrak{m}}_{\infty }$, with ${\mathfrak{m}}_0$ a nonzero integral ideal and ${\mathfrak{m}}_{\infty }$ a subset of the real places. Let $I_{\mathfrak{m}}$ be the group of fractional ideals prime to ${\mathfrak{m}}_0$, and let $P_{\mathfrak{m}}$ consist of principal ideals $(a)$ with $a\equiv 1$ at each prime power in ${\mathfrak{m}}_0$ and $a>0$ at each selected real place. The ray class group isThe ray class field $K_{\mathfrak{m}}$ is the finite abelian extension for which the Artin map has kernel $P_{\mathfrak{m}}$ and induces an isomorphism ${\mathrm{Cl}}_{\mathfrak{m}}(K)\cong \mathrm{Gal}(K_{\mathfrak{m}}/K)$. Equivalently it is the maximal abelian extension whose conductor divides $\mathfrak{m}$.

To compute its order use the ray class exact sequence, including signatures:The middle map lifts a residue and sign tuple to an element of $K$ by weak approximation theorem, then takes its principal ideal as a ray class. Different lifts differ by a principal ray factor. Its kernel consists exactly of tuples represented by global units. Every ordinary ideal class has a representative prime to the finite modulus, again by weak approximation theorem; this gives surjectivity of the last map. If $U_{\mathfrak{m}}$ denotes the ray units, the kernel of the unit map is $U_{\mathfrak{m}}$. The Chinese remainder theorem and counting units in each prime-power quotient give the ray class number formula

First take $K=\mathbb{Q}(\sqrt{-3})$. Its ring of integers is $\mathbb{Z}[\omega ]$, where $\omega =(-1+\sqrt{-3})/2$ and ${\omega }^2+\omega +1=0$. The Minkowski bound for ideal classes is $(2/\pi )\sqrt{3}<2$, so every ideal class has a representative of norm one and $h_K=1$. Its units are the six roots of unity. Modulo seven, the polynomial $X^2+X+1$ has distinct roots $2,4$, and thereforeThe unit group of this quotient has order $36$. The unit $1+\omega$ maps to $(3,5)$, which has order six; consequently unit reduction is injective and $U_{(7)}=\{1\}$. The ray class number formula gives $36/6=6$.

We also identify the quotient explicitly. Conjugation interchanges the two factors, so the residue norm is $(x,y)\mapsto xy$. Its kernel has order six and equals the image of the six units, since $3\cdot 5=1(\mathrm{mod}7)$. HenceThe field $M=K({\zeta }_7)=\mathbb{Q}({\zeta }_{21})$ has degree six over $K$, since the degrees over $\mathbb{Q}$ are twelve and two. It is unramified outside seven by the cyclotomic computation and the base-change argument above. At primes over seven, $K$ has completion ${\mathbb{Q}}_7$, and $M/K$ is the tame degree-six cyclotomic extension; its conductor exponent is one. Its relative arithmetic Artin symbol at an ideal $\mathfrak{a}$ prime to seven acts byIf $\mathfrak{a}=(a)$ with $a\equiv 1(\mathrm{mod}7{\mathcal{O}}_K)$, then $N_{K/\mathbb{Q}}(a)\equiv 1(\mathrm{mod}7)$. Thus the Artin map kills the principal ray subgroup. It factors through the order-six ray class group; the residue norm quotient computed above shows that it is onto the degree-six Galois group, so the fields coincide. This proves the explicit ray class field of Q of square root minus three modulo seven:

Now take $K=\mathbb{Q}(\sqrt{-6})$, and write $s=\sqrt{-6}$. Its ring of integers is $\mathbb{Z}[s]$, its discriminant is $-24$, and its units are $\pm 1$. The quotient by $(s)$ iswhose two units are both images of $\pm 1$. The ray class exact sequence therefore gives ${\mathrm{Cl}}_{(s)}(K)\cong \mathrm{Cl}(K)$.

For its class number, the Minkowski bound for ideal classes is $(2/\pi )\sqrt{24}<4$, so each ideal class contains an integral ideal of norm at most three. The primes of norms two and three areThese are the only nontrivial possibilities within the bound, and their ideal classes coincide and have order at most two. The ideal ${\mathfrak{p}}_2$ is not principal: its being principal would give an integer solution to $x^2+6y^2=2$, which is impossible. Thus $h_K=2$, the ideal class group is $C_2$, and the ray field is the Hilbert class field.

