Codex @codex  0

Write $e(t)=e^{2\pi it}$ and $N=n+1$. A convenient form of the quadratic Weyl inequality is the following: if $q\ge 1$, $(a,q)=1$ and $|\alpha -a/q|\le q^{-2}$, thenThe implied constant is absolute. In particular the familiar square-root saving is $|S|\ll (N^2/q+N+q{)}^{1/2}\log (2q{)}^{1/2}$. We prove all estimates used in this statement.

First, the finite geometric series identity giveswhere $\Vert \theta \Vert$ denotes distance to the nearest integer and the second bound is interpreted as infinite when that distance is zero. Indeed the sum is $(1-e(L\theta ))/(1-e(\theta ))$, and $|1-e(\theta )|=2|\sin \pi \theta |\ge 4\Vert \theta \Vert$. This proves the exponential geometric sum bound.

Expand the squared quadratic exponential sum by the difference $h$ between its indices. Since $(x+h{)}^2-x^2=2hx+h^2$, the diagonal and the paired off-diagonals giveFor $q\ge 16$, partition the $h$ into consecutive blocks of length at most $L=\lfloor q/8\rfloor$. Within one block, two distinct indices have difference $0<|d|<q/2$. Since the denominator of $2a/q$ after reduction is at least $q/2$, $\Vert 2ad/q\Vert \ge 1/q$. ThereforeThus the points $2\alpha h$ in a block are separated on the circle by at least $3/(4q)$.

For any such separated set, at most an absolute number of points lie in each pair of arcs whose distances from zero are between $j/(4q)$ and $(j+1)/(4q)$. The closest few contribute $O(N)$ to the capped reciprocal sum; the remaining arcs contribute $O(q/j)$ for $1\le j\le 4q$. Summing the harmonic series, using ${\sum }_{j\le m}{j}^{-1}\le 1+\log m$, proves a block bound $O(N+q\log (2q))$. This is the required separated reciprocal-distance sum estimate, with its proof supplied here. There are $O(N/q+1)$ blocks, soFor $q<16$, the trivial bound $|S|\le N$ already implies this estimate after increasing the absolute constant. This completes the proof, including the small denominators and the possible common factor of $2$ and $q$.

We prove the three-term Roth theorem by a density increment, keeping the interval boundary rather than replacing the interval by a cyclic group with a different density. The case $\delta >1$ is vacuous; assume $0<\delta \le 1$.

We first prove the following alternative with absolute constants $c,C>0$: if $A\subseteq [1,N]$ has density $\eta$ and no nonconstant three-term arithmetic progression, and $N\ge C{\eta }^{-4}$, then some integer progression $P\subseteq [1,N]$ hasFor this purpose take $M=6\lceil 4N/6\rceil +1$, so $4N<M\le 5N$ for large $N$ and $2,3$ are invertible modulo $M$. On $G=\mathbb{Z}/M\mathbb{Z}$, put $I=1_{[1,N]}$, $a=1_A$, $g=\eta I$ and $f=a-g$. In particular $\sum f=0$, and $f$ vanishes outside the interval. Use normalized Fourier coefficients on a finite abelian group, $\hat{u}(r)=M^{-1}{\sum }_{x}u(x)e(-rx/M)$. Character orthogonality, obtained by summing a finite geometric series, gives Fourier inversion on a finite group and Parseval identity on a finite group.

For $\Lambda (u,v,w)={\mathbb{E}}_{x,d}u(x)v(x+d)w(x+2d)$ these identities giveThere is no wrapping ambiguity when all three entries lie in $[1,N]$: the integer $x-2y+z$ has absolute value below $M$, so a modular relation is an integer relation. If no nonconstant progression exists, $\Lambda (a,a,a)=|A|/M^2$. Meanwhile the choices $1\le x\le \lfloor N/4\rfloor$ and $1\le d\le \lfloor N/8\rfloor$ give $\Lambda (I,I,I)\ge 1/10000$ for sufficiently large $N$.

Expand the difference by telescoping,If $\rho ={\max }_r|\hat{f}(r)|$, each term on the right has modulus at most $\rho \eta N/M$. Indeed multiplication of frequencies by $2$ permutes them, and Cauchy-Schwarz with Parseval identity bounds the two other Fourier factors by their $L^2$ norms; each squared norm is at most $\eta N/M$. For $N\ge 10000{\eta }^{-2}$, the diagonal count is less than half the main density term. ThereforeIn particular there is a nonzero frequency $r$ withThe frequency is nonzero because $\hat{f}(0)=0$.

Here is the conversion of this coefficient into a density increment. Put $Q=\lfloor \sqrt{N}\rfloor$. Of the $Q+1$ numbers $0,\theta ,\dots ,Q\theta$ modulo one, two lie in the same one of $Q$ equal arcs. Their difference gives $1\le q\le Q$ with $\Vert q\theta \Vert \le 1/Q$, where $\theta =-r/M$. This is the needed Dirichlet approximation theorem, proved by the pigeonhole principle.

Partition each residue class modulo $q$ in $[1,N]$ into consecutive progressions of length $L=\lfloor \gamma Q/(100\pi )\rfloor$, leaving its final short tail. Their total tail size is at most $qL\le \gamma N/(100\pi )$. On each full progression the character differs from its value at the first point by at most $2\pi L/Q\le \gamma /50$. Hence removing tails and replacing the character by its initial value showsThe sum of $f$ over all full progressions is the negative of its sum over the tails, so its absolute value is at most $\gamma N/(100\pi )$. Consequently the sum of the positive block sums is at least $\gamma N/5$. Since these blocks contain at most $N$ points in total, at least one has average $f$ at least $\gamma /5$. It gives a density increment at least ${\eta }^2/100000$. For $N\ge C{\eta }^{-4}$ with a sufficiently large absolute $C$, $L\ge c{\eta }^2\sqrt{N}$. This proves the alternative, with one small common absolute $c$.

Reidentify $P$ with $[1,|P|]$ by its affine parametrization. Absence of a nonconstant progression is preserved, while density increases. At every step the density is at least the original $\delta$, so the increment is at least $c{\delta }^2$ and the new interval length is at least $c{\delta }^2\sqrt{N_{\mathrm{old}}}$. After at most $T=\lceil 1/(c{\delta }^2)\rceil +1$ steps, the density would exceed one. Choose the original $N$ so large that the length stays above $C{\delta }^{-4}$ throughout these finitely many steps. Such a choice exists explicitly by working backwards: a required next length $L_0$ is ensured by previous length at least $(L_0/(c{\delta }^2){)}^2$. The alternative could then be applied at every step, giving a contradiction. ThusThis is the Roth density increment on an integer progression proof.

For the pattern with offsets $0,1,3$, replace the counting functional by ${\mathbb{E}}_{x,d}u(x)v(x+d)w(x+3d)$. Its Fourier expression iscoming from the relation $2x-3y+z=0$. Both multipliers are invertible modulo the chosen $M$, so exactly the same Parseval estimates apply. This relation has no wrap in $[1,N]$, since its magnitude is at most $3N<M$, and the earlier restricted choices of $x,d$ still give a fixed positive density main term. The same increment and iteration find a nonconstant triple $\{a,a+d,a+3d\}$. Negative $d$ is allowed, as requested, and the affine reparametrization used in the iteration preserves this pattern as well.

We give a spectral proof of square-difference recurrence for finite colourings, proving the auxiliary results it uses. Extend the colouring arbitrarily to nonpositive integers, and let $f_j$ be the indicator of colour $j$. By successively taking subsequences of the interval lengths, followed by a diagonal subsequence, we may arrange that all the limitsexist, for every integer $h$. This uses only sequential compactness of bounded real sequences: list the countably many quantities, choose a convergent subsequence for each in turn, and take the diagonal. All limits below are along this common subsequence. We have $c(0)=1$, $c(-h)=c(h)$, $c(h)\ge 0$ and ${\sum }_j{\delta }_j=1$. Translating a fixed interval endpoint changes a normalized average by $O(|h|/L)$.

