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32 results · all verified · 21 also independently AI-judged
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Chain Conditions, Semisimple Modules and the Wedderburn–Artin Theorem

1 · Prerequisites

2 · Summary

Submodules, quotient modules, short exact sequences, direct sums, simple modules, projective modules, and the module isomorphism theorem provide the algebraic language for chain conditions. The published splitting lemma and free-module quotient construction are used to relate finite generation, exactness, complements, and projectivity for left modules over a unital ring.

Noetherian and Artinian conditions are related to finite length and Jordan-Holder factors before endomorphism rings and semisimple modules are developed. Schur's lemma and the finite decomposition of the regular module lead to the radical-free Wedderburn-Artin classification and its left-right consequences. The integrality results characterize integral elements through finite modules and apply the determinant trick to closure and rational algebraic integers.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Noetherian modules: every submodule is finitely generated

Definition

A left R-module M is Noetherian when every submodule of M is finitely generated (Generated submodule, cyclic and finitely generated modules, module basis and free module). This finite-generation definition is the convention; its equivalence with ACC and the maximal condition is proved in Finite generation, ACC, and maximal-condition characterizations of Noetherian modules.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Artinian modules by the descending chain condition

Definition

A left R-module M is Artinian when every descending chain M0M1 of submodules stabilizes: there is N such that Mn=MN for all nN. This is the descending chain condition.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-17Open item page →

Left and right Noetherian rings

Definition

A unital ring R is left Noetherian when its left regular module RR is Noetherian, and right Noetherian when the right regular module RR is Noetherian. Unqualified “Noetherian ring” means left Noetherian here; the side is stated whenever both notions occur.

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Left and right Artinian rings

Definition

A unital ring R is left Artinian when RR is Artinian, and right Artinian when RR is Artinian. Unqualified “Artinian ring” means left Artinian here.

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Composition series and length of a module

Definition

A composition series of a left R-module M is a finite chain 0=M0<M1<<Mn=M whose factors Mi/Mi1 are simple. If such a series exists, the length R(M) is its number n of factors; Jordan–Hölder theorem for modules proves independence of the chosen series. The zero module has the empty series and length 0.

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The opposite ring Rop

Definition

For a unital ring R, the opposite ring Rop has the same underlying abelian group, identity, and addition as R, with multiplication ab:=ba. Associativity and both distributive laws follow from those of R with the order reversed, and the same element 1 is a two-sided identity. Thus the displayed operations really form a unital ring, including when R is the zero ring.

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The endomorphism ring EndR(M) under addition and composition

Definition

For a left R-module M, define EndR(M):=HomR(M,M). Addition is pointwise and multiplication is composition, (fg)(m):=f(g(m)). The ring laws and the identity endomorphism are established in Module endomorphisms form a ring under pointwise addition and composition .

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-17Open item page →

Semisimple modules as direct sums of simple modules

Definition

A left R-module is semisimple when it is an internal direct sum of simple submodules, allowing the empty direct sum. Hence the zero module is semisimple.

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A semisimple ring as a ring whose left regular module is semisimple

Definition

A unital ring R is semisimple when its left regular module RR is semisimple. This is a left-module definition and uses no Jacobson radical. For the zero ring, the regular module is zero and hence semisimple; the Wedderburn-Artin theorem below is stated for nonzero rings.

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The socle as the sum of all simple submodules

Definition

For a left R-module M, the socle is Soc(M):={SM:S is simple}. Concretely it is the submodule generated by the union of all simple submodules. If there are none, the sum is 0, so the definition is always meaningful.

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Integral elements over a commutative ring and algebraic integers

Definition

Let AB be a homomorphism of commutative rings. An element bB is integral over A when it is a root of a monic polynomial in A[X]. The extension is integral when every element is integral. An algebraic integer is a complex number integral over Z.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Second isomorphism theorem for modules

Statement

For submodules L,NM, there is a canonical isomorphism L/(LN)(L+N)/N. See First isomorphism theorem for modules: M/kerfimf.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For every module homomorphism f:MN, there is a module isomorphism M/kerf  imf, given by m+kerff(m). (First isomorphism theorem for modules: M/kerfimf).

[L2]

For submodules A,BM, the intersection AB and the sum A+B:={a+b:aA, bB} are submodules of M. (The one-step submodule criterion; intersections and sums of submodules are submodules).

[L3]

For NM, the additive cosets m+N form the quotient module M/N under the well-defined scalar action r(m+N):=rm+N. (Quotient module M/N with scalar multiplication on additive cosets).

Proof

technique · direct
1.1

For submodules L,NM, map L(L+N)/N.

L1L2L3givenalgebra
2.1

Its kernel is LN, and it is surjective by the definition of L+N; the first isomorphism theorem gives L/(LN)(L+N)/N.

step 1.1givenalgebra
3.1

The coincident and zero cases are admitted and give equalities rather than exceptions. For L=0 both sides are 0, since 0/(0N)=0 and (0+N)/N=N/N=0; for N=0 both sides are L; and for L=N both sides are 0. This proves the stated claim.

step 2.1givenalgebra
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Third isomorphism theorem for modules

Statement

If NLM, then L/N is a submodule of M/N and (M/N)/(L/N)M/L. See First isomorphism theorem for modules: M/kerfimf.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For every module homomorphism f:MN, there is a module isomorphism M/kerf  imf, given by m+kerff(m). (First isomorphism theorem for modules: M/kerfimf).

[L2]

For NM, the additive cosets m+N form the quotient module M/N under the well-defined scalar action r(m+N):=rm+N. (Quotient module M/N with scalar multiplication on additive cosets).

[L3]

Let f:MP be a module homomorphism and let NM satisfy Nkerf. There is a unique module homomorphism fˉ:M/NP such that fˉ(m+N)=f(m), equivalently f=fˉπ. (A module homomorphism vanishing on N factors uniquely through M/N).

Proof

technique · direct
1.1

For NLM, send m+N to m+L.

L1L2L3givenalgebra
2.1

If m+N=m+N, then mmNL, so the images modulo L agree. The map is surjective, and its kernel consists exactly of the cosets m+N with mL, namely L/N; the first isomorphism theorem gives the displayed isomorphism.

step 1.1givenalgebra
3.1

The two coincident cases are admitted by NLM and hold. For N=L the submodule L/N is zero and the isomorphism reads (M/N)/0M/N=M/L; for L=M it is L/N=M/N and the isomorphism reads (M/N)/(M/N)=0M/M=0. This proves the stated claim.

step 2.1givenalgebra
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Correspondence theorem for submodules of a quotient module

Statement

For NM, inverse image and quotient induce mutually inverse inclusion-preserving bijections between submodules of M/N and submodules of M containing N. They preserve sums, intersections, and successive quotients. See Quotient module M/N with scalar multiplication on additive cosets.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For NM, the additive cosets m+N form the quotient module M/N under the well-defined scalar action r(m+N):=rm+N. (Quotient module M/N with scalar multiplication on additive cosets).

[L2]

Let f:MP be a module homomorphism and let NM satisfy Nkerf. There is a unique module homomorphism fˉ:M/NP such that fˉ(m+N)=f(m), equivalently f=fˉπ. (A module homomorphism vanishing on N factors uniquely through M/N).

[L3]

If NLM, then L/N is a submodule of M/N and (M/N)/(L/N)M/L. (Third isomorphism theorem for modules).

Proof

technique · direct
1.1

Inverse image and quotient give mutually inverse inclusion-preserving bijections between submodules of M/N and submodules of M containing N.

L1L2L3givenalgebra
2.1

If PM/N, then π(π1P)=P by surjectivity of π; if LM contains N, then π1(L/N)=L. Direct calculation with inverse images gives preservation of sums and intersections, while [L3] identifies successive quotients. This proves the stated claim.

L3step 1.1givenalgebra
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Finite generation, ACC, and maximal-condition characterizations of Noetherian modules

Statement

For a left R-module M, the following are equivalent: every submodule is finitely generated; every ascending chain of submodules stabilizes; and every nonempty family of submodules has a maximal member. The implication from ACC to the maximal condition uses dependent choice; the other displayed implications are choice-free. See Noetherian modules: every submodule is finitely generated.

