Simple and Semisimple Modules: Definition and Equivalences

Q: Is a simple module semisimple?

Yes. If M is simple its only submodules are 0 and M, and both are direct summands (M = 0 ⊕ M). The converse fails in two ways: the zero module is semisimple but not simple, and Z/6 is semisimple of length 2.

Q: If M = _i ∈ I S_i, is every submodule of M one of the sub-sums?

No — only isomorphic to one. In S ⊕ S the diagonal (s,s) is a submodule isomorphic to S but distinct from both factors. The correct statement is that there is J ⊆ I with M = N ⊕ _j ∈ J S_j, so N ≅ _i J S_i.

Q: Does semisimple imply finite length?

No. _p prime Z/p is a semisimple Z-module of infinite length. A semisimple module has finite length precisely when it is finitely generated, and then the length is the number of simple summands.

Q: What replaces semisimplicity when it fails?

Two devices. The socle soc(M) picks out the largest semisimple part, and the radical measures the obstruction: for a module over a finite-dimensional algebra, M/(rad A)M is the largest semisimple quotient. Together they give the Loewy filtration, whose layers are semisimple even when M is not.

Executive Summary

Two definitions carry the whole of Wedderburn–Artin theory. A left $R$ -module $M$ is simple if $M \neq 0$ and its only submodules are $0$ and $M$ ; it is semisimple if every submodule of $M$ is a direct summand. The second definition is a splitting condition, stated without ever mentioning a simple module.

Lam's $(2.4)$ shows that the splitting condition is equivalent to two structural ones: $M$ is a direct sum of simple submodules, and $M$ is merely a sum of simple submodules. The last of these is by far the easiest to verify in practice, and it is the form used everywhere downstream — including in the proof that a ring is left semisimple as soon as its regular module is.

3Equivalent conditions in (2.4)

$(2.3)$ Simple submodule lemma

ZornTool behind both proofs

$0$ Semisimple but not simple

Overview

Throughout, $R$ is a ring with identity and modules are unital. We write $_{R} M$ for a left module and $M_{R}$ for a right module; every statement below has a right-handed mirror image obtained by reading in $R^{op}$ , and nothing on this page is genuinely one-sided.

The problem these definitions solve is decomposition. Given a module, one wants to break it into pieces that cannot be broken further and then to understand the pieces separately. Two things can go wrong: the pieces may not exist, and even when they do the module may fail to be their direct sum. Semisimplicity is exactly the hypothesis that rules out both failures.

M semisimple ⟺ \forall N \subseteq M \exists N' \subseteq M : M = N \oplus N'

(2.1b)

The definition. Note that $N'$ is required to exist, not to be unique — and it very rarely is.

Two conventions matter. The zero module is semisimple (vacuously: its only submodule is $0 = 0 \oplus 0$ ) but is not simple, because simple modules are required to be nonzero. And the empty sum and empty direct sum of submodules are both taken to be $0$ , which is what makes $(2.4)$ true in the degenerate case $M = 0$ .

The largest semisimple submodule of an arbitrary module is its socle, treated on The Socle of a Module and of a Ring; a ring all of whose modules are semisimple is treated on Semisimple Rings: Definition and Equivalent Characterisations.

Learning Objectives

State the definitions of simple and semisimple and identify the degenerate cases.
Prove that submodules and quotient modules of semisimple modules are semisimple.
Prove Lam $(2.3)$ : a nonzero semisimple module contains a simple submodule.
Prove all four implications in Lam $(2.4)$ and locate exactly where Zorn's Lemma is used.
Decide semisimplicity for $k [x]$ -modules of dimension $2$ by inspecting invariant subspaces.
Explain why complements exist but are not unique, and what uniqueness statement does hold.

Definitions

Definition(2.1)Simple and semisimple modules

Let $R$ be a ring and $M$ a left $R$ -module.

$M$ is simple (equivalently irreducible) if $M \neq 0$ and the only $R$ -submodules of $M$ are $0$ and $M$ .
$M$ is semisimple (equivalently completely reducible) if every $R$ -submodule of $M$ is an $R$ -module direct summand of $M$ .

