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Chain Conditions, Semisimple Modules and the Wedderburn–Artin Theorem
1 · Prerequisites
- Binary Operations, Monoids, Groups and Subgroups
- Construction of the Natural Numbers
- Construction of the Real Numbers via Cauchy Sequences
- Construction of the Real Numbers via Dedekind Cuts
- Cosets, Index and Lagrange's Theorem
- Countability and Uncountability
- Determinants of Matrices over a Commutative Ring
- Divisibility, Greatest Common Divisors and Bézout's Identity
- Finite Counting, Factorials and Binomial Coefficients
- Foundations of the Real Numbers for Analysis
- Free Modules, Exact Sequences, Projective and Injective Modules
- Group Actions, Orbits, Stabilisers and Cayley's Theorem
- Group Homomorphisms and the Isomorphism Theorems
- Modules, Submodules, Quotient Modules and the Isomorphism Theorems
- Normal Subgroups and Quotient Groups
- Order, Zorn's Lemma, and the Axiom of Choice
- Polynomial Rings, the Division Algorithm and Roots
- Relations, Functions, and Quotients
- Rings, Subrings, Integral Domains and Fields
- Roots, Rational Powers, and Classical Inequalities
- Symmetric Groups, Cycle Decomposition and the Sign Homomorphism
- Tensor Products of Modules
- The Determinant of a Linear Operator, Cofactors and Cramer's Rule
- The ZFC Axioms and the Basic Set Constructions
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
Noetherian modules: every submodule is finitely generated
Definition
A left -module is Noetherian when every submodule of 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.
Artinian modules by the descending chain condition
Definition
A left -module is Artinian when every descending chain of submodules stabilizes: there is such that for all . This is the descending chain condition.
Left and right Noetherian rings
Definition
A unital ring is left Noetherian when its left regular module is Noetherian, and right Noetherian when the right regular module is Noetherian. Unqualified “Noetherian ring” means left Noetherian here; the side is stated whenever both notions occur.
Left and right Artinian rings
Definition
A unital ring is left Artinian when is Artinian, and right Artinian when is Artinian. Unqualified “Artinian ring” means left Artinian here.
Composition series and length of a module
Definition
A composition series of a left -module is a finite chain whose factors are simple. If such a series exists, the length is its number of factors; Jordan–Hölder theorem for modules ↗ proves independence of the chosen series. The zero module has the empty series and length .
The opposite ring
Definition
For a unital ring , the opposite ring has the same underlying abelian group, identity, and addition as , with multiplication . Associativity and both distributive laws follow from those of with the order reversed, and the same element is a two-sided identity. Thus the displayed operations really form a unital ring, including when is the zero ring.
The endomorphism ring under addition and composition
Definition
For a left -module , define Addition is pointwise and multiplication is composition, . The ring laws and the identity endomorphism are established in Module endomorphisms form a ring under pointwise addition and composition ↗.
Semisimple modules as direct sums of simple modules
Definition
A left -module is semisimple when it is an internal direct sum of simple submodules, allowing the empty direct sum. Hence the zero module is semisimple.
A semisimple ring as a ring whose left regular module is semisimple
Definition
A unital ring is semisimple when its left regular module 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.
The socle as the sum of all simple submodules
Definition
For a left -module , the socle is Concretely it is the submodule generated by the union of all simple submodules. If there are none, the sum is , so the definition is always meaningful.
Integral elements over a commutative ring and algebraic integers
Definition
Let be a homomorphism of commutative rings. An element is integral over when it is a root of a monic polynomial in . The extension is integral when every element is integral. An algebraic integer is a complex number integral over .
Second isomorphism theorem for modules
Statement
For submodules , there is a canonical isomorphism See First isomorphism theorem for modules: .
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
For every module homomorphism , there is a module isomorphism given by . (First isomorphism theorem for modules: ).
For submodules , the intersection and the sum are submodules of . (The one-step submodule criterion; intersections and sums of submodules are submodules).
For , the additive cosets form the quotient module under the well-defined scalar action (Quotient module with scalar multiplication on additive cosets).
Proof
For submodules , map .
Its kernel is , and it is surjective by the definition of ; the first isomorphism theorem gives .
The coincident and zero cases are admitted and give equalities rather than exceptions. For both sides are , since and ; for both sides are ; and for both sides are . This proves the stated claim.
Third isomorphism theorem for modules
Statement
If , then is a submodule of and See First isomorphism theorem for modules: .
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
For every module homomorphism , there is a module isomorphism given by . (First isomorphism theorem for modules: ).
For , the additive cosets form the quotient module under the well-defined scalar action (Quotient module with scalar multiplication on additive cosets).
Let be a module homomorphism and let satisfy . There is a unique module homomorphism such that , equivalently . (A module homomorphism vanishing on factors uniquely through ).
Proof
For , send to .
If , then , so the images modulo agree. The map is surjective, and its kernel consists exactly of the cosets with , namely ; the first isomorphism theorem gives the displayed isomorphism.
The two coincident cases are admitted by and hold. For the submodule is zero and the isomorphism reads ; for it is and the isomorphism reads . This proves the stated claim.
Correspondence theorem for submodules of a quotient module
Statement
For , inverse image and quotient induce mutually inverse inclusion-preserving bijections between submodules of and submodules of containing . They preserve sums, intersections, and successive quotients. See Quotient module with scalar multiplication on additive cosets.
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
For , the additive cosets form the quotient module under the well-defined scalar action (Quotient module with scalar multiplication on additive cosets).
Let be a module homomorphism and let satisfy . There is a unique module homomorphism such that , equivalently . (A module homomorphism vanishing on factors uniquely through ).
If , then is a submodule of and (Third isomorphism theorem for modules).
Proof
Inverse image and quotient give mutually inverse inclusion-preserving bijections between submodules of and submodules of containing .
If , then by surjectivity of ; if contains , then . Direct calculation with inverse images gives preservation of sums and intersections, while [L3] identifies successive quotients. This proves the stated claim.
Finite generation, ACC, and maximal-condition characterizations of Noetherian modules
Statement
For a left -module , 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.
A left -module is Noetherian when every submodule of 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
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.
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.
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.
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.
DCC and minimal-condition characterizations of Artinian modules
Statement
For a left -module , 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.
A left -module is Artinian when every descending chain of submodules stabilizes: there is such that for all . This is the descending chain condition. (Artinian modules by the descending chain condition).
Proof
DCC gives a minimal member of every nonempty family by contradiction: under dependent choice, absence of a minimal member yields a strict descending chain.
Conversely, a nonstabilizing descending chain has no minimal member.
Repeated terms are what make step 2.1 correct rather than an equivocation: minimality of in the family means no member is properly contained in it, so for all , which is stabilization. For the zero module the only submodule is , every nonempty family is with minimal member , and every descending chain is constant, so both conditions hold. This proves the stated claim.
Noetherian and Artinian conditions are each exact in short exact sequences
Statement
In a short exact sequence , the module is Noetherian if and only if and 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.
A left -module is Noetherian when every submodule of 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).
A left -module is Artinian when every descending chain of submodules stabilizes: there is such that for all . This is the descending chain condition. (Artinian modules by the descending chain condition).
For submodules , there is a canonical isomorphism . (Second isomorphism theorem for modules).
For , inverse image and quotient induce mutually inverse inclusion-preserving bijections between submodules of and submodules of containing . They preserve sums, intersections, and successive quotients. (Correspondence theorem for submodules of a quotient module).
A short exact sequence is an exact sequence thus is injective, is surjective, and . (Exact sequences and short exact sequences of modules).
A module is Noetherian if and only if every ascending chain of submodules stabilizes. (Finite generation, ACC, and maximal-condition characterizations of Noetherian modules).
A module is Artinian if and only if every descending chain of submodules stabilizes. (DCC and minimal-condition characterizations of Artinian modules).
Proof
Identify with its image in . A chain in is a chain in , and the correspondence theorem lifts every chain in to a chain of submodules of containing ; hence both ACC and DCC pass from to and .
Conversely, for a chain in , the chains and stabilize when and have the relevant chain condition. If are beyond both stabilization indices and , equality of the images gives with ; equality of the intersections then puts , so . The same argument with the inclusions reversed handles descending chains.
Thus has ACC exactly when and do, and it has DCC exactly when and do. Facts [L6] and [L7] convert these chain statements into the asserted Noetherian and Artinian equivalences.
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.
In a short exact sequence , the module is Noetherian if and only if and are Noetherian; the same equivalence holds with “Artinian” in place of “Noetherian”. (Noetherian and Artinian conditions are each exact in short exact sequences).
Let be a unital ring and a family of left -modules (def-left-and-right-modules). Their direct product is the module with coordinatewise operations. The support of is , and the direct sum is the submodule (def-submodule). (The direct sum of an indexed family of modules).
Proof
Induct on the number of summands using the split short exact sequence for a binary direct sum.
Reflection follows because every summand is a submodule and a quotient.