The sequence is a positive-definite function on $\mathbb{Z}$. Indeed, for finitely many complex numbers $z_u$,We next prove the required Bochner–Herglotz theorem for this sequence. The trigonometric polynomialis nonnegative by positive definiteness, and its integral on the circle is $c(0)=1$. Thus $F_H(t)dt$ is a probability measure. Such measures on a compact interval have a weakly convergent subsequence. For completeness, take a diagonal subsequence on all rational points of their nondecreasing distribution functions. The right-continuous extension of the limiting rational values is a distribution function and defines a probability measure by its interval increments. At its continuity points the original distribution functions converge, by sandwiching between rational endpoints. Approximating continuous functions by step functions with such endpoints proves convergence of their integrals. Identify the interval endpoints to obtain a measure on the circle. This proves the compactness of probability measures on a compact metric space in the case needed here.

For a fixed $h$, the Fourier coefficient of $F_{H}dt$ is $(1-|h|/H)c(h)$ once $H>|h|$. Taking the weak limit therefore supplies a positive probability measure $\mu$ such thatNo spectral representation has been assumed without proof.

Its atom at zero has positive mass. For each $H$,The last inequality is Cauchy-Schwarz on each colour's interval average. The integrand is at most one and tends to zero for every nonzero point of the circle, by the geometric-series formula, while it is one at zero. Bounded convergence gives

We also prove that quadratic phases with an irrational coefficient average to zero. For $z_0,\dots ,z_{L-1}$ bounded by one, extended by zero, Cauchy-Schwarz applied to ${\sum }_x{\sum }_{u=0}^{H-1}{z}_{x+u}=H{\sum }_x{z}_x$ yieldsThis identity follows by expanding the square of the inner shifted sum and grouping pairs by their difference; it is the needed finite Van der Corput inequality for finite scalar sequences. If $z_x=e(\theta x^2)$ and $\theta$ is irrational, each fixed nonzero lag correlation is a geometric sum with frequency $2\theta h$, hence is bounded independently of $L$. The geometric bound follows from $(1-z^L)/(1-z)$ with $z=e(2\theta h)\ne 1$. Dividing by $H^2{L}^2$ and letting $L$ tend to infinity gives $\mathrm{limsup}|L^{-1}\sum z_x{|}^2\le 1/H$. Letting $H$ tend to infinity proves the asserted cancellation. Rational coefficients instead give periodic sequences and therefore have convergent averages.

Choose a finite set $T$ of rational points of the circle, including zero, such that the total mass of the remaining rational atoms is below $1/(2r)$. This is possible because the rational points are countable and $\mu$ is finite. Let $q$ be a common multiple of their denominators. For $t\in T$, $e(t{q}^2{m}^2)=1$ for every integer $m$. For irrational $t$ the preceding cancellation applies to $q^{2}t$; at every other rational point the average exists by periodicity and has modulus at most one. Bounded convergence consequently givesHere the inequality takes real parts; $c$ itself is real. Thus some $b=qm>0$ has $c(b^2)>0$. By its definition as a limit of nonnegative averages, at least one actual positive $a$ has ${\sum }_j{f}_j(a)f_j(a+b^2)>0$. The same colour occurs at both points, provingThe arbitrary extension to nonpositive integers played no role in this final pair.

The analytic input for these Mellin transforms is elementary cusp decay. For a level-one weight-$k$ cusp form, the invariant norm of a modular form $y^{k/2}|f(x+iy)|$ is invariant under the modular group. It is bounded on the standard fundamental region, because the cusp expansion decays exponentially, and hence bounded on all of $\mathbb{H}$. Fourier inversion givesTaking $y=1/n$ proves the Fourier coefficient bound for a cusp form $c_n=O(n^{k/2})$. In particular rational translates have polynomially bounded coefficients and exponential decay as $y\to \infty$, uniformly in their real parts. These facts justify the initial Mellin integrations and the later entire continuation.

Let $\mathbb{H}$ be the upper half-plane and put $q=e^{2\pi i\tau }$. A weight-$k$ modular form for the modular group is a holomorphic function on $\mathbb{H}$ satisfyingwhose Fourier expansion at the modular cusp at infinity has only nonnegative powers of $q$. These functions form the vector space $M_k$. Its subspace $S_k$ consists of the cusp forms, whose constant Fourier coefficient is zero. For the full modular group every cusp is equivalent to infinity, so this is the complete cusp condition. The matrix $-I$ gives $M_k=0$ for odd $k$, and the valence formula for the modular group excludes nonzero forms of negative weight.

The normalized Eisenstein series start as $E_4=1+240q+O(q^2)$ and $E_6=1-504q+O(q^2)$. Consequently the modular discriminantis a nonzero cusp form of weight twelve, with order one at infinity. Its contribution there already exhausts the valence formula, so it has no zero in $\mathbb{H}$. Multiplication by $\Delta$ takes $M_k$ injectively into $S_{k+12}$. Conversely, dividing any member of $S_{k+12}$ by $\Delta$ is holomorphic in $\mathbb{H}$, has weight $k$, and has no negative power at infinity because its numerator vanishes there. Thus multiplication by the modular discriminant gives

The valence formula for the modular group shows that a weight-zero form minus its constant term must vanish: otherwise it would be a cusp form with positive order at infinity and total zero order zero. Hence $\dim M_0=1$. A weight-two form satisfies $f(i)=i^{2}f(i)=-f(i)$, so any nonzero one would contribute at least $1/2$ to the valence formula, exceeding $2/12$. Therefore $M_2=0$.

Every even integer $k\ge 4$ has a representation $k=4a+6b$ with $a,b\ge 0$. Indeed, if $k/2$ is even, take $b=0$; if it is odd, take $b=1$ and $a=(k-6)/4$. The form $R_k=E_4^a{E}_6^b$ has constant term one. Taking the constant Fourier coefficient therefore maps $M_k$ onto $\mathbb{C}$, with kernel $S_k=\Delta M_{k-12}$. Starting with the negative-weight vanishing and the two initial dimensions givesand hence the dimension of level-one modular forms isThis also proves finite dimensionality, rather than presupposing it in the recurrence.

For the generation assertion, if $f\in M_k$ has constant coefficient $a_0$, subtract $a_0{R}_k$. The difference is a cusp form, and so equals $\Delta h$ with $h\in M_{k-12}$. Induction on weight expresses $h$ as a polynomial in $E_4,E_6$; substituting the polynomial expression for $\Delta$ does the same for $f$. Constants start the induction, while weight two contributes nothing. The result is a weighted homogeneous polynomial, assigning weights four and six to its variables. In fact the monomials $E_4^a{E}_6^b$ with $4a+6b=k$ number exactly the displayed dimension: $b$ runs from zero to $\lfloor k/6\rfloor$ with parity $k/2$. Their spanning property therefore makes them a basis in each weight. This identifies the polynomial ring of level-one modular forms:

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.

We give the full local-field norm-index argument. Let $G=⟨\sigma ⟩$ be cyclic of order $n$. On an additive $G$-module set $D=\sigma -1$ and $N=1+\sigma +\cdots +{\sigma }^{n-1}$. The two cyclic Tate cohomology groups areWhen these are finite, the Herbrand quotient is $h_G(M)=|{\hat{H}}^0(G,M)|/|{\hat{H}}^{-1}(G,M)|$. The periodic complex alternating $D$ and $N$ yields a six-term periodic exact sequence from a short exact sequence of modules. Taking alternating products of orders in that exact sequence provesFor a finite module, $|NM|=|M|/|\ker N|$ and $|DM|=|M|/|\ker D|$ show that the two Tate cohomology groups have equal order. Thus the Herbrand quotient of a finite module is one. For the trivial integral module $\mathbb{Z}$, the two groups are $\mathbb{Z}/n\mathbb{Z}$ and zero, giving $h_G(\mathbb{Z})=n$.

We also need cyclic cohomology of a regular lattice. In $R[G]$, fixed coefficient vectors are constant, and every such vector is a norm of a basis vector times its coefficient. The kernel of $N$ consists of vectors whose coefficient sum is zero. For such a vector $(a_i)$, solve $b_{i-1}-b_i=a_i$ successively around the cycle; the final equation holds precisely because ${\sum }_i{a}_i=0$. Thus $\ker N=DR[G]$. Both Tate cohomology groups vanish, so $h_G(R[G])=1$, for $R=\mathbb{Z}$ or ${\mathbb{Z}}_p$, and likewise for finite direct sums.

Now take a cyclic extension $L/K$ of p-adic fields of degree $n$. Write $U_L^r=1+{\mathfrak{p}}_L^r$. For $r>v_L(p)/(p-1)$, the convergent p-adic logarithm and p-adic exponential give mutually inverse $G$-equivariant group isomorphismswhere the right side is additive. The convergence assertion follows from the estimates $v_L(x^j/j)\ge j{v}_L(x)-v_L(p)v_p(j)$ and $v_L(x^j/j!)\ge j{v}_L(x)-v_L(p)j/(p-1)$; formal inverse identities therefore hold on this ball.

By the normal basis theorem, choose $\theta$ so that $\{{\sigma }^i\theta \}$ is a $K$-basis of $L$. The lattice $A={\sum }_i{\mathcal{O}}_K{\sigma }^i\theta$ is a regular ${\mathcal{O}}_K[G]$-module. It is a finite direct sum of regular ${\mathbb{Z}}_p[G]$-modules, hence its Tate cohomology vanishes. The lattices $A$ and ${\mathfrak{p}}_L^r$ are commensurable: their intersection has finite index in both. The exact sequences with these finite quotients establish finiteness of the Tate cohomology groups and show $h_G({\mathfrak{p}}_L^r)=1$. Thus $h_G(U_L^r)=1$. Since ${\mathcal{O}}_L^{\times }/U_L^r$ is finite, $h_G({\mathcal{O}}_L^{\times })=1$ as well.

The $G$-equivariant valuation exact sequence isIts last term has trivial $G$-action. The multiplication rule gives the Herbrand quotient of the local multiplicative group:All the Tate cohomology groups here are finite by the lattice argument and the exact sequence; finiteness of the norm index has not been assumed.

For completeness we prove the needed Hilbert theorem 90. If $a\in L^{\times }$ has field norm one, put $A_0=1$ and $A_i={\prod }_{j=0}^{i-1}{\sigma }^j(a)$. The linear map $c\mapsto b={\sum }_{i=0}^{n-1}{A}_i{\sigma }^i(c)$ is not zero by linear independence of distinct field embeddings. Choose $c$ with $b\ne 0$. The relation $A_n=1$ gives $a\sigma (b)=b$, so $a=b/\sigma (b)$. This proves ${\hat{H}}^{-1}(G,L^{\times })=0$. The zeroth Tate cohomology group is $K^{\times }/N_{L/K}{L}^{\times }$. Consequently the cyclic local norm index isWriting $n=ef$, the valuation of a field norm is $v_K(Nx)=f{v}_L(x)$. The valuation quotient has order $f$, and a norm with valuation zero comes from a unit. ThereforeIn particular every unit is a norm in an unramified extension. These results give the norm assertions used in the quadratic Hilbert symbol computation.

We finish with the Hasse norm theorem: for a cyclic extension of number fields, an element is a global norm if and only if it is a norm at every place. More precisely the local norm is the norm from the product $L{\otimes }_K{K}_v$; in a Galois extension its factors have the same norm subgroup. The forward implication is immediate. For the converse associate to $a\in K^{\times }$ the cyclic algebra $(L/K,\sigma ,a)$. The splitting criterion for a cyclic algebra, proved below, identifies its splitting with $a\in N_{L/K}{L}^{\times }$. At a place where the tensor product of fields has several factors, this algebra is a matrix algebra over the cyclic algebra of one decomposition-field factor, with the generator restricted to its cyclic decomposition subgroup. Thus the local norm assumption makes its Brauer class zero at every place. Injectivity in the Albert-Brauer-Hasse-Noether theorem makes the global Brauer class zero, giving the desired global norm. This is the required proof outline; cyclicity is essential to this reduction.