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✓ 19 results · all verified · 0 also independently AI-judged
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Ideals, Quotient Rings and the Isomorphism Theorems for Rings

1 · Prerequisites

2 · Summary

Rings, unital subrings, additive quotients, group kernels, and Zorn's lemma provide the starting point. Ideals are additive subgroups closed under multiplication by ambient ring elements, so they are exactly the subobjects that permit representative-independent multiplication of additive cosets. The canonical quotient map and the group factorisation result guide the corresponding ring constructions.

The page defines ideals, generated ideals, quotient rings, ideal sums and products, and prime and maximal ideals. It proves quotient-ring well-definedness and laws, the universal property, each ring isomorphism theorem, and the ideal correspondence. For commutative rings it characterises domain and field quotients, derives maximal-implies-prime, and obtains maximal ideals from Zorn's lemma.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

Left, right and two-sided ideals

Definition

Left, right and two-sided ideals.

Let R be a ring. An additive subgroup I≤(R,+) is a left ideal when ri∈I for every r∈R and i∈I, and a right ideal when ir∈I for every such r,i. A two-sided ideal, written I⊴R, is both a left and a right ideal. In a commutative ring these three notions agree.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Ideal criteria and intersections of ideals

Statement

Ideal criteria and intersections of ideals.

A nonempty subset I⊆R is a two-sided ideal exactly when it is closed under x−y and under rx,xr for all r∈R, x,y∈I. Any intersection of two-sided ideals is a two-sided ideal, with the empty intersection equal to R.

Facts & Assumptions

Given: A ring R and a subset I⊆R.

[L1]

A two-sided ideal is an additive subgroup closed under left and right multiplication by ring elements (Left, right and two-sided ideals).

Proof

technique · direct
1.1

Closure under x−y is exactly the additive subgroup criterion, and the two absorption conditions then give the ideal criterion.

L1L2L3givenalgebra
2.1

An intersection has subtraction closure and both absorption properties because each member ideal has them; for the empty family the intersection is R.

step 1.1L1L2L3givenalgebra
3.1

The criterion and intersection assertion follow.

step 2.1∎
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

The ideal generated by a subset and principal ideals

Definition

The ideal generated by a subset and principal ideals.

For S⊆R, define (S) to be the intersection of all two-sided ideals of R containing S. This family is nonempty because it contains R, and Ideal criteria and intersections of ideals shows that (S) is an ideal. For a∈R, the ideal ({a}) is written (a) and is called principal.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

In a commutative ring, (S) consists of finite sums ∑risi, and (a)=Ra

Statement

In a commutative ring, (S) consists of finite sums ∑risi, and (a)=Ra.

The empty sum is included and equals 0.

Facts & Assumptions

Given: A commutative ring R and a subset S⊆R.

[L1]

(S) is the intersection of ideals containing S (The ideal generated by a subset and principal ideals).

[L2]

The ideal criterion uses subtraction and absorption (Ideal criteria and intersections of ideals).

[L3]

Multiplication in a commutative ring commutes (Commutative ring).

Proof

technique · direct
1.1

Let J be the finite sums ∑risi; it contains S and is closed under subtraction and multiplication by arbitrary ring elements.

L1L2L3givenalgebra
2.1

Thus J is an ideal containing S, while every ideal containing S contains each such finite sum.

step 1.1L1L2L3givenalgebra
3.1

Hence J=(S); taking S={a} gives (a)=Ra.

step 2.1∎
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

The sum I+J and product IJ of two-sided ideals

Definition

The sum I+J and product IJ of two-sided ideals.

For two-sided ideals I,J⊴R, set

I+J:={i+j:i∈I, j∈J},IJ:={∑k=1mikjk:m≥0, ik∈I, jk∈J}.

The empty sum in the definition of IJ is 0.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The sum and product of two-sided ideals are two-sided ideals

Statement

The sum and product of two-sided ideals are two-sided ideals.

Facts & Assumptions

Given: Two-sided ideals I,J⊴R.

[L1]

I+J and IJ are the indicated elementwise and finite-sum sets (The sum I+J and product IJ of two-sided ideals).