Direct summand: $N \subseteq M$ is a direct summand if $M = N \oplus N'$ for some submodule $N'$ , i.e. $N + N' = M$ and $N \cap N' = 0$ .
Minimal submodule: A nonzero submodule containing no nonzero proper submodule — the same thing as a simple submodule.
Minimal left ideal: A simple submodule of the left regular module $_{R} R$ .
Composition series: A finite chain $0 = M_{0} ⊊ \dots ⊊ M_{n} = M$ with each $M_{i} / M_{i - 1}$ simple; $n$ is the length $ℓ (M)$ .
Essential submodule: $N \subseteq M$ meeting every nonzero submodule of $M$ nontrivially. In a semisimple module the only essential submodule is $M$ itself.

Semisimple is a property of a module. A ring called semisimple is one whose regular module has the property — see the companion page.

Core Concepts

Simple modules are cyclic quotients

If $M$ is simple and $0 \neq m \in M$ then $R m$ is a nonzero submodule, hence $R m = M$ : simple modules are cyclic. The surjection $R \to M$ , $r \mapsto r m$ , has kernel a left ideal $𝔪$ with $R / 𝔪 ≅ M$ simple, so $𝔪$ is a maximal left ideal. Conversely $R / 𝔪$ is simple for every maximal left ideal $𝔪$ .

Maximal left ideal

𝔪

Simple module

R / 𝔪

Every simple module, up to isomorphism

So the supply of simple modules is controlled by the maximal left ideals, and by Zorn's Lemma these exist whenever $R \neq 0$ . This is the point of contact with the Jacobson radical, which is precisely the intersection of the annihilators of all simple left modules.

Why complements behave

The definition of semisimplicity is a demand that complements exist. Existence is a strong condition, but it propagates well: it survives passage to submodules and quotients, and it is stable under arbitrary sums. The technical engine for all three is the modular law: if $P \subseteq N \subseteq M$ and $Q \subseteq M$ , then

N \cap (P + Q) = P + (N \cap Q) .

(mod)

Valid in the lattice of submodules of any module; the reason intersecting a decomposition with a submodule gives a decomposition.

Directness is automatic

The surprise in $(2.4)$ is the implication (3) $\Rightarrow$ (2): a module presented as an unstructured sum $M = \sum_{i \in I} M_{i}$ of simple submodules — with arbitrary overlaps and repetitions — is automatically a direct sum of some subfamily. One does not choose the subfamily by any formula; Zorn's Lemma produces it as a maximal independent set, in exact analogy with extracting a basis from a spanning set in linear algebra.

Isotypic components

Group the simple summands by isomorphism type. For each isomorphism class $[S]$ of simple modules let ${soc}_{S} (M)$ be the sum of all submodules of $M$ isomorphic to $S$ . Then a semisimple $M$ splits canonically as $M = ⨁_{[S]} {soc}_{S} (M)$ , and while the individual simple summands are not unique, the isotypic components are: they are defined without any choice.

Multiplicities are well defined. If $⨁_{i \in I} S_{i} ≅ ⨁_{j \in J} T_{j}$ with all $S_{i}, T_{j}$ simple, then for each isomorphism class the two index sets have the same cardinality.
By Schur's Lemma $D = {End}_{R} (S)$ is a division ring, and ${End}_{R} (S^{\oplus n}) ≅ M_{n} (D)$ — the first appearance of the matrix rings in the Wedderburn–Artin theorem.
A semisimple module is finitely generated iff it is a finite direct sum of simples iff it has finite length iff it is noetherian iff it is artinian.

Key Results

Remark(2.2)Heredity of semisimplicity

Let $M$ be a semisimple left $R$ -module. Then every submodule and every quotient module of $M$ is semisimple.

Proof

Submodules. Let $N \subseteq M$ and let $P \subseteq N$ . Since $M$ is semisimple, $M = P \oplus Q$ for some $Q \subseteq M$ . Intersecting with $N$ and using the modular law with $P \subseteq N$ gives $N = N \cap (P \oplus Q) = P \oplus (N \cap Q)$ . So $P$ is a direct summand of $N$ , and $N$ is semisimple.

