Uniqueness in the Wedderburn–Artin Decomposition

Q: How is (3.8) stronger than saying the factors are unique up to isomorphism?

It concludes B_i = C_i as subsets of R, not just B_i ≅ C_i. So there is genuinely only one way to write a ring as a direct sum of indecomposable ideals. Compare the left-module situation, where M_2(Q) splits into two minimal left ideals in infinitely many different ways, all of them isomorphic.

Q: Why does this give a second proof of Wedderburn-Artin?

Because the decomposition R = B_1 ⊕ … ⊕ B_r is obtained from the socle structure alone, with no endomorphism-ring computation. Each B_i is then simple and left artinian, so the classification of simple artinian rings as M_n(D) finishes the job. The two proofs are genuinely independent, and Lam presents both.

Q: How many ideals does a semisimple ring have?

Exactly 2^r, where r is the number of simple components: every ideal is the sum of the components it contains, so the ideal lattice is the Boolean lattice of subsets of 1,…,r. In particular a semisimple ring is simple exactly when r = 1.

Executive Summary

Wedderburn–Artin produces an isomorphism $R ≅ M_{n_{1}} (D_{1}) \times \dots \times M_{n_{r}} (D_{r})$ . This page explains in what sense that decomposition is unique, and it is a stronger sense than one might expect: the ideals $B_{1}, \dots, B_{r}$ of $R$ corresponding to the factors are uniquely determined as subsets of $R$ , not merely up to isomorphism.

Two lemmas do the work. Lemma $(3.8)$ says a decomposition of any ring into indecomposable ideals is unique. Lemma $(3.9)$ constructs the components intrinsically as isotypic sums of minimal left ideals, with no reference to Wedderburn–Artin — which yields a second, independent proof of the structure theorem.

As idealsStrength of the uniqueness

$B_{𝔄}$ Intrinsic construction

$2^{r}$ Total number of ideals

(3.8)–(3.9)Lam's numbering

Overview

Uniqueness statements in algebra come in three strengths, and confusing them is a common source of error. A decomposition can be unique as a list of subobjects, unique up to isomorphism of the pieces, or accompanied by no canonical isomorphism at all. The Wedderburn decomposition exhibits all three at once, at different levels.

R = B_{1} \oplus B_{2} \oplus \dots \oplus B_{r}, B_{i} ⊴ R indecomposable, B_{i} ≅ M_{n_{i}} (D_{i}),

(3.8)–(3.9)

The $B_{i}$ are unique as ideals; the pairs $(n_{i}, D_{i})$ are unique up to permutation and isomorphism; the isomorphism $B_{i} ≅ M_{n_{i}} (D_{i})$ is not canonical.

The reason the ideals themselves are canonical is that they have an intrinsic description with no choices in it: $B_{i}$ is the sum of all minimal left ideals of $R$ lying in one isomorphism class. Sums of all objects with a property are automatically invariant under every automorphism, which is what canonicity means.

By contrast, decompositions into left ideals are wildly non-unique: $M_{2} (ℚ)$ is a direct sum of two minimal left ideals in infinitely many ways, one for each pair of distinct points of $ℙ^{1} (ℚ)$ .

Learning Objectives

State $(3.8)$ with the correct notion of indecomposability — as an ideal, not as a ring or as a left ideal.
Prove $(3.8)$ using the correspondence between ideals of $R$ and products of ideals of the factors.
Prove both parts of $(3.9)$ for an arbitrary ring.
Show each isotypic component of a semisimple ring is a simple left artinian ring with identity $e_{i}$ .
Deduce that a semisimple ring with $r$ components has exactly $2^{r}$ two-sided ideals.
Compute the central idempotents of $ℚ S_{3}$ and verify orthogonality.

Definitions

Definition—Isotypic component of a minimal left ideal

Let $R$ be any ring and $𝔄$ a minimal left ideal of $R$ . Define

B_{𝔄} = \sum {𝔅 : 𝔅 a minimal left ideal of R, 𝔅 ≅ 𝔄 as left R -modules} .