[L2]

The ideal criterion is subtraction closure plus two-sided absorption (Ideal criteria and intersections of ideals).

Proof

technique · direct
1.1

Subtraction closure of I+J and of IJ follows by subtracting representatives and concatenating finite sums.

L1L2L3givenalgebra
2.1

Multiplying a representative on either side keeps it in I+J, and distributivity keeps each summand of a product in IJ.

step 1.1L1L2L3givenalgebra
3.1

The closure established in step 2.1 proves both claims.

step 2.1∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The kernel of a ring homomorphism is a two-sided ideal

Statement

The kernel of a ring homomorphism is a two-sided ideal.

Facts & Assumptions

Given: A ring homomorphism f:R→S.

[L1]

A ring homomorphism preserves addition, multiplication, 0, and 1 (Ring homomorphism: additive, multiplicative, and required to send 1 to 1).

[L3]

The group kernel is the inverse image of 0 in the additive groups (The kernel and image of a group homomorphism).

[L4]

A two-sided ideal is an additive subgroup with two-sided absorption (Left, right and two-sided ideals).

Proof

technique · direct
1.1

The additive-group kernel of f is an additive subgroup of R.

L1L2L3L4given
2.1

If x∈ker⁡f and r∈R, then f(rx)=f(r)f(x)=0 and f(xr)=0.

step 1.1L1L2L3L4givenalgebra
3.1

Thus the additive and absorption properties make ker⁡f a two-sided ideal.

step 2.1∎
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

The quotient ring R/I with (r+I)(s+I)=rs+I

Definition

The quotient ring R/I with (r+I)(s+I)=rs+I.

Let I⊴R. Since I is an additive subgroup of the abelian group (R,+), its additive cosets form the quotient group R/I (The quotient group G/N and coset product (gN)(hN)=ghN, Every subgroup of an abelian group is normal). Define

(r+I)(s+I):=rs+I.

The well-definedness and ring laws are established by Multiplication of additive cosets is well defined if and only if the additive subgroup is a two-sided ideal ↗ and For a two-sided ideal I, the additive cosets form a ring R/I with identity 1+I ↗.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Multiplication of additive cosets is well defined if and only if the additive subgroup is a two-sided ideal

Statement

Multiplication of additive cosets is well defined if and only if the additive subgroup is a two-sided ideal.

Let J≤(R,+). The rule (r+J)(s+J)=rs+J is independent of representatives exactly when J⊴R.

Facts & Assumptions

Given: A ring R and an additive subgroup J≤(R,+).

[L1]

The displayed rule is the proposed quotient multiplication (The quotient ring R/I with (r+I)(s+I)=rs+I).

[L2]

A two-sided ideal is an additive subgroup absorbing multiplication on both sides (Left, right and two-sided ideals).

[L4]

Coset equality is membership of a difference in the subgroup (x∈aH iff a−1x∈H, and aH=bH iff a−1b∈H).

[L5]

Additive cosets are available for an additive subgroup (The quotient group G/N and coset product (gN)(hN)=ghN).

Proof

technique · direct
1.1

If J is an ideal and r′=r+i, s′=s+j, then r′s′−rs=rj+is+ij∈J, so the product is well defined.

L1L2L3L4L5givenalgebra
2.1

Conversely, compare (r+J)(0+J) with (r+J)(j+J) and the reversed product; [L4] gives rj,jr∈J for all r,j.

step 1.1L1L2L3L4L5givenalgebra
3.1

Thus the rule is well defined exactly when J is a two-sided ideal.

step 2.1∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

For a two-sided ideal I, the additive cosets form a ring R/I with identity 1+I

Statement

For a two-sided ideal I, the additive cosets form a ring R/I with identity 1+I.

Facts & Assumptions

Given: A ring R and a two-sided ideal I⊴R.

[L1]

R/I has the stated coset addition and multiplication (The quotient ring R/I with (r+I)(s+I)=rs+I).

Proof

technique · direct
1.1

By [L3], coset addition has an abelian-group structure, and [L2] makes coset multiplication a defined operation.

