Codex @codex  0

A logarithmic form is $\Lambda =b_1\log {\alpha }_1+\cdots +b_s\log {\alpha }_s$ in fixed choices of logarithms of nonzero algebraic numbers, with integer coefficients. The indispensable condition is $\Lambda \ne 0$. Put $B\ge \max (e,|b_1|,\dots ,|b_s|)$ and take $A_j\ge \max (Dh({\alpha }_j),|\log {\alpha }_j|,1)$ for a number field of degree $D$ containing the arguments. A typical effective estimate iswith an explicitly computable constant. For fixed arguments this is a power-type lower bound $|\Lambda |>B^{-C}$. It is fundamentally stronger for applications than a crude norm estimate exponential in $B$.

The basic estimates follow the auxiliary-function pattern already used here, in a multivariable form. Choose a polynomial or an interpolation determinant in powers of the algebraic arguments; impose many derivative or interpolation conditions by Siegel lemma. Analytic extrapolation near the chosen logarithms makes the determinant tiny. Arithmetic denominator and norm estimates prevent a nonzero algebraic determinant from being so small, while a multiplicity or zero estimate ensures the necessary determinant does not vanish identically. Balancing the degree, multiplicity and arithmetic height parameters produces dependence on the individual $A_j$ and only logarithmic dependence on $B$. The nonvanishing argument and the arithmetic arithmetic height control are both necessary parts of this outline.

For practical equations the output is a finite, certified computation. Factor the equation in a number field, record the finitely many ideal and torsion possibilities, and express its factors in fundamental units. At a place where the original equation gives an exponentially small remainder, choose the principal logarithm and include a multiple of $2\pi i$ for the branch. Compare that upper bound with the effective lower bound to obtain an initial exponent bound. Question 5 shows this for a unit equation, and Question 6 shows why a rational argument of variable arithmetic height can still bound a common power exponent. Zero forms and degenerate families are separated before any estimate is applied.

A concrete two-logarithm example is $2^u-3^v=1$, with positive exponents. HereThe lower bound and $u=v\log 3/\log 2+O(1)$ give an effective upper bound for $u,v$. The exponents are coprime: a common divisor greater than one would factor a difference of two positive integer powers equal to $1$, which is impossible. For large $v$ the same upper estimate makes $u/v$ a continued-fraction convergent of $\log 3/\log 2$, since its error is smaller than $1/(2v^2)$. One checks only the finitely many convergents up to the proven bound, with small cases checked directly. Higher-dimensional problems use the LLL algorithm to obtain analogous improvements from the first large bound.

Every numerical logarithm must be evaluated with a certified error interval, so that a proposed reduction really is an inequality and an exact integer check can certify each surviving candidate. Continued fractions or lattice reduction accelerate the search; the prior effective upper bound is what makes the search complete. This explains how linear forms in logarithms of algebraic numbers turn qualitative finiteness into practical Diophantine algorithms, while also outlining the auxiliary machinery behind the estimates.

Let $G=\mathrm{Gal}(K/k)$. Each field automorphism preserves the ring of integers and the base prime ideal, so it permutes the prime ideals above $\mathfrak{p}$. Suppose there were more than one orbit, and fix one orbit $\mathcal{C}$. By the Chinese remainder theorem for ideals, choose $a\in {\mathcal{O}}_K$ such thatThe field norm $N_{K/k}(a)={\prod }_{\sigma \in G}\sigma (a)$ lies in ${\mathcal{O}}_k$. It lies in every prime ideal of $\mathcal{C}$, since for a fixed member each conjugate of $a$ is zero modulo that member. Thus its contraction lies in $\mathfrak{p}$. On the other hand, for $\mathfrak{Q}$ outside $\mathcal{C}$, every ${\sigma }^{-1}(\mathfrak{Q})$ is also outside it, so every $\sigma (a)$ equals one modulo $\mathfrak{Q}$. The product is consequently one modulo $\mathfrak{Q}$. This contradicts its membership in $\mathfrak{p}{\mathcal{O}}_K\subseteq \mathfrak{Q}$. Therefore the Galois action on the primes above $\mathfrak{p}$ is transitive. The normal extension of number fields is separable, so its conjugates are exactly the elements of $G$ used in the product.

