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Every finite-dimensional semisimple algebra is a finite product of matrix algebras over division algebras. In particular, every central simple algebra over $k$ is isomorphic to $M_r(D)$ for a finite-dimensional central division algebra $D$ over $k$. The integer $r$ and the isomorphism class of $D$ are unique.

Indeed, decompose the right regular module of a finite-dimensional semisimple algebra $A$ aswhere the $S_i$ are pairwise nonisomorphic simple modules. The Schur lemma makes $D_i={\mathrm{End}}_A(S_i)$ a division ring and gives ${\mathrm{Hom}}_A(S_i,S_j)=0$ for $i\ne j$. Left multiplication identifiesConversely, the column module over $M_n(D)$ is simple and the regular module is a direct sum of $n$ copies of it, so finite products of such matrix rings are semisimple. The simple-module decomposition also proves uniqueness up to permuting the factors.

The Brauer class of a central simple algebra is its equivalence class under adjoining matrix-algebra factors. Two algebras have the same class exactly when their central division-algebra representatives are isomorphic. The zero class consists of split matrix algebras; the inverse of $[A]$ is the class of the opposite algebra. For a cyclic algebra $(L/K,\sigma ,a)$, the class is zero exactly when $a$ is a field norm from $L$.

A central division algebra over $K$ is a finite-dimensional division algebra whose center is exactly $K$. It is a central simple algebra. Every Brauer group class has a unique central division algebra representative up to isomorphism.

For a cyclic extension $E/K$ of degree $m$ with generator $\sigma$ and $a\in K^{\times }$, the cyclic algebra is generated by $E$ and $u$ with $ub=\sigma (b)u$, $u^m=a$. It is a central simple algebra of degree $m$. Its Brauer class can also be written $(\chi ,a)$, where $\chi (\sigma )=1/m$.

For the function field $K$ of a curve over a finite full constant field $k$, extend constants to $\overline{k}$. The resulting curve function field is $C_1$ by the Tsen theorem, so every central simple algebra splits there. The finite collection of coefficients of a splitting isomorphism descends to some finite constant extension $k^{\prime }/k$. Regularity makes $K{k}^{\prime }/K$ cyclic, giving a finite cyclic splitting field.

A central simple algebra $A$ has dimension $d^2$ over its center. Its degree is this integer $d$. Over a splitting field it becomes $M_d$, so degree records the matrix size independently of the chosen splitting.

A splitting isomorphism of a central simple algebra with $M_n(L)$ transports its Galois action to $T_{\sigma }=c_{\sigma }{\sigma }_0$, giving a nonabelian first cohomology class. Conversely a cocycle gives a semilinear algebra action whose invariants descend to a central simple algebra. Equivalent cocycles give isomorphic fixed algebras. This gives a pointed-set bijection ${\mathrm{CSA}}_n(L/K)\cong H^1(\mathrm{Gal}(L/K),{\mathrm{PGL}}_n(L))$.

For an associative algebra $A$, its matrix algebra $M_n(A)$ consists of all $n$ by $n$ matrices over $A$, with ordinary matrix multiplication. If $A=k$ is a field, this is ${\mathrm{End}}_k(k^n)$ and is a central simple algebra. Full matrix algebras over division rings are the factors in the Artin–Wedderburn theorem.

Keep the nonzero $z$ of minimal two-factor tensor rank constructed in part (i). For every $a\in A$, its commutator with $a\otimes 1$ lies in $J$ and equalsIf this tensor were nonzero, its two-factor tensor rank would be at most $n-1$, contradicting minimality. Thus it is zero. The linear independence of $d_2,\dots ,d_n$ forces $a{c}_i=c_{i}a$ for every $a$ and $i\ge 2$. Since $Z(A)=k$, each $c_i$ is a scalar multiple of $1$.

But $1,c_2,\dots ,c_n$ are linearly independent. Hence $n=1$, so $z=1\otimes 1\in J$. A two-sided ideal containing the multiplicative identity is the entire ring; therefore $J=A{\otimes }_{k}B$.