L1L2L3L4given
2.1

Associativity and both distributive laws follow by applying the corresponding ring law in R to representatives; 1+I is a multiplicative identity.

step 1.1L1L2L3L4givenalgebra
3.1

Hence R/I is a ring with identity 1+I.

step 2.1∎
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The canonical projection R→R/I is a surjective ring homomorphism with kernel I

Statement

The canonical projection R→R/I is a surjective ring homomorphism with kernel I.

For I⊴R, π(r)=r+I has these properties.

Facts & Assumptions

Given: A ring R and a two-sided ideal I⊴R.

[L1]
[L2]

A ring homomorphism preserves addition, multiplication, and identity (Ring homomorphism: additive, multiplicative, and required to send 1 to 1).

[L4]

A kernel is the inverse image of the identity element (The kernel and image of a group homomorphism).

[L5]

The coset-equality criterion gives r+I=0+I exactly when −r∈I, hence exactly when r∈I because I is an additive subgroup (x∈aH iff a−1x∈H, and aH=bH iff a−1b∈H).

Proof

technique · direct
1.1

The identities π(r+s)=π(r)+π(s), π(rs)=π(r)π(s), and π(1)=1+I show that π is a ring homomorphism.

L1L2givenalgebra
2.1

By [L3] it is surjective; by [L4] and [L5], its kernel is exactly I.

step 1.1L3L4L5givenalgebra
3.1

Thus ker⁡π=I.

step 2.1∎
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

For every n∈N, the congruence-class ring Z/n is the quotient ring Z/nZ

Statement

For every n∈N, the congruence-class ring Z/n is the quotient ring Z/nZ.

This includes n=0 and n=1.

Facts & Assumptions

Given: A natural number n, viewed as a nonnegative integer.

[L1]

The canonical quotient map is a surjective ring homomorphism (The canonical projection R→R/I is a surjective ring homomorphism with kernel I).

[L4]

[a]n+[b]n=[a+b]n and [a]n[b]n=[ab]n (Addition and multiplication on Z/n by [a]n+[b]n=[a+b]n and [a]n[b]n=[ab]n).

[L5]

Z is a commutative ring with identity (The integers form a commutative ring).

Proof

technique · direct
1.1

Map a+nZ to [a]n; [L2] makes this an equality of additive quotient sets.

L1L2L3L4L5givenconstruct
2.1

The quotient product maps to [ab]n, exactly the modular product in [L4], and the identities agree.

step 1.1L1L2L3L4L5givenalgebra
3.1

Therefore the congruence-class ring is literally the quotient ring, including at n=0,1.

step 2.1∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring

Statement

A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring.

If I⊴R, f:R→S is a ring homomorphism, and I⊆ker⁡f, there is a unique ring homomorphism fˉ:R/I→S such that fˉ(r+I)=f(r).

Facts & Assumptions

Given: A two-sided ideal I⊴R and a ring homomorphism f:R→S with I⊆ker⁡f.

[L2]

A two-sided ideal is an additive subgroup (Left, right and two-sided ideals).

[L3]

Every subgroup of an abelian group is normal (Every subgroup of an abelian group is normal).

[L4]

The canonical quotient map is a surjective ring homomorphism (The canonical projection R→R/I is a surjective ring homomorphism with kernel I).

[L5]

A group homomorphism killing a normal subgroup factors uniquely through its quotient (A homomorphism that kills a normal subgroup factors uniquely through the quotient group).

[L6]

A ring homomorphism preserves addition, multiplication, and identity (Ring homomorphism: additive, multiplicative, and required to send 1 to 1).

Proof

technique · constructive
1.1

By [L1]--[L3], I is normal in the additive group of R; applying [L5] to the additive homomorphism underlying f defines fˉ(r+I)=f(r) and proves representative independence.

