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ArticlePublished 8 Aug 202621 min readBy KEVOS®

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Engineering Mathematics Advanced Local rings

Local Group Algebras

For a nontrivial finite group $G$ and a field $k$ , the group algebra $k G$ is local precisely when $char k = p > 0$ and $G$ is a $p$ -group — the one case where the augmentation ideal is nilpotent and the only simple module is trivial.

Page ID: KEVOS-ENG-MATH-NCR-0146
Taxonomy: ENG / ENG-MATH
Collection: noncommutative-rings-core
Source: (19.31), §19 (pp. 309–310)
Reviewed: 2026-08-08
Version: 1.0.0

Executive Summary

Group algebras are almost never local. If $char k$ does not divide $| G |$ , Maschke makes $k G$ semisimple, and a semisimple ring is local only if it is a division ring — which $k G$ is not once $G \neq 1$ , since $g$ of order $n$ gives $(g - 1) (g^{n - 1} + \dots + 1) = 0$ .

Exactly one family survives. For $k$ a field and $G$ a nontrivial finite group, $k G$ is local iff $char k = p > 0$ and $G$ is a $p$ -group. In that case the augmentation ideal is nilpotent, $k G / rad (k G) ≅ k$ , and $k G$ is completely primary with $(rad k G)^{| G |} = 0$ . Since $k G$ is artinian, $(19.19)$ makes locality equivalent to the regular module being indecomposable, so the whole question is about idempotents.

$p$ -groups onlyWhich

G

$char k = p$ Which

k

$k G / rad ≅ k$ Residue ring

$\leq | G |$ Nilpotency index

Overview

The augmentation map $ε : k G \to k$ , $\sum a_{g} g \mapsto \sum a_{g}$ , is always a surjective ring homomorphism, so its kernel $I (k G)$ is always a maximal ideal with residue field $k$ . Locality asks whether it is the only maximal one-sided ideal.

I (k G) = \ker ε = ⨁_{g \neq 1} k (g - 1), \dim_{k} I (k G) = | G | - 1 .

(A)

The augmentation ideal. It is always maximal; it equals $rad (k G)$ exactly in the local case.

This is the noncommutative source of local rings that matters most in practice. Restricting a $k G$ -module to a Sylow $p$ -subgroup lands it over a local ring, and the theorems of the previous pages — projectives are free, Krull–Schmidt uniqueness — become available. Dickson's divisibility theorem is the first payoff.

The result also survives replacing the field of coefficients by a local ring: for $(R, 𝔪)$ commutative local with $char (R / 𝔪) = p > 0$ and $G$ a finite $p$ -group, $R G$ is local $(19.11)$ . That version, with $R = ℤ_{p}$ or $ℤ_{(p)}$ , is the foundation of integral representation theory.

Learning Objectives

Prove nilpotence of $I (k G)$ for a finite $p$ -group in characteristic $p$ , by induction on $| G |$ .
Deduce $(19.10)$ : $k G$ is local and completely primary with residue ring $k$ .
Prove the converse: if $k G$ is local with $G \neq 1$ finite, then $char k = p$ and $G$ is a $p$ -group.
Connect locality with indecomposability of $k G$ as a module over itself using $(19.19)$ .
State $(19.11)$ and identify where Nakayama's Lemma is used in its proof.
Use locality of $k H$ for $H$ a Sylow $p$ -subgroup to derive divisibility constraints on dimensions.

Definitions

$k G$: The group algebra: $k$ -vector space with basis $G$ and multiplication extending that of $G$ . $\dim_{k} k G = | G |$ for $G$ finite.
Augmentation $ε$: The $k$ -algebra map $k G \to k$ with $ε (g) = 1$ for all $g \in G$ .
Augmentation ideal $I (k G)$: $\ker ε$ , spanned over $k$ by ${g - 1 : g \in G}$ ; a two-sided ideal of codimension $1$ .
Reduced augmentation ideal: For a coefficient ring $R$ with maximal ideal $𝔪$ : the kernel of $R G \to R \to R / 𝔪$ , generated by $𝔪$ together with all $g - 1$ .
$p$ -group: A finite group whose order is a power of the prime $p$ ; equivalently every element has $p$ -power order.
Principal indecomposable: An indecomposable direct summand of $k G$ as a right module over itself; the projective indecomposables.