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.
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.
A unital ring is left Noetherian when its left regular module is Noetherian, and right Noetherian when the right regular module is Noetherian. Unqualified “Noetherian ring” means left Noetherian here; the side is stated whenever both notions occur. (Left and right Noetherian rings).
Let be a left -module and . The submodule generated by is The family is nonempty because , and its intersection is a submodule by lem-submodule-criterion-sums-and-intersections. Thus is the smallest submodule of containing , just as def-generated-subgroup defines a generated subgroup. (Generated submodule, cyclic and finitely generated modules, module basis and free module).
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).
For every left -module , the free module on its underlying set admits a canonical surjection , determined by . Consequently . (Every module is a quotient of a free module).
In a short exact sequence , the module is Noetherian if and only if and are Noetherian. (Noetherian and Artinian conditions are each exact in short exact sequences).
Proof
A module generated by elements is a quotient of .
The left regular module is Noetherian, [L3] makes Noetherian, and [L5] makes its quotient Noetherian.
The case is admitted: a module generated by the empty set is , and is the zero module, so step 1.1 presents as a quotient of . 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.
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.
For a left -module , 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).
For a left -module , define Addition is pointwise and multiplication is composition, . The ring laws and the identity endomorphism are established in prop-endomorphisms-form-a-ring. (The endomorphism ring under addition and composition).
Proof
For a surjective endomorphism , the ascending chain stabilizes.
Choose with . For , surjectivity of gives with . Then , so and .
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.
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.
A composition series of a left -module is a finite chain whose factors are simple. If such a series exists, the length is its number of factors; thm-jordan-holder-theorem-for-modules proves independence of the chosen series. The zero module has the empty series and length . (Composition series and length of a module).
For a left -module , 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).
For a left -module , 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).
In a short exact sequence , the module is Noetherian if and only if and are Noetherian; the same equivalence holds with “Artinian” in place of “Noetherian”. (Noetherian and Artinian conditions are each exact in short exact sequences).
Proof
Let be a composition series [L1] and induct on that is Noetherian and Artinian. The zero module satisfies both conditions vacuously. A simple factor has only the submodules and itself, so every chain of its submodules stabilizes and it too satisfies both conditions. Applying [L4] to the short exact sequence carries both conditions from and the simple factor to . At this gives the forward implication, which uses no choice principle.
Conversely, assume dependent choice and let be Noetherian and Artinian. Every submodule of is Noetherian by [L4], so a nonzero submodule has a nonempty family of proper submodules, which by the maximal condition of [L2] has a maximal member — a maximal proper submodule of . Dependent choice applied to this relation, starting at , yields a chain in which is a maximal proper submodule of for as long as .
The chain of step 2.1 is strictly descending while its terms are nonzero, so the descending chain condition forces some . Since is maximal proper in , the quotient is nonzero and has no proper nonzero submodule, hence is simple. Reversing the chain gives , a composition series. For the empty chain is already the required series, so no choice is consumed in that case. This proves the stated claim.
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.
A composition series of a left -module is a finite chain whose factors are simple. If such a series exists, the length is its number of factors; thm-jordan-holder-theorem-for-modules proves independence of the chosen series. The zero module has the empty series and length . (Composition series and length of a module).
For submodules , there is a canonical isomorphism . (Second isomorphism theorem for modules).
If , then is a submodule of and . (Third isomorphism theorem for modules).
For , inverse image and quotient induce mutually inverse inclusion-preserving bijections between submodules of and submodules of containing . They preserve sums, intersections, and successive quotients. (Correspondence theorem for submodules of a quotient module).
Proof
Fix a composition series and prove by induction on that it has the asserted comparison with every other composition series of . We use simultaneously the elementary consequence that, for any , intersecting the fixed series with and deleting repetitions gives a composition series of : each remaining factor embeds in the corresponding simple factor and is therefore simple.
The case is . For , let and let be the penultimate term of a second series. If , the induction hypothesis applied in matches all lower factors, and the common top factor finishes.
Suppose . Since and are maximal proper submodules, . Put . The second isomorphism theorem gives so both quotients are simple.
By step 1.1, has a composition series. Appending gives a composition series of ending in . Compare it with using the induction hypothesis, whose fixed first series has length . It follows that the series of has length and that its factors together with are exactly the factors below in the fixed series.
Appending to the same series of gives a composition series of of length . 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 and with and . Hence the two original series have length and the same factors up to permutation. This also covers , when .
Module length is additive in short exact sequences
Statement
For a short exact sequence , the module has finite length if and only if and do, and then See Jordan–Hölder theorem for modules.
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
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).
A composition series of a left -module is a finite chain whose factors are simple. If such a series exists, the length is its number of factors; thm-jordan-holder-theorem-for-modules proves independence of the chosen series. The zero module has the empty series and length . (Composition series and length of a module).
For , inverse image and quotient induce mutually inverse inclusion-preserving bijections between submodules of and submodules of containing . They preserve sums, intersections, and successive quotients. (Correspondence theorem for submodules of a quotient module).
Proof
If and have composition series, lift the series of along and splice it above the series of . Correspondence identifies all lifted factors, so this is a composition series of with factors.
Conversely, let be a composition series. Put and let be the image of in . For each , the simple factor has submodule and corresponding quotient ; exactly one is that simple factor and the other is zero. Deleting repetitions therefore gives composition series of and , and their numbers of factors add to .
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 , , or , the relevant series is empty and the same count applies.
Module endomorphisms form a ring under pointwise addition and composition
Statement
For every left -module , pointwise addition and composition make a unital ring with identity . See The endomorphism ring under addition and composition.
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
For a left -module , define Addition is pointwise and multiplication is composition, . The ring laws and the identity endomorphism are established in prop-endomorphisms-form-a-ring. (The endomorphism ring under addition and composition).
For left -modules , the set of module homomorphisms is an abelian group under pointwise addition, with zero the zero homomorphism and inverse (def-module-homomorphism-kernel-image-and-cokernel, def-group). (The abelian group and maps induced by pre- and postcomposition).
For left -modules , a function is an -module homomorphism if and for all and (Module homomorphism and isomorphism, kernel, image and cokernel).
Proof
If then is again a module homomorphism, since and . So composition is a binary operation on .
Taking in [L2] makes an abelian group, with the zero homomorphism as neutral element and as the inverse of .
Composition is associative: for all , .
Both distributive laws hold. For all , , where the middle equality is additivity of ; and directly from pointwise addition.
The identity map satisfies and , so it lies in , and for every .
Steps 1.1 through 1.5 are exactly the axioms of a unital ring for . For the zero module the only map is , so has one element and is the one-element ring, in which the identity coincides with the zero element. This proves the stated claim.
Statement
For every unital ring , evaluation at identifies endomorphisms of the left regular module with right multiplications and gives a ring isomorphism See The opposite ring .
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
For a unital ring , the opposite ring has the same underlying abelian group, identity, and addition as , with multiplication . Associativity and both distributive laws follow from those of with the order reversed, and the same element is a two-sided identity. Thus the displayed operations really form a unital ring, including when is the zero ring. (The opposite ring ).
For every left -module , pointwise addition and composition make a unital ring with identity . (Module endomorphisms form a ring under pointwise addition and composition).
Let be a ring. A left -module is an abelian group with a scalar action , , satisfying A right -module has an action , , 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
Evaluate an endomorphism at .
Left linearity gives , so endomorphisms are right multiplications; composition reverses the order of their defining elements.
The map with is inverse to : it is a left -module homomorphism because , it satisfies by step 2.1, and . It is additive, and with the multiplication of [L1], so it is a ring isomorphism onto . For the zero ring , both and are the one-element ring and the correspondence is the unique map between them. This proves the stated claim.
Endomorphisms of a finite direct sum are matrices of Hom-groups
Statement
For and left -modules , endomorphisms of correspond to matrices with , and composition is matrix multiplication using composition in the entries. For , both sides are the one-element zero ring. See The endomorphism ring under addition and composition.
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
For a left -module , define Addition is pointwise and multiplication is composition, . The ring laws and the identity endomorphism are established in prop-endomorphisms-form-a-ring. (The endomorphism ring under addition and composition).
For left -modules , the set of module homomorphisms is an abelian group under pointwise addition, with zero the zero homomorphism and inverse (def-module-homomorphism-kernel-image-and-cokernel, def-group). (The abelian group and maps induced by pre- and postcomposition).
Let be left -modules and a left -module. For every family of homomorphisms , there is a unique homomorphism such that for every . It is given by For , this is the unique map . (Universal property of a direct sum of modules).
Proof
We use inclusions and projections to send to entries , and reconstruct by finite sums.
Composition becomes matrix multiplication because on a finite direct sum, so the entry of is which is the matrix product with composition in the entries; the sum is finite because is.
For the direct sum is the zero module, has one element by [L1], and the set of matrices also has exactly one element, so both sides are the one-element zero ring as the Statement records. For the matrix is the single entry , and the correspondence is the identity on . This proves the stated claim.