L1L2L3L5L6givenconstruct
2.1

Since [L4] gives (r+I)(s+I)=rs+I, one has fˉ((r+I)(s+I))=f(rs)=f(r)f(s), and fˉ(1+I)=f(1)=1; thus fˉ is a ring homomorphism.

step 1.1L4L6givenalgebra
3.1

The factor identity f=fˉ∘π holds by step 1.1, and any ring-homomorphic factor is additive, so the uniqueness in [L5] proves its uniqueness as a ring factor.

step 1.1step 2.1L5L6discharge-construct∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

First isomorphism theorem for rings: R/ker⁡f≅im⁡f

Statement

First isomorphism theorem for rings: R/ker⁡f≅im⁡f.

Facts & Assumptions

Given: A ring homomorphism f:R→S.

[L1]

A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring (A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring).

[L4]

The underlying group map is isomorphic modulo its kernel to its image (First isomorphism theorem for groups: G/ker⁡f≅im⁡f).

Proof

technique · direct
1.1

By [L2] and [L1], f induces a ring homomorphism fˉ:R/ker⁡f→im⁡f with fˉ(r+ker⁡f)=f(r).

L1L2L3L4L5givenconstruct
2.1

This map is surjective by the definition of image, and its additive kernel is trivial, hence it is injective by [L4].

step 1.1L1L2L3L4L5givenalgebra
3.1

Thus fˉ is a ring isomorphism.

step 2.1∎
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

If S is a subring and I is an ideal of R, then S+I is a subring, I is an ideal of S+I, and S∩I is an ideal of S

Statement

If S is a subring and I is an ideal of R, then S+I is a subring, I is an ideal of S+I, and S∩I is an ideal of S.

Facts & Assumptions

Given: A unital subring S⊆R and a two-sided ideal I⊴R.

[L1]

A subring contains the ambient identity and is closed under ring operations (Subring: a subset containing 1R and closed under addition, additive inverses and multiplication).

[L3]

An ideal is an additive subgroup with two-sided absorption (Left, right and two-sided ideals).

[L4]

Intersections of ideals and the ideal criterion are valid (Ideal criteria and intersections of ideals).

Proof

technique · direct
1.1

S+I contains 1R and is subtraction-closed; expanding (s+i)(t+j)=st+sj+it+ij proves multiplication closure.

L1L2L3L4givenalgebra
2.1

For s+i∈S+I and j∈I, both (s+i)j and j(s+i) lie in I, while S∩I is subtraction-closed and absorbed by S.

step 1.1L1L2L3L4givenalgebra
3.1

Hence S+I is a unital subring, I⊴S+I, and S∩I⊴S.

step 2.1∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Second isomorphism theorem for rings: S/(S∩I)≅(S+I)/I

Statement

Second isomorphism theorem for rings: S/(S∩I)≅(S+I)/I.

If S is a unital subring of R and I⊴R, then this is an isomorphism of unital rings.

Facts & Assumptions

Given: A unital subring S⊆R and a two-sided ideal I⊴R.

[L2]

The first ring isomorphism theorem identifies a ring modulo a kernel with its image (First isomorphism theorem for rings: R/ker⁡f≅im⁡f).

[L3]

The canonical quotient map is a surjective ring homomorphism (The canonical projection R→R/I is a surjective ring homomorphism with kernel I).

Proof

technique · direct
1.1

Restrict the quotient map S+I→(S+I)/I to ϕ:S→(S+I)/I, ϕ(s)=s+I.

L1L2L3givenconstruct
2.1

Its kernel is S∩I, and every (s+i)+I equals s+I, so its image is all of (S+I)/I.

step 1.1L1L2L3givenalgebra
3.1

The kernel and image computation gives S/(S∩I)≅(S+I)/I.

step 2.1∎
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

If I⊆J are ideals of R, then J/I is an ideal of R/I

Statement

If I⊆J are ideals of R, then J/I is an ideal of R/I.

Facts & Assumptions

Given: Two-sided ideals I⊆J⊴R.

[L1]

Ideals are additive subgroups with absorption (Left, right and two-sided ideals).

Proof

technique · direct
1.1

J/I={j+I:j∈J} is an additive subgroup of R/I because J is an additive subgroup.

L1L2L3givenalgebra
2.1

For r+I∈R/I and j+I∈J/I, both (r+I)(j+I)=rj+I and (j+I)(r+I)=jr+I lie in J/I.

step 1.1L1L2L3givenalgebra
3.1

Hence J/I is a two-sided ideal of R/I.

step 2.1∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Third isomorphism theorem for rings: (R/I)/(J/I)≅R/J

Statement

Third isomorphism theorem for rings: (R/I)/(J/I)≅R/J.

Facts & Assumptions

Given: Two-sided ideals I⊆J⊴R.

[L2]

A ring modulo a kernel is isomorphic to the image (First isomorphism theorem for rings: R/ker⁡f≅im⁡f).

[L3]

Proof

technique · direct
1.1

Define ϕ:R/I→R/J by ϕ(r+I)=r+J; it is a well-defined surjective ring homomorphism because I⊆J.

L1L2L3givenconstruct
2.1

The equality ϕ(r+I)=0+J holds exactly when r∈J, so ker⁡ϕ=J/I.

step 1.1L1L2L3givenalgebra
3.1

The kernel and image computation gives (R/I)/(J/I)≅R/J.

step 2.1∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Correspondence theorem: ideals of R/I correspond to ideals of R containing I

Statement

Correspondence theorem: ideals of R/I correspond to ideals of R containing I.

For I⊴R, J↦J/I and K↦π−1(K) are inverse inclusion-preserving bijections between ideals J of R containing I and ideals K of R/I.

Facts & Assumptions

Given: An ideal I⊴R and π:R→R/I.

[L1]
[L2]

Kernels of ring homomorphisms are ideals (The kernel of a ring homomorphism is a two-sided ideal).

[L3]

Ideals are additive subgroups with absorption (Left, right and two-sided ideals).

Proof

technique · direct
1.1

If J contains I, then J/I=π[J] is an ideal of R/I; if K is an ideal of R/I, then π−1(K) is an ideal containing I.

L1L2L3L4givenconstruct
2.1

Surjectivity gives π[π−1(K)]=K, and I⊆J gives π−1(π[J])=J; inclusion is preserved.

step 1.1L1L2L3L4givenalgebra
3.1

Therefore the two assignments give the claimed correspondence.

step 2.1∎
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

Prime ideals and maximal ideals in a commutative ring

Definition

Prime ideals and maximal ideals in a commutative ring.

Let R be a commutative ring. A proper ideal P⊊R is prime when ab∈P implies a∈P or b∈P. A proper ideal M⊊R is maximal when there is no proper ideal strictly between M and R; equivalently, it is a maximal element of the poset of proper ideals ordered by inclusion.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

R/P is an integral domain if and only if P is a prime ideal

Statement

R/P is an integral domain if and only if P is a prime ideal.

Here R is commutative and P⊴R is an ideal.

Facts & Assumptions

Given: A commutative ring R and a two-sided ideal P⊴R.

[L1]

A prime ideal is proper and satisfies ab∈P⇒a∈P or b∈P (Prime ideals and maximal ideals in a commutative ring).

[L2]

Products in R/P are (a+P)(b+P)=ab+P (For a two-sided ideal I, the additive cosets form a ring R/I with identity 1+I).

[L3]

An integral domain is a nonzero commutative ring without zero divisors (Zero divisor, and integral domain: a commutative ring with 1≠0 and no zero divisors).

[L4]

The canonical projection R→R/P has kernel P; equivalently, a+P=0+P exactly when a∈P (The canonical projection R→R/I is a surjective ring homomorphism with kernel I).

Proof

technique · direct
1.1

If P is prime, then P is proper, so 1+P≠0+P by [L4]; and (a+P)(b+P)=0+P gives ab∈P, hence a+P=0+P or b+P=0+P, while commutativity of R makes the quotient commutative.

L1L2L3L4givenalgebra
2.1

If R/P is a domain and ab∈P, then (a+P)(b+P)=0+P, so [L3] and [L4] give a∈P or b∈P; its nonzero identity gives 1∉P, hence P⊊R.

step 1.1L1L2L3L4givenalgebra
3.1

These implications prove the equivalence.

step 2.1∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

R/M is a field if and only if M is a maximal ideal

Statement

R/M is a field if and only if M is a maximal ideal.

Here R is commutative and M is an ideal of R.

Facts & Assumptions

Given: A commutative ring R and a proper ideal M⊴R.

[L1]

A maximal ideal has no proper intermediate ideal (Prime ideals and maximal ideals in a commutative ring).

[L2]

R/M is a quotient ring with its usual coset operations (For a two-sided ideal I, the additive cosets form a ring R/I with identity 1+I).

[L3]

A field is a commutative ring in which every nonzero element is invertible (Every field is a commutative ring with 1≠0; it is an integral domain, and it is a commutative division ring).

[L4]

The definition of field requires a multiplicative inverse for each nonzero element (Field).

[L5]

The ideal criterion verifies ideals by subtraction and absorption (Ideal criteria and intersections of ideals).

Proof

technique · direct
1.1

If M is maximal and a∉M, the set J={m+ra:m∈M,r∈R} is an ideal by the subtraction-and-absorption criterion, properly contains M, and hence is R; thus m+ra=1 for some m∈M,r∈R, giving (a+M)(r+M)=1+M.

L1L2L3L4L5givenalgebra
2.1

If R/M is a field and M⊊J⊴R, choose a∈J∖M; an inverse r+M of a+M gives ar−1∈M⊆J, while ar∈J, so 1∈J and J=R.

step 1.1L1L2L3L4L5givenchoose
3.1

Hence R/M is a field exactly when M is maximal.

step 2.1∎
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Every maximal ideal of a commutative ring is prime

Statement

Every maximal ideal of a commutative ring is prime.

Facts & Assumptions

Given: A maximal ideal M of a commutative ring R.

[L1]

R/M is a field when M is maximal (R/M is a field if and only if M is a maximal ideal).

[L2]

R/P is a domain exactly when P is prime (R/P is an integral domain if and only if P is a prime ideal).

Proof

technique · direct
1.1

By [L1], R/M is a field.

L1L2L3given
2.1

By [L3], that quotient is an integral domain.

step 1.1L1L2L3given
3.1

The domain conclusion of step 2.1 yields that M is prime.

step 2.1∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

In a nonzero commutative ring, every proper ideal is contained in a maximal ideal

Statement

Assume the Axiom of Choice (The Axiom of Choice).

In a nonzero commutative ring, every proper ideal is contained in a maximal ideal.

Facts & Assumptions

Given: A nonzero commutative ring R and a proper ideal I⊴R.

[L1]

A maximal ideal is a maximal proper ideal under inclusion (Prime ideals and maximal ideals in a commutative ring).

[L2]

Ideals are additive subgroups with multiplication absorption (Left, right and two-sided ideals).

[L3]

An ideal criterion and intersection closure are available (Ideal criteria and intersections of ideals).

[L4]

Assuming the Axiom of Choice, a nonempty poset in which every chain has an upper bound has a maximal element (Zorn's lemma).

[L5]

A chain is a subset linearly ordered by the ambient order (Chain in a poset).

[L6]

A maximal element has no strictly larger element in the poset (Maximal element and greatest element).

Proof

technique · direct
1.1

Let P be the poset of proper ideals of R containing I, ordered by inclusion.

L1L2givenconstruct
1.2

P is nonempty, because I is a proper ideal containing I.

givenL2
2.1

The empty chain of P has an upper bound in P: every member of P is vacuously above all of its members, and P is nonempty, so I is such an upper bound.

step 1.1step 1.2L5
2.2

A nonempty chain C⊆P has an upper bound in P: ⋃C is an ideal containing I, and it is proper, since 1∈⋃C would place 1 in some member of C, forcing that member to equal R and contradicting its properness.

step 1.1L2L3L5L7algebra
3.1

Every chain of P has an upper bound in P, and P is a nonempty poset, so Zorn's lemma yields a maximal element M of P.

step 1.2step 2.1step 2.2L4
4.1

M is a proper ideal containing I that is maximal among proper ideals of R, so M is a maximal ideal containing I.

step 3.1L1L6∎

5 · Examples, counterexamples and false statements

None yet.

Sources