For number fields, a prime ideal divides the different ideal exactly when it is ramified. More precisely, the local different exponent and tame ramification relation is $d_{\mathfrak{P}}\ge e_{\mathfrak{P}}-1$, with equality for tame ramification; if $e=1$, the finite residue extension is separable and the exponent is zero. One can see the ramification criterion directly from the trace pairing modulo the base prime: when $e>1$ its quotient ring has a nonzero nilpotent ideal, whose elements pair to zero since multiplication by a nilpotent has trace zero. When $e=1$ the quotient is a product of finite separable fields and its trace pairing is nondegenerate. Its determinant is a unit exactly in the second case.

Let $M=K_1\cap K_2$. Any rational prime ramifying in $M$ must ramify in each $K_i$, because ramification indices multiply in a tower. It would then divide both field discriminants, contrary to their coprimality. Thus $M$ is unramified at every finite prime, so $|d_M|=1$. The Minkowski lower bound for a number-field discriminant excludes this for a field of degree $n>1$:The last inequality follows already for the weakest signature bound $r_2\le n/2$: the expression starts above one at $n=2$ and increases with $n$. Consequently $M=\mathbb{Q}$.

For completeness, normality converts this intersection statement into the desired degree statement. Put $E=K_1{K}_2$. Restriction injects $\mathrm{Gal}(E/K_2)$ into $\mathrm{Gal}(K_1/\mathbb{Q})$. The fixed field of its image in $K_1$ is exactly $K_1\cap K_2$, because $E/K_2$ is Galois. The Galois correspondence therefore gives $[E:K_2]=[K_1:M]=[K_1:\mathbb{Q}]$. This proves coprime discriminants imply linear disjointness:

For a number field $k$, its ordinary Hilbert class field $H_k$ is the maximal finite abelian extension unramified at every finite prime and with every real place splitting completely. The Artin reciprocity map inducesThe image of an unramified prime ideal is its Frobenius automorphism. Thus a prime splits completely in $H_k$ precisely when its ideal class is trivial. This field packages ideal-class obstructions into splitting behavior. Ideals become principal in $H_k$ by the principal ideal theorem. Specifying the real-place convention matters for real fields: omitting it leads instead to the narrow class-field problem. Both illustrative fields below are imaginary quadratic, so there are no real places to discuss.

First consider $k=\mathbb{Q}(\sqrt{-23})$, whose quadratic discriminant is $-23$. The correspondence between ideal classes and reduced primitive positive definite binary quadratic forms lets us compute its class number without guessing a class field. A reduced form $(a,b,c)$ obeys $|b|\le a\le c$, $b^2-4ac=-23$, and $a\le \sqrt{23/3}<3$; a boundary case uses $b\ge 0$. For $a=1$ the only class is $(1,1,6)$; for $a=2$ the two possibilities are $(2,1,3)$ and $(2,-1,3)$. Hence $h_k=3$ and its ideal class group is cyclic of order three.

Take $g(X)=X^3-X-1$. Its only possible rational roots are $\pm 1$, neither a root, so it is irreducible. Its polynomial discriminant is $4-27=-23$, which is nonsquare. Thus its splitting field $H$ has Galois group $S_3$ and the quadratic fixed field of $A_3$ is $k$, generated by the square root of its discriminant. Therefore $H/k$ is cyclic of degree three and $H=k(\theta )$ for a root $\theta$ of $g$.

At a prime other than $23$, the separable residue polynomial splits over a finite residue-field extension; Hensel's lemma puts all its roots in an unramified local extension. At $23$ there is the exact factorizationThe simple factor lifts by Hensel's lemma. The remaining quadratic factor has discriminant valuation one, because the cubic discriminant is $-23$ and the resultant with the simple factor is a unit. Its local splitting field is ramified quadratic, with inertia group of order two. Its intersection with $A_3$ is trivial, so the relative extension $H/k$ is unramified at $23$ as well. There are no real places. An unramified abelian extension of degree $h_k=3$ must be all of $H_k$, givingThis is the Hilbert class field of Q of square root minus twenty-three.

The other alternative, $k=\mathbb{Q}(\sqrt{-30})$, has quadratic discriminant $-120$. Reduced forms have $a\le \sqrt{40}<7$ and even $b$. Testing $1\le a\le 6$, $|b|\le a$, and $c=(b^2+120)/(4a)\ge a$ gives exactlyThus $h_k=4$. Each class is its own inverse, because replacing $b$ by $-b$ leaves these representatives unchanged; consequently the ideal class group is $C_2\times C_2$.

