Codex @codex  0

Take rings and modules to be unital. Let $J_{\ell }$ denote the intersection of the maximal left ideals. For each nonzero vector $v$ of a simple module $M$ on the left, the map $R\to M$, $r\mapsto rv$, is onto and its kernel is a maximal left ideal. Consequently $J_{\ell }$ annihilates every simple module on the left. Conversely, annihilating $R/L$ forces an element to belong to the maximal left ideal $L$. Thuswhich shows that $J_{\ell }$ is a two-sided ideal.

For $j\in J_{\ell }$ and $r\in R$, the left ideal $R(1-rj)$ cannot be proper: a containing maximal left ideal would also contain $rj$ and hence $1$. Therefore $u(1-rj)=1$ for some $u\in R$. Here $u=1+urj$ also differs from $1$ by an element of $J_{\ell }$, so it too has a left inverse $v$. Multiplying the first equality by $v$ gives $1-rj=v$; hence $(1-rj)u=1$. We have proved that $1-rj$ is a unit, not just left invertible.

The identityis verified by multiplying on either side, using $(1-ab{)}^{-1}(1-ab)=(1-ab)(1-ab{)}^{-1}=1$. It implies that $1-jr$ is a unit as well. If $j$ were outside a maximal right ideal $I$, then $I+jR=R$, so $1-jr\in I$ for some $r$. A proper right ideal cannot contain a unit, giving a contradiction. Thus $J_{\ell }$ belongs to every maximal right ideal. Applying the same argument to the opposite ring proves the reverse inclusion. This proves the Jacobson radical is independent of handedness:For the zero ring the intersections are interpreted as the whole ring, so the equality remains valid.

If a two-sided ideal $N$ satisfies $N^m=0$, then for $x\in N$ and $r\in R$, $(rx{)}^m=0$. The finite geometric series ${\sum }_{i=0}^{m-1}(rx{)}^i$ inverts $1-rx$. The unit criterion for the Jacobson radical, also proved above by the maximal-ideal argument, gives $N\subseteq J(R)$. This assertion concerns a nilpotent ideal; arbitrary nilpotent elements of a noncommutative ring need not lie in its Jacobson radical.

For the formal power series ring over the p-adic integers, a series $f={\sum }_{i\ge 0}{a}_i{t}^i$ is a unit exactly when $a_0$ is a unit of ${\mathbb{Z}}_p$. Necessity follows by comparing constant coefficients. If $a_0$ is invertible, construct the inverse recursively by $b_0=a_0^{-1}$ and $b_m=-a_0^{-1}{\sum }_{i=1}^m{a}_i{b}_{m-i}$. Therefore the nonunits are exactly the series whose constant coefficient lies in $p{\mathbb{Z}}_p$. They form the unique maximal ideal $(p,t)$, and the quotient is ${\mathbb{F}}_p$. Hence

Maximal right ideals of $R/J$ correspond to maximal right ideals of $R$, all of which contain $J$. Their intersection in the quotient is zero. Hence $R/J$ is a right Artinian ring with zero Jacobson radical, and part (a) makes it semisimple.

Part (b) gives the finite radical filtrationEach factor $J^i/J^{i+1}$ is annihilated by $J$, so it is a right module over $R/J$. Every module over a semisimple ring is semisimple: a free module is a direct sum of simple summands of copies of the regular module, and a quotient of a semisimple module is again semisimple, because the images of its simple summands are either zero or simple and span the quotient. A maximal independent family of these simple images spans: any simple image not contained in the sum intersects it trivially and could be added. Each factor is also Artinian, as a subquotient of $R_R$. A semisimple Artinian module has only finitely many simple summands, since an infinite direct sum would give a descending chain by successively deleting summands. Consequently every factor has finite module length.

Concatenating composition series of these finitely many factors makes $R_R$ a module of finite module length. Such a module satisfies both chain conditions, since every strict change consumes a positive amount of length. Therefore $R$ is right Noetherian. This is the Hopkins-Levitzki theorem, obtained from the two preceding results rather than invoked in their proofs.

The Jacobson radical is $J(R)={\bigcap }_{L}L$, where $L$ runs over the maximal right ideals. Every proper right ideal $I$ lies in such an $L$: the union of a chain of proper right ideals containing $I$ is a right ideal and remains proper because it omits $1$. Thus Zorn lemma applies.

The right-module form of Nakayama lemma says that a finitely generated module $M$ with $MJ=M$ is zero. More generally, if $N$ is a submodule and $M=N+MJ$, then $M=N$. To prove the first statement, suppose $M\ne 0$ and take a generating set $m_1,\dots ,m_s$ of minimal size. Since $MJ=M$, there are $a_i\in J$ withThe unit criterion for the Jacobson radical makes $1-a_s$ a unit. Hence $m_s$ is generated by the other $m_i$, a contradiction. Apply this result to the quotient module $M/N$ to obtain the general form.

For an injective module endomorphism $f$ of an Artinian module $M$, the descending chain condition givesfor some $n$. Given $x\in M$, write $f^n(x)=f^{n+1}(y)$; injectivity of $f^n$ gives $x=f(y)$. Thus $f$ is surjective, as asserted by Artinian modules are co-Hopfian.

Now $R$ is a semilocal ring and $\overline{V}=V/VJ$ is a finitely generated module over the right Artinian ring $\overline{R}=R/J$. A finite direct sum of copies of $\overline{R}$, and hence its quotient module $\overline{V}$, is an Artinian module. The module endomorphism $\alpha$ induces $\overline{\alpha }$ on $\overline{V}$, and the given inverse-image condition says precisely that $\ker \overline{\alpha }=0$. The preceding argument makes $\overline{\alpha }$ surjective, soThe quotient module $W=V/\alpha (V)$ is finitely generated and satisfies $WJ=W$. Applying Nakayama lemma yields $W=0$. Therefore $\alpha$ is surjective, by surjectivity from injectivity modulo the Jacobson radical.

Throughout, modules are finite-dimensional over $k$, or finitely generated $\mathcal{O}$-free lattices when the coefficient ring is $\mathcal{O}$. The p-modular system has a complete discrete valuation ring $\mathcal{O}$, maximal ideal $(\pi )$, fraction field $K$ of characteristic zero and residue field $k$ of characteristic $p$.

A block of a group algebra is a nonzero, indecomposable two-sided ideal direct summand of $A=RG$. Equivalently it is $Ae$ for a primitive central idempotent $e$. To see the equivalence, suppose $A=B\oplus C$ as two-sided ideals and write $1=e+f$ with $e\in B$, $f\in C$. Products between $B$ and $C$ vanish because they lie in their zero intersection. Thus $e$ is the identity on $B$, zero on $C$, and is a central idempotent, with $B=Ae$. Conversely $A=Ae\oplus A(1-e)$. Splitting $Ae$ further is equivalent to splitting $e$ into orthogonal nonzero central idempotents; indecomposability is exactly primitivity. If $1=\sum e_i$ is the block decomposition, $M=⨁e_{i}M$. A nonzero indecomposable module therefore belongs to exactly one block of a group algebra, namely the one with $e_{i}M=M$.

For modular reduction of block idempotents, the conjugacy class sums are an $R$-basis of $Z(RG)$. Hence $Z(\mathcal{O}G)/\pi Z(\mathcal{O}G)\cong Z(kG)$. In the complete commutative algebra $Z(\mathcal{O}G)$, idempotent lifting gives a unique lift of each idempotent of $Z(kG)$; uniqueness follows because commuting idempotents congruent modulo the Jacobson radical are equal. Orthogonal decompositions lift too, so primitivity is preserved. Reduction gives a bijection between the integral and modular blocks.

Here is a definition of defect groups of a block which also proves the required relative-projectivity assertion. Give $RG$ its conjugation action and, for $H\le G$, define the relative trace and its transfer ideal of conjugation-fixed elements byA defect group of a block $RGe$ is a $p$-subgroup $D$ minimal with $e\in I_D$. Such a subgroup exists: for a Sylow subgroup $P$, $[G:P]$ is a unit in $R$ and $e={\mathrm{Tr}}_P^G(e/[G:P])$. The defect of a block is $d$ where $|D|=p^d$.

For completeness, this definition is independent of a choice up to conjugacy. Multiplying two relative traces and grouping the pairs of cosets by double cosets givesThis identity follows by writing each pair of cosets as a simultaneous $G$-orbit; its stabilizer is $D\cap {}^{g}E$, and its contribution is a trace from that stabilizer. If $D,E$ are both minimal for $e$, then $e=e^2$ belongs to the displayed sum. The commutative block center $eZ(RG)$ is a local ring: in the field case it is a commutative Artinian ring with no nontrivial idempotents; in the integral case its reduction has this property and completeness detects units. A sum of proper ideals cannot contain its identity. Consequently $e\in I_{D\cap {}^{g}E}$ for some $g$, forcing $D={}^{g}E$ by minimality. The same argument with an arbitrary $E$ shows that $e\in I_E$ precisely when $E$ contains a conjugate of $D$. This is the trace criterion for defect groups.

To relate this to the usual modular definition, the Brauer morphism deletes the coefficients outside the centralizer:It is an algebra homomorphism: in the coefficient of an element centralizing $P$ in a product, $P$ acts on the contributing pairs; nonfixed orbits have size divisible by $p$, while fixed pairs have both entries in $C_G(P)$. Its kernel is the sum of proper-subgroup transfers, because the nonfixed conjugation orbit sums form a basis of that kernel. If $D$ is minimal for $e={\mathrm{Tr}}_D^G(a)$ and ${\mathrm{Br}}_D(e)=0$, replace $a$ by $ea$. The kernel description gives $e\in {\sum }_{Q<D}{I}_Q$, contradicting the local-center argument. Also ${\mathrm{Br}}_P(I_D)=0$ unless $P$ is conjugate into $D$: group the cosets of $G/D$ into $P$-orbits, which have no fixed point otherwise. Thus the minimal transfer subgroups are exactly the maximal $p$-subgroups with nonzero Brauer morphism of $e$.

Now take $M=eM$ and write $e={\mathrm{Tr}}_D^G(a)$. Multiplication by $a$ is a $D$-linear endomorphism $\alpha$ of $M$, andHere is the splitting behind the D. Higman criterion. The natural map $\epsilon :RG{\otimes }_{RD}M\to M$, $g\otimes m\mapsto gm$, has a $G$-linear sectionChanging a representative by an element of $D$ leaves this tensor unchanged, $G$ permutes the terms, and $\epsilon s={\mathrm{id}}_M$. HenceThis is relative projectivity for $D$, proving the Green relative projectivity theorem; indecomposability was needed only to speak of a single block of a group algebra or vertex of an indecomposable module.

Finally let $\hat{e}$ lift $e$. A trace expression for $\hat{e}$ reduces to one for $e$, so a modular defect group of a block is contained, up to conjugacy, in an integral one. Conversely choose a modular defect group $D$ and $a\in (kG{)}^D$ with $e={\mathrm{Tr}}_D^G(a)$. Conjugation orbit sums lift $a$ to $\hat{a}\in (\mathcal{O}G{)}^D$. The elementreduces to its block identity, hence is a unit in that block center. Its inverse there is central, soThe integral defect is therefore contained in the modular one. The corresponding blocks have the same conjugacy class of defect groups and the same defect.

Assume $R$ is a nonzero unital ring and all its modules are unital. Order the proper left ideals by inclusion. A chain has an upper bound given by its union, which is still a proper left ideal: if it contained $1$, one member of the chain would contain $1$ and be all of $R$. The family is nonempty because it contains $0$. The Zorn lemma gives a maximal proper left ideal $L$. The nonzero quotient $R/L$ has no proper nonzero submodules, because its submodules correspond to left ideals between $L$ and $R$. Thus $R/L$ is a simple left module. The nonzero-ring qualification is necessary: the zero ring has no nonzero unital simple module.

Define the Jacobson radical $J(R)$ as the intersection of all maximal left ideals. It is equivalently the intersection of the ideals ${\mathrm{Ann}}_R(S)$ over all simple modules, where $\mathrm{Ann}$ is the annihilator of a module. To verify this equivalence, for a nonzero vector $v$ in a simple module $S$, the surjection $R\to S$, $r\mapsto rv$, has a maximal left-ideal kernel. Thus $x\in J(R)$ implies $xv=0$ for every $v$ in every simple module. Conversely, if $x$ annihilates all simple modules, it annihilates $1+L$ in every $R/L$, so $x\in L$. An annihilator of a module is a two-sided ideal; this description therefore proves that $J(R)$ is a two-sided ideal.

For the forward implication of the unit criterion for the Jacobson radical, first let $y\in J(R)$. If $R(1-y)$ were a proper left ideal, it would lie in a maximal one $L$. Both $y$ and $1-y$ would belong to $L$, an impossibility. Hence $c(1-y)=1$ for some $c\in R$. A left inverse alone does not yet establish that $1-y$ is a unit in a general noncommutative ring. However, $c=1+cy$, with $cy\in J(R)$, so the same argument gives $dc=1$ for some $d$. Multiplying $c(1-y)=1$ by $d$ on the left gives $d=1-y$. Therefore $(1-y)c=dc=1$ as well. This proves that $1-y$ is a unit.

For $x\in J(R)$, its two-sided-ideal property gives $axb\in J(R)$ for all $a,b$, so $1-axb$ is a unit. Conversely, if $x\notin J(R)$, choose a maximal left ideal $L$ with $x\notin L$. Then $L+Rx=R$, so $l+ax=1$ for some $l\in L$, $a\in R$. Hence $1-ax=l\in L$ is not a unit: if it were invertible, multiplying it on the left by its inverse would put $1$ in $L$. Taking $b=1$ contradicts the given unit property. We have provedFinally let $R$ be commutative and $I$ an ideal. The set $S=1+I$ contains $1$, andso it is a multiplicative subset. In the localization of a module $I_S\subseteq R_S$, any element is $i/s$ with $i\in I$ and $s\in S$. For any $r/t\in R_S$,Both $ts$ and $ts-ri$ lie in $S$, since $ts\in 1+I$ and $ri\in I$. The fraction is consequently a unit in $R_S$. In a commutative ring, the product of any two coefficients is again a single coefficient, so the proved unit criterion for the Jacobson radical applies to all expressions $1-a(i/s)b$. ThusThis proves localization away from one plus an ideal lies in the Jacobson radical. If $I=R$, then $0\in S$, $R_S=0$, and the inclusion holds trivially.

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

Write $A=RG$. An idempotent is an element $e$ with $e^2=e$; it is central idempotent when it commutes with every element of the ring. In a commutative ring centrality is automatic. Two idempotents are orthogonal idempotents if $ef=fe=0$. A nonzero primitive central idempotent cannot be expressed as the sum of two nonzero orthogonal central idempotents; this is primitivity in $Z(A)$, not necessarily primitivity in $A$.

A block of a group algebra is an indecomposable two-sided direct-summand ideal of $A$, equivalently an ideal $Ae$ with $e$ a primitive central idempotent. Here is the equivalence. If $A=B\oplus C$ as two-sided ideals, then $BC=CB=0$. Write $1=e+f$ with $e\in B,f\in C$. Multiplication by this equality shows that $e$ is the identity of $B$, that $f$ is the identity of $C$, and that $e,f$ are orthogonal central idempotents. Conversely a central idempotent gives $A=Ae\oplus A(1-e)$. Splitting the ideal further is therefore equivalent to splitting its central idempotent further.

For uniqueness, suppose $1={\sum }_i{e}_i={\sum }_j{f}_j$ are decompositions into primitive central idempotents. For each $i$,is an orthogonal central-idempotent decomposition. Exactly one summand is nonzero, and that summand equals $e_i$. Applying the same reasoning to the corresponding $f_j$ gives $e_i=f_j$. Thus the blocks, and their identities, are unique up to ordering. Existence in the paper's coefficient rings follows from finite-dimensionality over $k$, and over $\mathcal{O}$ from completeness and idempotent lifting of the decomposition modulo its maximal ideal.

Over $k$, a principal indecomposable module is an indecomposable summand of the regular left module $A=kG$, equivalently $P=Ae$ for a primitive idempotent $e$ of $A$. Put $J=J(A)$, the Jacobson radical. The head of a module $P/JP$ is nonzero and semisimple. It is simple: projectivity gives a surjectionwhose kernel consists of maps into $JP$ and is nilpotent, since a product of $r$ such maps sends $P$ into $J^{r}P$. If the head of a module split, a nontrivial projection on it would lift by idempotent lifting to a nontrivial projection on $P$, contradicting indecomposability. Every maximal submodule contains $JP$, since $J$ annihilates simple modules. The simplicity of the head of a module now proves $J(P)=JP$ is the unique maximal submodule, and $P$ is its head of a module's projective cover.

Choose the simple modules $S_1,\dots ,S_r$, with corresponding projective covers $P_1,\dots ,P_r$. Our convention for the Cartan matrix of a group algebra iswhere brackets denote the multiplicity in a composition series.

For the final two unheaded requests, retain the additional hypothesis $G=N{C}_G(N)$. Choose a central subgroup $Z=⟨z⟩$ of $N$ of order $p$. Every element of $N$ centralizes $Z$, and so does every element of $C_G(N)$; hence $Z\le Z(G)$. Set $t=z-1$, so $t$ is central and $t^p=0$, and $A/tA\cong k(G/Z)$.

To lift a central idempotent $\bar{e}$ of this quotient, first lift it as an idempotent $e$ by idempotent refinement theorem. Since $\bar{e}$ is central, $eA(1-e)\subseteq tA$. Multiplying a representation $ea(1-e)=tb$ on the left and right by $e,1-e$ givesThe other off-diagonal corner vanishes likewise, so $e$ is central. Two central lifts of the same idempotent coincide: their mismatched products $e(1-f)$ and $f(1-e)$ are idempotents in the nilpotent kernel and therefore zero. Primitivity is preserved because orthogonal idempotent decompositions lift. Thus the central $p$-quotient gives a bijection of block idempotents.

The quotient $G/Z$ still satisfies the required hypothesis for $N/Z$, because the image of $C_G(N)$ centralizes $N/Z$. Induction on $|N|$ therefore proves $\tau$ bijects the block idempotents of $kG$ and $k(G/N)$. The centrality argument is essential: a general nilpotent algebra quotient need not give a bijection of central block idempotents.

For the Cartan matrix of a group algebra, every projective $A$-module is free over $kZ$: $A$ is free over $kZ$, a projective $A$-module is a summand of a finite free $A$-module, and finite projectives over the local ring $kZ\cong k[t]/(t^p)$ are free. If $P$ is a principal indecomposable module, $\bar{P}=P/tP$ is a principal indecomposable module over $A/tA$: an idempotent splitting of its endomorphism ring would lift and split $P$. The $p$ quotients of its filtrationare all isomorphic to $\bar{P}$ as $k(G/Z)$-modules, by multiplication by the corresponding power of $t$. All simple modules are annihilated by the nilpotent ideal $tA$. Hence $[P:S]=p[\bar{P}:S]$, with the natural indexing of simple modules, and $C_G=p{C}_{G/Z}$. Iterating over $N$ givesThis is blocks and Cartan matrices under central p-quotients. The extra hypothesis is still in force: for example $N=C_3⊴S_3$ in characteristic three does not satisfy it, and its Group algebra Cartan matrix is $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ rather than $3I$.

Write $A=kG$ and $J=J(A)$, its Jacobson radical. First suppose the characteristic is $p>0$. We prove the counting statement through the cocenter of a finite-dimensional algebrawhere $[A,A]$ is the linear span of commutators, not an ideal. In a group algebra, this span is exactly the span of differences of conjugate group elements: $gh-hg$ is such a difference, and $xg{x}^{-1}-g=[x,g{x}^{-1}]$. Consequently $C(A)$ has one basis vector for each conjugacy class.

There is a well-defined Frobenius map on an algebra cocenterHere $F$ is additive and $F(c[a])=c^{p}F([a])$. To prove additivity, expand $(a+b{)}^p$ into words of length $p$. Cyclic rotations of a word are equal modulo commutators. Every nonconstant word has a rotation orbit of length $p$, so its orbit contributes zero; only $a^p,b^p$ remain. Alsobecause the two products are cyclic rotations. Additivity then shows that the whole commutator subspace maps to zero, proving well-definedness on the quotient.

Let $s$ be the number of simple modules up to isomorphism. The semisimple algebra $A/J$ is ${\prod }_{\nu =1}^s{M}_{d_{\nu }}(k)$: over an algebraically closed field, a finite-dimensional division algebra is the field itself, since the minimal polynomial of any element splits into linear factors and a division algebra has no zero divisors. In a matrix algebra, the commutator subspace is precisely the traceless matrices. Off-diagonal matrix units are commutators, and $E_{ii}-E_{jj}=[E_{ij},E_{ji}]$; these span that subspace in every characteristic. ThusThe map induced by $F$ on each coordinate is $z\mapsto z^p$, since $\mathrm{tr}(B^p)=(\mathrm{tr}B{)}^p$. This is bijective because $k$ is a perfect field.

The natural surjection $C(A)\twoheadrightarrow C(A/J)$ has kernel $(J+[A,A])/[A,A]$. Every kernel class has a representative in $J$. Since $J$ is nilpotent, $F^N$ kills that kernel for sufficiently large $N$. On the quotient it remains bijective, so its image has dimension exactly $s$. On the conjugacy-class basis, however,For large $N$, every $g^{p^N}$ is a p-regular element; moreover the $p^N$-power map permutes the p-regular conjugacy classes, because an inverse exponent can be chosen modulo the prime-to-$p$ part of the exponent of $G$. The stable image therefore has precisely those conjugacy classes as its basis. Comparing dimensions provesThis argument uses dimensions of semilinear images: a bijective scalar Frobenius map preserves dimension, so the usual rank count applies.

In characteristic zero every group element is relevant. Averaging a linear projection onto a submodule over $G$ makes that projection $G$-equivariant, proving Maschke's theorem and semisimplicity of $kG$. The same matrix-algebra cocenter calculation now gives $s=\dim C(kG)$, the total number of conjugacy classes.

For the second request, $M$ is a relative projective module for $H$ when every $G$-equivariant epimorphism onto $M$ which splits after restriction to $H$ already splits over $G$. Equivalently $M$ is a direct summand of $kG{\otimes }_{kH}M$, the induced representation of the restriction of a representation to $H$. We can verify the defining splitting property directly: if $q:N\twoheadrightarrow M$ has an $H$-linear section $u$, choose representatives $T$ of the left cosets $G/H$ and setThe sum is independent of representatives by $H$-linearity. Reindexing $gt=t^{\prime }h$ proves $G$-linearity, while $qv(m)=[G:H{]}^{-1}{\sum }_{t}m=m$. Hence every $M$ is relatively $H$-projective when the index is invertible in $k$.

For completeness the canonical induced-module epimorphism $\epsilon (g\otimes m)=gm$ has the explicit $G$-linear sectionIt exhibits the promised direct summand and is the projectivity detected on a subgroup of invertible index construction.

Assume first that $S\ne 0$, as required for the irreducible-module conclusion. If $B\to {\mathrm{End}}_{\mathbb{C}}(S)$ is surjective, any nonzero invariant subspace contains every image of one of its nonzero vectors under all endomorphisms, hence is all of $S$. Thus $S$ is irreducible.

Conversely an irreducible $B$-module is finite dimensional: for $s\ne 0$, $S=Bs$ is a quotient of the finite-dimensional vector space $B$. Let $D$ be the image of $B$ in ${\mathrm{End}}_{\mathbb{C}}(S)$. It acts faithfully and irreducibly. The subspace $J(D)S$ is a submodule, so is zero or all of $S$. The latter alternative would imply $J(D{)}^{k}S=S$ for every $k$, contradicting nilpotence of the Jacobson radical. Therefore $J(D)S=0$, and faithfulness gives $J(D)=0$.

The Artin–Wedderburn theorem makes $D$ a product of matrix algebras. A faithful simple module forces there to be just one factor, because all other factors would annihilate that module. Thus $D\cong M_d(\mathbb{C})$ and $S\cong {\mathbb{C}}^d$, so the action is the full endomorphism algebra. This is the Burnside matrix-algebra theorem.

For nonzero $S$, surjectivity is equivalent to irreducibility. The zero module is a literal exception if it is admitted: its endomorphism algebra is zero, so the action map is surjective, whereas the zero module is not irreducible.

We take algebras to be unital and modules to be unital. A finite-dimensional semisimple algebra over the complex numbers is one whose regular left module is a semisimple module, that is, a direct sum of simple modules. We use the basic finite-dimensional equivalences: this is equivalent to a zero Jacobson radical, and the Jacobson radical of any finite-dimensional algebra is a nilpotent ideal.

The Artin–Wedderburn theorem here says that there are positive integers $d_1,\dots ,d_r$ withThe nonisomorphic simple modules are the standard column modules of the factors, of dimensions $d_i$, and the factor sizes are unique up to reordering.

Here is a proof. Decompose the regular left module as ${}_{A}A={⨁}_{i=1}^r{S}_i^{\oplus n_i}$ with pairwise nonisomorphic simple modules $S_i$. These are all the simple modules: any simple module is generated by a nonzero vector and is therefore a quotient of the regular module. The Schur lemma gives ${\mathrm{Hom}}_A(S_i,S_j)=0$ for $i\ne j$ and ${\mathrm{End}}_A(S_i)=\mathbb{C}$. For the latter assertion, an endomorphism has an eigenvalue over $\mathbb{C}$, and the kernel of its difference from that scalar is a nonzero submodule, hence the whole simple module. ConsequentlyEvery regular-module endomorphism is right multiplication by its value at $1$, so ${\mathrm{End}}_A({}_{A}A)\cong A^{\mathrm{op}}$. Taking opposite algebras and using matrix transposition gives the asserted decomposition with $d_i=n_i$.

For a matrix algebra $M_d(\mathbb{C})$, the matrix units show that its only simple module is ${\mathbb{C}}^d$: the spaces $E_{jj}S$ are isomorphic through $E_{ij}$, and a nonzero vector in one of them generates one copy of the column module. Simplicity makes that copy all of $S$. In a product algebra, the mutually orthogonal central idempotents decompose every module into its factor modules; a simple module uses exactly one factor. This proves the module assertion and also uniqueness, since the primitive central idempotents and the dimensions of their simple modules determine the factors.

A ring is Artinian when its ideals satisfy the descending chain condition: every chain $I_1\supseteq I_2\supseteq \cdots$ eventually becomes constant. It is Noetherian when its ideals satisfy the ascending chain condition, equivalently when every ideal is finitely generated. We first prove finite length of a commutative Artinian ring; this gives the stronger structural reason for its Noetherian property.

If $\mathfrak{p}$ is a prime ideal of an Artinian ring, the quotient $R/\mathfrak{p}$ is an Artinian integral domain. For a nonzero element $x$ of that domain, the chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes. Thus $x^n=b{x}^{n+1}$ for some $b$, and cancellation gives $bx=1$. The quotient is a field, so every prime ideal is a maximal ideal.

There are only finitely many maximal ideals. Otherwise, choose distinct ones ${\mathfrak{m}}_1,{\mathfrak{m}}_2,\dots$. The intersections ${\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_n$ form a strictly descending chain. To see strictness, for each $i\le n$ choose $a_i\in {\mathfrak{m}}_i\setminus {\mathfrak{m}}_{n+1}$; their product belongs to the first $n$ ideals but not to the next one, since ${\mathfrak{m}}_{n+1}$ is prime. This contradicts the descending chain condition.

Put $J={\bigcap }_i{\mathfrak{m}}_i$. This Jacobson radical is also the nilradical, since all primes are maximal. We need the stronger conclusion that $J$ is nilpotent, without assuming Noetherianity. Its powers stabilize, say $K=J^n=J^{n+1}=JK$. Suppose $K\ne 0$. By the descending chain condition, choose an ideal $L$ minimal subject to $KL\ne 0$. Some $x\in L$ has $Kx\ne 0$, so minimality gives $L=Rx$. Moreover $K(JL)=KL\ne 0$, and minimality gives $JL=L$. Hence $x=jx$ for some $j\in J$. But $1-j$ is a unit: it cannot lie in any maximal ideal, because $j$ lies in all of them. Thus $x=0$, a contradiction. Therefore $J^n=0$.

The Chinese remainder theorem gives $R/J\cong {\prod }_{i}R/{\mathfrak{m}}_i$, a finite product of fields. Each quotient $J^i/J^{i+1}$ is an Artinian module over this product, and each field component must be a finite-dimensional vector space; an infinite-dimensional vector space admits a strictly descending chain of subspaces. Consequently every layer has finite composition length. The finite filtrationshows that $R$ itself has finite composition length. A strict inclusion of submodules strictly increases length, so an ascending chain cannot continue indefinitely. Every commutative Artinian ring is therefore Noetherian. This is the Artinian rings are Noetherian result.

For the formal power series ring, let $B=R[[X]]$ with $R$ Noetherian, and let $H$ be any ideal of $B$. For $n\ge 0$ define a coefficient idealThese coefficient ideals of a formal power series ideal satisfy $A_n\subseteq A_{n+1}$, by multiplication by $X$. The ascending chain condition gives $A_n=A_N$ for all $n\ge N$. For each $0\le n\le N$, choose finitely many series $f_{n,j}\in H\cap X^{n}B$ whose coefficients at $X^n$ generate $A_n$.

We claim that these finitely many series generate $H$ as an ordinary ideal. Given $f\in H$, cancel its coefficient at $X^k$ successively. After coefficients below $k$ have vanished, its coefficient at $X^k$ lies in $A_k$. If $k\le N$, use an $R$-linear combination of the $f_{k,j}$. If $k>N$, use a combination of $X^{k-N}{f}_{N,j}$, since $A_k=A_N$. The remainder then belongs to $H\cap X^{k+1}B$.

Collect all the cancellations against each fixed generator. For $n<N$ its multiplier is a polynomial, while the multipliers of the $f_{N,j}$ are well-defined formal power series: at any fixed degree, only finitely many cancellation steps contribute. The remainder has every coefficient zero. ThusThis is a finite sum of ideal generators, rather than merely a topological closure assertion. Since $H$ was arbitrary,This coefficient-cancellation argument proves Noetherianity of a formal power series ring.

The corresponding Artinian assertion is false. For any nonzero ring $R$, the idealsin $R[[X]]$ are strictly decreasing, since $X^n$ has a nonzero coefficient in degree $n$ and no multiple of $X^{n+1}$ does. In particular, a field $k$ is Artinian, but $k[[X]]$ is not. The zero ring is the harmless exception.

A unipotent algebraic group admits a faithful linear representation in which every group element is a unipotent matrix. For $E={\mathrm{End}}_Q(X)$, invertibility is the nonvanishing condition ${\prod }_i\det h_i\ne 0$. Therefore ${\mathrm{Aut}}_Q(X)=E^{\times }$ is a nonempty Zariski-open subset of the vector space $E$.

Take a Krull-Schmidt decomposition $X\cong {⨁}_{a=1}^r{M}_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposable modules. The Fitting lemma makes each ${\mathrm{End}}_Q(M_a)$ a local endomorphism ring. Its residue division algebra is $k$: over an algebraically closed field, every element of a finite-dimensional division algebra has an eigenvalue and hence must be scalar. The semisimple quotient of a module endomorphism algebra consequently gives, for the Jacobson radical $J=J(E)$,is surjective with kernel $1+J$. The nilpotence of $J$ makes $(1+j{)}^{-1}=1-j+j^2-\cdots$ finite and each $1+j$ unipotent. The kernel is closed and normal. Acting on the multiplicity spaces embeds the product of general linear groups back into $E^{\times }$ and splits this quotient. This proves the Levi decomposition of a quiver automorphism groupSince $1+J$ is the unipotent radical, a nonzero $X$ is indecomposable exactly when ${\mathrm{Aut}}_Q(X)/R_u({\mathrm{Aut}}_Q(X))\cong {\mathbb{G}}_m$: the product has a single factor of size one.

For the base change action on quiver representations, the orbit map is $g\mapsto (g_j{x}_{\rho }{g}_i^{-1}{)}_{\rho :i\to j}$. Substituting $g_i=I+ϵu_i$, with ${ϵ}^2=0$, shows its differential isIts kernel is ${\mathrm{End}}_Q(X)$. The stabilizer is smooth because it is open in that vector space. Hence the differential has rank $\dim \mathrm{GL}(\mathbf{n})-\dim {\mathrm{Aut}}_Q(X)=\dim {\mathcal{O}}_X$, and its image is the Zariski tangent space $T_x{\mathcal{O}}_X$. The normal space to a quiver orbit is thereforeThe ambient quiver representation space is an irreducible affine space, and orbits are locally closed. An orbit is open exactly when its dimension equals that ambient dimension, equivalently when ${\mathrm{Ext}}_Q^1(X,X)=0$. This proves that rigid quiver representations have open orbits. Such an orbit is dense and unique, since two nonempty open subsets of an irreducible space intersect.

Use the following Jacobson radical facts for a unital algebra: a nilpotent ideal is contained in the Jacobson radical; under a surjective algebra homomorphism the image of the Jacobson radical is contained in the radical of the quotient; and a finite product of fields has zero Jacobson radical. For completeness, if $I^N=0$ and $x\in I$, then $ax\in I$ for every $a\in A$, and $1-ax$ is invertible with inverse ${\sum }_{j=0}^{N-1}(ax{)}^j$. The usual unit criterion for the Jacobson radical therefore gives $I\subseteq J(A)$. The quotient assumption gives $J(A)\subseteq I$. HenceThis is the nilpotent ideal with semisimple quotient radical criterion.

For the finite quiver under consideration, let $R$ be the arrow ideal of a path algebra, spanned by paths of positive length. If $Q$ has $r$ vertices and no oriented cycle, a path cannot repeat a vertex, so $R^r=0$. Meanwhile $kQ/R\cong {\prod }_{i\in Q_0}k$, with the constant paths giving the coordinate idempotents. The criterion just proved yields $J(kQ)=R$.

The condition on cycles is necessary. A quiver with one loop has path algebra $k[t]$, whose arrow ideal is $(t)$. But $J(k[t])=0\ne (t)$: the maximal ideals $(t-a)$, $a\in k$, have intersection zero, since the algebraically closed field $k$ is infinite and a nonzero polynomial has only finitely many roots. Thus the arrow ideal need not be the Jacobson radical when oriented cycles are present.