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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.

A central division algebra of degree $d>1$ has an anisotropic reduced norm of degree $d$ in $d^2>d$ variables. That contradicts the C1 field property. Thus only degree one is possible, and every central simple algebra is split.