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Every nonzero commutative Artinian ring is a finite product of Artinian local rings. The primary components of zero have distinct maximal radicals and are pairwise comaximal ideals, so the Chinese remainder theorem gives the product. Each local factor has nilpotent maximal ideal and finite composition length. These factors and the components of zero are unique.

The center of an indecomposable finite-dimensional block of a group algebra is a commutative Artinian local ring. Its quotient by its Jacobson radical is a field, giving the block's central character. Over a splitting field for finite group representations the residue field is the coefficient field; without this assumption one retains the residue field rather than pretending every central character is scalar-valued over $k$.

The Brauer morphism retains precisely the fixed group-basis elements under conjugation by $D$:It is unital and surjective because $k{C}_G(D)\subseteq (kG{)}^D$. For multiplicativity, fix $x\in C_G(D)$. Its coefficient in a product is ${\sum }_{gh=x}{a}_g{b}_h$. Conjugation by $D$ permutes these pairs and the summand is constant on each orbit. Nonsingleton orbits have p-divisible size and contribute zero. Singleton pairs have both entries in $C_G(D)$, so they give exactly the coefficient in ${\mathrm{Br}}_D(a){\mathrm{Br}}_D(b)$. This proves the ring-homomorphism assertion.

Orbit sums form a basis of $(kG{)}^D$. A nonfixed orbit sum of $g$ is ${\mathrm{Tr}}_{C_D(g)}^D(g)$, a proper-subgroup relative trace. Conversely any proper-subgroup trace has zero coefficient on a fixed basis element, since that coefficient is multiplied by $[D:Q]=0$ in $k$. ThusThere is no assertion that this whole kernel is nilpotent.

Brauer first main theorem bijects the modular blocks of $kG$ with defect $D$ and those of $k{N}_G(D)$ with defect $D$, matching their nonzero Brauer images. Here is a proof with its auxiliary ingredients.

Write $N=N_G(D)$, $C=C_G(D)$ and defineThese are ideals in the indicated commutative algebras, since an invariant multiplier passes through a relative trace. The Brauer morphism and relative trace identityfollows by letting $D$ act on $G/D$. Fixed cosets have representatives in $N$; projection onto $C$ gives equal contributions on each other orbit, whose p-divisible size makes the contribution zero. Thus ${\mathrm{Br}}_D$ surjects $I_D(G)$ onto $J_D$.

To prove the trace criterion for defect groups, choose a p-subgroup $E$ of least order with a given block idempotent $b$ in $I_E(G)$. Such a choice exists because a Sylow $P$ gives $b={\mathrm{Tr}}_P^G(b/[G:P])$. Write $b={\mathrm{Tr}}_E^G(a)$ and replace $a$ by $ba$. If ${\mathrm{Br}}_E(b)=0$, the kernel formula and trace transitivity give $b\in {\sum }_{Q<E}{I}_Q(G)$. But $bZ(kG)$ is an Artinian local ring, and a finite sum of proper ideals cannot contain its identity. Multiplication by $b$ therefore forces $b\in I_Q(G)$ for some smaller $Q$, a contradiction. Hence ${\mathrm{Br}}_E(b)\ne 0$.

If $F$ is not conjugate into $E$, its action on $G/E$ has no fixed coset and the same orbit argument gives ${\mathrm{Br}}_F(I_E(G))=0$. Every subgroup with nonzero Brauer image of $b$ is therefore conjugate into $E$. Its maximal such subgroups are exactly the conjugates of $E$, proving the definition's conjugacy assertion andAmong block idempotents in this ideal, the Brauer image at $D$ survives precisely when their defect is $D$.

We also use primitive idempotents under a surjection of commutative Artinian ideals. Decompose the ambient algebras into local factors. The component of an ideal in a local factor is either the whole factor, possessing its identity, or a proper ideal in its nilpotent radical, possessing no nonzero idempotent. A nonzero quotient of a local algebra is local. A surjection of these ideals consequently bijects its surviving primitive idempotents with those of the target. Entire factors may be killed; the whole kernel need not be nilpotent.

Apply this result to $I_D(G)\to J_D$. Its surviving primitive idempotents are exactly the modular blocks of $G$ with defect $D$. Apply it also to $I_D(N)\to J_D$, where the target is identical since $N_N(D)=N$ and $C_N(D)=C$. Its surviving primitive idempotents are exactly the modular blocks of $N$ with defect $D$. Matching images in $J_D$ proves Brauer first main theorem, in both directions and with uniqueness.

For the last construction, the intended hypothesis is that $D$ is a defect group of the source modular block $b$. Put $L=N_H(D)$ and use the theorem inside $H$ to obtain the corresponding modular block $c$ of $kL$. The centralizer condition gives $C_H(D)=C_G(D)$, so ${\mathrm{Br}}_D$ maps $Z(kG)$ into $Z(kL)$. Let ${\omega }_c:Z(kL)\to F_c$ be the central character of a block, retaining its residue field $F_c$ rather than assuming it equals $k$. The unital compositesends exactly one global block idempotent to one, because their orthogonal idempotent images sum to one in a field. Define $b^G$ to be this unique modular block. This is block induction, independent of the chosen $H$-conjugate of $D$. It needs no containment of $N_G(D)$ in $H$, and its defect may be larger than $D$. With that containment it agrees with the first-main-theorem correspondence.

The printed PDF does not require $D$ to be a defect group of $b$. Its literal conditions do not always define the usual induced modular block. For example, over an algebraically closed field of characteristic three, take $G=A_5$, $H=A_4$ as a point stabilizer, and $D=⟨(123)⟩$. Then $D{C}_G(D)=D\le H$. The ordinary degree-three irreducible of $A_4$ gives a defect-zero modular block $b$ on reduction. The two distinct ordinary degree-three irreducibles of $A_5$ give distinct defect-zero modular blocks $B_{+}$ and $B_{-}$. Both restrict to that same character of $A_4$, with values $3,-1,0$ on its identity, double transpositions and three-cycles.

The standard defect-zero lifting theorem says that these reductions are simple projective modules and their block algebras are full matrix algebras. Thus both module restrictions $B_{+}{\downarrow }_{H\times H}$ and $B_{-}{\downarrow }_{H\times H}$ contain the same modular block bimodule $b$; indeed each module restriction is isomorphic to it here. The actual defect of $b$ is 1, whose centralizer is $G$, not a subgroup of $H$. The unrelated subgroup $D=C_3$ cannot select between the two modular blocks. This is block induction requires the block's defect centralizer; the qualified construction above answers the intended request.

The Brauer first main theorem says that, for a fixed p-subgroup $D$ and $N=N_G(D)$, there is a bijection between blocks of a group algebra of $kG$ with defect group of a block $D$ and blocks of $kN$ with that defect group. Correspondents have the same nonzero image under ${\mathrm{Br}}_D$. The integral version follows by modular reduction of block idempotents.

Here is a proof using relative traces. Put $C=C_G(D)$ andThe latter ambient algebra is commutative because invariance under $N$ includes invariance under $C$. The Brauer morphism and relative trace identity isIndeed $D$ acts on $G/D$. Its fixed cosets are exactly $N/D$. On a nonfixed orbit the terms have the same Brauer image, so the orbit contributes zero in characteristic $p$. The fixed cosets give the right-hand side. Since every element of $kC$ is $D$-fixed, this proves that ${\mathrm{Br}}_D:I_D^G\twoheadrightarrow J_D$ is a surjective algebra homomorphism of commutative ideals. The same argument for $N$ has exactly the same target.

We use the following elementary fact about primitive idempotents under a surjection of commutative Artinian ideals. Under a surjection $f:I\twoheadrightarrow J$ of ideals in finite-dimensional commutative algebras, the primitive ambient idempotents lying in $I$ and having nonzero image correspond bijectively to the primitive idempotents in $J$. To prove it, decompose the source ambient algebra into Artinian local rings $A_i$. Either $I\cap A_i=A_i$, or it is a proper, hence nilpotent, ideal. Proper components contribute no idempotents to the image. A surviving identity $e_i$ has corner $f(e_i)J=f(A_i)$, a local quotient of $A_i$, so is primitive. The surviving identities are orthogonal and exhaust the idempotents of the target. Notice that this proof does not assert that the whole kernel is nilpotent: it can kill entire block components.

By the trace criterion for defect groups proved above, primitive block idempotents in $I_D^G$ have defect conjugate into $D$. Among them, the Brauer morphism is nonzero precisely for defect $D$, again by that criterion and its Brauer characterization. The identical reasoning applies to $I_D^N$. Thus both sets are identified with the primitive idempotents of $J_D$, proving the bijection and its uniqueness.

For the more general subgroup, $D$ is understood to be a defect group of a block $b$ of $kH$. This is the implicit hypothesis needed to apply the theorem; it is not repeated in the printed sentence. Without it the printed uniqueness assertion is false: take $G=A_5$, $H=A_4$, $p=3$ and $D=C_3$. The centralizer of $D$ in $A_5$ is $D$, but the unique three-dimensional defect-zero block of $k{A}_4$ is the bimodule restriction of each of the two three-dimensional defect-zero blocks of $k{A}_5$. The two ordinary degree-three characters both restrict to the degree-three character of $A_4$, with values $3,-1,0$ on its identity, involutions and elements of order three. Defect-zero reduction gives the same simple projective $k{A}_4$-module, so both restricted block bimodules are this same matrix block. This illustrates why block induction requires the block's defect centralizer. Write $e_b$ for its block idempotent and $K=N_H(D)$. Since $D{C}_G(D)\le H$, we have $D{C}_G(D)\le K\le N_G(D)$. Apply the theorem inside $H$ to obtain the unique corresponding block $\beta$ of $kK$. Its center of an associative algebra has a residue field $F_{\beta }$, and the quotient map is its central character of a block ${\omega }_{\beta }:Z(kK)\to F_{\beta }$. No assumption that $k$ is algebraically closed is needed here. DefineThe Brauer image is in $(k{C}_G(D){)}^{N_G(D)}\subseteq Z(kK)$, so this is a unital algebra homomorphism. Among the orthogonal primitive central idempotents of $kG$, exactly one has image $1$. The block it selects is $b^G$. If $H$ contains $N_G(D)$ this is the inverse of the bijection just proved. For arbitrary $H$, the resulting block can have a strictly larger defect group. Block induction is transitive on admissible intermediate subgroups. For the normalizer chain used in the proof below, the required transitivity will be verified directly with Brauer images.

To prove the bimodule characterization, we explicitly state the standard permutation-module fact used here. For a permutation module, the Brauer quotient of a module at $P$ has as basis the permutation-basis points fixed by $P$. An indecomposable summand has nonzero quotient at $P$ exactly when its vertex of an indecomposable module contains a conjugate of $P$. For summands with vertex exactly $P$, taking this quotient, with its $N(P)/P$-action, detects the summand and detects which orthogonal idempotent projection acts as its identity. These facts hold for trivial source modules; the fixed-basis description follows by deleting proper transfers, and the detection assertion is the permutation form of the first-main-theorem argument, applied to their endomorphism algebra.

As an $H\times H$ permutation module, with $(h_1,h_2)g=h_{1}g{h}_2^{-1}$, the group algebra decomposes asThe block bimodule $b$ is an indecomposable module with vertex of an indecomposable module $\Delta D=\{(d,d):d\in D\}$. None of the other double-coset modules can contain $b$: a basis element fixed by $\Delta D$ centralizes $D$, hence lies in $C_G(D)\subseteq H$, whereas those double cosets lie outside $H$. Thus $b$ occurs exactly once as an indecomposable summand of this restriction, in the $kH$ term. Decomposing $kG$ into its global block bimodules therefore puts this occurrence in precisely one of them, say $B$.

Under the fixed-basis description, the quotient of $kG$ at $\Delta D$ is $k{C}_G(D)$, and multiplication by $e_B$ induces multiplication by ${\mathrm{Br}}_D(e_B)$. The quotient of $b$ is ${\mathrm{Br}}_D^H(e_b)k{C}_G(D)=e_{\beta }k{C}_G(D)$. Consequently the detecting projection for this occurrence of $b$ satisfieswhich is equivalent to ${\omega }_b^G(e_B)=1$. Thus $B=b^G$, proving the requested unique-block direct-summand characterization. Here the restriction is to $H\times H$; the extra subscript in the converted TeX is a transcription error, not a different subgroup.

A local ring is a nonzero ring with one maximal ideal $\mathfrak{m}$. Write $\kappa =R/\mathfrak{m}$ for its residue field. Its Krull dimension $\dim R$ is the supremum of lengths of strict chains of prime ideals; in a local ring chains may be extended to end at $\mathfrak{m}$.

For the local Hilbert-function convention, define the Hilbert–Samuel growth dimensionEquivalently it is the order of the pole at $z=1$ of the Hilbert series of the associated graded ringThis finite standard graded $\kappa$-algebra is generated by $\mathfrak{m}/{\mathfrak{m}}^2$. The Hilbert-Serre theorem makes the cumulative Hilbert function eventually polynomial, which establishes the definition. For an Artinian local ring, the polynomial is constant and nonzero, so $d(R)=0$.

For every prime chain of length $s$, the prime-chain lower bound for local length proved in Question 4 gives $H_R(t)\ge c{t}^s$. A polynomial of degree $d(R)$ cannot satisfy this when $s>d(R)$. Thus the requested inequality isFor clarity, the embedding dimension is the different invariantThe equality follows from the Nakayama lemma. Applying Question 4 to $I=\mathfrak{m}$ gives $\dim R\le e(R)$, while the polynomial-algebra surjection below also gives $d(R)\le e(R)$. In particular, growth dimension and embedding dimension should be kept distinct.

A regular local ring is a Noetherian local ring with $e(R)=\dim R$. Let this common value be $e$ and choose a minimal generating set $x_1,\dots ,x_e$ of $\mathfrak{m}$. The initial forms give a surjective graded ring homomorphismFor $e=0$, the Nakayama lemma gives $\mathfrak{m}=0$, so $R$ is a field. Assume $e>0$. If the kernel contained a nonzero homogeneous polynomial $F$ of degree $q>0$, the cumulative Hilbert function of the target would be bounded by that of $\kappa [T_1,\dots ,T_e]/(F)$. Multiplication by $F$ is injective in the polynomial integral domain, so this bound isBut $\dim R=e$ provides a prime chain of length $e$, and Question 4 gives $H_R(t)\ge c{t}^e$. This is a contradiction. Hence the associated graded ring of a regular local ring isan integral domain.

Finally the Krull intersection theorem gives ${\bigcap }_j{\mathfrak{m}}^j=0$. One can see the needed separatedness directly: for the finitely generated ideal $K={\bigcap }_j{\mathfrak{m}}^j$, the Artin-Rees lemma gives $K=\mathfrak{m}K$, and the Nakayama lemma gives $K=0$. Therefore each nonzero $a\in R$ has finite $\mathfrak{m}$-adic order $r$, with nonzero initial form in ${\mathfrak{m}}^r/{\mathfrak{m}}^{r+1}$. For nonzero $a,b$, their initial forms have nonzero product in the graded integral domain. It follows that $ab\ne 0$, and indeed its order is the sum of their orders. Thus $R$ is an integral domain.

Write $A=kG$, $N=N_G(D)$ and $C=C_G(D)$. The Brauer morphism is the coefficient projectionwhere the superscript $D$ denotes invariance under conjugation. This is an algebra homomorphism: for $c\in C$, the pairs $(x,y)$ with $xy=c$ are permuted by $D$; their coefficient products are constant on each orbit. Every nonfixed orbit has size divisible by $p$ and contributes zero. The fixed pairs are exactly $x,y\in C$, giving the coefficient of $c$ in the product of the projections. It is surjective since $kC\subset A^D$.

We need two trace facts. The kernel isIndeed, a basis of $A^D$ consists of conjugation-orbit sums. Nonfixed orbit sums are traces from their proper stabilizers, while fixed basis elements are exactly the elements of $C$. Traces from proper subgroups have zero coefficient at every fixed element, proving the reverse inclusion. Second, Brauer morphism and relative trace givesOne can see this by letting $D$ act on the cosets $G/D$: fixed cosets are exactly $N/D$, and all other orbit contributions vanish after projection. More generally, ${\mathrm{Br}}_E^G({\mathrm{Tr}}_D^G(A^D))=0$ unless $E$ is conjugate into $D$.

Set $I_D={\mathrm{Tr}}_D^G(A^D)$ and $J_D={\mathrm{Tr}}_D^N(kC)=(kC{)}_D^N$. The first is an ideal of the center of an associative algebra $Z(A)$; the second is an ideal of $(kC{)}^N$, which is commutative because $C\subset N$. The displayed trace identity and surjectivity of the Brauer projection give a surjective algebra homomorphismThe ideals can lack identities, so we justify the required idempotent argument. A finite-dimensional commutative algebra is a product of Artinian local rings. Its ideal intersects each local factor either in the whole factor, containing its identity, or in a proper nilpotent ideal. Under a surjection of such ideals, the proper parts have nilpotent image. Each nonzero image of a local-factor identity has a local corner algebra, a quotient of that factor, and is therefore primitive. If $u$ is the sum of these nonzero images, then $j-uj$ lies in the nilpotent image of the proper-factor ideals for every $j$ in the target; an idempotent has $j-uj=0$. Its components in the local corners are consequently zero or the corner identities. These orthogonal images account for every primitive idempotent in the target, because what remains is nilpotent. Thus primitive idempotents under a surjection of commutative Artinian ideals apply: the primitive idempotents of $J_D$ correspond exactly to primitive central idempotents $b\in I_D$ with ${\mathrm{Br}}_D^G(b)\ne 0$.

To identify these with the required blocks, we prove the trace criterion for defect groups. Choose a subgroup $D_0$ of smallest order with $b\in I_{D_0}$. Such a subgroup exists: for a Sylow $p$-subgroup $P$, $b={\mathrm{Tr}}_P^G(b/[G:P])$. If ${\mathrm{Br}}_{D_0}^G(b)=0$, write $b={\mathrm{Tr}}_{D_0}^G(a)$ and replace $a$ by $ba$. The kernel formula and transitivity of trace would giveMultiplying by $b$ gives the identity of the local algebra $bZ(A)$ in a sum of the ideals $b{I}_Q$. If all were proper, they would lie in its unique maximal ideal; hence some $b{I}_Q$ contains $b$, contradicting minimality. Therefore ${\mathrm{Br}}_{D_0}^G(b)\ne 0$.

The general trace-vanishing assertion shows that any $E$ with nonzero Brauer image is conjugate into $D_0$. Conversely, every conjugate of $D_0$ has nonzero image. Hence the maximal such subgroups, each called a defect group of a block, are precisely the conjugates of $D_0$. It also follows, using trace transitivity, that $b\in I_D$ if and only if a defect group is conjugate into $D$. Combining this with ${\mathrm{Br}}_D^G(b)\ne 0$ forces equality of the subgroup orders, so the idempotents singled out above are exactly those with defect group $D$.

We have proved the desired bijectionFinally, Brauer first main theorem states that these blocks correspond bijectively to the blocks of $k{N}_G(D)$ with defect group $D$. Apply the same argument to $N$: its normalizer of $D$ is itself and its centralizer of $D$ is still $C$. The target ideal is therefore the same $J_D$. Matching the two bijections proves the theorem and gives the Brauer correspondence, characterized by equal nonzero Brauer images. Notice that we never assume the entire kernel on $I_D$ is nilpotent; it may kill whole blocks of smaller defect.

Let $\mathfrak{m}$ be the maximal ideal and $k=A/\mathfrak{m}$ the finite residue field. The maximal ideal of an Artinian local ring is nilpotent, so ${\mathfrak{m}}^r=0$ for some $r$. Every quotientis both an Artinian $A$-module and a vector space over $k$. An Artinian vector space is finite-dimensional, hence each quotient is a finite set. The finite filtrationtherefore proves that the underlying set of $A$ is finite. This is the Finiteness criterion for an Artinian local ring.

Let $A,B$ be finite-dimensional commutative algebras, let $I\subset A$ and $J\subset B$ be ideals, and let $f:I\to J$ be a surjective algebra homomorphism. The ideals need not have identities. Then $f$ gives a bijection between the primitive idempotents of $A$ lying in $I$ with nonzero image and the primitive idempotents of $J$.

Decompose $A$ into its commutative Artinian local rings $A_i$, with identities $e_i$. Each $I\cap A_i$ is either all of $A_i$, in which case $e_i\in I$, or a proper ideal and therefore a nilpotent ideal. The sum of the latter ideals is nilpotent, so its image contributes no idempotents. Each nonzero $f(e_i)$ has local corner algebra $f(e_i)J=f(A_i)$, a quotient of $A_i$, so is primitive. These idempotents are orthogonal, and the remaining part of $J$ is nilpotent; hence they are exactly its primitive idempotents. This statement does not require the entire kernel of $f$ to be nilpotent.

Let $(R,\mathfrak{m})$ be an Artinian local ring. If ${\dim }_{R/\mathfrak{m}}\mathfrak{m}/{\mathfrak{m}}^2=1$, then every ideal of $R$ is principal. The Nakayama lemma gives $\mathfrak{m}=(x)$, and nilpotence gives a finite chainFor a nonzero ideal $J$, choose the largest $r$ for which $J\subseteq (x^r)$ and take $y\in J\setminus (x^{r+1})$. Writing $y=x^{r}u$, one has $u\notin \mathfrak{m}$, so $u$ is a unit and $J=(x^r)=(y)$.