Throughout $G$ is a finite group. The theory for infinite groups is different: for infinite $G$ the group algebra is not artinian and the radical is a hard invariant.

Core Concepts

Why characteristic $p$ makes $g - 1$ nilpotent

If $g$ has order $p^{a}$ and $char k = p$ , then inside the commutative subalgebra $k ⟨ g ⟩$ the freshman's dream applies:

(g - 1)^{p^{a}} = g^{p^{a}} - 1 = 0 .

(B)

Valid because $k ⟨ g ⟩$ is commutative of characteristic $p$ , so the binomial coefficients $(\binom{p^{a}}{i})$ vanish for $0 < i < p^{a}$ .

So every generator of $I (k G)$ is nilpotent. That alone does not make the ideal nilpotent — a sum of nilpotents need not be nilpotent — but combined with the fact that a nontrivial $p$ -group has nontrivial centre it does, by induction.

Why any other prime destroys locality

Suppose $g \in G$ has order $q$ with $q \neq char k$ and $q$ prime. Then $k ⟨ g ⟩ ≅ k [x] / (x^{q} - 1)$ is semisimple by Maschke, commutative, and therefore a finite product of fields — in particular it has no nonzero nilpotent element. So $g - 1 \neq 0$ is not nilpotent, and it lies in the augmentation ideal. Any argument that forces $rad (k G) = I (k G)$ therefore fails.

g

of order

q \neq p

k ⟨ g ⟩

reduced

g - 1

not nilpotent

rad (k G) \neq I (k G)

Nilpotent radical: completely primary, not just local

In the local case $rad (k G) = I (k G)$ is nilpotent with $(rad k G)^{| G |} = 0$ , so $k G$ is completely primary. The exact nilpotency index is not $| G |$ in general — for $G = C_{2} \times C_{2}$ over $𝔽_{2}$ it is $3$ — and is computed by Jennings' theory of dimension subgroups. Over a local coefficient ring rather than a field the radical need not be nilpotent at all.

Key Results

Lemma—Nilpotence of the augmentation ideal

Let $k$ be a field of characteristic $p > 0$ and let $G$ be a finite $p$ -group. Then the augmentation ideal $I (k G)$ is nilpotent.

Proof

Induct on $| G |$ . If $| G | = 1$ then $I (k G) = 0$ .

Let $| G | > 1$ . A nontrivial finite $p$ -group has nontrivial centre, so choose a central element $z$ of order $p$ and set $Z = ⟨ z ⟩ ⊴ G$ . Since $z$ is central, $N = k G (z - 1)$ is a two-sided ideal, and by $(B)$ , $N^{p} = k G (z - 1)^{p} = 0$ .

The surjection $k G ↠ k [G / Z]$ has kernel exactly $N$ and carries $I (k G)$ onto $I (k [G / Z])$ . By induction $I (k [G / Z])^{m} = 0$ for some $m$ , so $I (k G)^{m} \subseteq N$ and therefore $I (k G)^{m p} \subseteq N^{p} = 0$ .

Theorem(19.10)Group algebras of

p

-groups

Let $k$ be a field of characteristic $p > 0$ and $G$ a finite $p$ -group. Then $rad (k G) = I (k G)$ , $(rad k G)^{| G |} = 0$ , and $k G / rad (k G) ≅ k$ . In particular $k G$ is an artinian local ring — indeed completely primary.

Proof

By the Lemma $I (k G)$ is nilpotent, hence a nil ideal, hence $I (k G) \subseteq rad (k G)$ by $(4.11)$ . Since $k G / I (k G) ≅ k$ is a field, $I (k G)$ is a maximal ideal, so $rad (k G) \subseteq I (k G)$ as well and the two coincide. The residue ring $k G / rad (k G) ≅ k$ is a division ring, so $k G$ is local by $(19.1) (3)$ . The sharper bound $(rad k G)^{| G |} = 0$ is $(8.8)$ .

Theorem(19.4 Ex.)Exactly when a group algebra is local

Let $k$ be a field and $G$ a nontrivial finite group, $R = k G$ . The following are equivalent.

$R$ is a local ring.
$R_{R}$ is an indecomposable $R$ -module.
$R / rad R$ is a simple ring.
$char k = p > 0$ and $G$ is a $p$ -group.

The hypothesis $G \neq 1$ is needed only for (4): if $G = 1$ then $k G = k$ is local in every characteristic.

Proof

**(4) $\Rightarrow$ (1)** is $(19.10)$ .

**(1) $\Leftrightarrow$ (2).** $k G$ is finite-dimensional over $k$ , hence right artinian, and idempotents of $End (R_{R}) ≅ R$ correspond to decompositions of $R_{R}$ . So this is exactly $(19.19)$ : for a nonzero right artinian ring, local $\Leftrightarrow$ no nontrivial idempotents $\Leftrightarrow$ $R_{R}$ indecomposable.

**(1) $\Rightarrow$ (3)** is trivial, a division ring being simple.

**(3) $\Rightarrow$ (1).** $R / rad R$ is semisimple; if it is simple then $R$ has a unique simple module $S$ up to isomorphism and $R / rad R ≅ M_{n} (D)$ with $D = {End}_{R} (S)$ and $\dim_{k} S = n \cdot \dim_{k} D$ . The trivial module $k_{triv}$ is a simple $R$ -module with $\dim_{k} = 1$ and ${End}_{R} (k_{triv}) = k$ , so $S ≅ k_{triv}$ , forcing $n = 1$ and $D = k$ . Hence $R / rad R ≅ k$ is a division ring and $R$ is local.

**(1) $\Rightarrow$ (4).** Assume $R = k G$ is local. Since $k G / I (k G) ≅ k$ , the augmentation ideal is a maximal left ideal, and by uniqueness $rad (k G) = I (k G)$ . As $k G$ is artinian its radical is nilpotent, so every $g - 1$ with $g \in G$ is nilpotent.

Suppose some $g \in G$ had order divisible by a prime $q \neq char k$ ; replacing $g$ by a suitable power we may assume $g$ has order exactly $q$ . Then $q$ is invertible in $k$ , so $k ⟨ g ⟩ ≅ k [x] / (x^{q} - 1)$ is semisimple by Maschke and commutative, hence a product of fields, hence has no nonzero nilpotent element. But $g - 1 \neq 0$ is nilpotent — a contradiction. Therefore every element of $G$ has order a power of $char k$ . Since $G \neq 1$ this forces $char k = p > 0$ , and $G$ is then a $p$ -group by Cauchy's theorem.

Proposition(19.11)Group rings over a local coefficient ring

Let $(R, 𝔪)$ be a commutative local ring whose residue field $k = R / 𝔪$ has characteristic $p > 0$ , and let $G$ be a finite $p$ -group. Then $A = R G$ is a local ring with $A / rad A ≅ k$ , and $rad A$ is the reduced augmentation ideal generated by $𝔪$ together with all $g - 1$ , $g \in G$ .

Proof

Let $V$ be any simple left $A$ -module. Being simple, $V$ is cyclic over $A$ , hence finitely generated over $R$ . By Nakayama's Lemma $(4.22)$ , $𝔪 V \neq V$ . But $𝔪 V$ is an $A$ -submodule of $V$ (as $𝔪$ is central in $A$ ), so simplicity gives $𝔪 V = 0$ .

Hence $V$ is a simple module over $A / 𝔪 A ≅ k G$ . Since $G$ is a $p$ -group and $char k = p$ , $(19.10)$ says $k G$ has exactly one simple module, the trivial one, so $G$ acts trivially on $V$ .

Therefore the ideal $ℑ$ generated by $𝔪$ and all $g - 1$ annihilates every simple left $A$ -module, giving $ℑ \subseteq rad A$ . Since $A / ℑ ≅ k$ is a field, $ℑ$ is maximal, so $rad A = ℑ$ and $A / rad A ≅ k$ is a division ring; $A$ is local by $(19.1) (3)$ .

Remark—Noncommutative coefficients

$(19.11)$ remains true when $R$ is a noncommutative local ring: the same reduction sends the problem to the case of a division ring of coefficients, where the argument behind $(8.8)$ again shows that $G$ acts trivially on every simple module. The result is central to the theory of integral representations of finite groups.

Corollary(19.30)Dickson's divisibility theorem

Let $k$ be a field of characteristic $p > 0$ , $G$ a finite group, $H$ a Sylow $p$ -subgroup, and $U$ a principal indecomposable right $k G$ -module. Then $| H |$ divides $\dim_{k} U$ . The proof restricts $U$ to $k H$ , which is local by $(19.10)$ , and uses that finitely generated projectives over a local ring are free $(19.29)$ .

Theorem(19.31)Multiplicities in the regular module

Let $k$ be a splitting field for $G$ of characteristic $p > 0$ and $R = k G$ . Then every irreducible right $R$ -module occurs as a composition factor of $R_{R}$ with multiplicity divisible by the $p$ -part of $| G |$ .

The multiplicity of an irreducible $V$ as a composition factor of $R_{R}$ equals the $k$ -dimension of a principal indecomposable left $k G$ -module; applying $(19.30)$ on the left gives the divisibility. The identification of multiplicities with dimensions of principal indecomposables belongs to the block theory of §25 and is not proved here.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Quotient by a central subgroup of order $p$

A nontrivial $p$ -group has nontrivial centre, so $k G$ always has a central square-zero-like ideal $k G (z - 1)$ with $k G / N ≅ k [G / Z]$ . Induction on $| G |$ then works. This is the standard induction for $p$ -group algebras.

Move 2

Test nilpotence inside a cyclic subalgebra

To show $g - 1$ is not nilpotent, restrict to $k ⟨ g ⟩$ , which is a quotient of $k [x]$ and completely understood: semisimple and reduced when the order of $g$ is invertible in $k$ .

Move 3

Identify the unique simple module

The trivial module is always simple of dimension one. If a group algebra has only one simple module, it must be the trivial one, and the Wedderburn quotient collapses to $k$ . This converts simple quotient into division ring quotient.

The asymmetry between the two directions is worth noting. Sufficiency ( $p$ -group $\Rightarrow$ local) is a nilpotence computation. Necessity is a reduction argument: it uses that $k G$ is artinian, so the radical is nilpotent, and then tests one element at a time inside a cyclic subalgebra.

Worked Example

The Klein four group in characteristic $2$

Let $k = 𝔽_{2}$ and $G = ⟨ g ⟩ \times ⟨ h ⟩ ≅ C_{2} \times C_{2}$ . Setting $x = g - 1$ and $y = h - 1$ , we have $x^{2} = g^{2} - 2 g + 1 = g^{2} + 1 = 0$ and likewise $y^{2} = 0$ , and $x, y$ commute. Counting dimensions gives

k G ≅ 𝔽_{2} [x, y] / (x^{2}, y^{2}), rad (k G) = (x, y), \dim_{k} k G = 4 .

(E.1)

The radical has $k$ -basis ${x, y, x y}$ , so $\dim_{k} rad (k G) = 3$ and $k G / rad (k G) ≅ 𝔽_{2}$ . Its powers are $(rad)^{2} = k x y$ of dimension $1$ and $(rad)^{3} = 0$ : the nilpotency index is $3$ , comfortably below the guaranteed bound $| G | = 4$ .

Cross-checks. The units are exactly $1 + rad (k G)$ , so $| U (k G) | = 2^{3} = 8$ out of $16$ elements. The only idempotents are $0$ and $1$ , as $(19.19)$ requires. And by $(19.29)$ every finitely generated projective $k G$ -module is free, so projective $k G$ -modules have $k$ -dimension divisible by $4$ — in particular the trivial module is not projective.

Three group algebras that are not local

$ℚ C_{2} ≅ ℚ \times ℚ$ , via $e = \frac{1}{2} (1 + g)$ . Characteristic $0$ , so Maschke applies and the algebra is semisimple with two blocks. Not local.
$𝔽_{2} C_{3} ≅ 𝔽_{2} [x] / (x^{3} - 1) ≅ 𝔽_{2} [x] / (x - 1) \times 𝔽_{2} [x] / (x^{2} + x + 1) ≅ 𝔽_{2} \times 𝔽_{4}$ . Here $p = 2$ but $G$ is a $3$ -group, so the hypotheses of $(19.10)$ fail on the group side. Not local.
$𝔽_{3} S_{3}$ : here $p = 3$ divides $| G | = 6$ but $G$ is not a $3$ -group. The quotient is $𝔽_{3} S_{3} / rad ≅ 𝔽_{3} \times 𝔽_{3}$ , of dimension $2$ , so there are two simple modules and two principal indecomposables, each of dimension $3$ . Not local, but the Sylow subalgebra $𝔽_{3} A_{3} ≅ 𝔽_{3} [t] / (t^{3})$ is.

A local group ring that is not artinian

Take $R = ℤ_{(2)}$ , a commutative local ring with $𝔪 = 2 ℤ_{(2)}$ and residue field $𝔽_{2}$ of characteristic $2$ , and $G = C_{2} = ⟨ g ⟩$ . By $(19.11)$ , $A = ℤ_{(2)} C_{2}$ is local with $rad A = (2, g - 1)$ and $A / rad A ≅ 𝔽_{2}$ .

Process and Workflow

Is my group ring local?

Coefficients a field, $char k ∤ | G |$ No (for

G \neq 1

). Maschke makes

k G

semisimple, and

k G

is not a division ring because

g

of order

n

gives zero divisors.

Coefficients a field of characteristic $p$ , $G$ a $p$ -groupYes, and completely primary:

rad (k G) = I (k G)

with

(rad)^{| G |} = 0

and residue ring

k

.

Coefficients a field of characteristic $p$ , $G$ not a $p$ -groupNo. Some

g \in G

has order divisible by a prime

q \neq p

, and

g - 1

is a non-nilpotent element of the augmentation ideal; equivalently the semisimple quotient is not a division ring.

Coefficients a local ring $(R, 𝔪)$ with residue characteristic $p$ , $G$ a $p$ -groupYes by

(19.11)

, with

rad = (𝔪, g - 1 : g \in G)

— but do not expect the radical to be nilpotent.

Compare $char k$ with $| G |$ If the characteristic does not divide the order, stop: semisimple, not local.

Test whether $G$ is a $p$ -groupA single element of order prime to

p

is enough to rule out locality.

Identify the radicalIn the local case it is the augmentation ideal, of codimension

1

. Otherwise compute it, or use the ideal generated by

O_{p} (G)

as a first approximation.

Exploit localityProjectives become free, Krull–Schmidt applies with local endomorphism rings, and restriction to a Sylow

p

-subgroup becomes a usable tool.

Comparison and Classification

Is $k G$ local?
$k$	$G$	$k G / rad$	Local?
$𝔽_{p}$	$C_{p^{n}}$	$𝔽_{p}$	yes
$𝔽_{2}$	$C_{2} \times C_{2}$	$𝔽_{2}$	yes
$k$ , $char k = p$	any finite $p$ -group	$k$	yes
$ℚ$	$C_{2}$	$ℚ \times ℚ$	no
$𝔽_{2}$	$C_{3}$	$𝔽_{2} \times 𝔽_{4}$	no
$𝔽_{3}$	$S_{3}$	$𝔽_{3} \times 𝔽_{3}$	no
$ℂ$	any $G \neq 1$	$\prod_{i} M_{n_{i}} (ℂ)$	no
any $k$	trivial group	$k$	yes

Properties of $k G$ across the three regimes
	$char k ∤ \| G \|$	$p ∣ \| G \|$ , $G$ not a $p$ -group	$G$ a $p$ -group, $char k = p$
Semisimple	yes	no	no
Local	no	no	yes
$rad$ nilpotent	yes (it is $0$ )	yes	yes
Unique simple module	no	no	yes
Nontrivial idempotents	yes	yes	no
Projectives are free	no	no	yes

Properties of $k G$ across the three regimes

The first column assumes $G \neq 1$ ; for the trivial group every entry degenerates to the field case.

Relationship Map

Group algebras $k G$ , $G$ finitealways artinian, always with augmentation ideal

I (k G)

$char k = p$ divides $| G |$ modular case;

rad (k G) \neq 0

$O_{p} (G) \neq 1$ the ideal generated by

{h - 1 : h \in O_{p} (G)}

lies in the radical

$G$ is a $p$ -group

rad (k G) = I (k G)

,

k G

local and completely primary

$k G$ local — $k$ a field, $G \neq 1$ finite
- equivalent to
  - $k G$ has no nontrivial idempotent (19.19)
  - $k G$ indecomposable as a right module over itself
  - $k G / rad$ is a simple ring
  - $char k = p$ and $G$ is a $p$ -group
- gives
  - projectives are free (19.29)
  - Dickson's divisibility (19.30)
  - multiplicity divisibility over a splitting field (19.31)
- extends to
  - $R G$ for $(R, 𝔪)$ local, $char (R / 𝔪) = p$ (19.11)
  - noncommutative local coefficient rings
  - $ℤ_{p} G$ in integral representation theory

Applications and Industry Use

Applications here means where this structure is used — inside mathematics and in the engineering and computing disciplines that consume it.

Modular representation theory

Restriction to a Sylow subgroup

The main technical use of $(19.10)$ : $k H$ is local for $H$ a Sylow $p$ -subgroup, so restricting a projective $k G$ -module to $H$ gives a free module. Dickson's theorem and much of the theory of vertices and sources start here.

Integral representation theory

$ℤ_{p} G$ for a $p$ -group

$(19.11)$ makes $ℤ_{p} G$ local for $G$ a $p$ -group, which is what allows lattices over $ℤ_{p} G$ to be studied by reduction modulo $p$ and lifting — the standard technique for integral representations.

Cohomology of groups

Filtration by radical powers

For a $p$ -group in characteristic $p$ the radical filtration of $k G$ is the filtration by powers of the augmentation ideal, and its associated graded algebra is computed by Jennings' theorem — the entry point to modular group cohomology.

Computer algebra

Meataxe and module decomposition

When $k G$ is local there is nothing to split: the regular module is indecomposable and decomposition algorithms terminate immediately. That base case is what recursive splitting routines over general $k G$ reduce to.

The honest framing: locality of $k G$ is a rare event, and its value lies in being the extreme opposite of the semisimple case. Between the two extremes sits everything interesting in modular representation theory, and both endpoints are used as reference points.

Computational Notes

Computational notes cover algorithms, cost and library behaviour rather than manufacturing process.

Deciding locality of $k G$ for an explicitly given finite group is trivial computationally: factor $| G |$ and compare with $char k$ . No algebra computation is needed, which is unusual for a ring-theoretic property.
Computing $rad (k G)$ in the local case costs nothing — it is the codimension-one augmentation ideal — while in the general modular case it requires a radical computation on an algebra of dimension $| G |$ .
The nilpotency index of $I (k G)$ for a $p$ -group is not read off from $| G |$ ; Jennings' theorem expresses it through the dimension subgroup series, and implementations in GAP compute it from the Jennings series rather than by multiplying ideals.
For large $p$ -groups the dimension $| G |$ makes dense linear algebra impractical; practical systems work with the Jennings filtration and with basis-free representations of the augmentation ideal.

Failure Modes and Common Mistakes

Do not assume $rad (k G)$ equals the augmentation ideal outside the $p$ -group case; in general it is the smaller ideal $rad (k G) ⊊ I (k G)$ , and for $char k ∤ | G |$ it is zero.
Do not read Dickson's theorem as constraining all indecomposable modules; it constrains the principal — that is, projective — indecomposables.
Do not apply $(19.31)$ without the splitting-field hypothesis; over a non-splitting field the multiplicities involve the endomorphism division rings of the irreducibles.
Do not confuse $O_{p} (G)$ , the largest normal $p$ -subgroup, with a Sylow subgroup. The ideal generated by ${h - 1 : h \in O_{p} (G)}$ lies in $rad (k G)$ , but the corresponding statement for a non-normal Sylow subgroup is false.

Historical Notes and Lessons Learned

1898MaschkeMaschke proves that $k G$ is semisimple when $char k$ does not divide $| G |$ — the theorem that confines local group algebras to the modular case.
1900sDicksonDickson initiates modular representation theory and proves the divisibility of the dimensions of principal indecomposables by the order of a Sylow $p$ -subgroup.
1935–1950sBrauerBrauer builds modular character theory and block theory, in which the $p$ -group case functions as the extreme local example and defect groups measure the distance from it.
1941JenningsJennings determines the associated graded algebra of the augmentation ideal filtration for a $p$ -group in characteristic $p$ , pinning down the nilpotency index.
1950s–1960sIntegral representationsThe study of lattices over $ℤ_{p} G$ and $ℤ G$ turns $(19.11)$ into a working tool: locality of the $p$ -adic group ring is what makes reduction and lifting arguments available.

The historical pattern is instructive. Maschke's theorem is a statement about when representation theory is easy; the local case is a statement about when it is maximally hard in a controlled way. Almost every technique in modular representation theory is a way of interpolating between the two by restricting to $p$ -subgroups, where locality is restored.

Quick Reference

Criterion

k G

local

\Leftrightarrow

char k = p > 0

and

G

a

p

-group (

G \neq 1

)

Radical

rad (k G) = I (k G)

, of codimension

1

Residue ring

k G / rad (k G) ≅ k

Nilpotency

(rad k G)^{| G |} = 0

; exact index by Jennings

Key identity

(g - 1)^{p^{a}} = g^{p^{a}} - 1 = 0

for

g

of order

p^{a}

Coefficient rings

R G

local for

(R, 𝔪)

local with

char (R / 𝔪) = p

Consequencef.g. projectives free;

| H | ∣ \dim_{k} U

for principal indecomposables

ObstructionAn element of order prime to

char k

yields a non-nilpotent

g - 1

Numbered results used on this page
Reference	Statement	Hypotheses
(8.8)	$G$ acts trivially on every simple $k G$ -module	$char k = p$ , $G$ a finite $p$ -group
(19.10)	$k G$ local, $rad = I (k G)$ , $(rad)^{\| G \|} = 0$	$char k = p$ , $G$ a finite $p$ -group
(19.11)	$R G$ local with residue ring $k$	$(R, 𝔪)$ local, $char k = p$ , $G$ a $p$ -group
(19.19)	Local $\Leftrightarrow$ no nontrivial idempotent	$R \neq 0$ right artinian
(19.29)	f.g. projectives are free	$R$ local
(19.30)	$\| H \|$ divides $\dim_{k} U$	$U$ principal indecomposable, $H$ Sylow $p$
(19.31)	Multiplicity divisible by the $p$ -part of $\| G \|$	$k$ a splitting field, $char k = p$

Frequently Asked Questions

Why does

k G

fail to be local as soon as some element has order prime to the characteristic?

If $g$ has prime order $q \neq char k$ then $k ⟨ g ⟩ ≅ k [x] / (x^{q} - 1)$ is semisimple and commutative, hence a product of fields with no nonzero nilpotents, so $g - 1$ is a non-nilpotent element of the augmentation ideal. But if $k G$ were local its radical would be the augmentation ideal and, $k G$ being artinian, would be nilpotent.

Is

k G

local when

char k = p

divides

| G |

but

G

is not a

p

-group?

No. $𝔽_{3} S_{3}$ is the smallest instructive example: it is not semisimple, but its radical has codimension $2$ and the semisimple quotient $𝔽_{3} \times 𝔽_{3}$ supplies a nontrivial idempotent. What survives is that the Sylow subalgebra $𝔽_{3} A_{3}$ is local, and that is what the theory uses.

Does the group have to be finite?

Yes for the statements here. For infinite groups $k G$ is not artinian, the augmentation ideal has infinite codimension in no useful sense, and determining $rad (k G)$ is a deep problem — for many infinite groups it is not even known whether the radical vanishes.

What is the exact nilpotency index of the augmentation ideal?

$(19.10)$ guarantees only that it is at most $| G |$ . The exact value is determined by Jennings' theory via the dimension subgroup series; for $C_{2} \times C_{2}$ over $𝔽_{2}$ it is $3$ , while for the cyclic group $C_{p^{n}}$ over $𝔽_{p}$ it is exactly $p^{n} = | G |$ .

Why does

(19.11)

need Nakayama's Lemma?

To show $𝔪 V \neq V$ for a simple module $V$ . Simplicity makes $V$ cyclic over $A = R G$ and therefore finitely generated over $R$ , which is exactly the hypothesis Nakayama needs. Once $𝔪 V \neq V$ , simplicity upgrades it to $𝔪 V = 0$ and the problem drops to the residue field.

Is a local group ring always completely primary?

Over a field, yes: $k G$ is finite-dimensional, so its radical is nilpotent. Over a local coefficient ring, no: $ℤ_{(p)} C_{p}$ is local, but its radical contains $p$ and is not nil.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §19 (pp. 294–310), especially (19.10), (19.11), (19.30), (19.31) and Exercise 19.4; also §8 for (8.8).
D. S. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, 1977 (radicals of group rings, augmentation ideals and the modular case).
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley, 1981 (modular representations, principal indecomposables and integral representations).
S. A. Jennings, “The structure of the group ring of a p-group over a modular field”, Transactions of the American Mathematical Society 50 (1941), 175–185.
J. L. Alperin, Local Representation Theory, Cambridge Studies in Advanced Mathematics 11, Cambridge University Press, 1986.
L. E. Dickson, “On the group defined for any given field by the multiplication table of any given finite group”, Transactions of the American Mathematical Society 3 (1902), 285–301.

AI Suggested Questions

State and prove Jennings' theorem on the graded algebra associated with the augmentation ideal filtration.
How is $rad (k G)$ described when $G$ has a normal $p$ -subgroup but is not itself a $p$ -group?
What is known about $rad (k G)$ for an infinite group $G$ , and what does the semiprimitivity problem for group algebras say?
Work out the principal indecomposable modules of $𝔽_{2} A_{4}$ and check Dickson's divisibility explicitly.
How does the theory of vertices and sources use locality of $k H$ for $p$ -subgroups $H$ ?
For which finite groups $G$ and fields $k$ is $k G$ semiperfect but not local, and what are the principal indecomposables?
Explain how $(19.11)$ underpins reduction-and-lifting arguments for lattices over $ℤ_{p} G$ .

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