Quotients. Let $N \subseteq M$ . By semisimplicity $M = N \oplus N'$ , and the projection $M \to N'$ induces $M / N ≅ N'$ . As $N'$ is a submodule of $M$ it is semisimple by the previous paragraph, hence so is $M / N$ .

Lemma(2.3)Existence of a simple submodule

Every nonzero semisimple left $R$ -module $M$ contains a simple submodule.

Proof

Fix $0 \neq m \in M$ . By $(2.2)$ the cyclic submodule $R m$ is semisimple, so we may replace $M$ by $R m$ and assume $M = R m$ .

Let $Σ$ be the set of submodules $N \subseteq M$ with $m \notin N$ . It is nonempty ( $0 \in Σ$ since $m \neq 0$ ) and the union of a chain in $Σ$ again omits $m$ , so Zorn's Lemma yields a maximal $N \in Σ$ . Since $m \notin N$ we have $N \neq M$ , and by semisimplicity $M = N \oplus N'$ with $N' \neq 0$ .

We claim $N'$ is simple. Let $0 \neq N'' \subseteq N'$ . Then $N ⊊ N \oplus N''$ , so maximality of $N$ forces $m \in N \oplus N''$ and hence $M = R m \subseteq N \oplus N''$ . Intersecting $N' = N' \cap (N \oplus N'')$ with the modular law (using $N'' \subseteq N'$ ) gives $N' = N'' \oplus (N' \cap N) = N''$ . So $N'$ has no nonzero proper submodule and is simple.

Theorem(2.4)Three characterisations of semisimplicity

For a left $R$ -module $M$ the following are equivalent.

$M$ is semisimple: every submodule of $M$ is a direct summand.
$M$ is the direct sum of a family of simple submodules.
$M$ is the sum of a family of simple submodules.

Here the sum and the direct sum of the empty family are both $0$ , so the statement is correct for $M = 0$ as well.

Proof

**(1) $\Rightarrow$ (3).** Let $M_{1} \subseteq M$ be the sum of all simple submodules of $M$ . By hypothesis $M = M_{1} \oplus M_{2}$ for some $M_{2}$ . If $M_{2} \neq 0$ then $M_{2}$ is semisimple by $(2.2)$ , so by $(2.3)$ it contains a simple submodule $S$ . But then $S \subseteq M_{1}$ by definition of $M_{1}$ , whence $S \subseteq M_{1} \cap M_{2} = 0$ , contradicting $S \neq 0$ . Therefore $M_{2} = 0$ and $M = M_{1}$ .

**(3) $\Rightarrow$ (1).** Write $M = \sum_{i \in I} M_{i}$ with each $M_{i}$ simple, and let $N \subseteq M$ be a submodule. Consider the subsets $J \subseteq I$ satisfying (a) the sum $\sum_{j \in J} M_{j}$ is direct, and (b) $N \cap \sum_{j \in J} M_{j} = 0$ . Both conditions involve only finitely many indices at a time, so the family of such $J$ is closed under unions of chains; it is nonempty because $J = \emptyset$ qualifies. Zorn's Lemma gives a maximal such $J$ . Put

M' : = N + \sum_{j \in J} M_{j} = N \oplus (⨁_{j \in J} M_{j}) .

(2.4a)

It suffices to prove $M' = M$ , for then $N$ is a direct summand of $M$ with complement $⨁_{j \in J} M_{j}$ . Since $M = \sum_{i} M_{i}$ , it is enough to show $M_{i} \subseteq M'$ for every $i \in I$ . Suppose not, say $M_{i} ⊈ M'$ . Then $M' \cap M_{i}$ is a proper submodule of the simple module $M_{i}$ , so $M' \cap M_{i} = 0$ and

M' + M_{i} = N \oplus (⨁_{j \in J} M_{j}) \oplus M_{i},

(2.4b)

which shows that $J \cup {i}$ also satisfies (a) and (b) — note $i \notin J$ because $M_{i} ⊈ M'$ . This contradicts the maximality of $J$ . Hence $M' = M$ .