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.
Let be a left -module and . The submodule generated by is The family is nonempty because , and its intersection is a submodule by lem-submodule-criterion-sums-and-intersections. Thus is the smallest submodule of containing , just as def-generated-subgroup defines a generated subgroup. (Generated submodule, cyclic and finitely generated modules, module basis and free module).
Let be a left -module. A subset is a submodule when it is a subgroup of the additive group of and is closed under scalars: The operations on are the restrictions of those of . Write when the ring and module are understood. (Submodule of a module).
Assume the Axiom of Choice (def-axiom-of-choice). Let be a nonempty poset in which every chain has an upper bound. Then 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
Let be generated over by [L1] and let be the set of proper submodules of [L2], ordered by inclusion. It is nonempty, since is a proper submodule of the nonzero module .
Let be a chain. If is empty, is an upper bound. Otherwise put ; any two elements of lie in a common member of the chain, so is a submodule. If , each generator would lie in some member of , and the largest of those finitely many members would contain every and hence equal , contradicting properness. So and is an upper bound for .
Every chain in the nonempty poset therefore has an upper bound in , so Zorn's lemma [L3] — which assumes the Axiom of Choice, as the Statement does — gives a maximal element of , that is, a maximal proper submodule of .
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.
Equivalent characterizations of semisimple modules
Statement
Assuming the Axiom of Choice, for a module the following are equivalent: is a direct sum of simple submodules; is the sum of its simple submodules; and every submodule of has a complementary submodule. See Semisimple modules as direct sums of simple modules.
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
A left -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).
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).
Assume the Axiom of Choice (def-axiom-of-choice). Let be a nonempty poset in which every chain has an upper bound. Then 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).
Let be left -modules and a left -module. For every family of homomorphisms , there is a unique homomorphism such that for every . It is given by For , this is the unique map . (Universal property of a direct sum of modules).
The socle is the sum of all simple submodules of . (The socle as the sum of all simple submodules).
Proof
A direct sum of simple submodules is plainly their sum. Conversely, if is a sum of simple submodules, Zorn's lemma applied to independent families of them gives a maximal direct sum ; if , a simple submodule not contained in meets trivially, contradicting maximality.
Given and a direct-sum decomposition into simples, use Zorn to choose a maximal sum with . If some were not contained in , simplicity would give , so adjoining it would contradict maximality. Hence every , and therefore .
Conversely suppose every submodule has a complement. By [L5], choose with . If , choose . The cyclic module has a maximal proper submodule by [L2]. Let complement in . Then , and is a nonzero simple submodule of , contrary to . Hence and is a sum of simples. The zero module is the empty sum.
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.
A composition series of a left -module is a finite chain whose factors are simple. If such a series exists, the length is its number of factors; thm-jordan-holder-theorem-for-modules proves independence of the chosen series. The zero module has the empty series and length . (Composition series and length of a module).
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
Suppose is a sum of simple submodules and start with . If , some simple is not contained in , so simplicity gives and . Intersecting a fixed -factor composition series of with and deleting repetitions gives a composition series of with at most factors. Since already has the -factor series obtained by adding the one at a time, [L2] gives . Thus after at most finite choices the process reaches , proving that a sum of simples is a finite direct sum without any choice axiom.
Now write a finite direct-sum decomposition and let . Process the finitely many in order, maintaining a sum with : add exactly when . In that case simplicity gives , so the invariant persists. At the end every , whence . This proves that the direct-sum condition implies the complement condition without Zorn.
Conversely, suppose every submodule of the finite-length module has a complement, and induct on a fixed composition-series length. If , let be the penultimate term of such a series. A complement gives with simple. The complement property passes to : for , if , then . The induction hypothesis makes a finite direct sum of simples, hence so is . 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.
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.
Assuming the Axiom of Choice, for a module the following are equivalent: is a direct sum of simple submodules; is the sum of its simple submodules; and every submodule of has a complementary submodule. (Equivalent characterizations of semisimple modules).
For submodules , there is a canonical isomorphism . (Second isomorphism theorem for modules).
Proof
A submodule inherits the complement property by intersecting a complement in the ambient module.
For a quotient, complement the kernel and identify the quotient with that semisimple complement.
The extreme submodules are admitted and give nothing new: the zero submodule is the empty direct sum, hence semisimple, and its quotient is semisimple by hypothesis; the whole submodule is semisimple by hypothesis and its quotient is again the empty direct sum. This proves the stated claim.