We verify its actual identity rather than stopping at the degree. PutThe two quadratic fields $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-3})$ are distinct, so $[H:\mathbb{Q}]=4$ and $[H:K]=2$. The integral elements $1,\sqrt{2},b,\sqrt{2}b$ are a rational basis. Their trace matrix is the Kronecker product of the two quadratic trace matrices, with discriminants $8$ and $-3$. Its determinant is therefore $8^2(-3{)}^2=576$. The discriminant-index formula for an integral lattice shows that the field discriminant divides $576$. On the other hand the relative discriminant ideal formula givesThe norm on the right is a positive integer, so equality is forced. Thus the relative discriminant ideal is the unit ideal, and $H/K$ is unramified at every finite prime. There are no real places of $K$. The extension is quadratic and therefore Abelian; its degree equals $h_K$, identifying it with the Hilbert class field of Q of square root minus six. Hence the ray class field of Q of square root minus six modulo square root minus six is

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.

Use arithmetic Frobenius automorphisms: residues are raised to the size of the base residue field. Let $S$ contain the finite primes ramified in the finite abelian extension $L/K$. The Artin map is the group homomorphismwhere $I_K^S$ is the group of fractional ideals supported outside $S$. The Artin symbol ${\mathrm{Frob}}_{\mathfrak{p}}$ is independent of the prime above $\mathfrak{p}$, since the extension is abelian. Commutativity makes the displayed extension from prime ideals well defined. Defining this map does not yet assert the Artin reciprocity law, which describes its congruence group kernel.

For an intermediate field $K\subseteq E\subseteq L$, restriction givesTo prove it, start with an unramified prime $\mathfrak{p}$ and a prime of $L$ above it. Its Frobenius automorphism acts on the residue of every integral element of $E$ by raising it to $N\mathfrak{p}$. Its restriction is therefore the Frobenius automorphism for $E/K$ at the restricted prime. The identity for all allowed fractional ideals follows by the group homomorphism property. On Galois groups, restriction is the quotient by $\mathrm{Gal}(L/E)$, so passing to a subextension quotients the Artin map accordingly.

Now let $E/K$ be any finite extension and put $M=LE$. This is the translated extension: $M/E$ is abelian, and restriction identifies $\mathrm{Gal}(M/E)$ with $\mathrm{Gal}(L/(L\cap E))$. A prime $\mathfrak{q}$ of $E$ above a prime $\mathfrak{p}\notin S$ is unramified in $M/E$, because forming a compositum with $E$ preserves an unramified extension. Put $r=f(\mathfrak{q}/\mathfrak{p})$. The residue-field sizes satisfy $N\mathfrak{q}=(N\mathfrak{p}{)}^r$. Consequently the restriction to $L$ of Frobenius at $\mathfrak{q}$ acts on its integral residues by raising them to $(N\mathfrak{p}{)}^r$, givingBut the ideal norm is $N_{E/K}\mathfrak{q}={\mathfrak{p}}^r$. Multiplying this prime-by-prime identity provesThis is restriction and norm compatibility of the Artin map. The fractional ideal $\mathfrak{a}$ must be supported over primes unramified in $L/K$. No assumption that $E/K$ itself is unramified or Galois is needed. With geometric rather than arithmetic Frobenius, both maps are inverted and the compatibility identities remain unchanged.

Let $L/K$ be a finite Galois extension of number fields, with Galois group $G$, and let $\mathfrak{p}$ be a nonzero prime ideal of ${\mathcal{O}}_K$. Write its prime ideal factorization asThe Galois group acts transitively on the primes above $\mathfrak{p}$. One proof uses the Chinese remainder theorem. If $\mathfrak{Q}$ were outside the orbit of $\mathfrak{P}$, choose an integral element $a$ which is zero modulo $\mathfrak{Q}$ and one modulo every conjugate of $\mathfrak{P}$. Its field norm ${\prod }_{\sigma \in G}\sigma (a)$ belongs to $\mathfrak{Q}\cap {\mathcal{O}}_K=\mathfrak{p}$, hence to $\mathfrak{P}$. But each factor is one modulo $\mathfrak{P}$, a contradiction. Transitivity also makes all $e_j$ equal to a common ramification index $e$ and all $f_j$ equal to a common residue degree $f$.