**(3) $\Rightarrow$ (2).** Run the same argument with $N = 0$ : it produces $J \subseteq I$ with $M = ⨁_{j \in J} M_{j}$ .

**(2) $\Rightarrow$ (3)** is immediate. This closes the cycle.

Corollary—Complements inside a decomposition

If $M = ⨁_{i \in I} S_{i}$ with each $S_{i}$ simple and $N \subseteq M$ is any submodule, then there is $J \subseteq I$ with $M = N \oplus ⨁_{j \in J} S_{j}$ ; consequently $N ≅ ⨁_{i \in I ∖ J} S_{i}$ . In particular every submodule and every quotient of $M$ is isomorphic to a direct sum of a subfamily of the $S_{i}$ — though it need not equal one.

Corollary—Sums, quotients and extensions

An arbitrary sum of semisimple submodules is semisimple, and an arbitrary direct sum of semisimple modules is semisimple: each is a sum of simple submodules, so $(2.4) (3)$ applies. Extensions, by contrast, are not: there is a nonsplit short exact sequence $0 \to ℤ / 2 \to ℤ / 4 \to ℤ / 2 \to 0$ of $ℤ$ -modules with semisimple ends and non-semisimple middle.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Zorn on an omission condition

To manufacture a simple submodule, maximise a submodule subject to missing a fixed element $m$ . The complement of a maximal such submodule is forced to be simple. This is the whole of $(2.3)$ .

Move 2

Zorn on an independence condition

To extract a direct sum from a plain sum, maximise a subfamily subject to independence plus disjointness from $N$ . Both conditions are of finite character, so chains cause no trouble.

Move 3

Simplicity turns non-containment into zero intersection

If $S$ is simple and $S ⊈ M'$ , then $S \cap M' = 0$ — there is no middle ground. This dichotomy converts a failure of covering into a fresh independent summand.

Move 3 is the reason the argument terminates at all: without simplicity, $M' \cap M_{i}$ could be a nonzero proper submodule and no contradiction with maximality would follow. It is worth noticing that the proof of $(2.4)$ uses the simplicity of the $M_{i}$ only through this dichotomy.

A fourth, lighter technique is the reduction to cyclic submodules: any module is the sum of its cyclic submodules, so a property that (i) holds for cyclic modules and (ii) is preserved under sums, holds for all modules. Both $(2.3)$ and the ring-level theorem $(2.5)$ use this reduction, and it is how one avoids ever having to handle an arbitrary module directly.

Worked Example

Two-dimensional modules over $k [x]$

Let $k$ be a field and $V = k^{2}$ . Giving $V$ the structure of a $k [x]$ -module is the same as choosing a matrix $A \in M_{2} (k)$ for $x$ to act by, and the $k [x]$ -submodules of $V$ are exactly the $A$ -invariant subspaces. So semisimplicity of $V$ can be decided by pure linear algebra.

Case 1: distinct rational eigenvalues

Take $k = ℚ$ and $A = (\begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix})$ . The invariant subspaces are $0$ , the two eigenlines, and $V$ . Then

V ≅ ℚ [x] / (x - 1) \oplus ℚ [x] / (x - 2),

(E.1)

Semisimple of length $2$ , with two non-isomorphic simple summands; not simple.

Case 2: irreducible characteristic polynomial

Take $k = ℚ$ and $A = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ , with characteristic polynomial $x^{2} + 1$ , irreducible over $ℚ$ . There is no rational eigenvalue, hence no invariant line, so $V$ is simple as a $ℚ [x]$ -module: $V ≅ ℚ [x] / (x^{2} + 1) ≅ ℚ (i)$ . Extending scalars to $ℂ$ destroys simplicity — over $ℂ$ the same matrix has eigenvalues $\pm i$ and $V$ splits into two lines. Simplicity is not preserved by field extension; semisimplicity here is.

Case 3: a Jordan block

Take $A = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ over any field $k$ . Solving $(A - I) v = 0$ gives $v_{2} = 0$ , so the only invariant line is $L = k e_{1}$ . Since $L$ has no complement among invariant subspaces, $V$ is not semisimple; it is indecomposable of length $2$ , with $soc (V) = L$ and $V / L ≅ k$ .