Under Choice, the socle is the largest semisimple submodule
Statement
Assuming the Axiom of Choice, for every module , is semisimple and contains every semisimple submodule of . See The socle as the sum of all simple submodules.
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
For a left -module , the socle is Concretely it is the submodule generated by the union of all simple submodules. If there are none, the sum is , so the definition is always meaningful. (The socle as the sum of all simple submodules).
Assuming the Axiom of Choice, for a module the following are equivalent: is a direct sum of simple submodules; is the sum of its simple submodules; and every submodule of has a complementary submodule. (Equivalent characterizations of semisimple modules).
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
The sum of all simple submodules is semisimple by the sum characterization.
Every semisimple submodule is itself a sum of simple submodules of the ambient module, hence lies in the socle.
If has no simple submodule, the defining sum of [L1] is empty, so ; the zero module is the empty direct sum and hence semisimple, and the only semisimple submodule of such an is itself, which it contains. This proves the stated claim.
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.
A left -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).
Let be a left -module and . The submodule generated by is The family is nonempty because , and its intersection is a submodule by lem-submodule-criterion-sums-and-intersections. Thus is the smallest submodule of containing , just as def-generated-subgroup defines a generated subgroup. (Generated submodule, cyclic and finitely generated modules, module basis and free module).
Proof
In a direct-sum decomposition, each generator has finite support; the union of those finite supports contains every generator and hence the whole module.
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 . For a module generated by the empty set, , which is the empty direct sum and semisimple by [L1]. This proves the stated claim.
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.
A left -module is simple if and its only submodules are and . Equivalently, has no proper nonzero submodule. (Simple module: a nonzero module with no proper nonzero submodule).
For a left -module , define Addition is pointwise and multiplication is composition, . The ring laws and the identity endomorphism are established in prop-endomorphisms-form-a-ring. (The endomorphism ring under addition and composition).
For every left -module , pointwise addition and composition make a unital ring with identity . (Module endomorphisms form a ring under pointwise addition and composition).
Proof
The kernel and image of a homomorphism between simple modules are each zero or whole.
A nonzero homomorphism is therefore injective and surjective.
Applied to a nonzero endomorphism, its inverse is linear, so the endomorphism ring is a division ring.
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.
Equivalent module-theoretic characterizations of semisimple rings
Statement
Assuming the Axiom of Choice, for a unital ring the following are equivalent: is semisimple; every left -module is semisimple; every short exact sequence of left -modules splits; and every left -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.
A unital ring is semisimple when its left regular module 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).
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).
For every left -module , the free module on its underlying set admits a canonical surjection , determined by . Consequently . (Every module is a quotient of a free module).
In a short exact sequence a section of is a homomorphism with , and a retraction of is a homomorphism with . (Split short exact sequences, sections, and retractions).
For a short exact sequence the following are equivalent: 1. has a section ; 2. has a retraction ; 3. there is an isomorphism with and . (The splitting lemma for short exact sequences of modules).
A left -module is projective if it has the lifting property for epimorphisms: whenever is a surjective module homomorphism and is a module homomorphism, there exists a module homomorphism such that (def-module-homomorphism-kernel-image-and-cokernel, def-injection-surjection-bijection). (Projective modules and the lifting property).
For a left -module , assertions 1 to 3 below are equivalent without choice. Under the Axiom of Choice, they are also equivalent to assertion 4: 1. is projective; 2. every short exact sequence splits; 3. takes every short exact sequence to a short exact sequence; 4. is a direct summand of a free module. (Equivalent characterizations of projective modules).
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
If is semisimple, every free left module, being a direct sum of copies of , is semisimple; every module is a quotient of a free module, so every left module is semisimple.
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.
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.
Matrix rings over division rings are semisimple
Statement
Let be a division ring and . On the set of arrays over , use entrywise addition and the product
These operations make a ring , and this ring is semisimple. More precisely, its left regular module is the direct sum of the simple column ideals for . See A semisimple ring as a ring whose left regular module is semisimple.
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
A unital ring is semisimple when its left regular module 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).
A division ring is a ring with in which every nonzero element has a two-sided multiplicative inverse. (Division ring: a ring with in which every nonzero element is a unit).
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).
Finite sums in a commutative monoid are independent of the chosen enumeration, and the empty sum is . (A finite sum in a commutative monoid indexed by an arbitrary finite set).
A left -module is simple if and its only submodules are and . Equivalently, has no proper nonzero submodule. (Simple module: a nonzero module with no proper nonzero submodule).
Proof
Entrywise addition makes the arrays an abelian group. Associativity of multiplication follows by expanding both and and reassociating the finite double sum; the two distributive laws follow entrywise from those of . The matrix is a two-sided identity. Thus the displayed operations make the unital ring without any commutativity assumption on .
Let . Every matrix has the unique decomposition , and because the two sides have disjoint possible nonzero columns. Hence
Sending a matrix in to its -th column identifies that left ideal with under left matrix multiplication. If , choose with . For any , the matrix whose only possibly nonzero column is column , with entry , sends to . Thus every nonzero submodule of is all of , so each column ideal is simple.
The decomposition in step 2.1 is therefore a finite direct sum of simple left modules, so [L1] makes semisimple. For it is the single simple column ideal, and the hypothesis excludes an empty decomposition.
Wedderburn–Artin theorem for semisimple rings
Statement
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
For left -modules , endomorphisms of correspond to matrices with , and composition is matrix multiplication using composition in the entries. (Endomorphisms of a finite direct sum are matrices of Hom-groups).
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).
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).
For a division ring and , matrices with product form a semisimple ring whose left regular module is the direct sum of its simple column ideals. (Matrix rings over division rings are semisimple).
The opposite ring has the same addition and identity as and multiplication . (The opposite ring ).
A unital ring is semisimple when its left regular module 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
If is semisimple, its cyclic left regular module is a finite direct sum of simple modules. Grouping isomorphic summands gives with pairwise nonisomorphic and positive .
Schur's lemma gives for and makes a division ring. Hence the endomorphism-matrix theorem gives .
Since , taking opposites gives . The opposite is again a division ring, and entrywise transpose is a ring isomorphism because reversing both the matrix product and the entry product gives in the target. Thus with .
Conversely, each 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.
The Statement assumes that is nonzero, so the decomposition has at least one factor; no empty-product convention is asserted. This proves the stated claim.
Simple modules over a product of matrix rings over division rings
Statement
Let , let every , let every be a division ring, and put . Then every simple left -module is supported on exactly one factor and is isomorphic to that factor's column module . These column modules give all simple left -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.
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).
If is a division ring and , then the left regular module of is the direct sum of its simple column ideals . (Matrix rings over division rings are semisimple).
Proof
Write for the central idempotent that is in factor and elsewhere. For a simple left -module , every is a submodule and . Hence some is nonzero and therefore equals ; then for . Thus is supported on exactly one factor .
Choose . The map , , is surjective because its image is a nonzero submodule. By [L2], is a direct sum of simple column ideals . At least one restriction is nonzero, so [L1] makes it an isomorphism. Hence .
Conversely each column module is simple by [L2]. Modules supported on different factors cannot be isomorphic, because the corresponding acts as the identity on one and as zero on the other. For a fixed factor all column ideals are isomorphic to by [L2]. This proves the classification, including the one-factor case .
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.
Let be a nonzero unital ring. Then is semisimple if and only if for positive integers and division rings . (Wedderburn–Artin theorem for semisimple rings).
For , , and division rings , every simple left module over is supported on exactly one factor and is isomorphic to that factor's column module ; these give all isomorphism classes. (Simple modules over a product of matrix rings over division rings).
For a division ring and , the left regular module of is the direct sum of the simple column ideals . (Matrix rings over division rings are semisimple).
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).
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
In a decomposition , the simple left module supported on factor is its column module . Distinct coordinate idempotents show that the are pairwise nonisomorphic as -modules, and [L3] decomposes the regular module with exactly copies of .
Let commute with the matrix action and put . For , the matrix unit annihilates , so it annihilates ; hence for a unique . Since , one has , and additivity then gives for every column . Thus the endomorphisms are precisely right scalar multiplications, composition reverses the scalar order, and . Therefore is determined by the simple-module type with the orientation fixed.
Jordan–Hölder [L5] makes the simple-module types and their multiplicities in invariant. Hence the pairs are determined up to reordering and division-ring isomorphism. This proves the stated claim.
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.
A unital ring is semisimple when its left regular module 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).
Let be a nonzero unital ring. Then is semisimple if and only if for positive integers and division rings . (Wedderburn–Artin theorem for semisimple rings).
For a unital ring , the opposite ring has the same underlying abelian group, identity, and addition as , with multiplication . Associativity and both distributive laws follow from those of with the order reversed, and the same element is a two-sided identity. Thus the displayed operations really form a unital ring, including when is the zero ring. (The opposite ring ).
For a division ring and , the ring is semisimple and its left regular module is a finite direct sum of simple column ideals. (Matrix rings over division rings are semisimple).
Proof
First suppose that is semisimple on the left. If is the zero ring, its right regular module is also the zero module and is semisimple. Otherwise [L2] gives .
By [L4], it suffices to identify the opposite factors as matrix rings over division rings. Each is a division ring, and entrywise transpose defines a ring isomorphism because with the products interpreted in the indicated rings. Hence . By [L4], each factor's left regular module is semisimple, and the regular module of the finite product is their finite direct sum. Thus is left semisimple, equivalently is right semisimple.
Conversely, if is right semisimple, then is left semisimple. Applying steps 1.1–2.1 to makes it right semisimple, which is exactly left semisimplicity of . This also retains the zero-ring case.
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.
A unital ring is semisimple when its left regular module 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).
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).
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).
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
The left regular module is cyclic and semisimple, hence a finite direct sum of simples and therefore has finite length.
The finite simple-factor chain in step 1.1 is a composition series, so [L3] gives left Noetherian and Artinian.
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.
Integrality and finite-module characterizations for one element
Statement
Let be commutative rings with , and let . The following are equivalent: is integral over ; is finitely generated as an -module; and there exists a faithful -module that is finitely generated over , where faithful means that implies for . See Integral elements over a commutative ring and algebraic integers.
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
Let be a homomorphism of commutative rings. An element is integral over when it is a root of a monic polynomial in . The extension is integral when every element is integral. An algebraic integer is a complex number integral over . (Integral elements over a commutative ring and algebraic integers).
Let be a left -module and . The submodule generated by is The family is nonempty because , and its intersection is a submodule by lem-submodule-criterion-sums-and-intersections. Thus is the smallest submodule of containing , just as def-generated-subgroup defines a generated subgroup. (Generated submodule, cyclic and finitely generated modules, module basis and free module).
For a commutative ring , , and , . (For every positive-sized square matrix over a commutative ring, ).
Proof
If satisfies a monic equation of degree , every power with is an -linear combination of ; hence is finite over . Taking itself gives a faithful -module finite over .
Conversely, let a faithful -module be generated over by . Write with , so the matrix annihilates the generating column.
Multiplying by the adjugate shows that the value annihilates every . It therefore annihilates , so faithfulness makes it zero. The formal polynomial is monic of degree and evaluates at to this element, giving a monic relation for over .
The case cannot occur: then , so faithfulness would force , contradicting . Hence , and the determinant in step 3.1 is a nonzero monic polynomial of positive degree. This proves the stated claim.
Integral elements over a nonzero base ring form a subring
Statement
Let be commutative rings with . The elements of integral over form a subring of . See Integrality and finite-module characterizations for one element.
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
Let be commutative rings with , and let . Then is integral over if and only if is finitely generated as an -module, if and only if there exists a faithful -module that is finitely generated over . (Integrality and finite-module characterizations for one element).
Let be a homomorphism of commutative rings. An element is integral over when it is a root of a monic polynomial in . The extension is integral when every element is integral. An algebraic integer is a complex number integral over . (Integral elements over a commutative ring and algebraic integers).
Proof
If are integral, is finite over , and the monic equation for over is also one over ; multiplying finite generating sets shows that is finite over .
For each , the -module is faithful and finite over , so the finite-module criterion makes integral.
The elements and are roots of the monic polynomials and , respectively; step 2.1 supplies closure under addition, multiplication, and additive inverses, including coincident elements. Thus the integral elements form a unital subring of .
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.
Let be a homomorphism of commutative rings. An element is integral over when it is a root of a monic polynomial in . The extension is integral when every element is integral. An algebraic integer is a complex number integral over . (Integral elements over a commutative ring and algebraic integers).
Let with . If a reduced rational number , where , , and , is a root of , then . (Rational root theorem).
with the operations of def-rat-operations is a field: a commutative ring with in which every nonzero element has a multiplicative inverse. (The rationals form a field).
The map is injective and preserves addition, multiplication, and order. Composing with lem-nat-embeds-int embeds in ; we write for throughout. (The integers embed in the rationals).
Proof
We write a rational algebraic integer in lowest terms with .
It is a root of a monic integer polynomial, so the rational-root theorem makes , hence .
Conversely every integer satisfies the monic polynomial .
Both are admitted and neither is exceptional. The polynomial of step 3.1 is monic with integer coefficients for every integer , giving at and for ; and in step 2.1 the lowest-terms representation covers as , where the denominator is already . This proves the stated claim.
5 · Examples, counterexamples and false statements
None yet.
Sources
Standard references
Recommended treatments; not extraction sources.