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For the preliminary definitions, a block is $Re$ for a primitive central idempotent $e$, and a module belongs to it when $eM=M$. For a semisimple algebra, the Artin–Wedderburn theorem identifies its blocks with the factors $M_{n_i}(D_i)$ in its product decomposition. Each factor is a block of a finite-dimensional algebra.

We prove Ext separation of finite-length modules. The extension hypothesis is ${\mathrm{Ext}}^1(S,T)={\mathrm{Ext}}^1(T,S)=0$ for simples in opposite sets. First, for a simple $S\in {\mathcal{C}}_1$ and a module $N$ of finite composition length with factors in ${\mathcal{C}}_2$, we have ${\mathrm{Ext}}^1(S,N)=0$. Induct on the length of $N$: for $0\to N^{\prime }\to N\to T\to 0$ with $T$ simple, the long exact sequence of the Ext functor containswhose outer groups are zero. The same argument works with the two sets interchanged.

Now induct on the length of $M$, with $M=0$ immediate. Take a simple quotient in $0\to N\to M\to S\to 0$. By induction, $N=N_1\oplus N_2$ with factors in the respective sets. Suppose $S\in {\mathcal{C}}_1$; the other case is symmetric. Quotienting by $N_1$ givesThis splits by the preceding Ext vanishing. Let $U_1$ be the inverse image in $M$ of the chosen complementary copy of $S$. Then $U_1\cap N_2=0$, $U_1+N_2=M$, and $0\to N_1\to U_1\to S\to 0$. Thus $U_1$ has only factors in ${\mathcal{C}}_1$, while $U_2=N_2$ has only factors in ${\mathcal{C}}_2$.

If two finite-length modules have factors in disjoint sets, any homomorphism between them is zero: a nonzero image would, by the Jordan–Hölder theorem, have a simple factor belonging to both sets. For any submodule $V$ of $M$ with factors in ${\mathcal{C}}_1$, projection onto $U_2$ is consequently zero, so $V\subset U_1$. The analogous argument applies to $U_2$. HenceIn particular both summands are preserved by every endomorphism of $M$.

If $P$ is a prime ideal of the commutative Artinian ring $R$, then $R/P$ is an Artinian domain. For $0\ne x\in R/P$, the descending chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes, so $x^n=x^{n+1}a$ for some $n$; cancellation gives $xa=1$. Thus $R/P$ is a field and $P$ is maximal.

There are only finitely many maximal ideals. Otherwise, distinct maximal ideals $P_i$ are pairwise comaximal and their finite products give the strictly descending chaincontradicting the Artinian condition. This proves the statement about the prime ideals of a commutative Artinian ring.

The nilradical $N$ is the ideal of all nilpotent elements, equivalently the intersection of all prime ideals. Its powers stabilize, say $N^r=N^{r+1}=I$. If $I\ne 0$, choose an ideal $J$ minimal subject to $IJ\ne 0$, then choose $x\in J$ with $Ix\ne 0$. Since $I^2=I$, minimality gives $Rx=Ix$, so $x=ax$ for some $a\in I$. But $a$ is nilpotent, hence $1-a$ is a unit, contradicting $(1-a)x=0$. Therefore the nilradical of a commutative Artinian ring is nilpotent.

If the maximal ideals are $P_1,\dots ,P_s$, the Chinese remainder theorem givesa finite product of fields. Each layer $N^j/N^{j+1}$ is an Artinian module over this product and therefore finite-dimensional. Since $N$ is nilpotent, these finitely many layers show that every ideal of $R$ is finitely generated. This proves the Artinian commutative ring is Noetherian theorem.

Now assume $R$ is local with maximal ideal $P$ and ${\dim }_{R/P}P/P^2=1$. The Nakayama lemma gives $P=(x)$. Since $P$ is nilpotent, its powers form a finite chain ending in zero. For a nonzero ideal $J$, choose the largest $r$ with $J\subseteq (x^r)$ and take $y\in J\setminus (x^{r+1})$. Write $y=x^{r}u$; then $u\notin P$, so $u$ is a unit. Henceand every ideal of $R$ is principal.

Finally, the Artin–Wedderburn theorem says that a finite-dimensional semisimple $k$-algebra has the formwhere each $D_i$ is a finite-dimensional division ring over $k$; the factors are unique up to permutation and isomorphism. By the assumed complete reducibility, write the right regular module aswith pairwise nonisomorphic simple right modules $S_i$. The Schur lemma says that $D_i={\mathrm{End}}_T(S_i)$ is a division ring and that homomorphisms between distinct $S_i$ vanish. Left multiplication and the endomorphism ring of the displayed direct sum therefore giveConversely, the regular module of $M_n(D)$ is a direct sum of $n$ simple column modules, proving that every algebra on the right is semisimple. Uniqueness follows from uniqueness of the simple summands and their multiplicities.

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.

A nonzero two-sided ideal of the classical right ring of quotients contains a nonzero element of the original ring: multiply a nonzero fraction by its regular right denominator. If the original ring is prime, the contractions of any two such ideals have nonzero product, so the quotient ring is prime too. The Goldie theorem makes it semisimple Artinian, and the Artin–Wedderburn theorem expresses it as a finite product of matrix rings over division rings. Primeness excludes more than one factor.

Repeated coordinates contribute zero. Sorting a distinct tuple and subtracting the staircase gives a partition of $m$, and its $m!$ orderings have identical squared summands. The beta-set hook-product identity turns each summand into $(\dim S^{\lambda }/m!{)}^2$. Summing and using the Artin–Wedderburn theorem for $\mathbb{C}{S}_m$ gives $m!{\sum }_{\lambda ⊢m}(\dim S^{\lambda }{)}^2/(m!{)}^2=1$. This is a normalized group algebra dimension identity.