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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:

Put $\alpha =44^{1/11}$ and $L=\mathbb{Q}(\alpha )$. The Eisenstein criterion at $11$ shows that $X^{11}-44$ is an irreducible polynomial, so $[L:\mathbb{Q}]=11$. Its derivative is $11X^{10}$, and the polynomial discriminant isThe discriminant-index formula for an integral lattice implies that the field discriminant has no prime divisor outside $\{2,11\}$. By the discriminant obstruction to ramification, every other prime is unramified.

To prove that neither candidate disappears through an index correction, let $\mathfrak{p}$ lie above $p\in \{2,11\}$ and normalize its valuation to have value group $\mathbb{Z}$. If $e=e_{\mathfrak{p}/p}$, the relation ${\alpha }^{11}=44$ givesIn both cases $11\mid e$. The fundamental identity for prime decomposition, ${\sum }_{\mathfrak{p}\mid p}{e}_{\mathfrak{p}/p}{f}_{\mathfrak{p}/p}=11$, now forces exactly one prime ideal above each candidate, with $e=11$ and $f=1$. This is ramification forced by a coprime root valuation; at $2$ it avoids an invalid direct application of the Eisenstein criterion, since $4\mid 44$. Hence

Write $e=e(\mathfrak{q}/\mathfrak{p})$, $f=f(\mathfrak{q}/\mathfrak{p})$ and $g$ for the number of primes above $\mathfrak{p}$. The Galois group acts transitively on those primes, with stabilizer $D_{\mathfrak{q}}$. The fundamental identity for prime decomposition and the orbit-stabilizer theorem therefore giveIt remains to establish that the residue action has image of order $f$, rather than assume this in the desired equality.

Use the decomposition group of a prime and local Galois group identification $D_{\mathfrak{q}}=\mathrm{Gal}(F_{\mathfrak{q}}/K_{\mathfrak{p}})$. A generator of the finite field extension $k_{\mathfrak{q}}/k_{\mathfrak{p}}$ has a separable monic minimal polynomial of an algebraic element. Lift its coefficients to ${\mathcal{O}}_{K_{\mathfrak{p}}}$. The Hensel lemma lifts each of its distinct residue roots to $F_{\mathfrak{q}}$. One such lift has field degree at most $f$, while its residue generates the degree-$f$ finite field extension, so its field degree is exactly $f$ and its ramification index of a prime ideal is one. It therefore generates an unramified extension of degree $f$, and sending it to each of the other lifts gives its $f$ embeddings. Because $F_{\mathfrak{q}}/K_{\mathfrak{p}}$ is a Galois extension, these embeddings extend to its automorphisms. Their reductions realize every automorphism of $k_{\mathfrak{q}}/k_{\mathfrak{p}}$. Thus the residue action is surjective, and

Now let $G$ be cyclic of order ${\ell }^2$, and let $H=\mathrm{Gal}(F/L)$ be its unique subgroup of order $\ell$. In a tower of Galois extensions, inertia maps onto inertia in the intermediate extension. This also follows from the just-proved order formula: its image has order $e(F/K)/e(F/L)=e(L/K)$, using multiplicativity of the ramification index of a prime ideal in a tower. If $\mathfrak{p}$ ramifies in $L/K$, this image in $G/H$ is nontrivial. Every proper subgroup of $G$ is contained in $H$, so inertia upstairs must be $G$. HenceThe fundamental identity for prime decomposition then also forces a single prime above $\mathfrak{p}$ and residue-field degree one.

Use the Hilbert class field theorem stated in part (a), and let $H_L$ be the Hilbert class field of $L$. Put $F=M\cap H_L$ inside a common algebraic closure.

Any intermediate extension of $M/L$ is totally ramified at the given prime. To verify this even when $M/L$ is not Galois, there is a unique prime of $F$ above $\mathfrak{p}$, its residue field embeds into the residue field at the unique prime of $M$ and therefore equals $k_{\mathfrak{p}}$, and the fundamental identity for prime decomposition gives ramification index $[F:L]$. On the other hand $F/L$ is a subextension of $H_L/L$, so it is unramified there. Its ramification index is consequently both $[F:L]$ and $1$, forcingBecause $H_L/L$ is Galois, this intersection identity givesThus $M{H}_L/M$ is abelian. We use the standard local fact that unramified extensions remain unramified under finite base extension: after base change each factor is obtained by lifting an extension of the residue field. Since $H_L/L$ is unramified at every finite prime, $M{H}_L/M$ is also unramified at every finite prime. There are no real places of $M$, since every embedding of $M$ restricts to an embedding of the totally imaginary field $L$. Hence there is no infinite-place obstruction.

The maximality of $H_M$ now implies $M{H}_L\subseteq H_M$. Taking degrees over $M$ proves the class number divisibility under total ramification:The argument uses the degree of an unramified class-field extension, and does not assume that extension of ideals induces an injection of ideal class groups.

A prime is completely split in an extension of degree $n$ exactly when there are $n$ primes above it. Indeed, the fundamental identity for prime decomposition is ${\sum }_{\mathfrak{q}\mid \mathfrak{p}}{e}_{\mathfrak{q}}{f}_{\mathfrak{q}}=n$, and every summand is a positive integer. Thus $n$ primes means that every ramification index and residue degree equals one, and conversely.

By part (a), the number of primes of $L$ above $\mathfrak{p}$ is the number of right $D$-orbits on $H\backslash G$. This set has $[G:H]=[L:K]$ elements. It has that many orbits exactly when every orbit is a singleton, equivalently whenThis condition says $gD{g}^{-1}\subseteq H$ for every $g$, orthe subgroup core.

Because $M$ is the Galois closure of $L/K$, it is generated by all the fields $g(L)$. The fixed field of $gH{g}^{-1}$ is $g(L)$, so the fixed field of the core is their compositum, namely $M$. The Galois correspondence therefore gives ${\mathrm{Core}}_G(H)=1$. HenceIn the Galois extension $M/K$, the number of primes above $\mathfrak{p}$ is $[G:D]$, so $D=1$ is equivalent to complete splitting in $M$. This proves that complete splitting is preserved by Galois closure:No prior unramifiedness assumption is needed: complete splitting already forces all ramification indices to be one, and the decomposition-group argument includes the ramified case.

The prime ideal factorization of $pB$ iswhich defines the ramification index of a prime ideal $e_{q/p}=v_q(pB)$. The residue-field degree isLocalizing at $p$ makes $B_p$ a free $R_p$-module of rank $[L:K]$, so $B/pB$ has dimension $[L:K]$ over the residue field $R/p$. The Chinese remainder theorem and the filtration by powers of each $q$ decompose this vector space into $e_{q/p}$ successive quotients isomorphic to $B/q$, each of dimension $f_{q/p}$. Therefore the fundamental identity for prime decomposition gives

Suppose ${\alpha }^m=a\in {\mathbb{Q}}^{\times }$ in a number field and $\mathfrak{p}$ lies above $p$. Its normalized integral valuation satisfies $m{v}_{\mathfrak{p}}(\alpha )=e_{\mathfrak{p}/p}{v}_p(a)$. If $m$ and $v_p(a)$ are coprime, then $m\mid e_{\mathfrak{p}/p}$. In a degree-$m$ field the fundamental identity for prime decomposition forces a unique prime, ramification index $m$, and residue degree one. This detects total ramification of a prime even when the displayed defining polynomial is not Eisenstein at $p$.

// Target: algebra.bigb