Case 4: scalar action

$A = I$ makes every line invariant, so $V ≅ S \oplus S$ with $S = k [x] / (x - 1)$ : isotypic, semisimple, and possessing infinitely many different complements to any given line when $k$ is infinite. This is the standard warning that complements are not unique.

The four cases at a glance ( $V = k^{2}$ as a $k [x]$ -module)
Action of $x$	Minimal polynomial	Invariant lines	Simple?	Semisimple?	$soc (V)$
$diag (1, 2)$ over $ℚ$	$(x - 1) (x - 2)$	exactly $2$	no	yes	$V$
rotation, over $ℚ$	$x^{2} + 1$	none	yes	yes	$V$
Jordan block $J_{2} (1)$	$(x - 1)^{2}$	exactly $1$	no	no	$k e_{1}$
identity, $k$ infinite	$x - 1$	infinitely many	no	yes	$V$

The general statement behind the table: for $k$ a field, a finite-dimensional $k [x]$ -module is semisimple iff the minimal polynomial of the acting matrix is squarefree — over $ℚ$ , $x^{2} + 1$ and $(x - 1) (x - 2)$ qualify while $(x - 1)^{2}$ does not.

Process and Workflow

A practical route for deciding whether a given module is semisimple.

Is the module $M$ semisimple?

$R$ is a semisimple ringThen yes, unconditionally — every module over a left semisimple ring is semisimple, by

(2.5)

. Check the ring first; it is usually the cheapest test.

$R$ is a finite-dimensional algebraCompute

rad R

and test whether

(rad R) M = 0

. This is necessary and sufficient, and reduces to a linear-algebra kernel computation.

$M$ is finite lengthList the simple submodules and check whether their sum is all of

M

; equivalently, check

soc (M) = M

. Zorn is not needed in this case.

General $M$ Exhibit

M

as a sum of simple submodules and invoke

(2.4) (3)

, or produce one submodule with no complement to refute semisimplicity.

Comparison and Classification

Concrete modules and where they sit
Module	Over	Simple	Semisimple	Indecomposable	Finite length
$ℤ / p$	$ℤ$	yes	yes	yes	yes, $ℓ = 1$
$ℤ / 6$	$ℤ$	no	yes	no	yes, $ℓ = 2$
$ℤ / 4$	$ℤ$	no	no	yes	yes, $ℓ = 2$
$ℤ$	$ℤ$	no	no	yes	no
$ℚ$	$ℤ$	no	no	yes	no
$⨁_{p} ℤ / p$	$ℤ$	no	yes	no	no
$k^{n}$ (natural)	$M_{n} (k)$	yes	yes	yes	yes, $ℓ = 1$
$k [x] / (x^{2})$	$k [x]$	no	no	yes	yes, $ℓ = 2$

Which properties are inherited by which constructions
	Submodules	Quotients	Direct sums	Extensions
Simple	no	no	no	no
Semisimple	yes	yes	yes	no
Finite length	yes	yes	finite only	yes
Indecomposable	no	no	no	no
Noetherian	yes	yes	finite only	yes

Which properties are inherited by which constructions

The single row worth memorising is the second: semisimplicity is closed under everything except extension, and the failure under extension is exactly what the Jacobson radical measures.

Relationship Map

SimpleSemisimple

soc (M) = M

Every submodule splits off

The last three nodes are equivalent; only the first implication is strict. Below, the properties a single module can carry, arranged by logical strength.

Semisimple module $M$ — $M = \sum {simple submodules}$
- implies
  - every submodule is a direct summand
  - every submodule and quotient is semisimple
  - $M$ has no nonzero superfluous submodule
  - $rad (M) = 0$ , where $rad (M)$ is the intersection of maximal submodules
- is implied by
  - $M$ simple
  - $M$ a sum of simple submodules
  - $M$ a module over a semisimple ring
  - $M$ a $k G$ -module with $char k ∤ | G |$ , $G$ finite
- does not imply
  - finitely generated
  - artinian or noetherian
  - indecomposable
  - simple

In the other direction, an arbitrary module $M$ has a largest semisimple submodule $soc (M)$ and a smallest submodule with semisimple quotient when $R$ is nice enough; the first always exists and is the subject of The Socle of a Module and of a Ring.