We have proved the stronger result that $A{\otimes }_{k}B$ is a simple ring. It has finite dimension $({\dim }_{k}A)({\dim }_{k}B)$, so part (b) provesPart (c)(ii) also gives $Z(A{\otimes }_{k}B)=k$, so this proves that a tensor product of central simple algebras is again a central simple algebra.

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.

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.

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.

The full idelic statement identifies the group kernel as $K^{\times }{N}_{L/K}{J}_L$ and induces $J_K/(K^{\times }{N}_{L/K}{J}_L)\cong \mathrm{Gal}(L/K)$. It explains why the local norm-residue pairings must glue globally. The global Brauer theorem supplies, in addition, the injectivity and realization statements needed to compute all algebra classes; the principal-product identity alone would not prove those statements. For cyclic algebras its injectivity yields the Hasse norm theorem, as in Question 3. In degree two, an invariant is either zero or $1/2$, so the zero-sum law becomes the Hilbert reciprocity law ${\prod }_v(a,b{)}_v=1$. For example, a quaternion algebra over $\mathbb{Q}$ can be ramified at precisely a prescribed finite set of places, including possibly the real place, if and only if that set has even cardinality; the invariants uniquely determine its Brauer class. This illustrates how arithmetic reciprocity becomes a concrete constraint on global algebra structure.

We first prove the simple Artinian case of the Artin–Wedderburn theorem. This part does not need finite dimension over $k$. Choose a minimal nonzero left ideal $L$, using the descending chain condition. Simplicity gives $LA=A$. Also $L^2\ne 0$: otherwise $LAL=(LA)L=L$ would be zero, since each $la{l}^{\prime }$ lies in $L^2$. Choose $x\in L$ for which right multiplication $r_x:L\to L$ is nonzero. By simplicity of $L$, it is an automorphism. Choose $e\in L$ with $ex=x$. Then $(e^2-e)x=0$, and injectivity of $r_x$ gives $e^2=e$. Thus $L=Ae$.

Right multiplication by $e$ splits the left regular module as $A=Ae\oplus A(1-e)$. Repeat within the remaining left ideal: if it contains a minimal left ideal $Af$ generated by an idempotent, the map $v\mapsto vf$ splits off $Af$. Each nonzero remaining complement is a strictly smaller left ideal, so the Artinian condition terminates the process. Therefore the regular module is a finite direct sum of minimal left ideals.

All these simple summands are isomorphic. For primitive idempotents $e,f$, simplicity gives $AeA=A$, so $fAe\ne 0$; otherwise $fAeA=fA$ would be zero. A nonzero element of $fAe$ defines a nonzero homomorphism $Af\to Ae$, which is an isomorphism between simple modules. Thus $A\cong L^{\oplus n}$ as a left module. Put $\Delta ={\mathrm{End}}_A(L)$, a division algebra by the Schur lemma. Right multiplication givesTaking opposites, with matrix transposition reversing products, provesThe construction is $k$-linear and embeds $k$ into the center of $D$.

A central simple algebra is a finite-dimensional simple $k$-algebra whose center is exactly $k$. The Brauer group consists of such algebras modulo matrix stabilization: $A,B$ represent the same class if $M_r(A)\cong M_s(B)$ for some positive integers $r,s$. Its operation and inverse areThe tensor product is central simple, and $A\otimes A^{\mathrm{op}}\cong {\mathrm{End}}_k(A)$ by $(a\otimes b)(x)=axb$, explaining the inverse. Both facts can be checked after a splitting field of a central simple algebra extension, where they are matrix-algebra identities, and descended by faithful scalar extension. Every class has a unique central division-algebra representative, by the regular-module description above.

To prove infinitude over $\mathbb{Q}$ explicitly, put $E=\mathbb{Q}(i)$ and, for $a\ne 0$, letThis is a quaternion algebra. It is central simple: the center is $\mathbb{Q}$, and after adjoining $i$ its generators have the faithful matrix modelThese generate $M_2(E)$; consequently a nonzero proper two-sided ideal over $\mathbb{Q}$ is impossible. The quaternion conjugation givesA nonzero element has an inverse exactly when this norm is nonzero. Thus $A_a$ is a division algebra when $a$ is not an $E/\mathbb{Q}$ norm; if $a$ is a norm it has a nonzero zero divisor and is $M_2(\mathbb{Q})$ by the classification just proved.

