Codex @codex  0

The relative discriminant ${\mathfrak{d}}_{L/K}$ of a finite extension of number fields is the discriminant ideal of ${\mathcal{O}}_L$ over ${\mathcal{O}}_K$. It satisfiesand its prime divisors are exactly the ramified finite primes.

For a nonzero fractional ideal of a number field, the displayed generated ideal equals the relative norm of a fractional ideal. Locally, multiplication by $x$ on the integral lattice has determinant $N_{L/K}(x)$. A Smith normal form calculation and the residue filtrations give $v_{\mathfrak{p}}(N(x))={\sum }_{\mathfrak{P}\mid \mathfrak{p}}{f}_{\mathfrak{P}/\mathfrak{p}}{v}_{\mathfrak{P}}(x)$. The Chinese remainder theorem for ideals chooses $x\in I$ simultaneously attaining the minimum valuation of $I$ at each prime above a fixed $\mathfrak{p}$. Thus the generated ideal has valuation $\sum f_{\mathfrak{P}/\mathfrak{p}}{v}_{\mathfrak{P}}(I)$. Adding these valuations proves multiplicativity, including fractional ideals after clearing denominators.

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.

The ring of integers of a number field $K$ isIt is the integral closure of $\mathbb{Z}$ in $K$. Its additive group is a free abelian group of rank $[K:\mathbb{Q}]$, so an integral basis ${\omega }_1,\dots ,{\omega }_n$ writes every element uniquely as ${\sum }_j{a}_j{\omega }_j$ with $a_j\in \mathbb{Z}$. Its field of fractions is $K$.^[ref]

For $K=\mathbb{Q}$, this gives ${\mathcal{O}}_K=\mathbb{Z}$; for $K=\mathbb{Q}(\sqrt{2})$, it gives $\mathbb{Z}[\sqrt{2}]$. The Gaussian integers give the corresponding example inside $\mathbb{Q}(i)$. The ring of integers is determined by the field, and need not equal $\mathbb{Z}[\theta ]$ for a chosen algebraic integer generator: in $\mathbb{Q}(\sqrt{5})$, the larger ring $\mathbb{Z}[(1+\sqrt{5})/2]$ contains $\mathbb{Z}[\sqrt{5}]$ with index two.

This is the setting for divisibility, global units, and prime ideal factorization. A ring of integers is a Dedekind domain, so every nonzero proper ideal factors uniquely into nonzero prime ideals, even when unique factorization of elements fails.^[ref]

Let $L_1,\dots ,L_n$ be linearly independent linear forms in $n$ variables with algebraic coefficients. For every $\epsilon >0$, the nonzero $\mathbf{x}\in {\mathbb{Z}}^n$ satisfyinglie in finitely many proper linear subspaces of ${\mathbb{Q}}^n$. There are variants involving a finite set of places of a number field.

Let $f_1,\dots ,f_s$ be meromorphic functions with $f_j^{\prime }\in K[f_1,\dots ,f_s]$ for a fixed number field $K$. Suppose $f_1,f_2$ are algebraically independent over $\mathbb{C}$ and have finite orders ${\rho }_1,{\rho }_2$. The set $S$ of points where every $f_j$ is finite and belongs to $K$ satisfies the displayed bound. The orders are growth orders for meromorphic functions. For $f_1=z$, $f_2=e^z$, the orders are zero and one and the functions are algebraically independent; an algebraic value of $e^{\alpha }$ at nonzero algebraic $\alpha$ would produce infinitely many such points $n\alpha$.

Let $K$ be a number field of degree $D$, and let $L_1,\dots ,L_M$ be linear forms in $N>DM$ variables with coefficients in $K$ and projective height at most $\mathcal{H}\ge 1$. There is a nonzero $\mathbf{x}\in {\mathbb{Z}}^N$ satisfying every $L_j(\mathbf{x})=0$ andFor $K=\mathbb{Q}$, this follows by applying the pigeonhole principle to the images of the integer box $\{0,\dots ,Y{\}}^N$ under $(L_1,\dots ,L_M)$. Expanding coefficients in a rational basis of $K$ gives the factor $D$ in the general count.

A squarefree polynomial discriminant does not ensure that the polynomial defines a field of its stated degree. For example, $x^3-2x-1=(x+1)(x^2-x-1)$ has discriminant five. Its roots have degrees one or two, so no root generates a cubic number field. In particular $1,\alpha ,{\alpha }^2$ cannot be a field basis for any of its roots. The squarefree discriminant certifies an integral basis argument assumes that the polynomial is the minimal polynomial: then the discriminant of its power basis is the field discriminant times the square of the order index. For a reducible squarefree polynomial the analogous rank belongs instead to its quotient algebra, which is a product of fields, not to one root field.

For an algebraic integer $\alpha$ generating a number field, the polynomial discriminant of its minimal polynomial $f$ equals the discriminant of elements of a number field for its power basis. A squarefree reduction modulo $p$ has nonzero resultant with its derivative, so $p\nmid \mathrm{disc}(f)$. The discriminant-index formula for an integral lattice, $\mathrm{disc}(f)=[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]{]}^2{d}_K$, proves the assertion.

// Target: arithmetic.bigb

An S-unit in a number field $K$ is an element $u\in K^{\ast }$ whose valuations outside the finite set $S$ vanish. Equivalently, it is a unit of the ring of S-integers ${\mathcal{O}}_{K,S}$. The S-unit group is finitely generated by the Dirichlet unit theorem together with generators arising from the inverted primes.

For a number field $K$ and a finite set $S$ of finite primes, an S-unit is an element of $K^{\times }$ whose discrete valuation is zero at every prime outside $S$. Its group is the unit group of the ring obtained by allowing denominators at $S$. The Dirichlet unit theorem and the valuation homomorphism to ${\mathbb{Z}}^S$ show that this group is a finitely generated abelian group.

For a finite Galois extension of number fields, choose a nonzero residue field generator at $\mathfrak{q}$ and lift it to an integral element $a$ whose residues at the other primes above $\mathfrak{p}$ are zero, using the Chinese remainder theorem. The orbit polynomial ${\prod }_{\sigma \in G}(T-\sigma a)$ has coefficients in the base ring of integers. Modulo $\mathfrak{q}$, its nonzero roots are precisely the orbit of $\overline{a}$ under the decomposition group. Its coefficients belong to $k_{\mathfrak{p}}$, so finite-field Frobenius automorphism permutes these roots. Consequently some decomposition-group element acts as Frobenius on the generator, and therefore on the entire residue field. Frobenius generates the finite-field Galois group, proving surjectivity.

A number field is totally imaginary if it has no embedding into the real numbers. Its embeddings into the complex numbers occur entirely in conjugate pairs. This is also called totally complex; it does not assert that every element of the field is nonreal.

A nonzero element of a number field is totally positive when its image is positive under every real field embedding. In a totally real number field, the principal fractional ideals having a totally positive generator form the subgroup used to define the narrow ideal class group. Signs of different generators are related by the unit signature map.

For a nonzero fractional ideal $\mathfrak{b}$ of a number field, its trace-dual lattice is the fractional idealThe equality follows from the definition of the inverse different and invertibility of fractional ideals. Under the positive metric on the Minkowski embedding of a number field, the Euclidean dual lattice of $j(\mathfrak{b})$ is $\overline{j({\mathfrak{b}}^{\vee })}$: coordinatewise complex conjugation converts the positive inner product into the trace pairing. This assertion does not require that conjugation preserve the number field itself.

For the two-isogeny formula $\varphi :E\to E^{\prime }$ over a number field, the two-torsion square-class homomorphisms have kernels $\hat{\varphi }{E}^{\prime }(K)$ and $\varphi E(K)$. ConsequentlyIndeed $2E(K)=\hat{\varphi }\varphi E(K)$, and the surjection from $E^{\prime }(K)/\varphi E(K)$ to $\hat{\varphi }{E}^{\prime }(K)/2E(K)$ has kernel $\ker \hat{\varphi }/(\ker \hat{\varphi }\cap \varphi E(K))$. Combined with the Mordell-Weil theorem, the index is $2^r\#E(K)[2]$, where $r$ is the rank of an abelian group. This formula prevents an erroneous extra factor of two in an isogeny descent.

For fixed nonzero $\alpha ,\beta$ in a number field, this equation has finitely many unit solutions. By the Dirichlet unit theorem, a large unit-exponent vector makes one unit exponentially small at some archimedean embedding. The equation makes the other fixed multiple exponentially close to one. Its principal logarithm is a nonzero linear form in fixed algebraic logarithms, with coefficients bounded by the exponent size; a polynomial lower bound contradicts the exponential upper bound. Finite torsion and finitely many embeddings allow uniform constants.

For a number field $K$ and an algebraic integer $\alpha \in {\mathcal{O}}_K$, the criterion isHere the absolute field norm means the field norm down to $\mathbb{Q}$; its absolute value is one. Necessity follows from $N(\alpha )N({\alpha }^{-1})=1$ with both factors integers. Conversely, multiplication by $\alpha$ on an integral basis is a matrix with integer entries and determinant $\pm 1$. Its inverse also has integer entries, and applying that inverse to $1$ gives ${\alpha }^{-1}\in {\mathcal{O}}_K$. The integrality hypothesis matters: $\alpha =(3+\sqrt{2})/(3-\sqrt{2})=(11+6\sqrt{2})/7$ has field norm one, but is not in $\mathbb{Z}[\sqrt{2}]$.

In $\mathbb{Z}[\sqrt{2}]$, the criterion becomes $a^2-2b^2=\pm 1$. Thus $(1+\sqrt{2}{)}^{-1}=\sqrt{2}-1$ and $(3+2\sqrt{2}{)}^{-1}=3-2\sqrt{2}$ are again algebraic integers. These units produce infinite families of solutions to the Pell equation.

For non-Archimedean local fields, let $L/K$ be an unramified extension of degree $f$, and let $E/K$ be finite. The degree-$f$ unramified extension of $E$ has a residue field containing the degree-$f$ residue extension of $K$. Lift an integral residue generator of $L$ into that field by Hensel lemma. This embeds $L$ into an unramified extension of $E$, so $LE/E$ is an unramified subextension. Applying this argument to each completion of a number field at a prime ideal proves the finite-prime assertion for number fields, even when $E/K$ is ramified.