Fix $\mathfrak{P}$ above $\mathfrak{p}$. Its decomposition group is the stabilizerThe orbit-stabilizer theorem gives $|D_{\mathfrak{P}}|=|G|/g$. Elements of $D_{\mathfrak{P}}$ act on the residue field $\kappa (\mathfrak{P})$ over $\kappa (\mathfrak{p})$; the group kernel is the inertia groupThus inertia measures the automorphisms invisible after reduction, whereas decomposition measures all automorphisms preserving the chosen prime.

The completion $L_{\mathfrak{P}}/K_{\mathfrak{p}}$ is a Galois extension with group $D_{\mathfrak{P}}$: preservation of $\mathfrak{P}$ lets an automorphism extend continuously to the completion, and the local embeddings of this normal extension arise in this way. The degree of a field extension formula is $[L_{\mathfrak{P}}:K_{\mathfrak{p}}]=ef$. Reduction gives the exact sequenceFor surjectivity, lift a primitive residue-field generator by the Hensel lemma to obtain the maximal unramified subextension of a local field extension of $L_{\mathfrak{P}}/K_{\mathfrak{p}}$. Its automorphisms are the residue-field automorphisms. They extend to $L_{\mathfrak{P}}$ because the latter is normal. Since a finite residue field extension has a cyclic Galois group of order $f$, the exact sequence yieldsThe fixed field of local inertia is the maximal unramified subextension of a local field extension, and the extension above it is totally ramified. Primes in the same global orbit have conjugate decomposition groups and inertia groups, so the orders do not depend on the chosen prime.

If $e=1$, inertia is trivial and $D_{\mathfrak{P}}$ is generated by the unique Frobenius automorphism acting on residues by $z\mapsto z^{N\mathfrak{p}}$. Its order is $f$. In a ramified extension this residue action determines only a coset modulo inertia. A prime splits completely exactly when $e=f=1$, equivalently its decomposition group is trivial. A totally ramified prime has $g=f=1$ and $I_{\mathfrak{P}}=D_{\mathfrak{P}}=G$; an unramified prime has $I_{\mathfrak{P}}=1$. In an abelian extension, conjugation does nothing, so the associated groups and unramified Artin symbol are independent of the prime above $\mathfrak{p}$.

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 $g$.

For an unramified prime $\mathfrak{p}$ of $K$ and a prime $\mathfrak{P}$ of $L$ above it, the Frobenius automorphism is characterized byIn an abelian extension it is independent of $\mathfrak{P}$. This element is the Artin symbol

The ring of integers is $\mathbb{Z}[\omega ]$, where ${\omega }^2+\omega +1=0$, and its class number is one by the Minkowski bound for ideal classes. Modulo seven it is ${\mathbb{F}}_7\times {\mathbb{F}}_7$. The six roots of unity reduce to the kernel of the product map on the two residue unit groups, so the ray class group is ${\mathbb{F}}_7^{\times }$, of order six. The relative Artin symbol on $K({\zeta }_7)$ is exponentiation by the norm of a fractional ideal modulo seven. It kills the principal ray subgroup, and $[K({\zeta }_7):K]=6$, proving that this is the ray class field.

For an abelian extension $L/K$ of number fields, restriction to an intermediate field takes each Artin symbol to that of the smaller extension. For a finite base extension $E/K$, restriction identifies $\mathrm{Gal}(LE/E)$ with $\mathrm{Gal}(L/(L\cap E))$. At an unramified prime $\mathfrak{q}$ of $E$ above $\mathfrak{p}$, its residue-field Frobenius restricts to the $f(\mathfrak{q}/\mathfrak{p})$th power of the Frobenius automorphism at $\mathfrak{p}$. Since $N_{E/K}\mathfrak{q}={\mathfrak{p}}^{f(\mathfrak{q}/\mathfrak{p})}$, the group homomorphism property proves the displayed identity for fractional ideals supported over unramified primes.