Facts & Assumptions

Given: The hypotheses and objects in the Statement. The adopted axiom of dependent choice is assumed for the one direction identified in the Statement; it is not cited as a forward dependency.

[L1]

A left R-module M is Noetherian when every submodule of M is finitely generated (def-generated-cyclic-finitely-generated-and-free-modules). This finite-generation definition is the convention; its equivalence with ACC and the maximal condition is proved in thm-equivalent-characterizations-of-noetherian-modules. (Noetherian modules: every submodule is finitely generated).

Proof

technique · direct
1.1

We prove that finite generation of every submodule implies ACC by taking the union of a chain and locating a finite generating set in one stage.

L1givenalgebra
2.1

Assuming the adopted dependent-choice axiom in Facts, ACC implies the maximal condition: if a nonempty family had no maximal member, recursively choose a strict ascending chain in it.

step 1.1givenalgebra
3.1

Choice-free: for a submodule N, the maximal condition applied to its finitely generated submodules gives a maximal L; if L is proper in N, adjoining one element of N minus L contradicts maximality.

step 2.1givenalgebra
4.1

Thus the finite-generation condition, ACC, and the maximal condition are equivalent. Only the recursive construction in step 2.1 uses the adopted dependent-choice axiom. This proves the stated claim.

step 3.1givenalgebra
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

DCC and minimal-condition characterizations of Artinian modules

Statement

For a left R-module M, DCC is equivalent to the condition that every nonempty family of submodules has a minimal member. The implication from DCC to the minimal condition uses dependent choice. See Artinian modules by the descending chain condition.

Facts & Assumptions

Given: The hypotheses and objects in the Statement. The adopted axiom of dependent choice is assumed for the one direction identified in the Statement; it is not cited as a forward dependency.

[L1]

A left R-module M is Artinian when every descending chain M0M1 of submodules stabilizes: there is N such that Mn=MN for all nN. This is the descending chain condition. (Artinian modules by the descending chain condition).

Proof

technique · direct
1.1

DCC gives a minimal member of every nonempty family by contradiction: under dependent choice, absence of a minimal member yields a strict descending chain.

L1givenalgebra
2.1

Conversely, a nonstabilizing descending chain has no minimal member.

step 1.1givenalgebra
3.1

Repeated terms are what make step 2.1 correct rather than an equivocation: minimality of Mk in the family {Mn} means no member is properly contained in it, so Mn=Mk for all nk, which is stabilization. For the zero module the only submodule is 0, every nonempty family is {0} with minimal member 0, and every descending chain is constant, so both conditions hold. This proves the stated claim.

step 2.1givenalgebra
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

Noetherian and Artinian conditions are each exact in short exact sequences

Statement

In a short exact sequence 0NMQ0, the module M is Noetherian if and only if N and Q are Noetherian; the same equivalence holds with “Artinian” in place of “Noetherian”. See Noetherian modules: every submodule is finitely generated.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A left R-module M is Noetherian when every submodule of M is finitely generated (def-generated-cyclic-finitely-generated-and-free-modules). This finite-generation definition is the convention; its equivalence with ACC and the maximal condition is proved in thm-equivalent-characterizations-of-noetherian-modules. (Noetherian modules: every submodule is finitely generated).

[L2]

A left R-module M is Artinian when every descending chain M0M1 of submodules stabilizes: there is N such that Mn=MN for all nN. This is the descending chain condition. (Artinian modules by the descending chain condition).

[L3]

For submodules L,NM, there is a canonical isomorphism L/(LN)(L+N)/N.. (Second isomorphism theorem for modules).

[L4]

For NM, inverse image and quotient induce mutually inverse inclusion-preserving bijections between submodules of M/N and submodules of M containing N. They preserve sums, intersections, and successive quotients. (Correspondence theorem for submodules of a quotient module).

[L5]

A short exact sequence is an exact sequence 0NiMpQ0; thus i is injective, p is surjective, and imi=kerp. (Exact sequences and short exact sequences of modules).

[L6]

A module is Noetherian if and only if every ascending chain of submodules stabilizes. (Finite generation, ACC, and maximal-condition characterizations of Noetherian modules).

[L7]

A module is Artinian if and only if every descending chain of submodules stabilizes. (DCC and minimal-condition characterizations of Artinian modules).

Proof

technique · direct
1.1

Identify N with its image in M. A chain in N is a chain in M, and the correspondence theorem lifts every chain in Q=M/N to a chain of submodules of M containing N; hence both ACC and DCC pass from M to N and Q.

L1L2L3L4L5L6L7givenalgebra
2.1

Conversely, for a chain (Mi) in M, the chains (MiN) and ((Mi+N)/N) stabilize when N and Q have the relevant chain condition. If MiMj are beyond both stabilization indices and xMj, equality of the images gives yMi with xyN; equality of the intersections then puts xyMi, so xMi. The same argument with the inclusions reversed handles descending chains.

step 1.1givenalgebra
3.1

Thus M has ACC exactly when N and Q do, and it has DCC exactly when N and Q do. Facts [L6] and [L7] convert these chain statements into the asserted Noetherian and Artinian equivalences.

L6L7step 1.1step 2.1
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17Open item page →

Finite direct sums preserve and reflect Noetherian and Artinian conditions

Statement

A finite direct sum is Noetherian if and only if every summand is Noetherian, and it is Artinian if and only if every summand is Artinian. The empty direct sum is included. See Noetherian and Artinian conditions are each exact in short exact sequences.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

In a short exact sequence 0NMQ0, the module M is Noetherian if and only if N and Q are Noetherian; the same equivalence holds with “Artinian” in place of “Noetherian”. (Noetherian and Artinian conditions are each exact in short exact sequences).

[L2]

Let R be a unital ring and (Mi)iI a family of left R-modules (def-left-and-right-modules). Their direct product is the module iIMi with coordinatewise operations. The support of m=(mi) is {iI:mi0}, and the direct sum is the submodule iIMi={miIMi:supp(m) is finite} (def-submodule). (The direct sum of an indexed family of modules).

Proof

technique · direct
1.1

Induct on the number of summands using the split short exact sequence for a binary direct sum.

L1L2givenalgebra
2.1

Reflection follows because every summand is a submodule and a quotient.

step 1.1givenalgebra
3.1

Both extremes of the induction are admitted. The empty direct sum is the zero module, whose only chains of submodules are constant, so it satisfies both conditions while the "every summand" side is vacuously true; a single summand makes the direct sum that summand, so the equivalence is an identity and supplies the base of the induction in step 1.1. This proves the stated claim.

step 1.1step 2.1givenalgebra
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Finitely generated modules over a left Noetherian ring are Noetherian

Statement

Every finitely generated left module over a left Noetherian ring is Noetherian. See Left and right Noetherian rings.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A unital ring R is left Noetherian when its left regular module RR is Noetherian, and right Noetherian when the right regular module RR is Noetherian. Unqualified “Noetherian ring” means left Noetherian here; the side is stated whenever both notions occur. (Left and right Noetherian rings).

[L2]

Let M be a left R-module and SM. The submodule generated by S is SR:={NM:SN}. The family is nonempty because MM, and its intersection is a submodule by lem-submodule-criterion-sums-and-intersections. Thus SR is the smallest submodule of M containing S, just as def-generated-subgroup defines a generated subgroup. (Generated submodule, cyclic and finitely generated modules, module basis and free module).

[L3]

A finite direct sum is Noetherian if and only if every summand is Noetherian, and it is Artinian if and only if every summand is Artinian. The empty direct sum is included. (Finite direct sums preserve and reflect Noetherian and Artinian conditions).

[L4]

For every left R-module M, the free module R(M) on its underlying set admits a canonical surjection εM:R(M)M, determined by εM(em)=m. Consequently MR(M)/kerεM. (Every module is a quotient of a free module).

[L5]

In a short exact sequence 0NMQ0, the module M is Noetherian if and only if N and Q are Noetherian. (Noetherian and Artinian conditions are each exact in short exact sequences).

Proof

technique · direct
1.1

A module generated by n elements is a quotient of Rn.

L1L2L3L4givenalgebra
2.1

The left regular module is Noetherian, [L3] makes Rn Noetherian, and [L5] makes its quotient M Noetherian.

L1L3L5step 1.1given
3.1

The case n=0 is admitted: a module generated by the empty set is 0, and R0 is the zero module, so step 1.1 presents M=0 as a quotient of 0. The zero module has only the constant chain of submodules and is Noetherian, so the argument needs no separate base case. This proves the stated claim.

step 1.1step 2.1givenalgebra
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Every surjective endomorphism of a Noetherian module is injective

Statement

Every surjective endomorphism of a Noetherian module is injective, hence an automorphism. See Finite generation, ACC, and maximal-condition characterizations of Noetherian modules.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a left R-module M, the following are equivalent: every submodule is finitely generated; every ascending chain of submodules stabilizes; and every nonempty family of submodules has a maximal member. The implication from ACC to the maximal condition uses dependent choice; the other displayed implications are choice-free. (Finite generation, ACC, and maximal-condition characterizations of Noetherian modules).

[L2]

For a left R-module M, define EndR(M):=HomR(M,M). Addition is pointwise and multiplication is composition, (fg)(m):=f(g(m)). The ring laws and the identity endomorphism are established in prop-endomorphisms-form-a-ring. (The endomorphism ring EndR(M) under addition and composition).

Proof

technique · direct
1.1

For a surjective endomorphism f, the ascending chain kerfkerf2 stabilizes.

L1L2givenalgebra
2.1

Choose n with kerfn=kerfn+1. For xkerf, surjectivity of fn gives y with fn(y)=x. Then fn+1(y)=0, so ykerfn+1=kerfn and x=fn(y)=0.

step 1.1givenalgebra
3.1

The zero module is admitted: its only endomorphism is the identity, which is injective, so the conclusion holds there and the argument of step 2.1 is not vacuous by accident. This proves the stated claim.

step 2.1givenalgebra
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

A module has a composition series if and only if it is Noetherian and Artinian, the converse using dependent choice

Statement

A module with a composition series is both Noetherian and Artinian. Conversely, assuming dependent choice, a module that is both Noetherian and Artinian has a composition series. The zero module has the empty composition series. See Composition series and length of a module.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A composition series of a left R-module M is a finite chain 0=M0<M1<<Mn=M whose factors Mi/Mi1 are simple. If such a series exists, the length R(M) is its number n of factors; thm-jordan-holder-theorem-for-modules proves independence of the chosen series. The zero module has the empty series and length 0. (Composition series and length of a module).

[L2]

For a left R-module M, the following are equivalent: every submodule is finitely generated; every ascending chain of submodules stabilizes; and every nonempty family of submodules has a maximal member. The implication from ACC to the maximal condition uses dependent choice; the other displayed implications are choice-free. (Finite generation, ACC, and maximal-condition characterizations of Noetherian modules).

[L3]

For a left R-module M, DCC is equivalent to the condition that every nonempty family of submodules has a minimal member. The implication from DCC to the minimal condition uses dependent choice. (DCC and minimal-condition characterizations of Artinian modules).

[L4]

In a short exact sequence 0NMQ0, the module M is Noetherian if and only if N and Q are Noetherian; the same equivalence holds with “Artinian” in place of “Noetherian”. (Noetherian and Artinian conditions are each exact in short exact sequences).

Proof

technique · direct
1.1

Let 0=M0<<Mn=M be a composition series [L1] and induct on i that Mi is Noetherian and Artinian. The zero module M0 satisfies both conditions vacuously. A simple factor Mi/Mi1 has only the submodules 0 and itself, so every chain of its submodules stabilizes and it too satisfies both conditions. Applying [L4] to the short exact sequence 0Mi1MiMi/Mi10 carries both conditions from Mi1 and the simple factor to Mi. At i=n this gives the forward implication, which uses no choice principle.

L1L4givenalgebra
2.1

Conversely, assume dependent choice and let M be Noetherian and Artinian. Every submodule of M is Noetherian by [L4], so a nonzero submodule N has a nonempty family of proper submodules, which by the maximal condition of [L2] has a maximal member — a maximal proper submodule of N. Dependent choice applied to this relation, starting at M, yields a chain M=N0>N1> in which Nk+1 is a maximal proper submodule of Nk for as long as Nk0.

step 1.1L2L4givenalgebra
3.1

The chain of step 2.1 is strictly descending while its terms are nonzero, so the descending chain condition forces some Nr=0. Since Nk+1 is maximal proper in Nk, the quotient Nk/Nk+1 is nonzero and has no proper nonzero submodule, hence is simple. Reversing the chain gives 0=Nr<<N0=M, a composition series. For M=0 the empty chain is already the required series, so no choice is consumed in that case. This proves the stated claim.

L1L3step 2.1givenalgebra
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17Open item page →

Jordan–Hölder theorem for modules

Statement

Any two composition series of a module have the same length, and their simple factors agree up to permutation and isomorphism. See Composition series and length of a module.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A composition series of a left R-module M is a finite chain 0=M0<M1<<Mn=M whose factors Mi/Mi1 are simple. If such a series exists, the length R(M) is its number n of factors; thm-jordan-holder-theorem-for-modules proves independence of the chosen series. The zero module has the empty series and length 0. (Composition series and length of a module).

[L2]

For submodules L,NM, there is a canonical isomorphism L/(LN)(L+N)/N.. (Second isomorphism theorem for modules).

[L3]

If NLM, then L/N is a submodule of M/N and (M/N)/(L/N)M/L.. (Third isomorphism theorem for modules).

[L4]

For NM, inverse image and quotient induce mutually inverse inclusion-preserving bijections between submodules of M/N and submodules of M containing N. They preserve sums, intersections, and successive quotients. (Correspondence theorem for submodules of a quotient module).

Proof

technique · direct
1.1

Fix a composition series 0=M0<<Mn=M and prove by induction on n that it has the asserted comparison with every other composition series of M. We use simultaneously the elementary consequence that, for any CM, intersecting the fixed series with C and deleting repetitions gives a composition series of C: each remaining factor embeds in the corresponding simple factor Mi/Mi1 and is therefore simple.

L1L2L3L4givenalgebra
2.1

The case n=0 is M=0. For n>0, let A=Mn1 and let B be the penultimate term of a second series. If A=B, the induction hypothesis applied in A matches all lower factors, and the common top factor finishes.

step 1.1given
3.1

Suppose AB. Since A and B are maximal proper submodules, A+B=M. Put C=AB. The second isomorphism theorem gives A/CM/B,B/CM/A, so both quotients are simple.

L2step 1.1step 2.1givenalgebra
4.1

By step 1.1, C has a composition series. Appending A gives a composition series of A ending in A/C. Compare it with 0=M0<<Mn1=A using the induction hypothesis, whose fixed first series has length n1. It follows that the series of C has length n2 and that its factors together with A/C are exactly the factors below M/A in the fixed series.

L3L4step 1.1step 3.1given
5.1

Appending B to the same series of C gives a composition series of B of length n1. Using this as the fixed first series, the induction hypothesis compares it with the lower part of the second series. The isomorphisms in step 3.1 exchange the two top simple factors A/C and M/A with M/B and B/C. Hence the two original series have length n and the same factors up to permutation. This also covers n=1, when C=0.

step 2.1step 3.1step 4.1given
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Module length is additive in short exact sequences

Statement

For a short exact sequence 0NMQ0, the module M has finite length if and only if N and Q do, and then R(M)=R(N)+R(Q). See Jordan–Hölder theorem for modules.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Any two composition series of a module have the same length, and their simple factors agree up to permutation and isomorphism. (Jordan–Hölder theorem for modules).

[L2]

A composition series of a left R-module M is a finite chain 0=M0<M1<<Mn=M whose factors Mi/Mi1 are simple. If such a series exists, the length R(M) is its number n of factors; thm-jordan-holder-theorem-for-modules proves independence of the chosen series. The zero module has the empty series and length 0. (Composition series and length of a module).

[L3]

For NM, inverse image and quotient induce mutually inverse inclusion-preserving bijections between submodules of M/N and submodules of M containing N. They preserve sums, intersections, and successive quotients. (Correspondence theorem for submodules of a quotient module).