No hypothesis on $R$ is needed for the definition or for $(3.9)$ below; if $R$ has no minimal left ideal — as with $ℤ$ or the Weyl algebra $A_{1} (ℂ)$ — the construction is simply empty.

Indecomposable as an ideal: $B \neq 0$ and $B$ is not $C \oplus C'$ for nonzero ideals $C, C'$ of the ambient ring $R$ . Weaker than being indecomposable as a left $R$ -module.
Simple component: An indecomposable ideal appearing in the decomposition of a semisimple ring; equivalently a minimal two-sided ideal.
Orthogonal idempotents: Idempotents $e_{1}, \dots, e_{r}$ with $e_{i} e_{j} = 0$ for $i \neq j$ . Complete if in addition $e_{1} + \dots + e_{r} = 1$ .
Block: Synonym for simple component in the semisimple case; in general, the ideal generated by a centrally primitive idempotent.
$soc$: The socle: the sum of all simple submodules. For a semisimple ring $soc (_{R} R) = R$ , and the $B_{i}$ are its isotypic pieces.

Ideal means two-sided ideal throughout unless the words left or right appear.

Core Concepts

Ideal decompositions are idempotent decompositions

If $R = B_{1} \oplus \dots \oplus B_{r}$ with each $B_{i}$ an ideal, write $1 = e_{1} + \dots + e_{r}$ with $e_{i} \in B_{i}$ . For $b \in B_{i}$ we get $b = b e_{1} + \dots + b e_{r}$ , and $b e_{j} \in B_{i} \cap B_{j} = 0$ for $j \neq i$ , so $b = b e_{i}$ ; symmetrically $b = e_{i} b$ . Hence each $e_{i}$ is the identity element of $B_{i}$ , the $e_{i}$ are orthogonal idempotents summing to $1$ , and each is central in $R$ .

So decomposing a ring into ideals is the same as splitting $1$ into orthogonal central idempotents, and $B_{i}$ is indecomposable exactly when $e_{i}$ is centrally primitive. Every statement on this page can be translated into that language; The Block Decomposition of a Ring takes that route.

Why isotypic sums are two-sided

A minimal left ideal $𝔅$ need not be a right ideal. But right multiplication by a fixed $r \in R$ is a homomorphism of left modules $𝔅 \to 𝔅 r$ , so $𝔅 r$ is either $0$ or a homomorphic image of the simple module $𝔅$ , hence isomorphic to $𝔅$ and again a minimal left ideal. Either way $𝔅 r$ lands inside the same isotypic sum. Summing over all $𝔅$ of one type gives a right ideal, and it was a left ideal by construction.

Why distinct types annihilate each other

Suppose $𝔄 not ≅ 𝔄'$ are minimal left ideals and $𝔄 a' \neq 0$ for some $a' \in 𝔄'$ . Then $𝔄 a' \subseteq 𝔄'$ is a nonzero submodule, so it equals $𝔄'$ ; and $x \mapsto x a'$ is a nonzero left-module map $𝔄 \to 𝔄'$ , hence an isomorphism by Schur's Lemma. That contradicts $𝔄 not ≅ 𝔄'$ . So $𝔄 𝔄' = 0$ , and summing gives $B_{𝔄} B_{𝔄'} = 0$ .

Three levels of uniqueness

The decomposition of $R$ — how unique is it?
- The ideals $B_{i}$
  - unique as subsets of $R$ — no choices at all
  - characterised as the minimal two-sided ideals
  - permuted by any automorphism of $R$
- The pairs $(n_{i}, D_{i})$
  - unique up to permutation of the indices
  - $D_{i}$ unique up to ring isomorphism
  - $n_{i}$ literally unique once the ordering is fixed
- The isomorphism $B_{i} ≅ M_{n_{i}} (D_{i})$
  - not unique and not canonical
  - depends on a choice of basis of the simple module
  - the ambiguity is described by Skolem–Noether

Key Results

Lemma(3.8)Uniqueness of decompositions into indecomposable ideals

Let $R$ be a ring with identity possessing nonzero ideals $B_{1}, \dots, B_{r}$ and $C_{1}, \dots, C_{s}$ such that

R = B_{1} \oplus \dots \oplus B_{r} = C_{1} \oplus \dots \oplus C_{s},

where each $B_{i}$ and each $C_{j}$ is indecomposable as an ideal of $R$ , that is, not a direct sum of two nonzero ideals of $R$ . Then $r = s$ and, after a permutation of indices, $B_{i} = C_{i}$ for $1 \leq i \leq r$ — equality of ideals, not merely isomorphism.

Proof

Regarding each $B_{i}$ as a ring with identity $e_{i}$ , the decomposition gives a ring isomorphism $R ≅ B_{1} \times \dots \times B_{r}$ . Under a product decomposition, every ideal of $R$ has the form $I_{1} \times \dots \times I_{r}$ with $I_{i} ⊴ B_{i}$ : indeed for $I ⊴ R$ we have $I = ⨁_{i} e_{i} I$ because the $e_{i}$ are central and sum to $1$ .

Apply this to $C_{1}$ : it corresponds to $I_{1} \times \dots \times I_{r}$ , and $C_{1} = ⨁_{i} e_{i} C_{1}$ is a decomposition of $C_{1}$ into ideals of $R$ . Indecomposability of $C_{1}$ forces all but one summand to vanish; after permuting the $B$ 's we may assume $C_{1} = e_{1} C_{1} \subseteq B_{1}$ .

By the symmetric argument applied to $B_{1}$ inside the second decomposition, $B_{1} \subseteq C_{i}$ for some $i$ . Then $C_{1} \subseteq B_{1} \subseteq C_{i}$ , and since $C_{1} \neq 0$ and $C_{1} \cap C_{i} = 0$ for $i \neq 1$ , we must have $i = 1$ . Hence $C_{1} \subseteq B_{1} \subseteq C_{1}$ , that is $B_{1} = C_{1}$ .

Now $B_{2} \oplus \dots \oplus B_{r} = C_{2} \oplus \dots \oplus C_{s}$ , both being complements of the same ideal, and each summand is still indecomposable as an ideal of $R$ . Repeating the argument $r$ times, or inducting on $r$ , gives $r = s$ and $B_{i} = C_{i}$ after permutation.

Lemma(3.9)Isotypic components

Let $R$ be any ring with identity and let $𝔄$ be a minimal left ideal of $R$ . Then:

$B_{𝔄}$ , the sum of all minimal left ideals isomorphic to $𝔄$ , is a two-sided ideal of $R$ ;
if $𝔄, 𝔄'$ are minimal left ideals that are not isomorphic as left $R$ -modules, then $B_{𝔄} \cdot B_{𝔄'} = 0$ .

Proof

(1) $B_{𝔄}$ is a sum of left ideals, hence a left ideal. For the right-hand closure it suffices to show $𝔅 r \subseteq B_{𝔄}$ for every minimal left ideal $𝔅 ≅ 𝔄$ and every $r \in R$ . The map $𝔅 \to 𝔅 r$ , $x \mapsto x r$ , is a homomorphism of left $R$ -modules and is onto, so its image is either $0$ or isomorphic to the simple module $𝔅$ . In the second case $𝔅 r$ is a minimal left ideal isomorphic to $𝔄$ . In both cases $𝔅 r \subseteq B_{𝔄}$ .

(2) Since $B_{𝔄}$ and $B_{𝔄'}$ are sums of minimal left ideals of the respective types, it suffices to show $𝔄 𝔄' = 0$ for minimal left ideals $𝔄 not ≅ 𝔄'$ . Suppose $𝔄 a' \neq 0$ for some $a' \in 𝔄'$ . Then $𝔄 a'$ is a nonzero submodule of the simple module $𝔄'$ , so $𝔄 a' = 𝔄'$ . But $x \mapsto x a'$ is a nonzero homomorphism of left $R$ -modules $𝔄 \to 𝔄'$ between simple modules, hence an isomorphism by Schur's Lemma — contradicting $𝔄 not ≅ 𝔄'$ .

Theorem(3.9a)Intrinsic construction of the simple components

Let $R$ be a semisimple ring. Write $_{R} R = 𝔄_{1} \oplus \dots \oplus 𝔄_{n}$ with each $𝔄_{j}$ a minimal left ideal, indexed so that $𝔄_{1}, \dots, 𝔄_{r}$ are pairwise non-isomorphic and every $𝔄_{j}$ is isomorphic to exactly one of them. Put $B_{i} = B_{𝔄_{i}}$ . Then

R = B_{1} \oplus B_{2} \oplus \dots \oplus B_{r},

each $B_{i}$ is a simple left artinian ring, each is indecomposable as an ideal of $R$ , and every minimal left ideal of $R$ lies in exactly one $B_{i}$ . These are the simple components of $R$ , and by $(3.8)$ they are uniquely determined as ideals.

Proof

The sum is everything and is direct. Each $B_{i}$ is an ideal by $(3.9) (1)$ , and their sum contains every $𝔄_{j}$ , hence contains $R$ . For directness, note $B_{i} B_{j} = 0$ for $i \neq j$ by $(3.9) (2)$ ; if $x \in B_{i} \cap \sum_{j \neq i} B_{j}$ then writing $x = x e$ with $e$ the component of $1$ in $B_{i}$ and expanding shows $x \in B_{i} \cdot \sum_{j \neq i} B_{j} = 0$ .

** $B_{i}$ is left artinian.** $R$ is left artinian because it is semisimple, and $B_{i} ≅ R / ⨁_{j \neq i} B_{j}$ is a quotient ring of $R$ , hence left artinian.

** $B_{i}$ is simple.** Let $0 \neq I ⊴ B_{i}$ . Since $B_{j} B_{i} = 0$ for $j \neq i$ , $I$ is also an ideal of $R$ . As a nonzero left ideal of the semisimple ring $R$ , $I$ contains a minimal left ideal $𝔄$ of $R$ , and $𝔄 \subseteq B_{i}$ forces $B_{𝔄} = B_{i}$ . It remains to show every minimal left ideal $𝔅 ≅ 𝔄$ lies in $I$ , for then $B_{i} = B_{𝔄} \subseteq I$ .

Since $𝔄$ is a direct summand of $_{R} R$ , we have $𝔄 = R e$ for an idempotent $e \in 𝔄$ , so $a e = a$ for all $a \in 𝔄$ . Fix an isomorphism $ϕ : 𝔄 \to 𝔅$ of left $R$ -modules. For $a \in 𝔄$ , $ϕ (a) = ϕ (a e) = a ϕ (e)$ , so $𝔅 = ϕ (𝔄) = 𝔄 ϕ (e) \subseteq I R \subseteq I$ , using that $I$ is a right ideal. Hence $I = B_{i}$ and $B_{i}$ is simple.

Indecomposability. A simple ring has no decomposition into two nonzero ideals of itself, and ideals of $B_{i}$ are ideals of $R$ by the annihilation property, so $B_{i}$ is indecomposable as an ideal of $R$ .

Corollary(3.9b)A second proof of Wedderburn–Artin, and the ideal count

Let $R$ be semisimple with simple components $B_{1}, \dots, B_{r}$ . Then each $B_{i}$ , being simple and left artinian, is isomorphic to $M_{n_{i}} (D_{i})$ for a division ring $D_{i}$ and an integer $n_{i} \geq 1$ , both uniquely determined; this reproves $(3.5)$ without the endomorphism-ring computation. Moreover every ideal of $R$ is a direct sum of a subset of the $B_{i}$ , so $R$ has exactly $2^{r}$ two-sided ideals, and the $B_{i}$ are precisely the minimal nonzero ideals of $R$ .

Proof

The classification of simple left artinian rings as matrix rings over division rings, with uniqueness of $n$ and $D$ , is $(3.13)$ ; see Simple Artinian Rings and Minimal One-Sided Ideals. For the ideal count, let $I ⊴ R$ and let $e_{1}, \dots, e_{r}$ be the central idempotents with $e_{i}$ the identity of $B_{i}$ . Then $I = ⨁_{i} e_{i} I$ , and $e_{i} I$ is an ideal of the simple ring $B_{i}$ , hence $0$ or $B_{i}$ . So $I$ is the sum of the $B_{i}$ over some subset $S \subseteq {1, \dots, r}$ , and distinct subsets give distinct ideals.

Proof Techniques and Method

How these proofs work, and which moves to reuse.

Move 1

Mutual containment, not counting

To prove two decompositions coincide, show $C_{1} \subseteq B_{1}$ and $B_{1} \subseteq C_{i}$ , then use disjointness to force $i = 1$ . This gives equality on the nose, which a counting or Krull–Schmidt argument would not.

Move 2

Sum over everything of a type

Objects defined by 'the sum of all subobjects with property $P$ ' are automatically invariant under automorphisms and usually acquire extra structure for free — here, two-sidedness out of a one-sided definition.

Move 3

Split off an idempotent to move maps around

A direct summand $𝔄 = R e$ satisfies $a e = a$ , so any module map out of $𝔄$ is right multiplication by $ϕ (e)$ . This converts a module isomorphism into an inclusion of ideals.

Move 2 is the conceptual heart. The isotypic component is defined without reference to any decomposition, so it cannot depend on one — canonicity is built into the definition rather than proved afterwards. That is why $(3.9)$ can be stated for an arbitrary ring while $(3.8)$ needs a finite decomposition to exist.

Worked Example

The three blocks of $ℚ S_{3}$

By The Wedderburn–Artin Theorem, $ℚ S_{3} ≅ ℚ \times ℚ \times M_{2} (ℚ)$ . The corresponding central idempotents are computed from the characters. Write $c = (123) + (132)$ for the sum of the three-cycles and $t$ for the sum of the three transpositions.

e_{triv} = \frac{1}{6} (1 + c + t), e_{sgn} = \frac{1}{6} (1 + c - t), e_{std} = \frac{1}{3} (2 \cdot 1 - c) .

(E.1)

Each is a class function, hence central; they are orthogonal and sum to $1$ .

Check idempotency of the interesting one. From $c^{2} = ((123) + (132))^{2} = (132) + 1 + 1 + (123) = 2 \cdot 1 + c$ , we get

e_{std}^{2} = \frac{1}{9} (4 \cdot 1 - 4 c + c^{2}) = \frac{1}{9} (4 \cdot 1 - 4 c + 2 \cdot 1 + c) = \frac{1}{9} (6 \cdot 1 - 3 c) = e_{std} .

(E.2)

Orthogonality is equally quick: $e_{triv} g = e_{triv}$ for every $g \in S_{3}$ , so $e_{std} e_{triv} = \frac{1}{3} (2 e_{triv} - 2 e_{triv}) = 0$ ; and $e_{sgn} g = sgn (g) e_{sgn}$ with $sgn$ equal to $+ 1$ on three-cycles, giving $e_{std} e_{sgn} = 0$ . Finally $e_{triv} + e_{sgn} + e_{std} = \frac{1}{3} (1 + c) + \frac{1}{3} (2 \cdot 1 - c) = 1$ .

Simple components of $ℚ S_{3}$
Component	Identity element	Isomorphism type	$\dim_{ℚ}$
$B_{1}$	$e_{triv}$	$ℚ$	1
$B_{2}$	$e_{sgn}$	$ℚ$	1
$B_{3}$	$e_{std}$	$M_{2} (ℚ)$	4
Total	$1$	$ℚ S_{3}$	6

$B_{3}$ is the isotypic component of the standard module: the sum of all minimal left ideals isomorphic to the two-dimensional simple module. Its dimension $4 = n_{3}^{2} \dim_{ℚ} D_{3} = 2^{2} \cdot 1$ agrees, and the multiplicity of the standard module in the regular module is $n_{3} = 2$ .

The smallest interesting case: $ℤ / 6$

$ℤ / 6 ≅ 𝔽_{2} \times 𝔽_{3}$ with central idempotents $e_{1} = 3$ and $e_{2} = 4$ : indeed $3^{2} = 9 \equiv 3$ , $4^{2} = 16 \equiv 4$ , $3 \cdot 4 = 12 \equiv 0$ and $3 + 4 = 7 \equiv 1 (\mod 6)$ . The simple components are $B_{1} = {0, 3} ≅ 𝔽_{2}$ and $B_{2} = {0, 2, 4} ≅ 𝔽_{3}$ . Here $r = 2$ , so $(3.9 b)$ predicts $2^{2} = 4$ ideals — and indeed the ideals of $ℤ / 6$ are $(0), (2), (3), (1)$ .

Comparison and Classification

Which decompositions are unique, and in what sense
Decomposition of $R$ or of $_{R} R$	Unique as subobjects?	Unique up to isomorphism?	Governing result
Into indecomposable two-sided ideals	yes	yes	$(3.8)$
Into simple components (semisimple $R$ )	yes	yes	$(3.8)$ with $(3.9)$
Into isotypic components of $_{R} R$	yes	yes	$(3.9)$
Into minimal left ideals	no	yes, with multiplicities	Jordan–Hölder
Into indecomposable left ideals	no	yes	Krull–Schmidt (needs local endomorphism rings)
As $\prod_{i} M_{n_{i}} (D_{i})$ with a chosen isomorphism	no	yes	$(3.5)$ plus Skolem–Noether

The pattern: two-sided data is rigid, one-sided data is rigid only up to isomorphism. The reason is $(3.9) (2)$ — different isotypic types annihilate one another, which leaves no room for a decomposition to be moved, whereas isomorphic minimal left ideals can be interchanged freely.

Relationship Map

Any ring

(3.9)

applies: isotypic sums of minimal left ideals are ideals that kill each other

Rings with a finite ideal decomposition

(3.8)

applies: the indecomposable ideals are unique

Semisimple rings

R = B_{1} \oplus \dots \oplus B_{r}

with each

B_{i}

simple artinian

Simple artinian rings

r = 1

;

R ≅ M_{n} (D)

Division rings

r = 1

n = 1

You have two decompositions of a ring. What can you conclude?

Both into indecomposable two-sided idealsThey are the same decomposition, after reordering — literal equality of ideals, by

(3.8)

Both into minimal left idealsSame number of summands and the same multiset of isomorphism types, by Jordan–Hölder; the summands themselves may be completely different.

Both into indecomposable left ideals, $R$ semiperfect or of finite lengthKrull–Schmidt gives a bijection matching isomorphic summands; again not equality.

Both as products $\prod_{i} M_{n_{i}} (D_{i})$ Same

r

, same multiset of pairs

(n_{i}, D_{i})

up to isomorphism, by

(3.5)

— but the two isomorphisms with

R

can differ by any automorphism.

Computational Notes

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

The simple components are found through the centre: $Z (R) ≅ Z (D_{1}) \times \dots \times Z (D_{r})$ is a commutative semisimple algebra of dimension $\sum_{i} \dim_{k} Z (D_{i})$ , and its $r$ primitive idempotents are the $e_{i}$ . Splitting a commutative semisimple algebra is a factorisation problem, cheap over finite fields and dependent on polynomial factorisation over $ℚ$ .
For a group algebra $k G$ with $char k ∤ | G |$ , the central idempotents are computed from the irreducible characters by $e_{χ} = \frac{χ (1)}{| G |} \sum_{g} χ (g^{- 1}) g$ , which requires the character table but no linear algebra on $k G$ itself.
Once the $e_{i}$ are known, every subsequent computation splits into $r$ independent computations in the $B_{i} = e_{i} R$ , each a matrix algebra over a division ring — an embarrassingly parallel step.
Deciding whether two given central idempotents generate the same component is trivial (compare the idempotents); deciding whether two components are isomorphic as rings requires identifying $n_{i}$ and $D_{i}$ , which over $ℚ$ is the hard explicit isomorphism problem.

Failure Modes and Common Mistakes

Do not conclude from $B_{i} B_{j} = 0$ that the $B_{i}$ are nilpotent: $B_{i}^{2} = B_{i}$ , since $B_{i}$ has an identity element $e_{i}$ .
$(3.8)$ assumes both decompositions are finite. Without finiteness, mutual containment still works pairwise, but the induction terminating the argument does not.
The number of simple components $r$ is not the number of minimal left ideals — that is usually infinite — nor the composition length $\sum_{i} n_{i}$ of the regular module.
An automorphism of $R$ permutes the $B_{i}$ but need not fix them; only those preserving isomorphism classes of simple modules do. This matters when computing $Aut (R)$ .

Best Practices

State which uniqueness you are invoking: equality of ideals $(3.8)$ , isomorphism of pieces (Krull–Schmidt), or matching of numerical data $(3.5)$ .
Work with the central idempotents rather than with the ideals; they are easier to manipulate and make orthogonality an equation rather than a containment.
Verify a claimed decomposition by checking $e_{i}^{2} = e_{i}$ , $e_{i} e_{j} = 0$ and $\sum_{i} e_{i} = 1$ — three cheap identities that catch nearly every arithmetic slip.
When a construction must be canonical, define it as a sum over all objects of a type, as $(3.9)$ does; this eliminates dependence on choices at the outset.
Record the ground field. The number of simple components of $k G$ changes with $k$ , so 'the blocks of the group algebra' is ambiguous without it.

Quick Reference

$(3.8)$ decompositions into indecomposable ideals coincide, after reordering

$(3.9) (1)$

B_{𝔄}

is a two-sided ideal, in any ring

$(3.9) (2)$

𝔄 not ≅ 𝔄' \Rightarrow B_{𝔄} B_{𝔄'} = 0

Semisimple case

R = B_{1} \oplus \dots \oplus B_{r}

, each

B_{i}

simple artinian

Idempotents

1 = e_{1} + \dots + e_{r}

, central, orthogonal, primitive

Ideal countexactly

2^{r}

; the

B_{i}

are the minimal ideals

Uniqueness level

B_{i}

as ideals;

(n_{i}, D_{i})

up to permutation; isomorphism not canonical

Left analoguefails — minimal left ideals are not unique, only their isomorphism types

Centre

Z (R) = ⨁_{i} Z (B_{i})

, with

r

primitive idempotents

Worked components
Ring	$r$	Components	Ideals
$ℤ / 6$	2	$𝔽_{2}$ , $𝔽_{3}$	4
$ℚ S_{3}$	3	$ℚ$ , $ℚ$ , $M_{2} (ℚ)$	8
$ℝ Q_{8}$	5	$ℝ^{4}$ , $ℍ$	32
$M_{n} (D)$	1	itself	2
$ℚ C_{4}$	3	$ℚ$ , $ℚ$ , $ℚ (i)$	8

Frequently Asked Questions

How is (3.8) stronger than saying the factors are unique up to isomorphism?

It concludes $B_{i} = C_{i}$ as subsets of $R$ , not just $B_{i} ≅ C_{i}$ . So there is genuinely only one way to write a ring as a direct sum of indecomposable ideals. Compare the left-module situation, where $M_{2} (ℚ)$ splits into two minimal left ideals in infinitely many different ways, all of them isomorphic.

Does (3.9) need the ring to be semisimple?

No. Both parts hold in an arbitrary ring with identity; the construction is simply empty when there are no minimal left ideals. Semisimplicity is used only in $(3.9 a)$ , to know that the isotypic components exhaust the ring and that each is a simple artinian ring.

What is the relationship between simple components and blocks?

For a semisimple ring they are the same thing: the indecomposable ideals, equivalently the ideals generated by the centrally primitive idempotents. For a general ring, blocks are still defined by centrally primitive idempotents, but a block need not be a simple ring — in modular representation theory a block typically has a nonzero radical, and block theory studies precisely that.

Why does this give a second proof of Wedderburn-Artin?

Because the decomposition $R = B_{1} \oplus \dots \oplus B_{r}$ is obtained from the socle structure alone, with no endomorphism-ring computation. Each $B_{i}$ is then simple and left artinian, so the classification of simple artinian rings as $M_{n} (D)$ finishes the job. The two proofs are genuinely independent, and Lam presents both.

How many ideals does a semisimple ring have?

Exactly $2^{r}$ , where $r$ is the number of simple components: every ideal is the sum of the components it contains, so the ideal lattice is the Boolean lattice of subsets of ${1, \dots, r}$ . In particular a semisimple ring is simple exactly when $r = 1$ .

Can an automorphism of R permute the simple components?

Yes. An automorphism maps minimal left ideals to minimal left ideals and preserves the property of being an indecomposable ideal, so it permutes the $B_{i}$ ; it fixes each one only if it preserves the isomorphism classes of the simple modules. Components can only be swapped when they are isomorphic as rings. In $ℚ S_{3}$ the two one-dimensional components are interchanged by a ring automorphism of the abstract ring $ℚ \times ℚ \times M_{2} (ℚ)$ , but by no automorphism induced by an automorphism of $S_{3}$ , since every group automorphism fixes both the trivial and the sign representation.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §3, (3.8)–(3.10) (pp. 37–39).
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapter IV.
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981, §3 (blocks and central idempotents).
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §7 and §12 (Krull–Schmidt, socle).
D. S. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, 1977, Chapter 2.

AI Suggested Questions

Prove that a two-sided ideal of a semisimple ring is a direct sum of simple components, and deduce that every quotient of a semisimple ring is semisimple.
Compute the centrally primitive idempotents of $ℚ C_{4}$ and of $ℝ Q_{8}$ .
State the Krull–Schmidt theorem precisely and explain why it gives only isomorphism, not equality.
How does the block decomposition behave for a modular group algebra where the blocks are not simple?
Show that the simple components of a semisimple ring are exactly its minimal nonzero two-sided ideals.
Give an example of a ring with two genuinely different decompositions into indecomposable left ideals.
What is the automorphism group of $M_{n_{1}} (D_{1}) \times \dots \times M_{n_{r}} (D_{r})$ , and how does Skolem–Noether describe it?

Uniqueness in Wedderburn–Artin

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Ideal decompositions are idempotent decompositions

Why isotypic sums are two-sided

Why distinct types annihilate each other

Three levels of uniqueness

Key Results

Proof Techniques and Method

Mutual containment, not counting

Sum over everything of a type

Split off an idempotent to move maps around

Worked Example

The three blocks of $ℚ S_{3}$

The smallest interesting case: $ℤ / 6$

Comparison and Classification

Relationship Map

Computational Notes

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Ideal decompositions are idempotent decompositions

Why isotypic sums are two-sided

Why distinct types annihilate each other

Three levels of uniqueness

06Key Results

07Proof Techniques and Method

Mutual containment, not counting

Sum over everything of a type

Split off an idempotent to move maps around

08Worked Example

The three blocks of ℚS3

The smallest interesting case: ℤ/6

09Comparison and Classification

10Relationship Map

11Computational Notes

12Failure Modes and Common Mistakes

13Best Practices

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

The three blocks of $ℚ S_{3}$

The smallest interesting case: $ℤ / 6$

Comparison and Classification

Relationship Map

Computational Notes

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions