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The Jacobson radical $J=J(R)$ is the intersection of all maximal right ideals, equivalently the largest ideal annihilating every simple right module. The Artin–Wedderburn theorem givesbecause $k$ is algebraically closed.

The descending chain $J\supseteq J^2\supseteq \cdots$ stabilizes since $R$ is finite-dimensional. If $J^m=J^{m+1}$, Nakayama lemma applied to the finite right module $J^m$ gives $J^m=0$. Thus $J$ is nilpotent.

For $f\in {\mathrm{End}}_R(P_i)$, the Fitting lemma givesfor large $n$. Indecomposability makes one summand zero, so $f$ is either invertible or nilpotent. In the latter case $1-f$ is invertible. This is the criterion that ${\mathrm{End}}_R(P_i)$ is a local ring.

Let $M={⨁}_i{P}_i$. Reduction modulo $MJ$ definesSince $M/MJ={⨁}_i{S}_i$ and the $S_i$ are pairwise nonisomorphic simples,by Schur lemma. Arbitrary scalars on the direct summands lift to scalar identity maps on the $P_i$, so $\theta$ is surjective.

If $f\in \ker \theta$, then $f(M)\subseteq MJ$. A product of $r$ such maps sends $M$ into $M{J}^r$, so $\ker \theta$ is nilpotent. A nilpotent ideal lies in the Jacobson radical, while the semisimplicity of the quotient $k^n$ gives the reverse inclusion. HenceThis is exactly the definition of a basic algebra.

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.