Proof

technique · direct
1.1

If N and Q have composition series, lift the series of Q along MQ and splice it above the series of N. Correspondence identifies all lifted factors, so this is a composition series of M with R(N)+R(Q) factors.

L1L2L3givenalgebra
1.2

Conversely, let 0=M0<<Mn=M be a composition series. Put Ni=MiN and let Qi be the image of Mi in Q. For each i, the simple factor Mi/Mi1 has submodule Ni/Ni1 and corresponding quotient Qi/Qi1; exactly one is that simple factor and the other is zero. Deleting repetitions therefore gives composition series of N and Q, and their numbers of factors add to n.

L2L3givenalgebra
2.1

Jordan–Hölder makes all three lengths independent of the chosen series, so steps 1.1 and 1.2 prove both directions and the formula. If N=0, Q=0, or M=0, the relevant series is empty and the same count applies.

L1L2step 1.1step 1.2given
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Module endomorphisms form a ring under pointwise addition and composition

Statement

For every left R-module M, pointwise addition and composition make EndR(M) a unital ring with identity idM. See The endomorphism ring EndR(M) under addition and composition.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a left R-module M, define EndR(M):=HomR(M,M). Addition is pointwise and multiplication is composition, (fg)(m):=f(g(m)). The ring laws and the identity endomorphism are established in prop-endomorphisms-form-a-ring. (The endomorphism ring EndR(M) under addition and composition).

[L2]

For left R-modules M,N, the set HomR(M,N) of module homomorphisms is an abelian group under pointwise addition, with zero the zero homomorphism and inverse (f)(m)=f(m) (def-module-homomorphism-kernel-image-and-cokernel, def-group). (The abelian group HomR(M,N) and maps induced by pre- and postcomposition).

[L3]

For left R-modules M,N, a function f:MN is an R-module homomorphism if f(m+m)=f(m)+f(m) and f(rm)=rf(m) for all m,mM and rR (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct
1.1

If f,gEndR(M) then fg is again a module homomorphism, since (fg)(m+n)=f(g(m)+g(n))=(fg)(m)+(fg)(n) and (fg)(rm)=f(rg(m))=r(fg)(m). So composition is a binary operation on EndR(M).

L1L3givenalgebra
1.2

Taking N=M in [L2] makes (EndR(M),+) an abelian group, with the zero homomorphism as neutral element and (f)(m)=f(m) as the inverse of f.

L1L2
1.3

Composition is associative: for all m, (f(gh))(m)=f(g(h(m)))=((fg)h)(m).

L1algebra
1.4

Both distributive laws hold. For all m, (f(g+h))(m)=f(g(m)+h(m))=f(g(m))+f(h(m))=(fg+fh)(m), where the middle equality is additivity of f; and ((g+h)f)(m)=g(f(m))+h(f(m))=(gf+hf)(m) directly from pointwise addition.

L1L2L3algebra
1.5

The identity map idM satisfies idM(m+n)=m+n and idM(rm)=rm, so it lies in EndR(M), and fidM=f=idMf for every f.

L1L3algebra
2.1

Steps 1.1 through 1.5 are exactly the axioms of a unital ring for (EndR(M),+,,idM). For the zero module the only map 00 is id0, so EndR(0) has one element and is the one-element ring, in which the identity coincides with the zero element. This proves the stated claim.

step 1.1step 1.2step 1.3step 1.4step 1.5algebra
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EndR(RR)Rop

Statement

For every unital ring R, evaluation at 1 identifies endomorphisms of the left regular module with right multiplications and gives a ring isomorphism EndR(RR)Rop. See The opposite ring Rop.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a unital ring R, the opposite ring Rop has the same underlying abelian group, identity, and addition as R, with multiplication ab:=ba. Associativity and both distributive laws follow from those of R with the order reversed, and the same element 1 is a two-sided identity. Thus the displayed operations really form a unital ring, including when R is the zero ring. (The opposite ring Rop).

[L2]

For every left R-module M, pointwise addition and composition make EndR(M) a unital ring with identity idM. (Module endomorphisms form a ring under pointwise addition and composition).

[L3]

Let R be a ring. A left R-module is an abelian group (M,+,0M) with a scalar action R×MM, (r,m)rm, satisfying r(m+n)=rm+rn,(r+s)m=rm+sm,(rs)m=r(sm),1Rm=m. A right R-module has an action M×RM, (m,r)mr, with the analogous right-handed axioms. Unless “right” is stated, module means a unital left module. (Unital left and right modules over a ring; unqualified module means left module).

Proof

technique · direct
1.1

Evaluate an endomorphism at 1.

L1L2L3givenalgebra
2.1

Left linearity gives f(r)=rf(1), so endomorphisms are right multiplications; composition reverses the order of their defining elements.

step 1.1givenalgebra
3.1

The map aρa with ρa(x)=xa is inverse to ff(1): it is a left R-module homomorphism because ρa(rx)=rxa=rρa(x), it satisfies ρf(1)=f by step 2.1, and ρa(1)=a. It is additive, and ρaρb(x)=xba=ρba(x)=ρab(x) with the multiplication of [L1], so it is a ring isomorphism onto EndR(RR). For the zero ring 1=0, both EndR(RR) and Rop are the one-element ring and the correspondence is the unique map between them. This proves the stated claim.

step 2.1L1L2givenalgebra
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Endomorphisms of a finite direct sum are matrices of Hom-groups

Statement

For nN and left R-modules M1,,Mn, endomorphisms of jMj correspond to n×n matrices (fij) with fijHomR(Mj,Mi), and composition is matrix multiplication using composition in the entries. For n=0, both sides are the one-element zero ring. See The endomorphism ring EndR(M) under addition and composition.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a left R-module M, define EndR(M):=HomR(M,M). Addition is pointwise and multiplication is composition, (fg)(m):=f(g(m)). The ring laws and the identity endomorphism are established in prop-endomorphisms-form-a-ring. (The endomorphism ring EndR(M) under addition and composition).

[L2]

For left R-modules M,N, the set HomR(M,N) of module homomorphisms is an abelian group under pointwise addition, with zero the zero homomorphism and inverse (f)(m)=f(m) (def-module-homomorphism-kernel-image-and-cokernel, def-group). (The abelian group HomR(M,N) and maps induced by pre- and postcomposition).

[L3]

Let (Mi)iI be left R-modules and N a left R-module. For every family of homomorphisms fi:MiN, there is a unique homomorphism f:iIMiN such that fȷi=fi for every i. It is given by f((mi))=isupp(m)fi(mi). For I=, this is the unique map 0N. (Universal property of a direct sum of modules).

Proof

technique · direct
1.1

We use inclusions and projections to send f to entries fij=πifιj, and reconstruct f by finite sums.

L1L2L3givenalgebra
2.1

Composition becomes matrix multiplication because kιkπk=id on a finite direct sum, so the (i,j) entry of fg is πi(fg)ιj=πif(kιkπk)gιj=k(πifιk)(πkgιj)=kfikgkj, which is the matrix product with composition in the entries; the sum is finite because n is.

step 1.1L3algebra
3.1

For n=0 the direct sum is the zero module, EndR(0) has one element by [L1], and the set of 0×0 matrices also has exactly one element, so both sides are the one-element zero ring as the Statement records. For n=1 the matrix is the single entry f11=π1fι1=f, and the correspondence is the identity on EndR(M1). This proves the stated claim.

step 1.1step 2.1L1algebra
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Under Choice, every finitely generated nonzero module has a maximal proper submodule

Statement

Assuming the Axiom of Choice, every finitely generated nonzero module has a maximal proper submodule. See Generated submodule, cyclic and finitely generated modules, module basis and free module.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let M be a left R-module and SM. The submodule generated by S is SR:={NM:SN}. The family is nonempty because MM, and its intersection is a submodule by lem-submodule-criterion-sums-and-intersections. Thus SR is the smallest submodule of M containing S, just as def-generated-subgroup defines a generated subgroup. (Generated submodule, cyclic and finitely generated modules, module basis and free module).

[L2]

Let M be a left R-module. A subset NM is a submodule when it is a subgroup of the additive group of M and is closed under scalars: rR, nNrnN. The operations on N are the restrictions of those of M. Write NM when the ring and module are understood. (Submodule of a module).