Set $H=\mathbb{Q}(\sqrt{2},\sqrt{-3},\sqrt{5})$. The square classes $2,-3,5$ are independent, so $[H:\mathbb{Q}]=8$. Their product gives $\sqrt{-30}$, and $H/k$ is abelian of degree four. Its seven quadratic subfields have discriminantsFor a multiquadratic abelian extension, the conductor-discriminant formula takes the product of these quadratic character conductors, namely the absolute quadratic discriminants. Hence $|d_H|=120^4$. The relative discriminant tower formula givesThe integral relative discriminant ideal therefore has norm one and is the unit ideal. Thus $H/k$ is everywhere unramified, and its degree equals the class number. This proves the other explicit answer:It agrees with the Hilbert class field of Q of square root minus thirty. Both alternatives have been computed independently from their class numbers and ramification.

A central simple algebra over $K$ is a finite-dimensional associative $K$-algebra with no nonzero proper two-sided ideals and with center $K$. The Artin–Wedderburn theorem writes it uniquely as $M_r(D)$ with $D$ a central division algebra. After a finite separable splitting field extension it becomes $M_d$, so its dimension is $d^2$, and $d$ is its degree of a central simple algebra. Passing from $D$ to a matrix algebra over $D$ does not change its Brauer class.

The Brauer group consists of these classes, with product induced by tensor product. Extending to a common splitting field shows that a tensor product of central simple algebras is again central simple; its two-sided ideals and center descend from the corresponding full matrix algebra. The neutral class is $[K]$, and the inverse of $[A]$ is $[A^{\mathrm{op}}]$, becauseThe displayed map becomes the standard isomorphism for matrices over a splitting field, so it is already an isomorphism over $K$. Thus Brauer classes measure precisely the obstruction to being a split matrix algebra.

An explicit family is supplied by cyclic algebras. Let $L/K$ be cyclic of degree $n$, with generator $\sigma$, and let $a\in K^{\times }$. DefineThis is an associative algebra: form the skew polynomial ring with relation $ux=\sigma (x)u$ and quotient by the central relation $u^n-a$. Reduction of powers gives the displayed $n^2$-dimensional $K$-basis. An element commuting with every $x\in L$ has no $u^i$ coefficient for $i\ne 0$, since ${\sigma }^i$ is not the identity. Commuting also with $u$ forces the remaining coefficient into $K$. Hence the center is $K$.

Here is an explicit splitting construction. On $L^n$, with basis $e_0,\dots ,e_{n-1}$, putThen $UD(x)=D(\sigma x)U$ and $U^n=aI$. After extending scalars to $L$, the separable decomposition $L{\otimes }_{K}L\cong {\prod }_{j=0}^{n-1}L$ supplies all diagonal matrices, and their products with powers of $U$ supply all matrix units. ThereforeAny nonzero proper two-sided ideal of $A$ would extend to one of $M_n(L)$, which is impossible. Thus this construction really gives a central simple algebra, with splitting field $L$.

Its central arithmetic feature is the splitting criterion for a cyclic algebra:If $a=Nc$, let $L$ act by multiplication on the $K$-vector space $L$, and let $u$ act as $c\sigma$. Its nth power is multiplication by $Nc=a$, so this gives a representation $A\to {\mathrm{End}}_K(L)$. It is injective by simplicity and an isomorphism by equal dimension. Conversely, if $A\cong {\mathrm{End}}_K(V)$ with ${\dim }_{K}V=n$, its embedded field $L$ makes $V$ one-dimensional over $L$. The relation for $u$ says that $u$ is $\sigma$-semilinear. In an $L$-basis it must have the form $c\sigma$, whose nth power is $Nc$. The relation $u^n=a$ forces $a=Nc$.

More generally the crossed-product algebra of a Galois extension attaches an algebra to a multiplicative two-cocycle. Changing the cocycle by a coboundary rescales its basis and preserves the Brauer class. Descent of a split matrix algebra identifiesFor cyclic $L/K$, cyclic Tate cohomology identifies this second cohomology group with $K^{\times }/N{L}^{\times }$. In a crossed product, rescale successive cyclic basis elements to powers of a single $u$; the remaining datum is $u^n=a$, changed by a field norm when $u$ is rescaled. This explains why cyclic algebras give every class split by a cyclic extension, and why multiplication of their parameters is addition of their Brauer classes.

For a finite field $K={\mathbb{F}}_q$, every finite splitting field $L={\mathbb{F}}_{q^n}$ is cyclic over $K$. The field norm is the power map $x\mapsto x^{(q^n-1)/(q-1)}$ on its cyclic multiplicative group, and is onto ${\mathbb{F}}_q^{\times }$. Thus every relative Brauer class is zero, and every central simple algebra has zero class:

For a non-Archimedean local field $K$, the valuation classification of central division algebras shows that every Brauer class splits over a finite unramified extension. For the degree-$n$ unramified extension $L/K$ with arithmetic Frobenius $\sigma$, units are norms and norm valuations are multiples of $n$. Hence the cyclic relative Brauer group is $\mathbb{Z}/n\mathbb{Z}$, represented by $a={\pi }^r$. The compatible local Brauer invariant isThe union over all $n$ givesA class with invariant $r/n$ in lowest terms has a central division algebra representative of degree of a central simple algebra $n$. This is the local valuation classification used above; the explicit cyclic algebras exhibit all of its invariants. At the Archimedean places, $\mathrm{Br}(\mathbb{C})=0$, while $\mathrm{Br}(\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$: its nontrivial class is the quaternion algebra $(\mathbb{C}/\mathbb{R},\text{conjugation},-1)$, since complex norms are positive.

For a number field, the Albert-Brauer-Hasse-Noether theorem describes the Brauer group by the exact sequenceThus its classes are exactly finite-support families of local invariants with sum zero; real invariants are $0$ or $1/2$ and complex invariants are zero. Every global Brauer class has a cyclic splitting extension, so cyclic algebras also realize all global classes. If $\chi (\sigma )=1/n$, their local invariants are ${\chi }_v({\mathrm{rec}}_{K_v}(a))$, with ${\chi }_v$ restricted to the local decomposition subgroup. The sum-zero relation for a principal parameter is the invariant form of Artin reciprocity. Injectivity supplies the local-to-global splitting test, which yields the Hasse norm theorem for cyclic extensions as explained in the preceding solution. Here the local valuation classification and the global invariant theorem are the structural theorems used to describe the Brauer groups; the construction and norm-splitting calculation have been proved explicitly.

One form of the Theorem of the Cube is the cube principle: a line bundle $M$ on $E^3$ which is trivial on each of $E\times E\times \{O\}$, $E\times \{O\}\times E$, and $\{O\}\times E\times E$ is trivial. Here is a proof for a complex elliptic curve.

Since $E$ is a real two-dimensional torus, its integral cohomology has no torsion. The Künneth theorem decomposesEvery single-factor summand is detected on a face containing that factor, and every cross summand is detected on the corresponding two-factor face. Restriction of $M$ to those faces is trivial, so its First Chern class has zero component in every summand. Thus $c_1(M)=0$.

The holomorphic exponential sequence identifies the kernel of $c_1$ in the holomorphic Picard group with $H^1(E^3,\mathcal{O})/\mathrm{im}{H}^1(E^3,\mathbb{Z})$. First cohomology decomposes factorwise here: $H^1(E^3,\mathcal{O})={⨁}_j{p}_j^{\ast }{H}^1(E,\mathcal{O})$ and likewise for integral first cohomology. Analytically the former decomposition says that the harmonic representatives of $(0,1)$-cohomology on the flat torus are the constant forms $d{\overline{z}}_j$, one from each factor. Consequently this kernel is ${\mathrm{Pic}}^0(E{)}^3$, andfor degree-zero line bundles $M_j$ on $E$. Restrict to each coordinate axis. The hypothesis makes every $M_j$ trivial, so $M$ is trivial. This proves the cohomological proof of the cube theorem for elliptic curves; the holomorphic isomorphism is algebraic on these projective varieties by GAGA.

To state its usual addition-map consequence, write $m_I:E^3\to E$ for summation of the coordinates indexed by $I$ and $p_j=m_{\{j\}}$. For any line bundle $L$ on $E$,More canonically, include the inverse of the constant bundle with fibre $L_O$ in the alternating product. Its restriction to a zero-coordinate face has pairwise cancellations, leaving the trivial bundle. The cube principle proves its triviality, which gives the displayed isomorphism after choosing a trivialization of the constant fibre. Pullback along any triple of maps to $E$ gives the corresponding alternating identity for their seven nonempty sums.

A pointwise divisor version can also be seen directly from Question 1. For a divisor $D$ and fixed points $a,b,c$, the alternating translation divisorhas degree zero and point sum zero: translation by $t$ subtracts $\mathrm{deg}(D)t$ from its point sum, and all the alternating terms cancel. The principality criterion proves that this divisor is principal.

Over a number field $K$, use algebraic line bundles and maps defined over $K$, and use Galois-invariant divisors rather than arbitrary individual complex points. The geometric line-bundle cube identity itself is still exact over $K$. To see descent in this setting, first pass to an algebraic closure, or an embedding in $\mathbb{C}$. The alternating bundle becomes trivial there by the preceding proof. Choose its nowhere-zero section with value one in a chosen $K$-basis at $(O,O,O)$. Any Galois conjugate of that section differs by an invertible global function, hence a constant on the connected projective variety; its normalized value at the rational origin forces that constant to be one. The section therefore descends, giving a $K$-defined trivialization.

For the arithmetic formulation used with heights, the necessary modification is a uniformly bounded error. A chosen Weil height machine representative $h_L$ satisfiesuniformly for $P,Q,R\in E(K)$. Indeed, pullback and tensor product of line bundles correspond to composition and addition of heights only modulo bounded functions. This is the arithmetic cube identity for Weil heights. Arbitrary naive heights do not satisfy an exact numerical cube identity; the canonical height of an elliptic curve removes the bounded error, as explained in Question 4. This arithmetic qualification does not weaken the exact geometric theorem.

Let $K$ be a number field of degree $d$, and normalize its absolute values so that the product formula is ${\prod }_v|a{|}_v^{n_v}=1$ for $a\ne 0$, where $n_v=[K_v:{\mathbb{Q}}_v]$; at a complex place use the ordinary modulus with weight two. For $P=[a_0:\cdots :a_N]$, defineOnly finitely many factors differ from one. Rescaling all coordinates multiplies the local maxima by $|a{|}_v$, whose weighted product is one, so the projective height is well-defined. It is at least one by comparison with any nonzero coordinate. On enlarging $K$ to $L$, the identity ${\sum }_{w\mid v}[L_w:{\mathbb{Q}}_v]=[L:K][K_v:{\mathbb{Q}}_v]$ makes the normalized product unchanged. Thus this is an absolute height on algebraic projective points. For coprime rational integer coordinates it is simply their maximum absolute value.

For a projective variety $V/K$ and an embedding $i:V\hookrightarrow {\mathbb{P}}^N$, take $h_i=h\circ i$. Changing the basis of the same projective linear system changes the height by a bounded amount: the coordinate transformation and its inverse have fixed local coefficient bounds, equal to one outside finitely many places. If $L$ is ample and $L^m$ is very ample, set $h_L=m^{-1}{h}_i$ for an embedding from $L^m$, up to a bounded function. A general line bundle can be written $(L\otimes A)\otimes A^{-1}$ with both $A$ and $L\otimes A$ very ample, defining its height as the difference of their embedding heights.

This is the Weil height machine, withThe first relation follows from the Segre embedding, whose local maximum is the product of the two maxima; tensor powers similarly use the Veronese embedding. The second follows by pulling back the defining sections. These constructions make the choices equivalent modulo bounded functions. For a fixed homogeneous morphism with no base point, upper local height bounds come from its coefficients, and lower bounds from the Projective Nullstellensatz expressing coordinate powers in the ideal of its defining forms. The constants are one outside a finite set of places. This gives the uniform bounded errors and shows why heights belong to line-bundle classes rather than being independent of every projective embedding.

The Northcott lemma states that for fixed $B$ and $D$, the algebraic points of ${\mathbb{P}}^N$ with $H(P)\le B$ and $[\mathbb{Q}(P):\mathbb{Q}]\le D$ form a finite set. In particular the $K$-rational points of bounded projective height are finite. Here is a proof, including the degree bound.

First let $\alpha$ have degree $e\le D$ and $H(\alpha )=H([1:\alpha ])\le B$. Write its primitive minimal polynomial as $a_e{\prod }_{j=1}^e(T-{\alpha }_j)$, with positive integer $a_e$. The height-Mahler measure formula isIts finite-place part follows from multiplicativity of the nonarchimedean polynomial coefficient norm: a primitive integer polynomial has norm one at each prime, so factoring it bounds the product of the roots' local maxima by the inverse leading-coefficient norm. The archimedean part is the displayed product over conjugates. Combining them with the product formula gives the identity.

Expanding the polynomial now bounds each integer coefficient byThere are finitely many degrees and finitely many such integer coefficient lists, each with finitely many roots. Hence there are only finitely many possible $\alpha$. This proves the minimal-polynomial coefficient proof of Northcott finiteness for algebraic numbers.

On the projective chart $a_j\ne 0$, each ratio ${\alpha }_i=a_i/a_j$ has degree at most the point's field degree andThe equality uses the product formula. Each coordinate ratio therefore belongs to a finite set, and there are finitely many charts. This proves the projective Northcott theorem. Restricting to $V$ gives the result for a chosen embedding. For an ample line-bundle height, the comparison $m{h}_L=h_i+O(1)$ reduces its bounded-height sets to this case.

Both qualifications matter: a degree bound cannot be dropped, since infinitely many roots of unity have height one; and the finiteness assertion need not hold for a nonample line bundle, such as the trivial bundle's bounded height on a variety with infinitely many rational points.

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.

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.

An integer version of Siegel lemma is as follows. For $M<N$ homogeneous linear equations $Az=0$, with integer matrix entries bounded in absolute value by $A_0\ge 1$, there is a nonzero integer solution satisfyingPut the displayed integer bound equal to $Q$ and evaluate the equations at every vector in $\{0,\dots ,Q{\}}^N$. There are $(Q+1{)}^N$ inputs and at most $(2N{A}_{0}Q+1{)}^M$ outputs. But $2N{A}_{0}Q+1\le 2N{A}_0(Q+1)$ and $(Q+1{)}^{N-M}>(2N{A}_0{)}^M$, so the number of inputs is larger. Two distinct inputs have identical outputs; their difference proves the claim. This is the pigeonhole proof of integer Siegel lemma. Rational coefficients are handled by clearing denominators. Over a number field of degree $d$, expanding in a rational basis gives at most $dM$ rational equations, so the same construction applies when $N>dM$, with the height of the resulting integer matrix controlling the bound.

Siegel's original arithmetic definition starts with an entire series $E(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$, with all $a_n$ in a fixed number field. For every $ϵ>0$ there must be positive integers $d_n$ such that $d_n{a}_0,\dots ,d_n{a}_n$ are algebraic integers, and both $d_n$ and the absolute values of every conjugate of $a_n$ are $O_{ϵ}(n^{ϵn})$. The growth condition ensures convergence everywhere. The stronger, now customary strict E-function definition uses bounds $C^{n+1}$ for a fixed $C$ and also includes the requirement of a nonzero linear differential equation over $\overline{\mathbb{Q}}(z)$. In the original formulation, that differential condition is explicitly part of the system hypothesis in the theorem below. These definitions should be distinguished; the exponentials used here satisfy even the stronger conditions.

The Siegel–Shidlovsky theorem states that if a vector of E-functions satisfies $f^{\prime }=A(z)f$, with $A(z)$ a matrix of rational functions over $\overline{\mathbb{Q}}$, then for every nonzero algebraic $\xi$ which is not a pole of $A$,In particular, algebraic independence of the functions implies algebraic independence of the values. The exclusion of zero is necessary: their Taylor values there are algebraic.

Take rationally independent algebraic ${\gamma }_1,\dots ,{\gamma }_s$ and $f_j(z)=e^{{\gamma }_{j}z}$. Their coefficients are ${\gamma }_j^n$, their conjugates grow exponentially, and powers of a fixed common denominator clear all coefficients through degree $n$. They satisfy the nonsingular diagonal system $f_j^{\prime }={\gamma }_j{f}_j$.

To check functional independence, a polynomial relation among these exponentials, after clearing rational-function denominators, would be ${\sum }_m{P}_m(z)e^{{\lambda }_{m}z}=0$ with distinct ${\lambda }_m={\sum }_j{m}_j{\gamma }_j$. Such an exponential polynomial cannot vanish identically. To isolate a nonzero term with exponent ${\lambda }_0$, apply ${\prod }_{\lambda \ne {\lambda }_0}(D-\lambda {)}^{\mathrm{deg}{P}_{\lambda }+1}$. All other terms vanish, while the remaining polynomial is acted on by operators $D+{\lambda }_0-\lambda$, each injective on polynomials because its nonzero constant preserves a nonzero leading coefficient. Thus the remaining term is nonzero, a contradiction.

The theorem at $\xi =1$ now makes $e^{{\gamma }_j}$ algebraically independent. For arbitrary distinct algebraic exponents ${\alpha }_i$, choose a rational basis of their span and divide its elements by a common denominator so that every ${\alpha }_i$ is an integer combination of the resulting ${\gamma }_j$. Then $e^{{\alpha }_i}$ are distinct Laurent monomials in $e^{{\gamma }_j}$. A linear relation among them becomes a nonzero polynomial relation after multiplying by a common monomial, contradicting independence. These particular E-functions therefore yield the full Lindemann–Weierstrass theorem.

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 $\mathfrak{p}$.

We give the detailed norm-index argument for a cyclic extension $L/K$ of p-adic fields; thus $K$ is a finite extension of ${\mathbb{Q}}_p$. Write $G=⟨\sigma ⟩$, $|G|=n$. For an additive $G$-module $M$, put $D=\sigma -1$ and $N=1+\sigma +\cdots +{\sigma }^{n-1}$. The two periodic Tate cohomology of a cyclic group groups and the Herbrand quotient arewhen these groups are finite. For multiplicative modules, $N$ is the product of conjugates and $D(z)=\sigma (z)/z$.

The periodic complex alternating $D$ and $N$ gives a six-term exact sequence for every short exact sequence of modules. Taking the alternating product of the finite group orders in that exact sequence provesFor a finite module, the counting fibers gives $|NM|=|M|/|\ker N|$ and $|DM|=|M|/|\ker D|$, so the two Tate groups have equal order. This proves the Herbrand quotient of a finite module is one. It follows that replacing a module by a finite-index submodule does not change its quotient, whenever the cohomology groups are finite.

Two basic calculations drive the norm computation. For the trivial module $\mathbb{Z}$, $D=0$ and $N$ is multiplication by $n$, so $h_G(\mathbb{Z})=n$. For a regular lattice $R[G]$, with $R={\mathbb{Z}}_p$, invariants are constant coefficient vectors and every such vector is a norm: put its constant coefficient in just one coordinate before applying $N$. The group kernel of $N$ consists of the coefficient vectors of sum zero. Successively taking partial sums solves the cyclic difference equation $Dx=y$ for every such vector $y$. Thus both Tate groups vanish and $h_G(R[G])=1$. This is cyclic cohomology of a regular lattice.

Let $U_L={\mathcal{O}}_L^{\times }$ and $U_L^r=1+{\mathfrak{P}}_L^r$. For sufficiently large $r$, the p-adic logarithm and p-adic exponential are inverse $G$-equivariant isomorphismsTo make the range explicit, $r>v_L(p)/(p-1)$ suffices: the higher terms in the logarithm and exponential have greater valuation than their leading terms, their series converge there, and the identities $\log (\exp x)=x$ and $\exp (\log (1+x))=1+x$ remain valid. Thus a deep multiplicative unit group becomes an additive ${\mathbb{Z}}_p$-lattice in $L$.

By the normal basis theorem, $L\cong K[G]$ as a $K[G]$-module. As a ${\mathbb{Q}}_p[G]$-module it is therefore ${\mathbb{Q}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$. Transport the regular lattice $M_0={\mathbb{Z}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$ into $L$. Both $M_0$ and ${\mathfrak{P}}_L^r$ are full $G$-stable ${\mathbb{Z}}_p$-lattices, so their intersection has finite index in each. The exact sequences with finite quotients show that the Tate groups of ${\mathfrak{P}}_L^r$ are finite and thatSince $U_L/U_L^r$ is finite, the exact-sequence formula for the Herbrand quotient gives $h_G(U_L)=1$. Finally, the valuation sequencehas trivial $G$-action on $\mathbb{Z}$. Hence the Herbrand quotient of the local multiplicative group is

For completeness, the denominator vanishes by Hilbert theorem 90. If $N_{L/K}a=1$, put $c_0=1$ and $c_j=a\sigma (a)\cdots {\sigma }^{j-1}(a)$. Independence of distinct field automorphisms lets us choose $t\in L$ with $b={\sum }_{j=0}^{n-1}{c}_j{\sigma }^j(t)\ne 0$. Since $c_n=1$, shifting the sum gives $\sigma (b)=b/a$. Thus $a=b/\sigma (b)$, a multiplicative coboundary. Therefore ${\hat{H}}^{-1}(G,L^{\times })=1$. The invariant subgroup of $L^{\times }$ is $K^{\times }$, soThis proves the cyclic local norm index without assuming local reciprocity. If the extension has ramification index $e$ and residue degree $f$, the valuation formula $v_K(Nz)=f{v}_L(z)$ shows that the valuation contribution to the index is $f$. A norm is a unit only when $z$ is a unit, so the remaining contribution isIn particular, unit norms are surjective in an unramified cyclic extension. The norm subgroup is open: on deep units its logarithm is ${\mathrm{Tr}}_{L/K}({\mathfrak{P}}_L^r)$, a nonzero full lattice in $K$, and therefore contains a sufficiently deep principal-unit group. This connects the index calculation to the open finite-index norm subgroups classified by local class field theory.

To finish, the Hasse norm theorem for a cyclic extension of number fields states that a nonzero element is a global norm if and only if it is a norm at every finite and infinite completion. Here is its Brauer-theoretic proof outline. Associate to $a\in K^{\times }$ the cyclic algebra $A=(L/K,\sigma ,a)$, with relations $z\ell =\sigma (\ell )z$ and $z^n=a$. The splitting criterion for a cyclic algebra says that it splits exactly when $a=Nc$: if $a=Nc$, multiplication by $L$ and the operator $c\sigma$ realize it as ${\mathrm{End}}_K(L)$; conversely, in a split matrix representation, the underlying $n$-dimensional space is one-dimensional over $L$, so $z$ must act as $c\sigma$ and $z^n=Nc$.

Apply this criterion at every completion, using the restricted cyclic character and any one completion of $L$ above that place. If $a$ is a local norm everywhere, every localized Brauer class is zero. Injectivity in the Albert-Brauer-Hasse-Noether theorem forces $[A]=0$ in $\mathrm{Br}(K)$; the global splitting criterion gives $a=Nc$. Conversely, a global norm is a local norm after completion, since the norm on $L{\otimes }_K{K}_v$ is the product of the component norms, whose images coincide in the Galois case. The cyclic hypothesis is essential; the stated local-global principle does not hold for arbitrary extensions.

The Brauer group of a field consists of equivalence classes of central simple algebras, with addition induced by tensor product, zero represented by matrix algebras, and inverse represented by the opposite algebra. Localization sends a global algebra to its scalar extensions at the completions. For a non-Archimedean completion of a number field, the local Brauer invariant is an isomorphismNormalize it so that an unramified cyclic algebra of degree $m$, with arithmetic Frobenius generator and parameter $a$, has invariant $v(a)/m$. For a real completion, $\mathrm{Br}(\mathbb{R})=\mathbb{Z}/2$ with the Hamilton quaternion algebra mapped to $1/2$; for a complex completion the Brauer group is zero.

The global computation is the exact sequence in the Albert-Brauer-Hasse-Noether theorem:Thus a global Brauer class is determined uniquely by a finite-support family of local invariants. Such a family occurs if and only if its sum is zero; at real places the only allowed entries are $0,1/2$, and at complex places the entry is zero. Injectivity is a local-global splitting principle; the middle exactness is the reciprocity constraint; surjectivity of the sum follows already from any one finite local summand. These are separate assertions, not merely different descriptions of the sum formula.

For an explicit group description, choose one finite place $v_0$. Assign arbitrary finite-support invariants at all other finite places and arbitrary allowed invariants at the real places, then set the invariant at $v_0$ equal to minus their sum. This givesThe description by the entire invariant family is canonical; this displayed elimination of one coordinate depends on the choice of $v_0$.

The connection with Artin reciprocity is especially transparent for cyclic algebras. Let $\chi :\mathrm{Gal}(L/K)\hookrightarrow \mathbb{Q}/\mathbb{Z}$ describe a cyclic extension, with $\chi (\sigma )=1/n$. Write $(\chi ,a)$ for the class of $(L/K,\sigma ,a)$. The local reciprocity pairing isIt follows from the local fundamental class construction of the Local Artin map; on an unramified character it is precisely the normalization $v(a_v)/n$. The group kernel in the parameter is the local norm group, and restriction/corestriction give the norm compatibility of this pairing.

For a principal parameter $a\in K^{\times }$, the zero-sum identity for its global cyclic algebra therefore readsOn the other hand the idelic Artin reciprocity law isand evaluating this product by $\chi$ gives exactly that zero-sum identity. Conversely, testing every cyclic character of a finite abelian Galois group separates its elements, so the zero-sum identities for cyclic algebras imply this principal-idele reciprocity law for every finite abelian extension. At almost all places a unit and an unramified character pair trivially, so the sums and products are finite.