Applications and Industry Use

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

Representation theory

Complete reducibility

A representation of a group over a field is a module over the group algebra; it is completely reducible exactly when that module is semisimple. Maschke's theorem supplies the hypothesis in characteristic zero, and the whole character-theoretic apparatus presumes it.

Signal processing

Symmetry-adapted bases

Decomposing a symmetric linear system into isotypic components block-diagonalises the system matrix. This is the algebraic content of symmetry reduction in finite element analysis and of the fast Fourier transform for abelian groups.

Coding theory

Cyclic codes as minimal ideals

When $\gcd (n, q) = 1$ the algebra $𝔽_{q} [x] / (x^{n} - 1)$ is semisimple, so cyclic codes of length $n$ are direct sums of minimal ideals, each generated by a primitive idempotent. Semisimplicity is what makes the idempotent generator exist.

Computational algebra

Module chopping

The MeatAxe splits a module over a finite field into its composition factors. Detecting simplicity of the pieces is the primitive operation, and semisimplicity of the whole is checked by testing whether the radical acts as zero.

Honest summary: simple and semisimple modules are infrastructure. They are almost never the object of study; they are the vocabulary in which decomposition theorems in representation theory, coding theory and harmonic analysis are stated.

Computational Notes

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

Let $A$ be a finite-dimensional algebra over a field $k$ and $M$ a finite-dimensional $A$ -module, both given by structure constants and matrices.

Semisimplicity test. $M$ is semisimple iff $(rad A) M = 0$ : a semisimple module is killed by the radical, and conversely such an $M$ is a module over the semisimple ring $A / rad A$ . Given a basis $r_{1}, \dots, r_{m}$ of $rad A$ , the test is $m$ matrix multiplications.
Simplicity test. Over a finite field, Norton's irreducibility test inside the MeatAxe decides simplicity in $O (d^{3})$ field operations per attempt, where $d = \dim_{k} M$ , with randomised element selection; a failure returns a proper submodule.
Composition factors. Recursive chopping yields a composition series; the factors are unique by Jordan–Hölder, so the output is canonical even though the series is not.
Where it stops. For infinite-dimensional modules over finitely presented rings none of this is available; simplicity is not decidable in general, since the word problem for finitely presented rings is undecidable.

Failure Modes and Common Mistakes

Do not read $(2.4) (2)$ as saying that a submodule of $⨁_{i} S_{i}$ is a sub-sum $⨁_{i \in K} S_{i}$ . It is only isomorphic to one; the diagonal submodule of $S \oplus S$ is the standard counterexample.
Do not assume semisimple modules are finitely generated. $⨁_{p} ℤ / p$ over $ℤ$ is semisimple, neither noetherian nor artinian.
Do not conflate $rad (M) = 0$ with semisimplicity for a general module. Semisimple implies $rad (M) = 0$ ; the converse needs a hypothesis such as $M$ artinian. Over $ℤ$ , $rad (ℤ) = 0$ but $ℤ$ is not semisimple.
Do not forget the nonzero requirement in the definition of simple; a stray zero module inserted into a list of composition factors invalidates length counts.

Quick Reference

Simple

M \neq 0

and only submodules are

0, M

Semisimpleevery submodule is a direct summand

Equivalently

M = ⨁_{i} S_{i}

with

S_{i}

simple

Equivalently

M = \sum_{i} S_{i}

with

S_{i}

simple

Equivalently

soc (M) = M

Hereditysubmodules, quotients and sums stay semisimple

Failure modeextensions do not:

ℤ / 4

Simple modulesexactly the

R / 𝔪

𝔪

maximal left ideal

Which result to quote
You want	Use	Reference
A simple submodule	$M$ nonzero semisimple	(2.3)
Directness from a plain sum	$M = \sum S_{i}$	(2.4)(3) $\Rightarrow$ (2)
A complement for $N$	$M$ semisimple	(2.4)(3) $\Rightarrow$ (1)
Semisimplicity of $N \subseteq M$	$M$ semisimple	(2.2)
Semisimplicity of $M / N$	$M$ semisimple	(2.2)
All modules semisimple	$_{R} R$ semisimple	(2.5)

Frequently Asked Questions

Is a simple module semisimple?

