Prime Ideal Decomposition Ramification and the Different

Choose $0\mathrm{\ne }a\in \mathfrak{a}$. Multiplication by $a$ is an injective endomorphism of the free rank-$n$ lattice ${\mathcal{O}}_K$, with nonzero integral determinant.

Its image $(a)$ has finite index, and $(a)\subseteq \mathfrak{a}\subseteq {\mathcal{O}}_K$; thus ${\mathcal{O}}_K\mathrm{/}\mathfrak{a}$ is a quotient of the finite group ${\mathcal{O}}_K\mathrm{/}(a)$.

The quotient $R={\mathcal{O}}_K\mathrm{/}\mathfrak{a}$ is finite, so its finite ideal lattice supplies the finite list of maximal ideals; their inverse images are exactly the primes $\mathfrak{p}$ containing $\mathfrak{a}$. The finite free integral lattice makes ${\mathcal{O}}_K$ noetherian, its definition as an integral closure makes it integrally closed, and every nonzero prime is maximal because its quotient is a finite domain. Thus $({\mathcal{O}}_K{)}_{\mathfrak{p}}$ is a DVR. Consequently $\mathfrak{a}({\mathcal{O}}_K{)}_{\mathfrak{p}}={\mathfrak{p}}^{e_{\mathfrak{p}}}({\mathcal{O}}_K{)}_{\mathfrak{p}}$ for a unique least $e_{\mathfrak{p}}>0$.

Put $\mathfrak{b}=\prod {\mathfrak{p}}^{e_{\mathfrak{p}}}$. At every maximal ideal in the finite list, step 1.1 gives ${\mathfrak{a}}_{\mathfrak{p}}={\mathfrak{b}}_{\mathfrak{p}}$; at any other maximal ideal both localisations are the unit ideal. Hence $\mathfrak{a}=\mathfrak{b}$ (otherwise a maximal ideal containing the appropriate colon ideal gives a contradictory localisation). Distinct prime powers are comaximal, so the Chinese remainder theorem reassembles this finite product. Localising a second factorisation at the same finite primes forces the same exponents. The list is finite and each exponent is the least natural number with its property, so no Choice is used.

Multiplication by $\alpha$ is an injective integer matrix on the integral lattice, and its image is $(\alpha )$.

The index of the image of an injective integer matrix is the absolute determinant; that determinant is the field norm. Hence the quotient cardinality has the asserted value.

Factor both ideals over their finite union of prime supports. The DVR calculation in the factorisation proof makes ${\mathfrak{p}}^r\mathrm{/}{\mathfrak{p}}^{r+1}$ one-dimensional over ${\mathcal{O}}_K\mathrm{/}\mathfrak{p}$, so every successive quotient has cardinality $N\mathfrak{p}$.

Hence $N({\mathfrak{p}}^{r+s})=N({\mathfrak{p}}^r)N({\mathfrak{p}}^s)$ at every prime; multiply over the finite support to obtain the claim.

The finite domain ${\mathcal{O}}_K\mathrm{/}\mathfrak{P}$ is a finite field; its characteristic is a rational prime $p$ and its kernel on $\mathbb{Z}$ is $(p)$.

As a finite-dimensional vector space over ${\mathbb{F}}_p$, that field has $p^f$ elements for some $f\ge 1$, which is exactly $N\mathfrak{P}$.

Localise ${\mathcal{O}}_L$ at $\mathfrak{p}$. It is a finite torsion-free module over the DVR $({\mathcal{O}}_K{)}_{\mathfrak{p}}$, hence free of rank $[L:K]$. Filtering its reduction modulo $\mathfrak{p}$ by the finite prime-power factors, the $e(\mathfrak{P}\mathrm{/}\mathfrak{p})$ layers at $\mathfrak{P}$ are copies of its residue field.

Taking dimensions over ${\mathcal{O}}_K\mathrm{/}\mathfrak{p}$ gives the left side, while the free rank in step 1.1 gives $[L:K]$ for the same quotient.

Substitute the prime factorisation of $\mathfrak{p}{\mathcal{O}}_L$ into its extension to ${\mathcal{O}}_M$ and compare the exponent of $\mathfrak{Q}$.

The three residue fields form a finite tower, so the dimensions multiply. These are exactly the two asserted equalities.

Put $A={\mathcal{O}}_K$, $B={\mathcal{O}}_L$ and $k=A\mathrm{/}\mathfrak{p}$. Monic division by $F$ shows that the kernel of $A[X]\to B$, $X\mapsto \alpha$, is $(F)$: a remainder of degree less than $\mathrm{deg}F$ vanishing at $\alpha$ is zero by minimality over $K$. Monogeneity gives surjectivity, hence $B\mathrm{/}\mathfrak{p}B\simeq k[X]\mathrm{/}(\bar{F})$.

The maximal ideals of this quotient are exactly those generated by the ${\bar{g}}_i$; their inverse images in $B$ are ${\mathfrak{P}}_i=(\mathfrak{p},g_i(\alpha ))$. They are independent of the chosen lifts. Their residue fields are $k[X]\mathrm{/}({\bar{g}}_i)$, of degree $\mathrm{deg}{\bar{g}}_i$, and their contractions are $\mathfrak{p}$. Thus they are exactly the primes above $\mathfrak{p}$.

The local factorisation theorem Integral ideal factorisation in a number field, in ZF writes $\mathfrak{p}B={\prod }_i{\mathfrak{P}}_i^{a_i}$. In the DVR $B_{{\mathfrak{P}}_i}$ used in that theorem's proof, the maximal ideal of $B_{{\mathfrak{P}}_i}\mathrm{/}\mathfrak{p}{B}_{{\mathfrak{P}}_i}$ has nilpotency index exactly $a_i$. On the polynomial side of step 1.1, localisation at $({\bar{g}}_i)$ gives $k[X{]}_{({\bar{g}}_i)}\mathrm{/}({\bar{g}}_i^{e_i})$, whose maximal ideal has nilpotency index exactly $e_i$: its $e_i$-th power vanishes and its $(e_i-1)$-st power does not. Thus $a_i=e_i$, proving the ideal factorisation and the stated ramification and residue degrees.

Let $\mathfrak{P}\mid \mathfrak{p}$ have ramification index $e$. Comparing the valuations of the terms in the Eisenstein equation forces $n{v}_{\mathfrak{P}}(\alpha )=e$; in particular $e\le [L:K]\le n$.

The inequalities in step 1.1 are therefore equalities, so $v_{\mathfrak{P}}(\alpha )=1$ and $e=n$. The resulting factorisation is $\mathfrak{p}{\mathcal{O}}_L=(\mathfrak{p},\alpha {)}^n$; hence this is the unique prime factor and the prime is totally ramified.

Reducing its trace-pairing matrix modulo $p$, a nontrivial radical is equivalent to failure of the residue algebra to be a product of separable fields, hence to some ramification index exceeding $1$.

The determinant of that matrix is $d_K$, so the radical is nontrivial exactly when $p\mid d_K$. This proves both directions.

The field discriminant is a fixed nonzero integer.

It has only finitely many prime divisors, and the criterion identifies these exactly with the ramified rational primes.

In a $\mathbb{Z}$-basis of the full lattice $A$, nondegeneracy of trace gives a dual basis, so $A^{\vee }$ is again a full lattice and is stable under ${\mathcal{O}}_K$.

The condition $\mathrm{Tr}(xcA)\subseteq \mathbb{Z}$ is equivalent to $cx\in A^{\vee }$, proving $(cA{)}^{\vee }=c^{-1}{A}^{\vee }$ and the fractional-ideal claim.

Lagrange interpolation in the conjugates shows that the trace-dual lattice has $\mathbb{Z}$-basis $\frac{1}{f^{\mathrm{\prime }}(\alpha )},\frac{\alpha }{f^{\mathrm{\prime }}(\alpha )},\dots ,\frac{{\alpha }^{n-1}}{f^{\mathrm{\prime }}(\alpha )}\mathrm{.}$

Thus ${\mathcal{O}}_K^{\vee }=f^{\mathrm{\prime }}(\alpha {)}^{-1}{\mathcal{O}}_K$; taking its fractional-ideal inverse yields $(f^{\mathrm{\prime }}(\alpha ))$.

The dual lattice is obtained from the original basis by the inverse matrix $G^{-1}$, hence its index relative to ${\mathcal{O}}_K$ has absolute determinant $\mathrm{\mid }\det G\mathrm{\mid }$.

Since $\det G=d_K$ and the different is the inverse codifferent, its norm is that same positive index.

Put $d=v_{\mathfrak{P}}({\mathfrak{D}}_K)$. The prime $\mathfrak{P}$ divides the different exactly when $d>0$.

If $e=1$, then $p\nmid e$ and [L1] gives $d=e-1=0$. If $e>1$, then [L1] gives either $d=e-1>0$ or $d\ge e>0$, according as $p\nmid e$ or $p\mid e$. This proves the stated local equivalence.

That trace criterion first gives $v_{\mathfrak{P}}({\mathfrak{D}}_K)\ge e-1$. For the next power, the successive quotients ${\mathfrak{P}}^i\mathrm{/}{\mathfrak{P}}^{i+1}$ are all isomorphic to the residue field as modules, so the relevant trace is $e{\mathrm{Tr}}_{{\mathcal{O}}_K\mathrm{/}\mathfrak{P}\mathrm{/}{\mathbb{F}}_p}\mathrm{.}$

The trace of a finite separable field extension is not the zero map. Hence the display in step 1.1 vanishes identically exactly when $p\mid e$. Thus $p\nmid e$ gives exact exponent $e-1$, while $p\mid e$ gives exponent at least $e$, which are precisely the tame and wild cases.

Factor ${\mathfrak{D}}_K$ into its finite prime powers and apply norm multiplicativity, using $N\mathfrak{P}=p^{f(\mathfrak{P}\mathrm{/}p)}$ over $p$.

The norm-of-the-different identity makes the $p$-adic valuation of that product $v_p(d_K)$, which is the displayed sum.