[L3]

Assume the Axiom of Choice (def-axiom-of-choice). Let (P,) be a nonempty poset in which every chain has an upper bound. Then P has a maximal element (def-maximal-element). Note the hypothesis asks only for an upper bound, not a least upper bound, and the conclusion asserts only that a maximal element exists, never that a greatest one does. (Zorn's lemma).

Proof

technique · direct
1.1

Let M0 be generated over R by x1,,xn [L1] and let P be the set of proper submodules of M [L2], ordered by inclusion. It is nonempty, since 0 is a proper submodule of the nonzero module M.

L1L2givenalgebra
2.1

Let CP be a chain. If C is empty, 0P is an upper bound. Otherwise put U=C; any two elements of U lie in a common member of the chain, so U is a submodule. If U=M, each generator xi would lie in some member of C, and the largest of those finitely many members would contain every xi and hence equal M, contradicting properness. So UP and is an upper bound for C.

step 1.1givenalgebra
3.1

Every chain in the nonempty poset P therefore has an upper bound in P, so Zorn's lemma [L3] — which assumes the Axiom of Choice, as the Statement does — gives a maximal element of P, that is, a maximal proper submodule of M.

L3step 2.1given
4.1

Both hypotheses are load-bearing above: nonzeroness is what makes the poset of step 1.1 nonempty, and finite generation is what makes the union of a chain proper in step 2.1. This proves the stated claim.

step 1.1step 2.1step 3.1given
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Equivalent characterizations of semisimple modules

Statement

Assuming the Axiom of Choice, for a module M the following are equivalent: M is a direct sum of simple submodules; M is the sum of its simple submodules; and every submodule of M has a complementary submodule. See Semisimple modules as direct sums of simple modules.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A left R-module is semisimple when it is an internal direct sum of simple submodules, allowing the empty direct sum. Hence the zero module is semisimple. (Semisimple modules as direct sums of simple modules).

[L2]

Assuming the Axiom of Choice, every finitely generated nonzero module has a maximal proper submodule. (Under Choice, every finitely generated nonzero module has a maximal proper submodule).

[L3]

Assume the Axiom of Choice (def-axiom-of-choice). Let (P,) be a nonempty poset in which every chain has an upper bound. Then P has a maximal element (def-maximal-element). Note the hypothesis asks only for an upper bound, not a least upper bound, and the conclusion asserts only that a maximal element exists, never that a greatest one does. (Zorn's lemma).

[L4]

Let (Mi)iI be left R-modules and N a left R-module. For every family of homomorphisms fi:MiN, there is a unique homomorphism f:iIMiN such that fȷi=fi for every i. It is given by f((mi))=isupp(m)fi(mi). For I=, this is the unique map 0N. (Universal property of a direct sum of modules).

[L5]

The socle Soc(M) is the sum of all simple submodules of M. (The socle as the sum of all simple submodules).

Proof

technique · direct
1.1

A direct sum of simple submodules is plainly their sum. Conversely, if M is a sum of simple submodules, Zorn's lemma applied to independent families of them gives a maximal direct sum D; if DM, a simple submodule not contained in D meets D trivially, contradicting maximality.

L1L2L3L4L5givenalgebra
2.1

Given NM and a direct-sum decomposition M=iISi into simples, use Zorn to choose a maximal sum C=jJSj with CN=0. If some Si were not contained in N+C, simplicity would give Si(N+C)=0, so adjoining it would contradict maximality. Hence every SiN+C, and therefore M=NC.

L3step 1.1givenalgebra
3.1

Conversely suppose every submodule has a complement. By [L5], choose C with M=Soc(M)C. If C0, choose 0xC. The cyclic module Rx has a maximal proper submodule K by [L2]. Let D complement K in M. Then Rx=K(RxD), and RxDRx/K is a nonzero simple submodule of C, contrary to CSoc(M)=0. Hence C=0 and M is a sum of simples. The zero module is the empty sum.

L2L5step 2.1givenalgebra
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Choice-free semisimple characterizations for finite-length modules

Statement

For a finite-length module, the direct-sum, sum-of-simples, and complement characterizations of semisimplicity are equivalent without any choice principle. See Equivalent characterizations of semisimple modules.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A composition series of a left R-module M is a finite chain 0=M0<M1<<Mn=M whose factors Mi/Mi1 are simple. If such a series exists, the length R(M) is its number n of factors; thm-jordan-holder-theorem-for-modules proves independence of the chosen series. The zero module has the empty series and length 0. (Composition series and length of a module).

[L2]

Any two composition series of a module have the same length, and their simple factors agree up to permutation and isomorphism. (Jordan–Hölder theorem for modules).

Proof

technique · direct
1.1

Suppose M is a sum of simple submodules and start with D0=0. If DkM, some simple Sk is not contained in Dk, so simplicity gives SkDk=0 and Dk+1=DkSk. Intersecting a fixed n-factor composition series of M with Dk and deleting repetitions gives a composition series of Dk with at most n factors. Since Dk already has the k-factor series obtained by adding the Sj one at a time, [L2] gives kn. Thus after at most n finite choices the process reaches M, proving that a sum of simples is a finite direct sum without any choice axiom.

L1L2givenalgebra
2.1

Now write a finite direct-sum decomposition M=i=1tSi and let NM. Process the finitely many Si in order, maintaining a sum C with CN=0: add Si exactly when Si≰N+C. In that case simplicity gives Si(N+C)=0, so the invariant persists. At the end every SiN+C, whence M=NC. This proves that the direct-sum condition implies the complement condition without Zorn.

step 1.1givenalgebra
3.1

Conversely, suppose every submodule of the finite-length module M has a complement, and induct on a fixed composition-series length. If M0, let A be the penultimate term of such a series. A complement S gives M=AS with SM/A simple. The complement property passes to A: for LA, if M=LD, then A=L(AD). The induction hypothesis makes A a finite direct sum of simples, hence so is M. Together with step 1.1 and the trivial direct-sum-to-sum implication, this proves all three equivalences, including lengths zero and one, without Choice.

L1step 1.1step 2.1givenalgebra
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Under Choice, submodules and quotients of semisimple modules are semisimple

Statement

Assuming the Axiom of Choice, every submodule and every quotient of a semisimple module is semisimple. See Equivalent characterizations of semisimple modules.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Assuming the Axiom of Choice, for a module M the following are equivalent: M is a direct sum of simple submodules; M is the sum of its simple submodules; and every submodule of M has a complementary submodule. (Equivalent characterizations of semisimple modules).

[L2]

For submodules L,NM, there is a canonical isomorphism L/(LN)(L+N)/N.. (Second isomorphism theorem for modules).

Proof

technique · direct
1.1

A submodule inherits the complement property by intersecting a complement in the ambient module.

L1L2givenalgebra
2.1

For a quotient, complement the kernel and identify the quotient with that semisimple complement.

step 1.1givenalgebra
3.1

The extreme submodules are admitted and give nothing new: the zero submodule is the empty direct sum, hence semisimple, and its quotient M/0M is semisimple by hypothesis; the whole submodule M is semisimple by hypothesis and its quotient M/M=0 is again the empty direct sum. This proves the stated claim.

step 1.1step 2.1givenalgebra
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Under Choice, the socle is the largest semisimple submodule

Statement

Assuming the Axiom of Choice, for every module M, Soc(M) is semisimple and contains every semisimple submodule of M. See The socle as the sum of all simple submodules.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a left R-module M, the socle is Soc(M):={SM:S is simple}. Concretely it is the submodule generated by the union of all simple submodules. If there are none, the sum is 0, so the definition is always meaningful. (The socle as the sum of all simple submodules).

[L2]

Assuming the Axiom of Choice, for a module M the following are equivalent: M is a direct sum of simple submodules; M is the sum of its simple submodules; and every submodule of M has a complementary submodule. (Equivalent characterizations of semisimple modules).

[L3]

Assuming the Axiom of Choice, every submodule and every quotient of a semisimple module is semisimple. (Under Choice, submodules and quotients of semisimple modules are semisimple).

Proof

technique · direct
1.1

The sum of all simple submodules is semisimple by the sum characterization.

L1L2L3givenalgebra
2.1

Every semisimple submodule is itself a sum of simple submodules of the ambient module, hence lies in the socle.

step 1.1givenalgebra
3.1

If M has no simple submodule, the defining sum of [L1] is empty, so Soc(M)=0; the zero module is the empty direct sum and hence semisimple, and the only semisimple submodule of such an M is 0 itself, which it contains. This proves the stated claim.

L1step 1.1step 2.1givenalgebra
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A finitely generated semisimple module is a finite direct sum of simple modules

Statement

Every finitely generated semisimple module is a finite direct sum of simple modules. See Semisimple modules as direct sums of simple modules.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A left R-module is semisimple when it is an internal direct sum of simple submodules, allowing the empty direct sum. Hence the zero module is semisimple. (Semisimple modules as direct sums of simple modules).

[L2]

Let M be a left R-module and SM. The submodule generated by S is SR:={NM:SN}. The family is nonempty because MM, and its intersection is a submodule by lem-submodule-criterion-sums-and-intersections. Thus SR is the smallest submodule of M containing S, just as def-generated-subgroup defines a generated subgroup. (Generated submodule, cyclic and finitely generated modules, module basis and free module).

Proof

technique · direct
1.1

In a direct-sum decomposition, each generator has finite support; the union of those finite supports contains every generator and hence the whole module.

L1L2givenalgebra
2.1

The finite subfamily located in step 1.1 already consists of simple summands of the original internal direct sum, so no summand is zero and no index repeats; its internal sum is therefore a finite direct sum of simple modules, and by step 1.1 it is all of M. For a module generated by the empty set, M=R=0, which is the empty direct sum and semisimple by [L1]. This proves the stated claim.

step 1.1L1L2givenalgebra
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Schur's lemma for simple modules

Statement

A nonzero homomorphism between simple modules is an isomorphism. Consequently the endomorphism ring of a simple module is a division ring. See Simple module: a nonzero module with no proper nonzero submodule.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A left R-module M is simple if M0 and its only submodules are 0 and M. Equivalently, M has no proper nonzero submodule. (Simple module: a nonzero module with no proper nonzero submodule).

[L2]

For a left R-module M, define EndR(M):=HomR(M,M). Addition is pointwise and multiplication is composition, (fg)(m):=f(g(m)). The ring laws and the identity endomorphism are established in prop-endomorphisms-form-a-ring. (The endomorphism ring EndR(M) under addition and composition).

[L3]

For every left R-module M, pointwise addition and composition make EndR(M) a unital ring with identity idM. (Module endomorphisms form a ring under pointwise addition and composition).

Proof

technique · direct
1.1

The kernel and image of a homomorphism between simple modules are each zero or whole.

L1L2L3givenalgebra
2.1

A nonzero homomorphism is therefore injective and surjective.

step 1.1givenalgebra
3.1

Applied to a nonzero endomorphism, its inverse is linear, so the endomorphism ring is a division ring.

step 2.1givenalgebra
4.1

The excluded case is genuinely excluded rather than overlooked: the zero homomorphism between nonzero simple modules is not an isomorphism, which is why the hypothesis asks for a nonzero one, and it is the zero element of the endomorphism ring of step 3.1 rather than a non-invertible unit. Contrapositively, if two simple modules are not isomorphic then every homomorphism between them is zero. This proves the stated claim.

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Equivalent module-theoretic characterizations of semisimple rings

Statement

Assuming the Axiom of Choice, for a unital ring R the following are equivalent: RR is semisimple; every left R-module is semisimple; every short exact sequence of left R-modules splits; and every left R-module is projective. See A semisimple ring as a ring whose left regular module is semisimple.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A unital ring R is semisimple when its left regular module RR is semisimple. This is a left-module definition and uses no Jacobson radical. For the zero ring, the regular module is zero and hence semisimple; the Wedderburn-Artin theorem below is stated for nonzero rings. (A semisimple ring as a ring whose left regular module is semisimple).

[L2]

Assuming the Axiom of Choice, every submodule and every quotient of a semisimple module is semisimple. (Under Choice, submodules and quotients of semisimple modules are semisimple).

[L3]

For every left R-module M, the free module R(M) on its underlying set admits a canonical surjection εM:R(M)M, determined by εM(em)=m. Consequently MR(M)/kerεM. (Every module is a quotient of a free module).

[L4]

In a short exact sequence 0AiBpC0, a section of p is a homomorphism s:CB with ps=idC, and a retraction of i is a homomorphism r:BA with ri=idA. (Split short exact sequences, sections, and retractions).

[L5]

For a short exact sequence 0AiBpC0, the following are equivalent: 1. p has a section s:CB; 2. i has a retraction r:BA; 3. there is an isomorphism Φ:ACB with Φ(a,0)=i(a) and p(Φ(a,c))=c. (The splitting lemma for short exact sequences of modules).

[L6]

A left R-module P is projective if it has the lifting property for epimorphisms: whenever q:EM is a surjective module homomorphism and f:PM is a module homomorphism, there exists a module homomorphism f~:PE such that qf~=f (def-module-homomorphism-kernel-image-and-cokernel, def-injection-surjection-bijection). (Projective modules and the lifting property).

[L7]

For a left R-module P, assertions 1 to 3 below are equivalent without choice. Under the Axiom of Choice, they are also equivalent to assertion 4: 1. P is projective; 2. every short exact sequence 0KEP0 splits; 3. HomR(P,) takes every short exact sequence to a short exact sequence; 4. P is a direct summand of a free module. (Equivalent characterizations of projective modules).

[L8]

Assuming the Axiom of Choice, a module is semisimple if and only if every submodule has a complementary submodule. (Equivalent characterizations of semisimple modules).

Proof

technique · direct
1.1

If RR is semisimple, every free left module, being a direct sum of copies of R, is semisimple; every module is a quotient of a free module, so every left module is semisimple.

L1L2L3L4L5L6L7L8givenalgebra
2.1

If every left module is semisimple, [L8] gives every submodule a complement, and the splitting lemma makes every short exact sequence split. Conversely, if every short exact sequence splits, the projective criterion makes every module projective; if every module is projective, each quotient map splits, so [L8] makes every module semisimple.

L5L7L8step 1.1givenalgebra
3.1

Applying the universal module condition to the left regular module recovers the first condition, and every clause is left-handed as asserted. This proves the stated claim.

step 2.1givenalgebra
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Matrix rings over division rings are semisimple

Statement

Let D be a division ring and n1. On the set of n×n arrays over D, use entrywise addition and the product

(AB)ij:=k=1naikbkj.

These operations make a ring Mn(D), and this ring is semisimple. More precisely, its left regular module is the direct sum of the simple column ideals Mn(D)ejjDn for 1jn. See A semisimple ring as a ring whose left regular module is semisimple.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A unital ring R is semisimple when its left regular module RR is semisimple. This is a left-module definition and uses no Jacobson radical. For the zero ring, the regular module is zero and hence semisimple; the Wedderburn-Artin theorem below is stated for nonzero rings. (A semisimple ring as a ring whose left regular module is semisimple).

[L2]

A division ring is a ring D with 10 in which every nonzero element has a two-sided multiplicative inverse. (Division ring: a ring with 10 in which every nonzero element is a unit).

[L3]

A ring has an abelian-group addition, an associative multiplication with identity, and both distributive laws; multiplication need not be commutative. (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L4]

Finite sums in a commutative monoid are independent of the chosen enumeration, and the empty sum is 0. (A finite sum in a commutative monoid indexed by an arbitrary finite set).

[L5]

A left R-module M is simple if M0 and its only submodules are 0 and M. Equivalently, M has no proper nonzero submodule. (Simple module: a nonzero module with no proper nonzero submodule).

Proof

technique · direct
1.1

Entrywise addition makes the arrays an abelian group. Associativity of multiplication follows by expanding both (AB)C and A(BC) and reassociating the finite double sum; the two distributive laws follow entrywise from those of D. The matrix I=(δij1D) is a two-sided identity. Thus the displayed operations make the unital ring Mn(D) without any commutativity assumption on D.

L2L3L4givenalgebra
2.1

Let ej=ejj. Every matrix has the unique decomposition A=j=1nAej, and Mn(D)eijiMn(D)ej=0 because the two sides have disjoint possible nonzero columns. Hence Mn(D)Mn(D)=j=1nMn(D)ej.

step 1.1L4givenalgebra
3.1

Sending a matrix in Mn(D)ej to its j-th column identifies that left ideal with Dn under left matrix multiplication. If 0vDn, choose k with vk0. For any wDn, the matrix whose only possibly nonzero column is column k, with entry aik=wivk1, sends v to w. Thus every nonzero submodule of Dn is all of Dn, so each column ideal is simple.

L2L5step 1.1step 2.1givenalgebra
4.1

The decomposition in step 2.1 is therefore a finite direct sum of simple left modules, so [L1] makes Mn(D) semisimple. For n=1 it is the single simple column ideal, and the hypothesis n1 excludes an empty decomposition.

L1step 2.1step 3.1given
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Wedderburn–Artin theorem for semisimple rings

Statement

Let R be a nonzero unital ring. Then R is semisimple if and only if Ri=1rMni(Di) for positive integers r,ni and division rings Di. See EndR(RR)Rop.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For every unital ring R, evaluation at 1 identifies endomorphisms of the left regular module with right multiplications and gives a ring isomorphism EndR(RR)Rop.. (EndR(RR)Rop).

[L2]

For left R-modules M1,,Mn, endomorphisms of jMj correspond to matrices (fij) with fijHomR(Mj,Mi), and composition is matrix multiplication using composition in the entries. (Endomorphisms of a finite direct sum are matrices of Hom-groups).

[L3]

Every finitely generated semisimple module is a finite direct sum of simple modules. (A finitely generated semisimple module is a finite direct sum of simple modules).

[L4]

A nonzero homomorphism between simple modules is an isomorphism. Consequently the endomorphism ring of a simple module is a division ring. (Schur's lemma for simple modules).

[L5]

For a division ring D and n1, matrices with product (AB)ij=kaikbkj form a semisimple ring whose left regular module is the direct sum of its simple column ideals. (Matrix rings over division rings are semisimple).

[L6]

The opposite ring Rop has the same addition and identity as R and multiplication ab:=ba. (The opposite ring Rop).

[L7]

A unital ring R is semisimple when its left regular module RR is semisimple. This is a left-module definition and uses no Jacobson radical. For the zero ring, the regular module is zero and hence semisimple; the Wedderburn-Artin theorem below is stated for nonzero rings. (A semisimple ring as a ring whose left regular module is semisimple).

Proof

technique · direct
1.1

If R is semisimple, its cyclic left regular module is a finite direct sum of simple modules. Grouping isomorphic summands gives RRi=1rSini with pairwise nonisomorphic Si and positive r,ni.

L1L2L3L4L5L6L7givenalgebra
2.1

Schur's lemma gives HomR(Sj,Si)=0 for ij and makes Ei=EndR(Si) a division ring. Hence the endomorphism-matrix theorem gives EndR(RR)iMni(Ei).

step 1.1givenalgebra
3.1

Since EndR(RR)Rop, taking opposites gives RiMni(Ei)op. The opposite Eiop is again a division ring, and entrywise transpose is a ring isomorphism Mni(Ei)opMni(Eiop),AAT, because reversing both the matrix product and the entry product gives (BA)T=ATBT in the target. Thus RiMni(Di) with Di=Eiop.

L5L6step 2.1givenalgebra
4.1

Conversely, each Mni(Di) is semisimple by its column-ideal decomposition, and a finite product is semisimple because its regular module is the finite direct sum of the factors' regular modules.

step 3.1givenalgebra
5.1

The Statement assumes that R is nonzero, so the decomposition has at least one factor; no empty-product convention is asserted. This proves the stated claim.

step 4.1givenalgebra
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Simple modules over a product of matrix rings over division rings

Statement

Let r1, let every ni1, let every Di be a division ring, and put R=i=1rMni(Di). Then every simple left R-module is supported on exactly one factor and is isomorphic to that factor's column module Dini. These column modules give all simple left R-module isomorphism classes, with one class for each factor. See Wedderburn–Artin theorem for semisimple rings.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A nonzero homomorphism between simple modules is an isomorphism. Consequently the endomorphism ring of a simple module is a division ring. (Schur's lemma for simple modules).

[L2]

If D is a division ring and n1, then the left regular module of Mn(D) is the direct sum of its simple column ideals Mn(D)ejjDn. (Matrix rings over division rings are semisimple).

Proof

technique · direct
1.1

Write ciR for the central idempotent that is 1 in factor i and 0 elsewhere. For a simple left R-module S, every ciS is a submodule and S=iciS. Hence some ciS is nonzero and therefore equals S; then cjS=cjciS=0 for ji. Thus S is supported on exactly one factor Ai=Mni(Di).

givenalgebra
2.1

Choose 0sS. The map AiS, aas, is surjective because its image is a nonzero submodule. By [L2], Ai is a direct sum of simple column ideals CjDini. At least one restriction CjS is nonzero, so [L1] makes it an isomorphism. Hence SDini.

L1L2step 1.1givenalgebra
3.1

Conversely each column module is simple by [L2]. Modules supported on different factors cannot be isomorphic, because the corresponding ci acts as the identity on one and as zero on the other. For a fixed factor all column ideals are isomorphic to Dini by [L2]. This proves the classification, including the one-factor case r=1.

L2step 1.1step 2.1givenalgebra
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Uniqueness of the Wedderburn–Artin factors

Statement

The division rings and matrix sizes in a Wedderburn-Artin decomposition of a nonzero semisimple ring are unique up to permutation and division-ring isomorphism. See Wedderburn–Artin theorem for semisimple rings.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let R be a nonzero unital ring. Then R is semisimple if and only if Ri=1rMni(Di) for positive integers r,ni and division rings Di. (Wedderburn–Artin theorem for semisimple rings).

[L2]

For r1, ni1, and division rings Di, every simple left module over iMni(Di) is supported on exactly one factor and is isomorphic to that factor's column module Dini; these give all isomorphism classes. (Simple modules over a product of matrix rings over division rings).

[L3]

For a division ring D and n1, the left regular module of Mn(D) is the direct sum of the n simple column ideals Mn(D)ejjDn. (Matrix rings over division rings are semisimple).

[L4]

A nonzero homomorphism between simple modules is an isomorphism. Consequently the endomorphism ring of a simple module is a division ring. (Schur's lemma for simple modules).

[L5]

Any two composition series of a module have the same length, and their simple factors agree up to permutation and isomorphism. (Jordan–Hölder theorem for modules).

Proof

technique · direct
1.1

In a decomposition RiMni(Di), the simple left module supported on factor i is its column module Si=Dini. Distinct coordinate idempotents show that the Si are pairwise nonisomorphic as R-modules, and [L3] decomposes the regular module with exactly ni copies of Si.

L1L2L3L4L5givenalgebra
2.1

Let f:SiSi commute with the matrix action and put v=f(e1). For k1, the matrix unit ekk annihilates e1, so it annihilates v; hence v=e1d for a unique dDi. Since ej=ej1e1, one has f(ej)=ej1f(e1)=ejd, and additivity then gives f(x)=xd for every column x. Thus the endomorphisms are precisely right scalar multiplications, composition reverses the scalar order, and EndR(Si)Diop. Therefore DiEndR(Si)op is determined by the simple-module type with the orientation fixed.

L3L4step 1.1givenalgebra
3.1

Jordan–Hölder [L5] makes the simple-module types and their multiplicities in RR invariant. Hence the pairs (ni,Di) are determined up to reordering and division-ring isomorphism. This proves the stated claim.

L5step 1.1step 2.1given
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Left and right semisimplicity of a ring agree

Statement

A unital ring is semisimple as a left regular module if and only if it is semisimple as a right regular module. See Wedderburn–Artin theorem for semisimple rings.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A unital ring R is semisimple when its left regular module RR is semisimple; the zero ring is semisimple because its regular module is zero. (A semisimple ring as a ring whose left regular module is semisimple).

[L2]

Let R be a nonzero unital ring. Then R is semisimple if and only if Ri=1rMni(Di) for positive integers r,ni and division rings Di. (Wedderburn–Artin theorem for semisimple rings).

[L3]

For a unital ring R, the opposite ring Rop has the same underlying abelian group, identity, and addition as R, with multiplication ab:=ba. Associativity and both distributive laws follow from those of R with the order reversed, and the same element 1 is a two-sided identity. Thus the displayed operations really form a unital ring, including when R is the zero ring. (The opposite ring Rop).

[L4]

For a division ring D and n1, the ring Mn(D) is semisimple and its left regular module is a finite direct sum of simple column ideals. (Matrix rings over division rings are semisimple).

Proof

technique · direct
1.1

First suppose that R is semisimple on the left. If R is the zero ring, its right regular module is also the zero module and is semisimple. Otherwise [L2] gives RiMni(Di).

L1L2given
2.1

By [L4], it suffices to identify the opposite factors as matrix rings over division rings. Each Diop is a division ring, and entrywise transpose defines a ring isomorphism Mni(Di)opMni(Diop),AAT, because (AB)T=BTAT with the products interpreted in the indicated rings. Hence RopiMni(Diop). By [L4], each factor's left regular module is semisimple, and the regular module of the finite product is their finite direct sum. Thus Rop is left semisimple, equivalently R is right semisimple.

L3L4step 1.1givenalgebra
3.1

Conversely, if R is right semisimple, then Rop is left semisimple. Applying steps 1.1–2.1 to Rop makes it right semisimple, which is exactly left semisimplicity of R. This also retains the zero-ring case.

L3step 1.1step 2.1givenalgebra
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Semisimple rings are left and right Noetherian and Artinian

Statement

Every semisimple ring is left and right Noetherian and left and right Artinian. See A semisimple ring as a ring whose left regular module is semisimple.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A unital ring R is semisimple when its left regular module RR is semisimple. This is a left-module definition and uses no Jacobson radical. For the zero ring, the regular module is zero and hence semisimple; the Wedderburn-Artin theorem below is stated for nonzero rings. (A semisimple ring as a ring whose left regular module is semisimple).

[L2]

Every finitely generated semisimple module is a finite direct sum of simple modules. (A finitely generated semisimple module is a finite direct sum of simple modules).

[L3]

A module with a composition series is both Noetherian and Artinian. The zero module has the empty composition series. (A module has a composition series if and only if it is Noetherian and Artinian, the converse using dependent choice).

[L4]

A unital ring is semisimple as a left regular module if and only if it is semisimple as a right regular module. (Left and right semisimplicity of a ring agree).

Proof

technique · direct
1.1

The left regular module is cyclic and semisimple, hence a finite direct sum of simples and therefore has finite length.

L1L2L3L4givenalgebra
2.1

The finite simple-factor chain in step 1.1 is a composition series, so [L3] gives left Noetherian and Artinian.

L3step 1.1given
3.1

By [L4], the right regular module is also semisimple; repeating steps 1.1–2.1 for right modules gives right Noetherian and Artinian. The zero ring is covered by its zero regular modules.

L1L2L3L4step 1.1step 2.1given
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Integrality and finite-module characterizations for one element

Statement

Let AB be commutative rings with A0, and let bB. The following are equivalent: b is integral over A; A[b] is finitely generated as an A-module; and there exists a faithful A[b]-module that is finitely generated over A, where faithful means that rM=0 implies r=0 for rA[b]. See Integral elements over a commutative ring and algebraic integers.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let AB be a homomorphism of commutative rings. An element bB is integral over A when it is a root of a monic polynomial in A[X]. The extension is integral when every element is integral. An algebraic integer is a complex number integral over Z. (Integral elements over a commutative ring and algebraic integers).

[L2]

Let M be a left R-module and SM. The submodule generated by S is SR:={NM:SN}. The family is nonempty because MM, and its intersection is a submodule by lem-submodule-criterion-sums-and-intersections. Thus SR is the smallest submodule of M containing S, just as def-generated-subgroup defines a generated subgroup. (Generated submodule, cyclic and finitely generated modules, module basis and free module).

[L3]

For a commutative ring R, n1, and AMn(R), Aadj(A)=adj(A)A=det(A)In.. (For every positive-sized square matrix over a commutative ring, Aadj(A)=adj(A)A=det(A)I).

Proof

technique · direct
1.1

If b satisfies a monic equation of degree d, every power bm with md is an A-linear combination of 1,b,,bd1; hence A[b] is finite over A. Taking A[b] itself gives a faithful A[b]-module finite over A.

L1L2L3givenalgebra
2.1

Conversely, let a faithful A[b]-module M be generated over A by m1,,mn. Write bmi=jaijmj with aijA, so the matrix bI(aij) annihilates the generating column.

step 1.1givenalgebra
3.1

Multiplying by the adjugate shows that the value det(bI(aij))A[b] annihilates every mi. It therefore annihilates M, so faithfulness makes it zero. The formal polynomial det(XI(aij))A[X] is monic of degree n and evaluates at b to this element, giving a monic relation for b over A.

L3step 2.1givenalgebra
4.1

The case n=0 cannot occur: then M=0, so faithfulness would force A[b]=0, contradicting A0. Hence n1, and the determinant in step 3.1 is a nonzero monic polynomial of positive degree. This proves the stated claim.

step 3.1givenalgebra
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Integral elements over a nonzero base ring form a subring

Statement

Let AB be commutative rings with A0. The elements of B integral over A form a subring of B. See Integrality and finite-module characterizations for one element.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let AB be commutative rings with A0, and let bB. Then b is integral over A if and only if A[b] is finitely generated as an A-module, if and only if there exists a faithful A[b]-module that is finitely generated over A. (Integrality and finite-module characterizations for one element).

[L2]

Let AB be a homomorphism of commutative rings. An element bB is integral over A when it is a root of a monic polynomial in A[X]. The extension is integral when every element is integral. An algebraic integer is a complex number integral over Z. (Integral elements over a commutative ring and algebraic integers).

Proof

technique · direct
1.1

If x,y are integral, A[x] is finite over A, and the monic equation for y over A is also one over A[x]; multiplying finite generating sets shows that A[x,y] is finite over A.

L1L2givenalgebra
2.1

For each z{x+y,xy,xy}, the A[z]-module A[x,y] is faithful and finite over A, so the finite-module criterion makes z integral.

step 1.1givenalgebra
3.1

The elements 0 and 1 are roots of the monic polynomials X and X1, respectively; step 2.1 supplies closure under addition, multiplication, and additive inverses, including coincident elements. Thus the integral elements form a unital subring of B.

L2step 2.1givenalgebra
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The rational algebraic integers are exactly the integers

Statement

A rational number is an algebraic integer if and only if it is an integer. See Integral elements over a commutative ring and algebraic integers.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

Let AB be a homomorphism of commutative rings. An element bB is integral over A when it is a root of a monic polynomial in A[X]. The extension is integral when every element is integral. An algebraic integer is a complex number integral over Z. (Integral elements over a commutative ring and algebraic integers).

[L2]

Let f=anxn++a1x+a0Z[x] with an0. If a reduced rational number r/s, where r,sZ, s>0, and gcd(r,s)=1, is a root of f, then ra0andsan.. (Rational root theorem).

[L3]

(Q,+,,0,1) with the operations of def-rat-operations is a field: a commutative ring with 10 in which every nonzero element has a multiplicative inverse. (The rationals form a field).

[L4]

The map j(k)=[(k,1)] is injective and preserves addition, multiplication, and order. Composing with lem-nat-embeds-int embeds N in Q; we write k for j(k) throughout. (The integers embed in the rationals).

Proof

technique · direct
1.1

We write a rational algebraic integer in lowest terms r/s with s>0.

L1L2L3L4givenalgebra
2.1

It is a root of a monic integer polynomial, so the rational-root theorem makes s1, hence s=1.

step 1.1givenalgebra
3.1

Conversely every integer satisfies the monic polynomial Xn.

step 2.1givenalgebra
4.1

Both are admitted and neither is exceptional. The polynomial Xn of step 3.1 is monic with integer coefficients for every integer n, giving X at n=0 and X+n for n<0; and in step 2.1 the lowest-terms representation covers 0 as 0/1, where the denominator is already 1. This proves the stated claim.

step 2.1step 3.1givenalgebra

5 · Examples, counterexamples and false statements

None yet.

Sources