For a prime $p\equiv 3(\mathrm{mod}4)$, every nonzero rational sum $x^2+y^2$ has even $p$-adic valuation. Divide out the minimum valuation of $x,y$; cancellation modulo $p$ is impossible because $-1$ is not a square modulo $p$. Therefore $p$ is not an $E/\mathbb{Q}$ norm, and $A_p$ is a nontrivial division-algebra Brauer class. This is the quaternion norm obstruction at primes congruent to three modulo four.

We also prove the product rule needed to distinguish classes. In $A_a\otimes A_b$, write the generators as $i_1,j_1,i_2,j_2$. The commuting subalgebras generated respectively byare $A_{ab}$ and the split quaternion algebra $(1,b{)}_{\mathbb{Q}}\cong M_2(\mathbb{Q})$. The latter has the matrix model $\mathrm{diag}(1,-1)$ and $\left(\begin{array}{cc}0& b\\ 1& 0\end{array}\right)$. Together the two subalgebras generate the whole tensor product, since they recover $i_2$ and $j_1$. Dimensions are sixteen on both sides, givingQuaternion conjugation identifies $A_a$ with its opposite, so these classes have order at most two. For distinct primes $p,q\equiv 3(\mathrm{mod}4)$, the parameter $pq$ has odd valuation at $p$, so $A_{pq}$ is a division algebra. Thus $[A_p]\ne [A_q]$. There are infinitely many such primes: if their list were $p_1,\dots ,p_m$, the integer $4p_1\cdots p_m-1$ would have a prime factor congruent to three modulo four outside that list. We concludeIndeed the same valuation argument for any nonempty finite product gives independent quaternion classes over rational numbers.

For $\mathbb{R}$, we prove the required division-algebra classification. Let $D$ be finite-dimensional and central over $\mathbb{R}$. If $D\ne \mathbb{R}$, choose a nonreal element; its irreducible real minimal polynomial is quadratic, so translating and rescaling it gives $i\in D$ with $i^2=-1$. The centralizer $D_{+}$ of $i$ is a finite-dimensional division algebra over $\mathbb{C}=\mathbb{R}(i)$, with that field central. Every element has a split minimal polynomial over $\mathbb{C}$, hence belongs to $\mathbb{C}$, so $D_{+}=\mathbb{C}$.

Conjugation by $i$ is an involution, giving $D=D_{+}\oplus D_{-}$, where $D_{-}=\{x:ix=-xi\}$. If $D_{-}=0$, then $D=\mathbb{C}$, contrary to its center being $\mathbb{R}$. Choose $j\ne 0$ in $D_{-}$. For any $x\in D_{-}$, $x{j}^{-1}$ commutes with $i$, so $D_{-}=\mathbb{C}j$. Also $j^2\in \mathbb{C}$ and commutes with $j$, forcing $j^2\in \mathbb{R}$. It cannot be positive, since $(j-\sqrt{j^2})(j+\sqrt{j^2})=0$, or zero, since $D$ is a division algebra. Rescale to $j^2=-1$. ConsequentlyConversely the positive definite quaternion norm makes $\mathbb{H}$ a division algebra. The two possible central division representatives are therefore $\mathbb{R}$ and $\mathbb{H}$. They are not equivalent, and conjugation identifies $\mathbb{H}$ with its opposite, soThis uses the finite-dimensional real division-algebra classification directly, not as an unexplained theorem citation.

Finally let $k_s$ be a separable closure and $\Gamma =\mathrm{Gal}(k_s/k)$. The cohomological description isHere $k_s^{\times }$ has the discrete topology and its natural Galois action. For a finite Galois splitting field $L$, identify $A{\otimes }_{k}L$ with a matrix algebra. Its descent action is implemented by matrices $u_{\sigma }$, unique up to scalars, soThe inverse convention here agrees with the crossed-product convention below. The semilinear operators $u_{\sigma }\sigma$ commute with the descended algebra $A$; their crossed product has class $[A{]}^{-1}$ by the commuting-factor decomposition of the full endomorphism algebra. Inverting its scalar cocycle therefore gives the class $[A]$. Associativity gives the two-cocycle identity, and changing the representatives changes $c$ by a coboundary. Conversely a two-cocycle defines the crossed-product algebra of a Galois extension, with $u_{\sigma }a=\sigma (a)u_{\sigma }$ and $u_{\sigma }{u}_{\tau }=c(\sigma ,\tau )u_{\sigma \tau }$. It is central simple and split by $L$. Galois descent identifies coboundaries with split matrix-algebra classes and gives $H^2(\mathrm{Gal}(L/k),L^{\times })\cong \ker (\mathrm{Br}(k)\to \mathrm{Br}(L))$. Every central simple algebra has a finite separable, hence finite Galois, splitting field, yielding the displayed limit.

Write $d=\mathrm{deg}A$ for the degree of a central simple algebra, so ${\dim }_{K}A=d^2$. Choose a finite Galois splitting field of a central simple algebra $E/K$ for $A$ and an isomorphism $\varphi :A{\otimes }_{K}E\to M_d(E)$. The reduced norm isThe existence of a finite separable splitting field is a standard structural property of a central simple algebra; passing to its Galois closure supplies $E$.

Every $E$-algebra automorphism of $M_d(E)$ is inner. Here is the matrix-unit argument for the inner automorphisms of a matrix algebra result. For an automorphism $h$, choose $w\ne 0$ in the image of $h(E_{11})$, and set $w_i=h(E_{i1})w$. Then $h(E_{ij})w_k={\delta }_{jk}{w}_i$. The $w_i$ are nonzero and independent, and their number is $d$, so they form a basis. Relative to that basis $h$ acts on each matrix unit in the usual way. Thus $h(T)=PT{P}^{-1}$ for some $P\in {\mathrm{GL}}_d(E)$. Determinants are unchanged by conjugation, proving independence of $\varphi$.

For $\sigma \in \mathrm{Gal}(E/K)$, compare $\varphi$ with the isomorphism obtained by applying $\sigma$ to matrix entries and ${\sigma }^{-1}$ to the scalar factor of $A\otimes E$. Their difference is again inner. For $a\in A$, this shows $\sigma (\det \varphi (a))=\det \varphi (a)$. The determinant therefore belongs to $K$. In fact, for a $K$-basis $a_1,\dots ,a_{d^2}$, the same comparison shows that all coefficients ofare fixed by the Galois group, so the reduced norm is a homogeneous polynomial of degree $d$ over $K$. This coefficient argument also applies over finite fields, where equality merely as functions would not identify polynomials.

Finally, two finite splitting fields embed into a common finite splitting extension. Determinant commutes with scalar extension, and over that common extension the two matrix identifications differ by an inner automorphism. Hence the resulting polynomials agree over $K$. The reduced norm is independent of both the splitting field and the matrix identification. It is multiplicative, and $\mathrm{Nrd}(a)=0$ exactly when $a$ is not invertible: a matrix with nonzero determinant is invertible after scalar extension, and invertibility descends by the invertibility of the $K$-linear multiplication map. In a division algebra the only element of reduced norm zero is zero.

Every Brauer class has a central division algebra representative $D$, and every central simple algebra is a matrix algebra over such a $D$. Put $d=\mathrm{deg}D$; its reduced norm is a homogeneous degree-$d$ polynomial on the $d^2$-dimensional vector space $D$.

If $d>1$, then $d^2>d$, so the $C_1$ property supplies a nonzero $x\in D$ with $\mathrm{Nrd}(x)=0$. But every nonzero element of a division algebra is invertible, and its reduced norm cannot be zero. Thus $d=1$, and a central division algebra of degree one is $K$ itself. Every central simple algebra over $K$ is therefore a matrix algebra over $K$, givingThis proves the trivial Brauer group of a C1 field. The essential ingredient is the anisotropy of a division algebra's reduced norm, not the norm polynomial of an arbitrary matrix algebra, which certainly can vanish on nonzero singular matrices.

Let $G=\mathrm{Gal}(L/K)$. A semilinear action on an $L$-vector space $V$ is an additive action with $\sigma (av)=\sigma (a)\sigma (v)$. The Galois descent of vector spaces theorem says that $V^G$ is a $K$-vector space and the natural mapis an isomorphism. In particular ${\dim }_K{V}^G={\dim }_{L}V$. Morphisms also descend: an $L$-linear equivariant map is the scalar extension of its restriction to invariant subspaces.

We prove that the invariant vectors span $V$ over $L$, without dividing by $|G|$. For $a\in L$ and $v\in V$, the vectoris invariant, because every $\tau \in G$ merely permutes the terms. Suppose their $L$-span were proper. There would be a nonzero $L$-linear functional $\lambda$ annihilating that span. For each fixed $v$, we would haveThe Artin independence theorem makes all coefficients zero, in particular $\lambda (v)=0$ for every $v$, a contradiction. To recall its elementary proof, take a nontrivial linear relation between distinct field automorphisms with the fewest nonzero coefficients. Replace the input $a$ by $ab$ and subtract one automorphism's value on $b$ times the original relation. Choose $b$ on which two participating automorphisms differ. The resulting nonzero relation has fewer terms, contradicting minimality.

Choose an $L$-basis $v_1,\dots ,v_m$ consisting of invariant vectors. If $v={\sum }_i{a}_i{v}_i$ is invariant, uniqueness of coordinates gives $\sigma (a_i)=a_i$ for every $\sigma$, so every $a_i$ lies in $K$. These vectors are therefore a $K$-basis of $V^G$, proving the displayed map is an isomorphism. Restriction and scalar extension are visibly inverse on equivariant morphisms, completing the descent theorem. The argument remains valid when the characteristic divides $|G|$.

Now put $B=M_n(L)$, and let ${\sigma }_0$ act on $B$ by applying $\sigma$ to each entry. The group ${\mathrm{Aut}}_L(B)$ is a $G$-group under $\sigma (h)={\sigma }_{0}h{\sigma }_0^{-1}$. Its nonabelian first cohomology consists of cocycles $c_{\sigma }$ satisfyingmodulo $c_{\sigma }^{\prime }=b{c}_{\sigma }\sigma (b{)}^{-1}$ for $b\in {\mathrm{Aut}}_L(B)$. This convention is equivalent to the usual one after replacing $b$ by its inverse. The matrix-automorphism calculation above identifies this coefficient group with ${\mathrm{PGL}}_n(L)$.

For a central simple algebra $A$ of degree $n$ split by $L$, choose $\varphi :A{\otimes }_{K}L\to B$. Transport the natural semilinear action to $T_{\sigma }=\varphi (1\otimes \sigma ){\varphi }^{-1}$ and setIt is $L$-linear and multiplicative, and $T_{\sigma }{T}_{\tau }=T_{\sigma \tau }$ gives the cocycle identity. Replacing $\varphi$ by $b\varphi$ changes $c$ by the displayed coboundary equivalence. Isomorphic $K$-algebras give the same class, so this defines the descent classification of central simple algebras map naturally.

Conversely, a cocycle defines a semilinear algebra action $T_{\sigma }=c_{\sigma }{\sigma }_0$. LetThe vector-space descent theorem gives $A_c{\otimes }_{K}L\cong B$, with the natural map also respecting multiplication and the identity. Hence ${\dim }_K{A}_c=n^2$. If $I$ is a nonzero two-sided ideal of $A_c$, then $I{\otimes }_{K}L$ is a nonzero ideal of $B$, so it is all of $B$. Dimensions force $I=A_c$, proving simplicity. If $z$ is central in $A_c$, it commutes with its $L$-span $B$, and thus is a scalar matrix with scalar $a\in L$. Twisted invariance of a scalar matrix means $\sigma (a)=a$, so $a\in K$. The center is exactly $K$. Thus $A_c$ is central simple, of degree $n$, and split by $L$.

Equivalent cocycles have conjugate semilinear actions: $T_{\sigma }^{\prime }=b{T}_{\sigma }{b}^{-1}$, so $b$ restricts to a $K$-algebra isomorphism between the fixed algebras. In the other direction, an isomorphism of fixed algebras extends to an $L$-algebra automorphism of $B$ and intertwines the two actions, giving exactly that equivalence relation. Starting from $A$ recovers the fixed algebra of $A\otimes L$, namely $A$; starting from a cocycle recovers its action. ThereforeThis is a bijection of pointed sets, with the split algebra corresponding to the trivial cocycle. Fixed-degree algebra classes do not carry the Brauer group's tensor-product group law, so a pointed-set interpretation is the appropriate meaning of the isomorphism here.

The Artin–Wedderburn theorem says that every central simple algebra $A$ over $k$ has the form $M_r(D)$ for a finite-dimensional central division algebra $D$, uniquely up to the evident data. If $A$ and $B$ are central simple, extend scalars to an algebraic closure $\bar{k}$. Both become full matrix algebras, hencefor suitable $a,b$. Any nonzero proper ideal of $A{\otimes }_{k}B$ would extend to one in this simple matrix algebra, and faithful flatness prevents it from vanishing or becoming the whole algebra. The same scalar-extension argument shows that the center is $k$. This proves the tensor product of central simple algebras theorem.

The Brauer group $\mathrm{Br}(k)$ consists of Morita equivalence classes of central simple $k$-algebras. Its product is $[A][B]=[A{\otimes }_{k}B]$, its identity is $[k]$, and $[A{]}^{-1}=[A^{\mathrm{op}}]$ because $A{\otimes }_k{A}^{\mathrm{op}}\cong {\mathrm{End}}_k(A)$ is a full matrix algebra.

Let $L/k$ be a Finite Galois extension with Galois group $G_L$, and let $\varphi :G_L\times G_L\to L^{\times }$ be a normalized two-cocycle. The crossed-product algebra of a Galois extension has underlying left $L$-vector spaceand multiplicationThe cocycle identity is exactly associativity. After scalar extension to $L$, the algebra acts by the twisted regular representation and becomes $M_{|G_L|}(L)$; Galois descent shows that it is central simple over $k$. If $\varphi$ is multiplied by the coboundary of a one-cochain $b_{\sigma }$, rescaling $u_{\sigma }$ by $b_{\sigma }$ gives an isomorphic algebra. Hence the cohomological construction of a Brauer class gives a well-defined map

It remains to show that every Brauer class is torsion. For a finite group $G$, restriction and corestriction on normalized bar cochains satisfyon cohomology: the first equality follows by summing the translated cochain over coset representatives, and each $g^{\ast }$ is the identity because an inner automorphism is cochain-homotopic to the identity. Restriction to the trivial subgroup is zero in positive degree, so this proves that finite-group cohomology is annihilated by the group order. In multiplicative notation, every $\alpha \in H^2(G_L,L^{\times })$ therefore satisfies ${\alpha }^{|G_L|}=1$.

By the permitted assumption, $[A]$ is the image of such an $\alpha$ for some $L$. Consequently $[A{]}^{|G_L|}=1$ in $\mathrm{Br}(k)$. By the definition of Brauer equivalence, this says that for some $n$,as required.

For a field of characteristic other than two and nonzero $a,b$, the quaternion algebra has basis $1,i,j,ij$ with the displayed relations. It is a four-dimensional central simple algebra. After adjoining a square root of $a$, diagonalizing $i$ and taking $j=\left(\begin{array}{cc}0& b\\ 1& 0\end{array}\right)$ identifies it with $M_2$. Quaternion conjugation fixes scalars and negates $i,j,ij$, giving norm $x_0^2-a{x}_1^2-b{x}_2^2+ab{x}_3^2$. An element is invertible exactly when its norm is nonzero. Thus the quaternion algebra is either a division algebra or a split matrix algebra. The familiar quaternion algebra is $(-1,-1{)}_{\mathbb{R}}$.

The reduced norm of an element of a central simple algebra of degree $d$ is its determinant after any splitting isomorphism to $M_d$. Inner automorphisms of a matrix algebra preserve that determinant, while Galois invariance descends its polynomial coefficients to the center. It is a multiplicative homogeneous degree-$d$ polynomial, vanishing exactly on noninvertible elements. On a central division algebra it is anisotropic.

An extension $E/K$ splits a central simple algebra $A$ of degree $d$ when $A{\otimes }_{K}E\cong M_d(E)$. A finite separable splitting field exists. Splitting is preserved by further scalar extension and is equivalent to the vanishing of the restricted Brauer group class.