Yes. If $M$ is simple its only submodules are $0$ and $M$ , and both are direct summands ( $M = 0 \oplus M$ ). The converse fails in two ways: the zero module is semisimple but not simple, and $ℤ / 6$ is semisimple of length $2$ .

Why is Zorn's Lemma needed, and can it be avoided?

It is needed twice: to produce a maximal submodule omitting a given element in $(2.3)$ , and to extract a maximal independent subfamily in $(2.4)$ . For modules of finite length both maximal objects can be found by descending induction on length, so the proofs are choice-free in that case. For arbitrary modules some form of choice is genuinely required — even the statement that every vector space has a basis is equivalent to the axiom of choice.

M = ⨁_{i \in I} S_{i}

, is every submodule of

M

one of the sub-sums?

No — only isomorphic to one. In $S \oplus S$ the diagonal ${(s, s)}$ is a submodule isomorphic to $S$ but distinct from both factors. The correct statement is that there is $J \subseteq I$ with $M = N \oplus ⨁_{j \in J} S_{j}$ , so $N ≅ ⨁_{i \notin J} S_{i}$ .

Does semisimple imply finite length?

No. $⨁_{p prime} ℤ / p$ is a semisimple $ℤ$ -module of infinite length. A semisimple module has finite length precisely when it is finitely generated, and then the length is the number of simple summands.

How is semisimplicity of a module related to semisimplicity of the ring?

A ring is left semisimple when its own left regular module is semisimple, and by $(2.5)$ that single condition forces every left module to be semisimple. So a semisimple ring is one for which the property is universal, while an individual module can be semisimple over a ring that is very far from semisimple — for instance $ℤ / p$ over $ℤ$ .

What replaces semisimplicity when it fails?

Two devices. The socle $soc (M)$ picks out the largest semisimple part, and the radical measures the obstruction: for a module over a finite-dimensional algebra, $M / (rad A) M$ is the largest semisimple quotient. Together they give the Loewy filtration, whose layers are semisimple even when $M$ is not.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §2, results (2.1)–(2.4), pp. 25–27.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §9 (semisimple modules and the socle).
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapter III.
N. Bourbaki, Algèbre, Chapitre VIII: Modules et anneaux semi-simples, Hermann, Paris, 1958.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Prove that a module is semisimple if and only if every cyclic submodule is semisimple.
State and prove the uniqueness of multiplicities in a semisimple decomposition when the index sets are infinite.
Show that ${End}_{R} (S_{1}^{n_{1}} \oplus \dots \oplus S_{r}^{n_{r}})$ is a product of matrix rings over division rings, and identify the division rings.
Give an example of a finite-dimensional algebra and a module that is semisimple over the algebra but not after extending the base field.
How does the notion of a simple module change for rings without an identity element?
Describe the lattice of submodules of a semisimple module of finite length and explain when it is a Boolean lattice.
Work through Maschke's theorem and identify exactly where the hypothesis that the group order is invertible is used.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Simple modules are cyclic quotients

Why complements behave

Directness is automatic

Isotypic components

06Key Results

07Proof Techniques and Method

Zorn on an omission condition

Zorn on an independence condition

Simplicity turns non-containment into zero intersection

08Worked Example

Two-dimensional modules over k[x]

Case 1: distinct rational eigenvalues

Case 2: irreducible characteristic polynomial

Case 3: a Jordan block

Case 4: scalar action

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Complete reducibility

Symmetry-adapted bases

Cyclic codes as minimal ideals

Module chopping

13Computational Notes

14Failure Modes and Common Mistakes

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

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Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Two-dimensional modules over $k [x]$

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions