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A block of a finite-dimensional associative algebra $R$ is a two-sided ideal $Re$ associated with a primitive idempotent $e$ in the center of an associative algebra. Its identity is $e$, and $R$ is the product of these blocks. An $R$-module belongs to this block if $eM=M$. For a semisimple algebra, the Artin–Wedderburn theorem says the blocks are its full matrix algebras over division rings. For a group algebra this specializes to a block of a group algebra.

For a nonzero finite-dimensional complex vector space $V$, a unital matrix algebra acting irreducibly on $V$ is the full endomorphism algebra. Its Jacobson radical annihilates $V$: otherwise the radical times $V$ is all of $V$, contradicting nilpotence. Faithfulness kills the radical, and the Artin–Wedderburn theorem then leaves one simple matrix factor acting on its standard column module. This is a matrix-algebra theorem, distinct from the orbit-counting Burnside lemma and the finite-group Burnside's theorem.

The matrix-block sizes in the Artin–Wedderburn theorem for $\mathbb{C}[G]$ recover the irreducible character degrees with their multiplicities. They determine the degree-square sum and the number $|G:G^{\prime }|$ of linear characters. Isaacs' character-degree criterion for p-nilpotency therefore detects each normal p-complement and applying it for every prime also detects nilpotent groups. The set of distinct degrees without multiplicities does not contain the same information.

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.

Write $o(\theta )$ for the order of the determinant character of an irreducible character $\theta$. An extension has the same degree and restricts exactly to $\theta$, rather than just containing it as a constituent.

Extension theorem. Thompson's coprime character extension theorem says that if $N◃G$, $\theta \in \mathrm{Irr}(N)$ is $G$-invariant, and $\gcd (\theta (1),[G:N])=1$, then $\theta$ extends to $G$ exactly when its determinant character extends to a linear character of $G$. In particular, if also $\gcd (o(\theta ),[G:N])=1$, there is a unique extension with determinant order coprime to the index. For a p-group quotient, the version used below gives a canonical extension whenever both the degree and the determinant order are prime to $p$. This theorem is an assertion of extendibility, not the false statement that every invariant irreducible character extends.

The p-nilpotency lemma. Definewhere $m_p$ is the largest power of $p$ dividing an integer $m$. Isaacs' character-degree criterion for p-nilpotency assertsEquivalently, $s_p(G{)}_p=h_p(G)$. We prove both the divisibility and the criterion.

Let $N=O^p(G)$ be the p-residual of a finite group, the least normal subgroup with p-group quotient. It can also be described as the subgroup generated by all p-prime-order elements: every homomorphism to a p-group kills those elements, and after quotienting by their generated subgroup the image of every element has p-power order, so the quotient is a p-group by Cauchy's theorem.

The residual is p-perfect, meaning $O^p(N)=N$. Indeed $O^p(N)$ is characteristic in $N$, hence normal in $G$, and $G/O^p(N)$ is an extension of the two p-groups $N/O^p(N)$ and $G/N$. Minimality of $N$ forces equality. Consequently $N/N^{\prime }$ has order prime to $p$, since otherwise it would have a quotient of order $p$. Every linear character of $N$, in particular every determinant character of $N$, therefore has order prime to $p$.

Let $\mathcal{T}$ consist of the $G$-invariant irreducible characters $\theta$ of $N$ of degree prime to $p$. Each extends to an irreducible character $\hat{\theta }$ of $G$ by the extension theorem. The irreducible characters of $G$ lying over $\theta$ are exactlyHere is the representation argument for this description. If a representation lies over the invariant $\theta$, its $N$-module is copies of the module for $\theta$. Identify it with that module tensored with its multiplicity space. Dividing the $G$-action by the fixed extending action on the first factor leaves an action on the multiplicity space, trivial on $N$. Irreducibility is equivalent to irreducibility of that quotient action, proving the displayed parametrization.

Every irreducible degree of the p-group $G/N$ is a power of $p$, since it divides the group order. Thus the p-prime-degree characters lying over $\theta$ are exactly those with $\nu$ linear, and there areof them. For the last equality, $G/N{G}^{\prime }$ is the largest abelian p-group quotient of $G$, hence the p-primary part of its abelianization. In particular all these characters have degree $\theta (1)$.

Conversely, for any $\chi \in \mathrm{Irr}(G)$ of p-prime degree, Clifford theory writes its restriction to $N$ as $e$ times the sum of a single orbit of irreducible characters. The orbit size divides $[G:N]$ and also divides $\chi (1)$, so it is one; the constituent $\theta$ is invariant and has p-prime degree. Hence $\theta \in \mathcal{T}$. The preceding extension parametrization now forces the quotient character to be linear. We have therefore obtained the exact identityThis proves the required divisibility without assuming p-nilpotency.

The p-group $G/N$ acts on $\mathrm{Irr}(N)$ by conjugation. Every nonfixed orbit has size divisible by $p$, and degrees are constant on each orbit. Fixed characters whose degrees are divisible by $p$ also contribute zero modulo $p$. The sum of squares of irreducible degrees consequently givesThe quotient is prime to $p$ exactly when $N$ has order prime to $p$. Since $G/N$ is a p-group, that is exactly the assertion that $N$ is a normal p-complement. Conversely, any normal p-complement equals $O^p(G)$ because it contains all p-prime-order elements and has p-group quotient. This finishes the lemma.

Detection from the group algebra. For finite groups $G$ and $H$ with isomorphic complex group algebras, the Artin–Wedderburn theorem givesThe simple matrix-block sizes and their multiplicities are determined by the algebra isomorphism. Hence $G$ and $H$ have identical irreducible degree multisets. They have the same order, from the degree-square sum, and the same number of linear characters, which equals $|G:G^{\prime }|$. Thus $s_p$ and $h_p$ coincide for every prime $p$. The lemma shows that one group is p-nilpotent exactly when the other is.

A finite group is nilpotent exactly when it has a normal p-complement for every prime dividing its order. For completeness, if all those complements exist, set $L_q={\bigcap }_{p\ne q}{K}_p$, where $K_p$ is the normal p-complement. The diagonal map $G/L_q\to {\prod }_{p\ne q}G/K_p$ is injective, so $[G:L_q]$ is prime to $q$. At the same time, containment in each $K_p$ removes every prime other than $q$ from $|L_q|$. Thus $L_q$ is a normal Sylow q-subgroup. Distinct normal Sylow subgroups commute, giving the direct-product decomposition. Conversely, the direct-product decomposition of a finite nilpotent group provides all the complements. We concludeThis is the assertion that the complex group algebra detects p-nilpotency and hence nilpotency. It needs the multiplicities of the degrees, not just their distinct values.

Normal-complement consequence. If $p$ divides every nonlinear irreducible degree, the p-prime-degree characters are exactly the linear characters. ThereforeThe lemma yields a normal p-complement immediately:This proves Thompson's character-degree normal complement theorem, including the abelian case where the hypothesis on nonlinear characters is vacuous.

Let $R$ be a commutative Artinian ring. We prove finite composition length first, without assuming the conclusion that $R$ is Noetherian.

Every prime ideal is maximal. Indeed, an Artinian integral domain is a field: for $a\ne 0$, stabilization of $(a^m)$ gives $a^m=a^{m+1}b$, and cancellation gives $ab=1$. There are only finitely many maximal ideals. To see this, choose a minimal finite intersection $L={\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_s$ using the descending chain condition. Minimality gives $L\subseteq \mathfrak{m}$ for any other maximal ideal $\mathfrak{m}$. Since ${\mathfrak{m}}_1\cdots {\mathfrak{m}}_s\subseteq L$, primeness forces one ${\mathfrak{m}}_i\subseteq \mathfrak{m}$, and maximality makes them equal. Thus the Jacobson radical $J$ is this finite intersection, and the Chinese remainder theorem gives

We also need nilpotence of $J$. Its powers stabilize; put $I=J^m$ at a stabilization stage, so $I^2=I$. If $I\ne 0$, choose an ideal $L$ minimal subject to $IL\ne 0$. Since $I(IL)=IL\ne 0$, minimality gives $IL=L$. Choose $a\in L$ with $Ia\ne 0$. Minimality applied to the subideal $Ra$ gives $L=Ra$. The equality $IL=L$ then yields $a=xa$ for some $x\in I\subseteq J$. But $1-x$ is a unit by the unit criterion for the Jacobson radical; hence $a=0$, a contradiction. Therefore $J^m=0$.

Each layer $J^i/J^{i+1}$ is an Artinian module over the finite product of fields $R/J$. Its components over those fields are finite-dimensional: an infinite-dimensional vector space has a strictly descending chain of subspaces. Each layer consequently has finite composition length. The finite filtrationproves that $R$ itself has finite composition length, and therefore satisfies the ascending chain condition on ideals. Every commutative Artinian ring is Noetherian. The zero ring satisfies the conclusion trivially.

The relevant Artin–Wedderburn theorem says that a semisimple ring which is a right Artinian ring has the formwhere the $D_i$ are division rings and the positive integers $n_i$ and division rings are unique up to permutation and isomorphism. The division rings need not be commutative or finite-dimensional over a field. Conversely every such finite product is semisimple and a right Artinian ring.

Here is a proof which also works if semisimplicity is initially expressed as $J(R)=0$. A nonzero right ideal contains a minimal nonzero right ideal $L$ by the descending chain condition; $L$ is a simple module. If $L^2=0$, then for every $l\in L$ and $r\in R$, $(lr{)}^2=0$, so $1-lr$ has inverse $1+lr$. The unit criterion for the Jacobson radical would give $L\subseteq J(R)=0$, a contradiction. Thus some $a\in L$ induces a nonzero map $L\to L$ by left multiplication. This map is an isomorphism, since $L$ is simple. Choose $e\in L$ with $ae=a$. Then $a(e^2-e)=0$ and injectivity on $L$ gives $e^2=e$. We obtain $L=eR$ and the splittingFor a right ideal $K$ containing $e$, the same argument splits $K=eR\oplus (K\cap (1-e)R)$. Repeatedly split a minimal right ideal out of the remaining summand. The remaining summands form a strictly descending chain of right ideals, so the process terminates. Thusfor pairwise nonisomorphic simple modules $S_i$.

A nonzero map between simple modules is an isomorphism: its kernel and image leave no alternatives. Thus $D_i={\mathrm{End}}_R(S_i)$ is a division ring, and ${\mathrm{Hom}}_R(S_i,S_j)=0$ for $i\ne j$. This is the needed Schur lemma. Taking endomorphisms of the direct sum gives ${\prod }_i{M}_{n_i}(D_i)$. Finally, left multiplication identifies $R$ with ${\mathrm{End}}_R(R_R)$: a right-module endomorphism sends $r$ to $f(1)r$, and composition corresponds to multiplication in $R$, not its opposite. This proves the displayed ring isomorphism.

For the converse, the right regular module of $M_n(D)$ is the direct sum of its $n$ row modules. Each row module is simple, since elementary matrices and nonzero scalar inverses send any nonzero row to any prescribed row. A finite product therefore has a finite semisimple regular module and is a right Artinian ring. The simple-module multiplicities and their endomorphism division rings recover the factors, establishing uniqueness.

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.

A right Ore localization at $S$ is a unital homomorphism $\iota :R\to Q$ such that every $\iota (s)$, $s\in S$, is a unit, every element of $Q$ is $\iota (r)\iota (s{)}^{-1}$, andEquivalently, the usual fraction construction has this zero criterion and is universal among homomorphisms that invert $S$. Injectivity is required only when the denominators are regular elements of a ring. If $0\in S$, every homomorphism inverting $S$ has zero target; this is the degenerate zero-ring localization. For a nonzero localization we require $0\notin S$.

The necessary and sufficient conditions are that $S$ be a right denominator set:

For necessity, write the element $\iota (s{)}^{-1}\iota (r)$ as $\iota (b)\iota (t{)}^{-1}$. Multiplying by $\iota (s)$ on the left and $\iota (t)$ on the right gives $\iota (rt-sb)=0$. The zero criterion supplies $u\in S$ with $(rt-sb)u=0$. Hence $r(tu)=s(bu)$, the right Ore condition in $R$, not merely an equality of images in $Q$. Also $sr=0$ makes $\iota (r)=0$ because $\iota (s)$ is invertible, and the zero criterion gives $rt=0$. This proves both necessities.

The sufficiency is the Ore theorem for denominator sets. Concretely, right fractions can be represented by $(r,s)$; a common-denominator equality has $su=tv\in S$ and $ru=r^{\prime }v$ for suitable $u,v\in R$. The two conditions make the induced fraction equivalence transitive and make addition and multiplication well defined; the zero fraction is exactly the kernel displayed above. The requested direction of the theorem was necessity, for which the complete argument has just been given.

For a right Noetherian domain, any two nonzero principal right ideals intersect. To see this without using the desired localization, suppose $bR\cap aR=0$ with $a,b\ne 0$. The sumis direct: reduce a finite relation modulo $aR$ to get its first summand zero, then cancel $a$ in the noncommutative domain and repeat. Every summand is nonzero, so the finite partial sums form a strictly ascending chain of right ideals, contradicting the ascending chain condition. A nonzero intersection $rR\cap sR$ thus provides $rt=sb\ne 0$, with $t\ne 0$. This is the right Ore condition for $S=R\setminus \{0\}$. Reversibility is automatic in a noncommutative domain, and the zero criterion makes the map injective. The localization is a division ring, since a nonzero $r{s}^{-1}$ has inverse $s{r}^{-1}$.

The Goldie theorem states that a semiprime ring has a semisimple Artinian classical right ring of quotients if and only if it has finite right uniform dimension and satisfies the ascending chain condition on right annihilators. A right Noetherian ring satisfies the annihilator condition because right annihilators are right ideals. Its right regular module has finite uniform dimension: an infinite direct sum of nonzero submodules would give a strictly ascending chain of finite partial sums. A prime ring is semiprime, so the theorem applies to a prime right Noetherian ring $R$.

To obtain a single matrix-ring factor, not just a finite product, prove that its quotient $Q$ is prime. A nonzero two-sided ideal $I\subseteq Q$ contains a nonzero fraction $r{s}^{-1}$ and hence the nonzero element $r=(r{s}^{-1})s\in I\cap R$. The same holds for any other nonzero ideal $K$. Since $R$ is prime, $(I\cap R)(K\cap R)\ne 0$, so $IK\ne 0$. Thus $Q$ is prime. The Artin–Wedderburn theorem gives a finite product of matrix rings over division rings; primeness excludes two different nonzero factors with zero product. This proves the prime classical quotient of a prime right Goldie ring:For the requested counterexample to Noetherianity, take $R=k[x_1,x_2,\dots ]$. It is a commutative integral domain and has its field of fractions as its classical right ring of quotients. But $(x_1)⊊(x_1,x_2)⊊\cdots$ is a strictly ascending ideal chain: evaluation at $x_1=\cdots =x_m=0$ shows $x_{m+1}$ is not in the preceding ideal. Therefore existence of a classical quotient does not imply right Noetherianity.

The ring form of the Artin–Wedderburn theorem states that the following conditions on a unital ring $R$ are equivalent: $R$ is a semisimple ring; $R$ satisfies the descending chain condition on left ideals and $J(R)=0$; andwhere $s$ is finite, each $n_i\ge 1$, and each $D_i$ is a division ring. The factors are unique up to permutation and isomorphism. The zero ring may be included as the empty product. This theorem also implies the corresponding right-handed statements.

First prove the implication from the chain condition and zero Jacobson radical. Among finite intersections of maximal left ideals, choose a minimal one, say $L=L_1\cap \cdots \cap L_t$. Such a minimal intersection exists by the descending chain condition. Intersecting with any other maximal left ideal cannot make $L$ smaller, so $L$ is contained in every maximal left ideal. Thus $L\subseteq J(R)=0$. It follows that the map of left modulesis injective. Each quotient is a simple module.

For completeness, a submodule $N$ of a finite direct sum $V={⨁}_j{T}_j$ of simple modules is itself a finite direct sum of simple modules and is a direct summand of $V$. Choose a sum $W$ of coordinate summands, maximal among those with $N\cap W=0$. For any omitted coordinate summand $T_j$, simplicity gives $T_j\cap (N+W)=0$ or $T_j\subseteq N+W$. In the first case $N\cap (W\oplus T_j)=0$, contradicting maximality; hence every coordinate summand lies in $N+W$. Therefore $V=N\oplus W$. Projection onto $N$ shows $N$ is a finite sum of images of the $T_j$, each image zero or simple. Add these images successively, discarding any image already in the sum: simplicity makes each new intersection zero. This yields a direct-sum decomposition. Applying this argument to the displayed injection proves that the left regular module is a finite sum of simple modules.

Conversely, if the regular module is a direct sum of simple modules, that sum is finite because $1$ lies in a finite subsum and generates the whole regular module. Every left ideal, being a submodule, has the preceding decomposition and splitting property. Its chain lengths are bounded by the number of simple summands, so the descending chain condition holds. The Jacobson radical annihilates every simple summand of the regular module, hence annihilates $1$ and is zero. This identifies the first two conditions.

Write the finite decomposition using pairwise nonisomorphic simple left modules:Every simple module is a quotient of $R$, by sending $1$ to any nonzero vector, so this list includes every simple isomorphism class. The Schur lemma gives ${\mathrm{Hom}}_R(S_i,S_j)=0$ for $i\ne j$, and $E_i={\mathrm{End}}_R(S_i)$ is a division ring: any nonzero endomorphism has zero kernel and full image by simplicity. ConsequentlyEvery endomorphism of the left regular module is right multiplication by its value on $1$. Composition reverses multiplication of those values, so the left side is $R^{\mathrm{op}}$. Taking the opposite ring, and using transpose to identify $M_n(E{)}^{\mathrm{op}}$ with $M_n(E^{\mathrm{op}})$, gives the theorem's decomposition with $D_i=E_i^{\mathrm{op}}$. This keeps the left/right conventions explicit.

Conversely, the standard column module $D^n$ over $M_n(D)$ is simple: a nonzero coordinate of a vector can be inverted to construct a matrix sending it to any prescribed vector. The left regular module decomposes as the sum of its $n$ column ideals, each isomorphic to $D^n$. A finite product of these matrix rings is therefore a semisimple ring. This proves all implications. For uniqueness, the isomorphism classes of simple modules and their multiplicities in the regular module determine the numbers $n_i$ and the division rings $E_i^{\mathrm{op}}$. The multiplicity can also be recovered as the dimension of ${\mathrm{Hom}}_R(S_i,R)$ as a right $E_i$-module, so it does not depend on the chosen decomposition.

Now let $A=\mathbb{C}[G]$ for the finite group $G$. It has dimension $|G|$, with the elements of $G$ as a basis, so its left ideals satisfy the descending chain condition by their dimensions. The permitted assumption $J(A)=0$ makes the Artin–Wedderburn theorem applicable. Its decomposition is a complex-algebra decompositionwith each $D_i$ finite-dimensional over the central copy of $\mathbb{C}$. Each such division algebra is just $\mathbb{C}$. Indeed, for $d\in D_i$, finite dimensionality gives a nonzero polynomial relation $p(d)=0$ with complex coefficients. Factor $p$ into linear factors over $\mathbb{C}$. Their product at $d$ is zero, so the absence of zero divisors forces $d-\lambda =0$ for some scalar root $\lambda$. Thus every element is scalar.

The pairwise nonisomorphic simple modules of ${\prod }_i{M}_{n_i}(\mathbb{C})$ are the column modules ${\mathbb{C}}^{n_i}$, with all other factors acting as zero. Hence ${\dim }_{\mathbb{C}}{S}_i=n_i$. Taking complex dimensions gives

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.

A partition of an integer $n$ is a finite sequence ${\lambda }_1\ge \cdots \ge {\lambda }_{\ell }>0$ with sum $n$; append zeros when comparing lengths. Its Young diagram has cells $(i,j)$ with $1\le j\le {\lambda }_i$. A Young tableau is a bijective filling of these cells by $1,\dots ,n$. Write $R_t,C_t$ for its row and column stabilizers, and fix the conventionPermutations act on entries on the left, with the rightmost factor acting first. This defines the Young symmetrizer used throughout. In the dictionary order on integer partitions, $\lambda >\mu$ means that at the first differing part ${\lambda }_i>{\mu }_i$; equality is allowed in $\ge$.

Suppose there is no row-column collision between $t$ of shape $\lambda$ and $u$ of shape $\mu$. Each column of $u$ contains at most one entry from each row of $t$. The first $k$ rows of $t$ therefore contain at mostentries, where ${\mu }_j^{\prime }$ is a column length. Thus ${\sum }_{i\le k}{\lambda }_i\le {\sum }_{i\le k}{\mu }_i$ for every $k$: $\mu$ dominates $\lambda$ in dominance order on partitions. If $\lambda \ge \mu$ in dictionary order on integer partitions, a first strictly larger part would contradict the corresponding prefix inequality. Hence $\lambda =\mu$.

All the bounds must now be equalities. Every column $j$ of $u$ contains exactly one entry from row $i$ of $t$ whenever $i\le {\lambda }_j^{\prime }$. Choose $r_0\in R_t$ sending that entry to the entry of $t$ in cell $(i,j)$. These prescriptions are bijections within the rows. The columns of $r_{0}u$ then have exactly the same sets of entries as the columns of $t$, so $r_{0}u=c_{0}t$ for some $c_0\in C_t$. ConsequentlyIf a collision was present instead, its two entries supply the first alternative. This proves the row-column collision lemma, including the prescribed order of the two stabilizer factors.

We next prove the Specht module classification. By Maschke's theorem, the group algebra $A=\mathbb{C}{S}_n$ is a semisimple algebra. Use the permitted basic quasi-idempotence of a Young symmetrizer,and put $e_t=h_t/H_{\lambda }$. The coefficient of $1$ in $h_t$ is one, because $R_t\cap C_t=\{1\}$, so $e_t\ne 0$.

For $\pi \in S_n$, consider $r_t\pi c_t$. A collision between the rows of $t$ and the columns of $\pi t$ gives a transposition $\tau \in R_t\cap \pi C_t{\pi }^{-1}$. Row symmetrization fixes $\tau$, whereas column antisymmetrization changes its sign, so $r_t\pi c_t=0$. If there is no collision, the proved lemma gives $\pi =rc$ with $r\in R_t,c\in C_t$, andIt follows that $h_t\pi h_t$ is zero or $\mathrm{sgn}(c)H_{\lambda }{h}_t$. Since permutations span $A$, $e_{t}A{e}_t=\mathbb{C}{e}_t$. In a semisimple algebra this means that $e_t$ is a primitive idempotent, so $A{e}_t=A{h}_t$ is an irreducible left module.

If $\lambda >\mu$ in dictionary order on integer partitions, the collision lemma applied to every $\pi u$ gives $r_{t}A{c}_u=0$, and therefore $h_{t}A{h}_u=0$. Sincethe two simple modules cannot be isomorphic. Conversely, Young tableaux of the same shape are related by a permutation, which conjugates their Young symmetrizers and gives isomorphic left ideals. Finally, the center of $\mathbb{C}{S}_n$ has the conjugacy-class sums as a basis. Conjugacy classes are indexed by cycle-type partitions, so its dimension is the number of partitions of $n$. The Artin–Wedderburn theorem gives exactly that many simple-module isomorphism classes. We have already produced one for each partition. Thus the $A{h}_{\lambda }$ form a complete set of pairwise nonisomorphic irreducible modules.

A standard Young tableau has its entries increasing from left to right in each row and from top to bottom in each column. For the convention $h_t=c_t{r}_t$, use column-reading order of standard Young tableaux: read columns from left to right, each from top to bottom, and compare the resulting words lexicographically. This makes the requested product direction explicit. The order and the symmetrizer multiplication convention must be chosen together.

We prove the vanishing claim. Suppose the rows of $t$ and the columns of $u$ have no collision. The argument in Question 1 shows that every column of $u$ contains one entry from each eligible row of $t$. Its first column thus selects one entry from every row. Each selected entry is at least the first entry of that row in $t$. Sorting the selected entries, as the standard tableau $u$ does, gives a column componentwise at least the first column of $t$: increasing individual entries cannot decrease any order statistic. If the columns are equal, equality of their sums forces every selected entry to be that row's first entry. Delete this common column and repeat. At the first differing column, its first differing entry in $u$ is therefore larger; otherwise $u=t$. Thus absence of a collision implies $u\ge t$ in column-reading order of standard Young tableaux.

If $t>u$, there must instead be a transposition in $R_t\cap C_u$. It fixes $r_t$ and negates $c_u$, giving $r_t{c}_u=0$. HenceThis is triangular vanishing of Young-symmetrizer products; it does not assert vanishing in the opposite order.

Normalize to $e_t=h_t/H_{\lambda }$. List all standard tableaux in increasing shape dictionary order on integer partitions, and within each shape in increasing column-reading order of standard Young tableaux. Then $e_i{e}_j=0$ for $i>j$, using Question 1 between different shapes and the result above within a shape. The left ideals $A{e}_i$ have an internal direct sum: if ${\sum }_i{z}_i=0$ with $z_i\in A{e}_i$, multiply on the right by $e_1$ to get $z_1=0$, since $z_1{e}_1=z_1$ and every later $z_i{e}_1=0$. Repeat with $e_2,e_3,\dots$. This proves directness without incorrectly treating all the idempotents as mutually orthogonal.