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Localisation of Modules and Support

1 · Prerequisites

2 · Summary

Ring localisation, exact sequences, tensor products, module homomorphisms, finitely generated modules, and finitely presented modules supply the background for extending scalars from $R$ to $S^{- 1} R$ . Using those ingredients, this development constructs module localisation, proves the tensor description $S^{- 1} M ≅ S^{- 1} R \otimes_{R} M$ , identifies the Hom-localisation map, and records the choice-dependent local tests for zero modules, morphisms, and exactness.

It then defines support and relates it to annihilators, short exact sequences, tensor products, direct sums, and further localisation. The final part defines the Jacobson radical, proves the determinant trick and the choice-dependent Nakayama consequences, characterises minimal generators over a local ring through the residue-field quotient, and ends with the principal-neighbourhood vanishing consequence for finite modules.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

Localisation of a module at a multiplicative subset

Definition

Let $R$ be a commutative ring, let $S \subseteq R$ be a multiplicative subset, and let $M$ be a left $R$ -module. On $M \times S$ , define

(m, s) \sim (n, t) ⟺ u (t m - s n) = 0 for some u \in S .

The localisation of $M$ at $S$ is the set $S^{- 1} M$ of equivalence classes for this relation, and the class of $(m, s)$ is written $m / s$ .

The proposed addition and $S^{- 1} R$ -scalar action are

\frac{m}{s} + \frac{n}{t} : = \frac{t m + s n}{s t}, \frac{a}{u} \cdot \frac{m}{s} : = \frac{a m}{u s} .

The canonical map is

λ_{M} : M ⟶ S^{- 1} M, m ⟼ \frac{m}{1} .

The relation, the addition formula, and the scalar action are justified by The module-fraction relation is an equivalence relation ↗, Addition of localised module fractions is independent of representatives ↗, and The localised scalar action is independent of representatives ↗.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

The module-fraction relation is an equivalence relation

Statement

For a commutative ring $R$ , a multiplicative subset $S \subseteq R$ , and a left $R$ -module $M$ , the relation on $M \times S$ from Localisation of a module at a multiplicative subset is an equivalence relation.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , and a left $R$ -module $M$ .

[L1]

In the localisation of a module, $(m, s) \sim (n, t)$ means that $u (t m - s n) = 0$ for some $u \in S$ , and $1 \in S$ while products of elements of $S$ stay in $S$ (Localisation of a module at a multiplicative subset).

Proof

technique · direct

1.1

Reflexivity holds because $1 (s m - s m) = 0$ , so $(m, s) \sim (m, s)$ for every $(m, s) \in M \times S$ .

L1algebra

1.2

Symmetry holds because $u (t m - s n) = 0$ implies $u (s n - t m) = 0$ , so $(m, s) \sim (n, t)$ implies $(n, t) \sim (m, s)$ .

L1algebra

1.3

If $(m, s) \sim (n, t)$ via $u$ and $(n, t) \sim (p, v)$ via $w$ , then $u w v (t m - s n) = 0$ and uws $(v n - t p) = 0$ ; adding these equalities gives $u w t (v m - s p) = 0$ , so $(m, s) \sim (p, v)$ .

L1algebra

2.1

Steps 1.1, 1.2, and 1.3 prove that $\sim$ is an equivalence relation.

step 1.1step 1.2step 1.3∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

Addition of localised module fractions is independent of representatives

Statement

If $m / s = m^{'} / s^{'}$ and $n / t = n^{'} / t^{'}$ in $S^{- 1} M$ , then

\frac{t m + s n}{s t} = \frac{t^{'} m^{'} + s^{'} n^{'}}{s^{'} t^{'}} .

So the addition formula of Localisation of a module at a multiplicative subset is independent of the chosen representatives.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , a left $R$ -module $M$ , and equalities $m / s = m^{'} / s^{'}$ and $n / t = n^{'} / t^{'}$ in $S^{- 1} M$ .

[L1]

In $S^{- 1} M$ , the equality $x / u = y / v$ means that $q (v x - u y) = 0$ for some $q \in S$ , and addition is defined by $(x / u) + (y / v) = (v x + u y) / (u v)$ (Localisation of a module at a multiplicative subset).

[L2]

The relation defining $S^{- 1} M$ is an equivalence relation (The module-fraction relation is an equivalence relation).

Proof

technique · direct

1.1

By [L1], choose $u, v \in S$ with $u (s^{'} m - s m^{'}) = 0$ and $v (t^{'} n - t n^{'}) = 0$ .

givenL1L2choose

2.1

Multiplying the first equality by $v t t^{'}$ and the second by $u s s^{'}$ and then adding gives $u v (s^{'} t^{'} (t m + s n) - s t (t^{'} m^{'} + s^{'} n^{'})) = 0$ .

step 1.1algebra

3.1

Step 2.1 is exactly the relation witnessing $(t m + s n) / (s t) = (t^{'} m^{'} + s^{'} n^{'}) / (s^{'} t^{'})$ , so the addition formula is independent of representatives.

step 2.1L1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

The localised scalar action is independent of representatives

Statement

If $a / u = a^{'} / u^{'}$ in $S^{- 1} R$ and $m / s = m^{'} / s^{'}$ in $S^{- 1} M$ , then

\frac{a m}{u s} = \frac{a^{'} m^{'}}{u^{'} s^{'}} .

So the scalar action of Localisation of a module at a multiplicative subset is independent of both ring-fraction and module-fraction representatives.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , a left $R$ -module $M$ , and equalities $a / u = a^{'} / u^{'}$ in $S^{- 1} R$ and $m / s = m^{'} / s^{'}$ in $S^{- 1} M$ .

[L1]

In $S^{- 1} M$ , the equality $x / r = y / t$ means that $q (t x - r y) = 0$ for some $q \in S$ , and the proposed scalar action is $(b / v) (x / r) = b x / (v r)$ (Localisation of a module at a multiplicative subset).

[L2]

In $S^{- 1} R$ , the equality $b / v = c / w$ means that $q (w b - v c) = 0$ for some $q \in S$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions).

Proof

technique · direct

1.1

By [L1] and [L2], choose $v, w \in S$ with $v (u^{'} a - u a^{'}) = 0$ and $w (s^{'} m - s m^{'}) = 0$ .

givenL1L2choose

2.1

Multiplying the module-fraction equality by $a u v$ and the ring-fraction equality by $u s m^{'} w$ , then adding, gives $u v w (u^{'} s^{'} a m - u s a^{'} m^{'}) = a u^{'} v w (s^{'} m - s m^{'}) + u s m^{'} v w (u^{'} a - u a^{'}) = 0.$ So $u v w (u^{'} s^{'} a m - u s a^{'} m^{'}) = 0$ .

step 1.1algebra

3.1

Step 2.1 is exactly the relation witnessing $a m / (u s) = a^{'} m^{'} / (u^{'} s^{'})$ , so the scalar action is independent of representatives.

step 2.1L1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Universal property of localisation for modules

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative, let $M$ be a left $R$ -module, and let $N$ be an $S^{- 1} R$ -module. Every $R$ -linear map $f : M \to N$ factors uniquely through the localisation map $λ_{M} : M \to S^{- 1} M$ by an $S^{- 1} R$ -linear map

\tilde{f} : S^{- 1} M ⟶ N, \tilde{f} (m / s) = (1 / s) f (m) .

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , a left $R$ -module $M$ , an $S^{- 1} R$ -module $N$ , and an $R$ -linear map $f : M \to N$ .

[L1]

In $S^{- 1} M$ , the localisation map is $λ_{M} (m) = m / 1$ , addition is $(m / s) + (n / t) = (t m + s n) / (s t)$ , and the scalar action is $(a / u) (m / s) = a m / (u s)$ (Localisation of a module at a multiplicative subset).

[L2]

The addition formula of $S^{- 1} M$ is independent of representatives (Addition of localised module fractions is independent of representatives).

[L3]

The scalar action of $S^{- 1} R$ on $S^{- 1} M$ is independent of representatives (The localised scalar action is independent of representatives).

[L4]

In $S^{- 1} R$ , every $s / 1$ with $s \in S$ is a unit with inverse $1 / s$ (The localisation relation is an equivalence relation and fraction arithmetic is well defined).

Proof

technique · direct

1.1

Define $\tilde{f} (m / s) : = (1 / s) f (m)$ . If $m / s = m^{'} / s^{'}$ , choose $u \in S$ with $u (s^{'} m - s m^{'}) = 0$ ; applying $f$ gives $(u / 1) ((s^{'} / 1) f (m) - (s / 1) f (m^{'})) = 0$ , and multiplying by the units $(u / 1)^{- 1} (s / 1)^{- 1} (s^{'} / 1)^{- 1}$ from [L4] gives $(1 / s) f (m) = (1 / s^{'}) f (m^{'})$ .

L1L4givenalgebra

2.1

For $m / s, n / t \in S^{- 1} M$ , $\tilde{f} ((m / s) + (n / t)) = (1 / s t) f (t m + s n) = (1 / s) f (m) + (1 / t) f (n) = \tilde{f} (m / s) + \tilde{f} (n / t)$ , and for $a / u \in S^{- 1} R$ one has $\tilde{f} ((a / u) (m / s)) = (1 / u s) f (a m) = (a / u) \tilde{f} (m / s)$ .

step 1.1L1L2L3algebra

2.2

For every $m \in M$ , $\tilde{f} (λ_{M} (m)) = \tilde{f} (m / 1) = f (m)$ .

step 1.1L1

3.1

If $g : S^{- 1} M \to N$ is $S^{- 1} R$ -linear and $g λ_{M} = f$ , then for every $m / s$ one has $g (m / s) = g ((1 / s) (m / 1)) = (1 / s) g (m / 1) = (1 / s) f (m) = \tilde{f} (m / s)$ , so $g = \tilde{f}$ .

L1L4step 2.2

4.1

Steps 1.1, 2.1, 2.2, and 3.1 prove the stated unique $S^{- 1} R$ -linear factorisation.

step 1.1step 2.1step 2.2step 3.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Localisation of modules is extension of scalars

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative, and let $M$ be a left $R$ -module. The map

Φ : (S^{- 1} R) \otimes_{R} M ⟶ S^{- 1} M, Φ ((a / s) \otimes m) = a m / s,

is an isomorphism of $S^{- 1} R$ -modules. Its inverse is

Ψ : S^{- 1} M ⟶ (S^{- 1} R) \otimes_{R} M, Ψ (m / s) = (1 / s) \otimes m .

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , and a left $R$ -module $M$ .

[L1]

A balanced pairing on $(S^{- 1} R) \times M$ induces a unique homomorphism from $(S^{- 1} R) \otimes_{R} M$ (Universal property of the tensor product for balanced maps into abelian groups, A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

[L2]

Over a commutative ring, a tensor product carries the scalar action $r (x \otimes m) = (r x) \otimes m$ (Over a commutative ring, $M \otimes_{R} N$ is an $R$ -module with $r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$ ).

[L3]

The localisation map $M \to S^{- 1} M$ is universal for maps into $S^{- 1} R$ -modules (Universal property of localisation for modules).

[L4]

In $S^{- 1} R$ , fraction arithmetic is well defined and every $s / 1$ with $s \in S$ is a unit with inverse $1 / s$ (The localisation relation is an equivalence relation and fraction arithmetic is well defined).

Proof

technique · direct

1.1

The pairing $q ((a / s), m) = a m / s$ is balanced because it is additive in each variable and $q (((a / s) r), m) = a r m / s = q (a / s, r m)$ for every $r \in R$ .

L1L4algebra

1.2

The map $i : M \to (S^{- 1} R) \otimes_{R} M$ , $i (m) = (1 / 1) \otimes m$ , is $R$ -linear, and every $s \in S$ acts invertibly on the target because $(s / 1)^{- 1} = 1 / s$ in $S^{- 1} R$ .

L2L4algebra

2.1

By [L1], step 1.1 induces a unique homomorphism $Φ : (S^{- 1} R) \otimes_{R} M \to S^{- 1} M$ with $Φ ((a / s) \otimes m) = a m / s$ .

step 1.1L1construct

2.2

By [L3], step 1.2 induces a unique $S^{- 1} R$ -linear map $Ψ : S^{- 1} M \to (S^{- 1} R) \otimes_{R} M$ with $Ψ (m / s) = (1 / s) \otimes m$ .

step 1.2L3construct

3.1

For every $m / s \in S^{- 1} M$ , $(Φ Ψ) (m / s) = Φ ((1 / s) \otimes m) = m / s$ .

step 2.1step 2.2

3.2

For every elementary tensor $(a / s) \otimes m$ , $(Ψ Φ) ((a / s) \otimes m) = Ψ (a m / s) = (1 / s) \otimes a m = (a / s) \otimes m$ by the tensor scalar action of [L2].

step 2.1step 2.2L2algebra

4.1

Steps 3.1 and 3.2 show that $Φ$ and $Ψ$ are inverse $S^{- 1} R$ -linear isomorphisms.

step 3.1step 3.2∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Localisation commutes with quotient modules and arbitrary direct sums

Statement

Let $R$ be a commutative ring and let $S \subseteq R$ be multiplicative.

For every submodule $N \leq M$ , there is a natural isomorphism

S^{- 1} (M / N) ≅ (S^{- 1} M) / (S^{- 1} N) .

For every family $(M_{i})_{i \in I}$ of left $R$ -modules, there is a natural isomorphism

S^{- 1} ⁣ (⨁_{i \in I} M_{i}) ≅ ⨁_{i \in I} S^{- 1} M_{i} .

Facts & Assumptions

Given: A commutative ring $R$ and a multiplicative subset $S \subseteq R$ .

[L1]

Localisation is naturally $(S^{- 1} R) \otimes_{R} -$ (Localisation of modules is extension of scalars).

[L2]

Tensoring a right-exact sequence with a fixed module preserves right exactness (Tensoring is right exact).

[L3]

Tensor products commute with arbitrary direct sums (Tensor products commute with arbitrary direct sums).

[L4]

A quotient module is the module of cosets $M / N$ (Quotient module $M / N$ with scalar multiplication on additive cosets).

Proof

technique · direct

1.1

For a submodule $N \leq M$ , the sequence $N \to M \to M / N \to 0$ is right exact, so [L2] and [L1] give a right-exact sequence $S^{- 1} N \to S^{- 1} M \to S^{- 1} (M / N) \to 0$ . Therefore $S^{- 1} (M / N)$ is the quotient of $S^{- 1} M$ by the image of $S^{- 1} N \to S^{- 1} M$ , namely by the submodule $S^{- 1} N$ .

L1L2L4

1.2

For a family $(M_{i})_{i \in I}$ , [L3] and [L1] give $S^{- 1} (⨁_{i} M_{i}) ≅ (S^{- 1} R) \otimes_{R} (⨁_{i} M_{i}) ≅ ⨁_{i} ((S^{- 1} R) \otimes_{R} M_{i}) ≅ ⨁_{i} S^{- 1} M_{i}$ .

L1L3

2.1

Thus $S^{- 1} (M / N) ≅ (S^{- 1} M) / (S^{- 1} N)$ naturally in $M$ and $N$ .

step 1.1

3.1

Steps 2.1 and 1.2 prove the quotient and direct-sum claims.

step 2.1step 1.2∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

A localised module fraction is zero exactly when one denominator kills its numerator

Statement

For $m \in M$ and $s \in S$ ,

\frac{m}{s} = 0 in S^{- 1} M ⟺ u m = 0 for some u \in S .

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , a left $R$ -module $M$ , an element $m \in M$ , and an element $s \in S$ .

[L1]

In $S^{- 1} M$ , the equality $m / s = n / t$ means that $u (t m - s n) = 0$ for some $u \in S$ (Localisation of a module at a multiplicative subset).

Proof

technique · direct

1.1

If $m / s = 0 / 1$ , then [L1] gives $u (1 m - s 0) = u m = 0$ for some $u \in S$ .

1.2

If $u m = 0$ for some $u \in S$ , then $u (1 m - s 0) = 0$ , so [L1] gives $m / s = 0 / 1 = 0$ in $S^{- 1} M$ .

2.1

Steps 1.1 and 1.2 prove the equivalence.

step 1.1step 1.2∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

Injective module maps remain injective after localisation

Statement

Let $f : M^{'} \to M$ be an injective $R$ -module homomorphism. Then the induced map

S^{- 1} f : S^{- 1} M^{'} ⟶ S^{- 1} M, (m^{'} / s) ⟼ f (m^{'}) / s,

is injective.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , left $R$ -modules $M^{'}, M$ , and an injective $R$ -module homomorphism $f : M^{'} \to M$ .

[L1]

A localised fraction is zero exactly when one element of $S$ kills its numerator (A localised module fraction is zero exactly when one denominator kills its numerator).

[L2]

A module homomorphism preserves scalar multiplication, so $f (r m^{'}) = r f (m^{'})$ for every $r \in R$ and $m^{'} \in M^{'}$ (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct

1.1

Suppose $(S^{- 1} f) (m^{'} / s) = 0$ . Then $f (m^{'}) / s = 0$ , so [L1] gives $u f (m^{'}) = 0$ for some $u \in S$ .

givenL1

2.1

By [L2], $u f (m^{'}) = f (u m^{'})$ , so injectivity of $f$ gives $u m^{'} = 0$ .

step 1.1L2

3.1

Applying [L1] again, step 2.1 gives $m^{'} / s = 0$ in $S^{- 1} M^{'}$ . Hence $S^{- 1} f$ is injective.

step 2.1L1∎

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

Surjective module maps remain surjective after localisation

Statement

Let $f : M \to M^{''}$ be a surjective $R$ -module homomorphism. Then the induced map

S^{- 1} f : S^{- 1} M ⟶ S^{- 1} M^{''}, (m / s) ⟼ f (m) / s,

is surjective.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , left $R$ -modules $M, M^{''}$ , and a surjective $R$ -module homomorphism $f : M \to M^{''}$ .

[L1]

A module homomorphism preserves scalar multiplication (Module homomorphism and isomorphism, kernel, image and cokernel).

[L2]

Elements of $S^{- 1} M^{''}$ are fractions $m^{''} / s$ with $m^{''} \in M^{''}$ and $s \in S$ (Localisation of a module at a multiplicative subset).

Proof

technique · direct

1.1

Let $m^{''} / s \in S^{- 1} M^{''}$ . Since $f$ is surjective, choose $m \in M$ with $f (m) = m^{''}$ .

givenL2choose

2.1

Then $(S^{- 1} f) (m / s) = f (m) / s = m^{''} / s$ , so $m^{''} / s$ lies in the image of $S^{- 1} f$ .

step 1.1L1

3.1

Since every element of $S^{- 1} M^{''}$ is hit, $S^{- 1} f$ is surjective.

step 2.1L2∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

Localisation of modules is exact

Statement

0 ⟶ M^{'} \overset{f}{\to} M \overset{g}{\to} M^{''} ⟶ 0

is a short exact sequence of $R$ -modules, then

0 ⟶ S^{- 1} M^{'} \overset{S^{- 1} f}{\to} S^{- 1} M \overset{S^{- 1} g}{\to} S^{- 1} M^{''} ⟶ 0

is a short exact sequence of $S^{- 1} R$ -modules.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , and a short exact sequence $0 \to M^{'} \overset{f}{\to} M \overset{g}{\to} M^{''} \to 0$ .

[L1]

Localisation is naturally $(S^{- 1} R) \otimes_{R} -$ (Localisation of modules is extension of scalars).

[L2]

Tensoring a right-exact sequence with a fixed module preserves right exactness (Tensoring is right exact).

[L3]

Injective module homomorphisms remain injective after localisation (Injective module maps remain injective after localisation).

[L4]

A short exact sequence is exact, with left map injective and right map surjective (Exact sequences and short exact sequences of modules).

Proof

technique · direct

1.1

By [L4], the tail $M^{'} \overset{f}{\to} M \overset{g}{\to} M^{''} \to 0$ is exact. Using [L1] to identify localisation with tensor product, [L2] gives an exact sequence $S^{- 1} M^{'} \overset{S^{- 1} f}{\to} S^{- 1} M \overset{S^{- 1} g}{\to} S^{- 1} M^{''} \to 0$ .

L1L2L4

1.2

The map $f$ is injective by [L4], so [L3] makes $S^{- 1} f$ injective.

L3L4

2.1

Steps 1.1 and 1.2 show that the localised sequence is short exact.

step 1.1step 1.2∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Localisation commutes with kernels images and cokernels

Statement

For every $R$ -module homomorphism $f : M \to N$ , localisation identifies

S^{- 1} (\ker f) ≅ \ker (S^{- 1} f), S^{- 1} (im f) ≅ im (S^{- 1} f), S^{- 1} (coker f) ≅ coker (S^{- 1} f) .

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , left $R$ -modules $M, N$ , and an $R$ -module homomorphism $f : M \to N$ .

[L1]

Localisation sends short exact sequences to short exact sequences (Localisation of modules is exact).

[L2]

The kernel, image, and cokernel of $f$ are the standard submodule and quotient constructions associated to $f$ (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct

1.1

The standard short exact sequences $0 \to \ker f \to M \to im f \to 0$ and $0 \to im f \to N \to coker f \to 0$ localise, by [L1], to short exact sequences $0 \to S^{- 1} (\ker f) \to S^{- 1} M \to S^{- 1} (im f) \to 0$ and $0 \to S^{- 1} (im f) \to S^{- 1} N \to S^{- 1} (coker f) \to 0$ .

L1L2

2.1

In the first localised sequence, the middle map is $S^{- 1} f$ restricted to $S^{- 1} M$ , so its kernel is exactly $S^{- 1} (\ker f)$ and its image is exactly the embedded copy of $S^{- 1} (im f)$ .

step 1.1L2

3.1

In the second localised sequence, the quotient by the image of $S^{- 1} f$ is therefore $S^{- 1} (coker f)$ , so $S^{- 1} (coker f) ≅ coker (S^{- 1} f)$ .

step 2.1L2

4.1

Steps 2.1 and 3.1 prove the kernel, image, and cokernel identifications.

step 2.1step 3.1∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Localisation commutes with finite intersections of submodules

Statement

Let $N_{1}, \dots, N_{r}$ be submodules of a left $R$ -module $M$ . Then

S^{- 1} ⁣ (⋂_{i = 1}^{r} N_{i}) = ⋂_{i = 1}^{r} S^{- 1} N_{i}

inside $S^{- 1} M$ .

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , a left $R$ -module $M$ , and submodules $N_{1}, \dots, N_{r} \leq M$ .

[L1]

Localisation identifies kernels with the kernels of localised maps (Localisation commutes with kernels images and cokernels).

[L2]

Localisation commutes with quotient modules and finite direct sums (Localisation commutes with quotient modules and arbitrary direct sums).

[L3]

The direct sum has its universal diagonal map into a family of targets (Universal property of a direct sum of modules, The direct sum of an indexed family of modules).

Proof

technique · direct

1.1

Let $δ : M \to ⨁_{i = 1}^{r} M / N_{i}$ be the diagonal map $δ (m) = (m + N_{1}, \dots, m + N_{r})$ . By construction, $\ker δ = ⋂_{i = 1}^{r} N_{i}$ .

L3algebra

2.1

By [L1], $S^{- 1} (⋂_{i} N_{i}) ≅ \ker (S^{- 1} δ)$ . By [L2], the codomain of $S^{- 1} δ$ identifies with $⨁_{i} (S^{- 1} M / S^{- 1} N_{i})$ , and under this identification $S^{- 1} δ$ is the diagonal map $S^{- 1} M \to ⨁_{i} (S^{- 1} M / S^{- 1} N_{i})$ .

L1L2step 1.1

3.1

An element of $S^{- 1} M$ lies in the kernel of that diagonal map exactly when its image in every quotient $S^{- 1} M / S^{- 1} N_{i}$ is zero, that is, exactly when it lies in every submodule $S^{- 1} N_{i}$ . Therefore $\ker (S^{- 1} δ) = ⋂_{i} S^{- 1} N_{i}$ .

step 2.1algebra

4.1

Combining steps 2.1 and 3.1 gives $S^{- 1} (⋂_{i} N_{i}) = ⋂_{i} S^{- 1} N_{i}$ inside $S^{- 1} M$ .

step 2.1step 3.1∎

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

There is a natural localisation map on Hom

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative, and let $M, N$ be left $R$ -modules. There is a natural $S^{- 1} R$ -linear map

θ_{M, N} : S^{- 1} ⁣ {Hom}_{R} (M, N) ⟶ {Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)

given by

θ_{M, N} (f / s) (m / u) = f (m) / (s u) .

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , and left $R$ -modules $M, N$ .

[L1]

For a commutative ring $A$ and $A$ -modules $X, Y$ , the group ${Hom}_{A} (X, Y)$ is an $A$ -module under pointwise scalar multiplication (Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module).

[L2]

The localisation map of a module is universal for maps into modules over the localised ring (Universal property of localisation for modules).

Proof

technique · direct

1.1

For each $R$ -linear map $f : M \to N$ , apply [L2] to the composite $M \overset{f}{\to} N \to S^{- 1} N$ to obtain a unique $S^{- 1} R$ -linear map $S^{- 1} f : S^{- 1} M \to S^{- 1} N$ with $(S^{- 1} f) (m / u) = f (m) / u$ .

L2construct

2.1

The assignment $f \mapsto S^{- 1} f$ is $R$ -linear because for $f, g \in {Hom}_{R} (M, N)$ , $r \in R$ , and $m / u \in S^{- 1} M$ one has $(S^{- 1} (f + g)) (m / u) = (f (m) + g (m)) / u$ and $(S^{- 1} (r f)) (m / u) = r f (m) / u = (r / 1) (f (m) / u)$ .

step 1.1L1algebra

3.1

By [L1], ${Hom}_{R} (M, N)$ is an $R$ -module, while ${Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)$ is an $S^{- 1} R$ -module; therefore [L2] applied to the $R$ -linear map of step 2.1 gives a unique $S^{- 1} R$ -linear map $θ_{M, N}$ with $θ_{M, N} (f / 1) = S^{- 1} f$ .

step 2.1L1L2construct

4.1

Since $θ_{M, N}$ is $S^{- 1} R$ -linear, $θ_{M, N} (f / s) = (1 / s) θ_{M, N} (f / 1)$ , so for $m / u \in S^{- 1} M$ one gets $θ_{M, N} (f / s) (m / u) = (1 / s) (f (m) / u) = f (m) / (s u)$ .

step 3.1algebra

5.1

Steps 3.1 and 4.1 produce the stated natural $S^{- 1} R$ -linear map.

step 3.1step 4.1∎

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-26Open item page →

The localised Hom map is an isomorphism for finite free sources

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative, let $M$ be a finite free $R$ -module, and let $N$ be a left $R$ -module. Then the natural map

θ_{M, N} : S^{- 1} ⁣ {Hom}_{R} (M, N) ⟶ {Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)

from There is a natural localisation map on Hom is an isomorphism.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , a finite free $R$ -module $M$ with basis $e_{1}, \dots, e_{n}$ , and a left $R$ -module $N$ .

[L1]

The natural map $θ_{M, N}$ satisfies $θ_{M, N} (f / s) (m / u) = f (m) / (s u)$ (There is a natural localisation map on Hom).

[L2]

A basis means that every element of $M$ has a unique finite linear combination in the basis elements (Generated submodule, cyclic and finitely generated modules, module basis and free module).

[L3]

A homomorphism from a finite free module is uniquely determined by its basis values, and every tuple of target values occurs as those basis values (For a commutative ring, ${Hom}_{R} (R^{n}, N) ≅ N^{n}$ ).

[L4]

A localised fraction is zero exactly when one element of $S$ kills its numerator (A localised module fraction is zero exactly when one denominator kills its numerator).

[L5]

The group ${Hom}_{R} (M, N)$ is an $R$ -module (Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module).

[L6]

In $S^{- 1} M$ , addition and scalar multiplication satisfy $m / u = \sum_{i} (a_{i} / u) (e_{i} / 1)$ whenever $m = \sum_{i} a_{i} e_{i}$ in $M$ (Localisation of a module at a multiplicative subset).

Proof

technique · direct

1.1

By [L2] and [L6], every element of $S^{- 1} M$ is an $S^{- 1} R$ -linear combination of $e_{1} / 1, \dots, e_{n} / 1$ , so two $S^{- 1} R$ -linear maps $S^{- 1} M \to S^{- 1} N$ that agree on all $e_{i} / 1$ are equal.

L2L6algebra

1.2

If $θ_{M, N} (f / s) = 0$ , then $f (e_{i}) / s = 0$ for each $i$ , so [L4] gives $u_{i} \in S$ with $u_{i} f (e_{i}) = 0$ . Let $u = u_{1} \dots u_{n}$ . Then $u f (e_{i}) = 0$ for every $i$ , hence $u f = 0$ by [L3]. Since ${Hom}_{R} (M, N)$ is an $R$ -module by [L5], [L4] applied there gives $f / s = 0$ . Thus $θ_{M, N}$ is injective.

L1L3L4L5choose

1.3

Let $φ : S^{- 1} M \to S^{- 1} N$ be $S^{- 1} R$ -linear. Write $φ (e_{i} / 1) = n_{i} / s_{i}$ with $n_{i} \in N$ and $s_{i} \in S$ , set $s = s_{1} \dots s_{n}$ , and set $m_{i} = (s / s_{i}) n_{i}$ . By [L3], there is a unique $R$ -linear map $f : M \to N$ with $f (e_{i}) = m_{i}$ for all $i$ .

L3choose

2.1

For each basis vector, $θ_{M, N} (f / s) (e_{i} / 1) = f (e_{i}) / s = (s / s_{i}) n_{i} / s = n_{i} / s_{i} = φ (e_{i} / 1)$ , so step 1.1 gives $θ_{M, N} (f / s) = φ$ . Thus $θ_{M, N}$ is surjective.

L1step 1.1step 1.3algebra

3.1

Steps 1.2 and 2.1 prove that $θ_{M, N}$ is an isomorphism.

step 1.2step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A finite presentation reduces localised Hom to the finite free case

Statement

Let

R^{m} \overset{α}{\to} R^{n} \overset{β}{\to} M ⟶ 0

be a finite presentation of an $R$ -module $M$ , and let $N$ be an $R$ -module. If the natural localisation maps for $R^{n}$ and $R^{m}$ are isomorphisms, then the natural localisation map

S^{- 1} ⁣ {Hom}_{R} (M, N) ⟶ {Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)

is an isomorphism as well.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , an $R$ -module $N$ , and a finite presentation $R^{m} \overset{α}{\to} R^{n} \overset{β}{\to} M \to 0$ .

[L1]

A finite presentation is an exact sequence $R^{m} \overset{α}{\to} R^{n} \overset{β}{\to} M \to 0$ , and equivalently $M ≅ R^{n} / im α$ (Finitely presented modules and finitely presented algebras).

[L2]

Localisation sends short exact sequences to short exact sequences (Localisation of modules is exact).

[L3]

The localisation map on Hom is natural in the source, with formula $θ_{X, N} (f / s) (x / u) = f (x) / (s u)$ (There is a natural localisation map on Hom).

Proof

technique · direct

1.1

Because $β$ is surjective in [L1], a homomorphism $h : R^{n} \to N$ factors through $M$ exactly when $h$ kills $im α$ , equivalently when $h \circ α = 0$ . Thus ${Hom}_{R} (M, N)$ is the kernel of $α^{*} : {Hom}_{R} (R^{n}, N) \to {Hom}_{R} (R^{m}, N)$ , $h \mapsto h \circ α$ .

L1algebra

1.2

By [L2], localising the presentation of [L1] gives an exact sequence $S^{- 1} R^{m} \overset{S^{- 1} α}{\to} S^{- 1} R^{n} \overset{S^{- 1} β}{\to} S^{- 1} M \to 0$ , so the same argument identifies ${Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)$ with the kernel of $(S^{- 1} α)^{*}$ .

L1L2algebra

1.3

Naturality from [L3] makes the square between $α^{*}$ and $(S^{- 1} α)^{*}$ commute.

L3algebra

2.1

If the two vertical maps in step 1.3 are isomorphisms, then they identify the kernel in step 1.1 with the kernel in step 1.2. Therefore the induced map on those kernels, namely $S^{- 1} ⁣ {Hom}_{R} (M, N) \to {Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)$ , is an isomorphism.

step 1.1step 1.2step 1.3algebra

3.1

Step 2.1 is exactly the reduction from a finite presentation to the finite free case.

step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

Localisation of Hom for finite and finitely presented modules

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative, and let $M, N$ be left $R$ -modules. The natural map

θ_{M, N} : S^{- 1} ⁣ {Hom}_{R} (M, N) ⟶ {Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)

is injective when $M$ is finitely generated, and it is an isomorphism when $M$ is finitely presented.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , and left $R$ -modules $M, N$ .

[L1]

The natural localisation map on Hom satisfies $θ_{M, N} (f / s) (m / u) = f (m) / (s u)$ (There is a natural localisation map on Hom).

[L2]

A finitely generated module has a finite generating set, and a finitely presented module admits a finite presentation by finite free modules (Generated submodule, cyclic and finitely generated modules, module basis and free module, Finitely presented modules and finitely presented algebras).

[L3]

A localised fraction is zero exactly when one denominator kills its numerator (A localised module fraction is zero exactly when one denominator kills its numerator).

[L4]

The group ${Hom}_{R} (M, N)$ is an $R$ -module (Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module).

[L5]

The natural localisation map on Hom is an isomorphism for finite free sources (The localised Hom map is an isomorphism for finite free sources).

[L6]

A finite presentation reduces the Hom-localisation comparison to the finite free case (A finite presentation reduces localised Hom to the finite free case).

Proof

technique · direct

1.1

Suppose that $M$ is generated by $m_{1}, \dots, m_{r}$ and that $θ_{M, N} (f / s) = 0$ . Then $f (m_{i}) / s = 0$ for each $i$ by [L1], so [L3] gives $u_{i} \in S$ with $u_{i} f (m_{i}) = 0$ . Let $u = u_{1} \dots u_{r}$ . Then $u f (m_{i}) = 0$ for every generator, hence $u f = 0$ as a homomorphism $M \to N$ . Since ${Hom}_{R} (M, N)$ is an $R$ -module by [L4], [L3] applied there gives $f / s = 0$ . Therefore $θ_{M, N}$ is injective whenever $M$ is finitely generated.

L1L2L3L4choose

1.2

If $M$ is finitely presented, [L2] supplies a finite presentation $R^{m} \to R^{n} \to M \to 0$ . The free modules $R^{m}$ and $R^{n}$ satisfy the isomorphism hypothesis of [L6] by [L5], so [L6] makes $θ_{M, N}$ an isomorphism.

L2L5L6

2.1

Steps 1.1 and 1.2 prove the injective and finitely presented claims.

step 1.1step 1.2∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Assuming the Axiom of Choice, local criteria for zero modules and for injective, surjective, and bijective maps

Statement

Assume the Axiom of Choice.

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

$M = 0$ if and only if $M_{p} = 0$ for every prime ideal $p$ , and this is equivalent to $M_{m} = 0$ for every maximal ideal $m$ .
For an $R$ -module homomorphism $f : M \to N$ , the map $f$ is injective, surjective, or bijective if and only if every prime localisation $f_{p}$ has the same property, and this is equivalent to checking every maximal localisation.

Facts & Assumptions

Given: A commutative ring $R$ , left $R$ -modules $M, N$ , and an $R$ -module homomorphism $f : M \to N$ .

[L1]

Localisation identifies kernels and cokernels: $S^{- 1} (\ker f) ≅ \ker (S^{- 1} f)$ and $S^{- 1} (coker f) ≅ coker (S^{- 1} f)$ (Localisation commutes with kernels images and cokernels).

[L2]

Every proper ideal of a nonzero commutative ring is contained in a maximal ideal (In a nonzero commutative ring, every proper ideal is contained in a maximal ideal).

[L3]

Every maximal ideal of a commutative ring is prime (Every maximal ideal of a commutative ring is prime).

[L4]

The annihilator of $m \in M$ is ${Ann}_{R} (m) = {r \in R : r m = 0}$ (Annihilators, torsion elements and the torsion subset of a module).

[L5]

A localised fraction is zero exactly when one denominator kills its numerator (A localised module fraction is zero exactly when one denominator kills its numerator).

[L6]

Kernels and cokernels are the standard constructions attached to a module homomorphism (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct

1.1

If $M = 0$ , then every localisation of $M$ is $0$ .

given

1.2

Suppose $M_{m} = 0$ for every maximal ideal $m$ and $m \in M$ is nonzero. Then ${Ann}_{R} (m)$ is a proper ideal by [L4], so [L2] gives a maximal ideal $m$ containing it. If $m / 1 = 0$ in $M_{m}$ , [L5] gives $t \notin m$ with $t m = 0$ , so $t \in {Ann}_{R} (m) \subseteq m$ , a contradiction. Hence $M_{m} \neq 0$ , contradicting the hypothesis. Therefore $M = 0$ iff all maximal localisations vanish.

L2L4L5choose

2.1

If all prime localisations vanish then all maximal localisations vanish by [L3], so step 1.2 gives $M = 0$ . Conversely, suppose $M_{p} \neq 0$ for some prime ideal $p$ and choose $m / s \neq 0$ in $M_{p}$ . Then no element outside $p$ annihilates $m$ , or else [L5] would give $m / s = 0$ ; hence ${Ann}_{R} (m) \subseteq p$ . By [L2] choose a maximal ideal $m$ containing ${Ann}_{R} (m)$ . The same zero-criterion argument as in step 1.2 gives $m / 1 \neq 0$ in $M_{m}$ , so maximal-local vanishing would fail. Thus prime-local vanishing and maximal-local vanishing are equivalent.

L2L3L4L5step 1.2choose

3.1

The map $f$ is injective iff $\ker f = 0$ . By [L1], this is equivalent to $\ker (f_{p}) = 0$ for every prime $p$ , and then by step 2.1 to $\ker (f_{m}) = 0$ for every maximal $m$ . So $f$ is injective iff all prime localisations, equivalently all maximal localisations, are injective.

L1L6step 2.1

3.2

The map $f$ is surjective iff $coker f = 0$ . By [L1], this is equivalent to $coker (f_{p}) = 0$ for every prime $p$ , and then by step 2.1 to $coker (f_{m}) = 0$ for every maximal $m$ . So $f$ is surjective iff all prime localisations, equivalently all maximal localisations, are surjective.

L1L6step 2.1

4.1

A map is bijective exactly when it is both injective and surjective, so step 3.1 and step 3.2 give the bijective criterion.

step 3.1step 3.2

5.1

Steps 1.2, 2.1, 3.1, 3.2, and 4.1 prove both claims.

step 1.2step 2.1step 3.1step 3.2step 4.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

Assuming the Axiom of Choice, a sequence of modules is exact exactly when all prime localisations are exact

Statement

Assume the Axiom of Choice.

Let

M^{'} \overset{f}{\to} M \overset{g}{\to} M^{''}

be a sequence of $R$ -module homomorphisms with $g \circ f = 0$ . Then the sequence is exact at $M$ if and only if, for every prime ideal $p$ , the localised sequence

M_{p}^{'} \overset{f_{p}}{\to} M_{p} \overset{g_{p}}{\to} M_{p}^{''}

is exact at $M_{p}$ . Equivalently, it suffices to check exactness at every maximal ideal.

Facts & Assumptions

Given: A commutative ring $R$ and a sequence $M^{'} \overset{f}{\to} M \overset{g}{\to} M^{''}$ of left $R$ -modules with $g \circ f = 0$ .

[L1]

Exactness at $M$ means $im f = \ker g$ (Exact sequences and short exact sequences of modules).

[L2]

Localisation identifies kernels and images, and it commutes with quotient modules (Localisation commutes with kernels images and cokernels, Localisation commutes with quotient modules and arbitrary direct sums).

[L3]

A module is zero exactly when all of its prime localisations are zero, equivalently all of its maximal localisations are zero (Assuming the Axiom of Choice, local criteria for zero modules and for injective, surjective, and bijective maps).

[L4]

The image and kernel are the standard submodule constructions attached to a module homomorphism (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct

1.1

Because $g \circ f = 0$ , one has $im f \subseteq \ker g$ , so the quotient module $H : = \ker g / im f$ is defined. By [L1], the original sequence is exact at $M$ if and only if $H = 0$ .

L1L4

2.1

For every prime ideal $p$ , [L2] gives $H_{p} ≅ (\ker g)_{p} / (im f)_{p} ≅ \ker (g_{p}) / im (f_{p})$ . Therefore the localised sequence is exact at $M_{p}$ if and only if $H_{p} = 0$ .

L2step 1.1

3.1

By [L3], $H = 0$ if and only if $H_{p} = 0$ for every prime ideal $p$ , and this is equivalent to $H_{m} = 0$ for every maximal ideal $m$ . Combining this with steps 1.1 and 2.1 gives the prime-local and maximal-local exactness criteria.

L3step 1.1step 2.1

4.1

Step 3.1 proves the theorem.

step 3.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-26Open item page →

Support of a module

Definition

Let $R$ be a commutative ring and let $M$ be a left $R$ -module. The support of $M$ is

{Supp}_{R} (M) : = {p : M_{p} \neq 0},

where $p$ ranges over the prime ideals of $R$ and $M_{p}$ denotes localisation at the multiplicative set $R ∖ p$ (Localisation at a prime ideal: $R_{p} = (R ∖ p)^{- 1} R$ ).

On this page, before the separate spectrum page is built, ${Supp}_{R} (M)$ is read simply as a set of prime ideals.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

A prime lies in the support exactly when some element has annihilator inside it

Statement

For a left $R$ -module $M$ and a prime ideal $p$ of $R$ ,

p \in {Supp}_{R} (M) ⟺ {Ann}_{R} (m) \subseteq p for some m \in M .

Facts & Assumptions

Given: A commutative ring $R$ , a left $R$ -module $M$ , and a prime ideal $p$ .

[L1]

The support condition $p \in {Supp}_{R} (M)$ means $M_{p} \neq 0$ , and localisation at $p$ uses denominators outside $p$ (Support of a module, Localisation at a prime ideal: $R_{p} = (R ∖ p)^{- 1} R$ ).

[L2]

The annihilator of $m \in M$ is ${Ann}_{R} (m) = {r \in R : r m = 0}$ (Annihilators, torsion elements and the torsion subset of a module).

[L3]

A localised fraction is zero exactly when one denominator kills its numerator (A localised module fraction is zero exactly when one denominator kills its numerator).

Proof

technique · direct

1.1

Suppose $p \in {Supp}_{R} (M)$ and choose $m / s \neq 0$ in $M_{p}$ . If $t \notin p$ satisfied $t m = 0$ , then [L3] would give $m / s = 0$ . Hence every element of ${Ann}_{R} (m)$ lies in $p$ , so ${Ann}_{R} (m) \subseteq p$ .

L1L2L3

1.2

Conversely, if ${Ann}_{R} (m) \subseteq p$ and $m / 1 = 0$ in $M_{p}$ , then [L3] gives $t \notin p$ with $t m = 0$ , so $t \in {Ann}_{R} (m) \subseteq p$ , a contradiction. Thus $m / 1 \neq 0$ , so $M_{p} \neq 0$ .

L1L2L3

2.1

Steps 1.1 and 1.2 prove the equivalence.

step 1.1step 1.2∎

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

The support of a cyclic quotient is its vanishing set

Statement

For an ideal $I$ of a commutative ring $R$ ,

{Supp}_{R} (R / I) = {p : I \subseteq p} .

Facts & Assumptions

Given: A commutative ring $R$ and an ideal $I ⊴ R$ .

[L1]

A prime ideal $p$ lies in ${Supp}_{R} (R / I)$ exactly when $(R / I)_{p} \neq 0$ (Support of a module).

[L2]

Localisation commutes with quotients: $(R / I)_{p} ≅ R_{p} / I R_{p}$ (Localisation commutes with quotient rings: $S^{- 1} R / S^{- 1} I ≅ {\overset{ˉ}{S}}^{- 1} (R / I)$ ).

[L3]

The local ring $R_{p}$ has maximal ideal $p R_{p}$ , and the units are exactly the fractions with numerator outside $p$ ( $R_{p}$ is local with unique maximal ideal $p R_{p}$ ).

Proof

technique · direct

1.1

Fix a prime ideal $p$ . By [L2], $(R / I)_{p} \neq 0$ exactly when $R_{p} / I R_{p} \neq 0$ .

L1L2

1.2

If $I \subseteq p$ , then every generator $i / 1$ of $I R_{p}$ lies in the maximal ideal $p R_{p}$ from [L3], so $I R_{p}$ is proper and the quotient is nonzero.

1.3

If $I ⊈ p$ , choose $i \in I ∖ p$ . Then $i / 1$ is a unit in $R_{p}$ by [L3], and it lies in $I R_{p}$ , so $I R_{p} = R_{p}$ and the quotient is zero.

L3choose

2.1

By steps 1.1, 1.2, and 1.3, $p$ lies in ${Supp}_{R} (R / I)$ exactly when $I \subseteq p$ .

step 1.1step 1.2step 1.3∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A finite module has the union of its generator-cyclic supports

Statement

If a left $R$ -module $M$ is generated by $m_{1}, \dots, m_{r}$ , then

{Supp}_{R} (M) = ⋃_{i = 1}^{r} {Supp}_{R} (R / {Ann}_{R} (m_{i})) .

Facts & Assumptions

Given: A commutative ring $R$ , a left $R$ -module $M$ , and generators $m_{1}, \dots, m_{r}$ of $M$ .

[L1]

A prime ideal $p$ lies in ${Supp}_{R} (M)$ exactly when ${Ann}_{R} (m) \subseteq p$ for some $m \in M$ (A prime lies in the support exactly when some element has annihilator inside it).

[L2]

For an ideal $I$ , the support of $R / I$ is the set of primes containing $I$ (The support of a cyclic quotient is its vanishing set).

[L3]

The generators $m_{1}, \dots, m_{r}$ generate every element of $M$ by finite $R$ -linear combinations (Generated submodule, cyclic and finitely generated modules, module basis and free module).

Proof

technique · direct

1.1

If $r = 0$ , then $M = 0$ because the empty list generates only the zero submodule. Both sides are therefore empty, so the formula holds. Hence assume $r \geq 1$ .

L3algebra

1.2

Suppose $p \in {Supp}_{R} (M)$ . By [L1], choose $m \in M$ with ${Ann}_{R} (m) \subseteq p$ , and write $m = \sum_{i} r_{i} m_{i}$ using [L3]. If no ${Ann}_{R} (m_{i})$ were contained in $p$ , choose $t_{i} \in {Ann}_{R} (m_{i}) ∖ p$ for every $i$ ; then $t = t_{1} \dots t_{r} \notin p$ because $p$ is prime, but $t m = 0$ , so $t \in {Ann}_{R} (m) \subseteq p$ , a contradiction. Thus ${Ann}_{R} (m_{i}) \subseteq p$ for some $i$ , and [L2] gives $p \in {Supp}_{R} (R / {Ann}_{R} (m_{i}))$ .

L1L2L3choose

1.3

Conversely, if $p \in {Supp}_{R} (R / {Ann}_{R} (m_{i}))$ for some $i$ , then [L2] gives ${Ann}_{R} (m_{i}) \subseteq p$ , and [L1] applied to the element $m_{i} \in M$ gives $p \in {Supp}_{R} (M)$ .

L1L2

2.1

Steps 1.1, 1.2, and 1.3 prove the union formula.

step 1.1step 1.2step 1.3∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

For a finite module, support is the set of primes containing the annihilator

Statement

If $M$ is a finitely generated left $R$ -module, then

{Supp}_{R} (M) = {p : {Ann}_{R} (M) \subseteq p} .

Facts & Assumptions

Given: A commutative ring $R$ and a finitely generated left $R$ -module $M$ .

[L1]

A prime ideal lies in ${Supp}_{R} (M)$ exactly when some element of $M$ has annihilator inside it (A prime lies in the support exactly when some element has annihilator inside it).

[L2]

If $m_{1}, \dots, m_{r}$ generate $M$ , then ${Supp}_{R} (M) = ⋃_{i} {Supp}_{R} (R / {Ann}_{R} (m_{i}))$ (A finite module has the union of its generator-cyclic supports).

[L3]

The annihilator of $M$ is ${Ann}_{R} (M) = {r \in R : r m = 0 for every m \in M}$ (Annihilators, torsion elements and the torsion subset of a module).

Proof

technique · direct

1.1

If $p \in {Supp}_{R} (M)$ , [L1] gives $m \in M$ with ${Ann}_{R} (m) \subseteq p$ . Since every element of ${Ann}_{R} (M)$ kills every element of $M$ , one has ${Ann}_{R} (M) \subseteq {Ann}_{R} (m) \subseteq p$ .

L1L3

1.2

Choose generators $m_{1}, \dots, m_{r}$ of $M$ . If ${Ann}_{R} (M) \subseteq p$ and no ${Ann}_{R} (m_{i})$ is contained in $p$ , choose $t_{i} \in {Ann}_{R} (m_{i}) ∖ p$ for every $i$ . Then $t = t_{1} \dots t_{r} \notin p$ , but $t$ annihilates every generator and hence all of $M$ , so $t \in {Ann}_{R} (M) \subseteq p$ , a contradiction. Thus ${Ann}_{R} (m_{i}) \subseteq p$ for some $i$ , and [L2] gives $p \in {Supp}_{R} (M)$ .

L2L3choose

2.1

Steps 1.1 and 1.2 prove the support-annihilator formula.

step 1.1step 1.2∎

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

Support in a short exact sequence is the union of the outer supports

Statement

0 ⟶ M^{'} ⟶ M ⟶ M^{''} ⟶ 0

is a short exact sequence of left $R$ -modules, then

{Supp}_{R} (M) = {Supp}_{R} (M^{'}) \cup {Supp}_{R} (M^{''}) .

Facts & Assumptions

Given: A commutative ring $R$ and a short exact sequence $0 \to M^{'} \to M \to M^{''} \to 0$ of left $R$ -modules.

[L1]

A prime ideal $p$ lies in ${Supp}_{R} (X)$ exactly when $X_{p} \neq 0$ (Support of a module).

[L2]

Localisation sends short exact sequences to short exact sequences (Localisation of modules is exact).

[L3]

In a short exact sequence, the left map is injective and the right map is surjective (Exact sequences and short exact sequences of modules).

Proof

technique · direct

1.1

Fix a prime ideal $p$ . By [L2], localising the given short exact sequence at $p$ gives $0 \to M_{p}^{'} \to M_{p} \to M_{p}^{''} \to 0$ .

2.1

In that localised sequence, $M_{p} = 0$ holds exactly when both $M_{p}^{'} = 0$ and $M_{p}^{''} = 0$ : if the middle term is zero then injectivity and surjectivity from [L3] force both outer terms to be zero, while if both outer terms are zero then exactness makes the middle term zero as well.

step 1.1L3

3.1

By [L1], step 2.1 says exactly that $p$ lies in ${Supp}_{R} (M)$ if and only if it lies in ${Supp}_{R} (M^{'})$ or in ${Supp}_{R} (M^{''})$ .

L1step 2.1

4.1

Since this holds for every prime ideal $p$ , the support identity follows.

step 3.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Support of a tensor product of finite modules is the intersection of the supports

Statement

If $M$ and $N$ are finitely generated left $R$ -modules, then

{Supp}_{R} (M \otimes_{R} N) = {Supp}_{R} (M) \cap {Supp}_{R} (N) .

Facts & Assumptions

Given: A commutative ring $R$ and finitely generated left $R$ -modules $M, N$ .

[L1]

For a finite module, the support is the set of primes containing its annihilator (For a finite module, support is the set of primes containing the annihilator).

[L2]

Localisation is naturally tensoring with the localised ring, so $(M \otimes_{R} N)_{p} ≅ M_{p} \otimes_{R_{p}} N_{p}$ (Localisation of modules is extension of scalars).

[L3]

The ring $R_{p}$ is local with maximal ideal $p R_{p}$ , and its residue field is $k (p) = R_{p} / p R_{p}$ ( $R_{p}$ is local with unique maximal ideal $p R_{p}$ , $R_{p} / p R_{p} ≅ Frac (R / p)$ is the residue field at $p$ ).

[L4]

Tensoring preserves surjections, and nonzero finite-dimensional vector spaces over a field have nonzero tensor product (Tensoring is right exact, $R^{m} \otimes_{R} R^{n} ≅ R^{m n}$ with the product basis, and $\dim_{F} (V \otimes_{F} W) = \dim_{F} V \dim_{F} W$ ).

[L5]

A finitely generated module admits finite generators, and for a positive-size square matrix $A$ over a commutative ring one has $A adj (A) = \det (A) I$ (Generated submodule, cyclic and finitely generated modules, module basis and free module, For every positive-sized square matrix over a commutative ring, $A adj (A) = adj (A) A = \det (A) I$ ).

Proof

technique · direct

1.1

Because $M$ and $N$ are finite, $M \otimes_{R} N$ is finite: if $m_{1}, \dots, m_{r}$ generate $M$ and $n_{1}, \dots, n_{s}$ generate $N$ , then the tensors $m_{i} \otimes n_{j}$ generate $M \otimes_{R} N$ . If $p \in {Supp}_{R} (M \otimes_{R} N)$ , then [L1] gives ${Ann}_{R} (M \otimes_{R} N) \subseteq p$ . Every element of ${Ann}_{R} (M)$ and every element of ${Ann}_{R} (N)$ annihilates every elementary tensor, so ${Ann}_{R} (M) + {Ann}_{R} (N) \subseteq {Ann}_{R} (M \otimes_{R} N)$ . Thus $p$ contains both annihilators, and [L1] gives $p \in {Supp}_{R} (M) \cap {Supp}_{R} (N)$ .

L1givenalgebra

1.2

Conversely, let $p \in {Supp}_{R} (M) \cap {Supp}_{R} (N)$ . By [L2], it is enough to prove $M_{p} \otimes_{R_{p}} N_{p} \neq 0$ . Put $A = R_{p}$ , $m = p A$ , and $k = A / m$ ; by [L3], $A$ is a local ring with residue field $k$ .

L2L3

1.3

If $X$ is a finite nonzero $A$ -module, then $X / m X \neq 0$ . Indeed, if $X = m X$ , choose generators $x_{1}, \dots, x_{t}$ of $X$ and coefficients $a_{i j} \in m$ with $x_{i} = \sum_{j} a_{i j} x_{j}$ . Writing $B = (a_{i j})$ and $x = (x_{1}, \dots, x_{t})^{T}$ , this says $(I - B) x = 0$ . By [L5], $\det (I - B) x = 0$ . The determinant has the form $1 - a$ with $a \in m$ , so $1 - a \notin m$ and therefore is a unit in the local ring $A$ . Hence $x = 0$ and $X = 0$ , a contradiction.

L3L5algebra

2.1

Apply step 1.3 to $M_{p}$ and $N_{p}$ . Since $p$ lies in both supports, these local modules are nonzero, so the $k$ -vector spaces $M_{p} / m M_{p}$ and $N_{p} / m N_{p}$ are nonzero. Tensoring the quotient maps with [L4] gives a surjection $M_{p} \otimes_{A} N_{p} ↠ (M_{p} / m M_{p}) \otimes_{A} (N_{p} / m N_{p})$ , and the target is the same as the tensor product over $k$ , hence nonzero by [L4]. Therefore $M_{p} \otimes_{A} N_{p} \neq 0$ .

L4step 1.3

3.1

Step 2.1 and [L2] give $p \in {Supp}_{R} (M \otimes_{R} N)$ . Together with step 1.1, this proves the support-intersection formula.

L2step 1.1step 2.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

Support of an arbitrary direct sum is the union of the supports

Statement

For any family $(M_{i})_{i \in I}$ of left $R$ -modules,

{Supp}_{R} ⁣ (⨁_{i \in I} M_{i}) = ⋃_{i \in I} {Supp}_{R} (M_{i}) .

Facts & Assumptions

Given: A commutative ring $R$ and a family $(M_{i})_{i \in I}$ of left $R$ -modules.

[L1]

A prime ideal $p$ lies in the support of a module exactly when the localisation at $p$ is nonzero (Support of a module).

[L2]

Localisation commutes with arbitrary direct sums (Localisation commutes with quotient modules and arbitrary direct sums).

Proof

technique · direct

1.1

Fix a prime ideal $p$ . By [L2], ${(⨁_{i} M_{i})}_{p} ≅ ⨁_{i} (M_{i})_{p}$ . This direct sum is nonzero exactly when at least one summand $(M_{i})_{p}$ is nonzero.

2.1

By [L1], step 1.1 says exactly that $p$ lies in the support of $⨁_{i} M_{i}$ if and only if it lies in the support of some $M_{i}$ .

L1step 1.1

3.1

Since this holds for every prime ideal $p$ , the support of the direct sum is the union of the supports.

step 2.1∎

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Support under localisation is restriction to primes disjoint from the denominator set

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative, and let $M$ be a left $R$ -module. Under the prime-ideal correspondence

{p \subseteq R : p \cap S = \emptyset} ⟷ Spec (S^{- 1} R), p ⟼ S^{- 1} p,

the support of $S^{- 1} M$ corresponds exactly to

{p \in {Supp}_{R} (M) : p \cap S = \emptyset} .

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , and a left $R$ -module $M$ .

[L1]

A prime ideal lies in the support of a module exactly when the localisation at that prime is nonzero (Support of a module).

[L2]

Prime ideals of $S^{- 1} R$ correspond to prime ideals of $R$ disjoint from $S$ (Ideals of $S^{- 1} R$ correspond to $S$ -saturated ideals of $R$ , and prime ideals correspond to primes disjoint from $S$ ).

[L3]

Localising twice is the same as localising once at the multiplicative set generated by the two denominator sets (Localising twice is localising once at the multiplicative set generated by both denominator sets).

Proof

technique · direct

1.1

Let $q$ be a prime ideal of $S^{- 1} R$ , and let $p \subseteq R$ be its contraction. By [L2], $p \cap S = \emptyset$ and $q = S^{- 1} p$ .

2.1

The localisation $(S^{- 1} M)_{q}$ is obtained by localising $M$ first at $S$ and then at the complement of $q$ . By [L3], this is the same as localising $M$ once at the multiplicative subset $R ∖ p$ , so $(S^{- 1} M)_{q} ≅ M_{p}$ .

step 1.1L3

3.1

By [L1], step 2.1 gives $q \in {Supp}_{S^{- 1} R} (S^{- 1} M)$ if and only if $p \in {Supp}_{R} (M)$ . Together with step 1.1, this proves the stated description of support under localisation.

L1step 1.1step 2.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

The Jacobson radical of a ring

Definition

For a commutative ring $R$ , the Jacobson radical is

J (R) : = ⋂_{m maximal} m .

When $R = 0$ , there are no maximal ideals, and on this page the empty intersection is taken to be $0$ , so $J (0) = 0$ .

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Assuming the Axiom of Choice, an element lies in the Jacobson radical exactly when one minus any multiple is a unit

Statement

Assume the Axiom of Choice.

For a commutative ring $R$ and an element $x \in R$ ,

x \in J (R) ⟺ 1 - r x is a unit for every r \in R .

Facts & Assumptions

Given: A commutative ring $R$ and an element $x \in R$ .

[L1]

The Jacobson radical is the intersection of the maximal ideals, with $J (0) = 0$ (The Jacobson radical of a ring).

[L2]

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

[L3]

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

Proof

technique · direct

1.1

If $R = 0$ , then $J (R) = 0$ by [L1], the only element is $x = 0$ , and $1 - r x = 0$ is the identity element of the zero ring, hence a unit. So the statement holds in this case.

L1given

1.2

Assume $R \neq 0$ and first suppose $x \in J (R)$ . Fix $r \in R$ . If $1 - r x$ lay in a maximal ideal $m$ , then [L1] gives $x \in m$ , hence also $r x \in m$ , so $1 = (1 - r x) + r x \in m$ , impossible. Thus $1 - r x$ lies in no maximal ideal.

L1L3algebra

1.3

Conversely, suppose $1 - r x$ is a unit for every $r \in R$ and that $x \notin J (R)$ . By [L1], choose a maximal ideal $m$ with $x \notin m$ . Then the ideal $m + (x)$ strictly contains $m$ , so maximality from [L3] gives $m + (x) = R$ . Thus $1 = a + r x$ for some $a \in m$ and $r \in R$ , so $1 - r x = a \in m$ . But an element of a proper ideal cannot be a unit, contradicting the hypothesis. Therefore $x \in J (R)$ .

L1L3algebra

2.1

Still under the hypothesis of step 1.2, if $1 - r x$ were not a unit, then the principal ideal it generates would be proper, so [L2] would place it in a maximal ideal, contradicting step 1.2. Hence $1 - r x$ is a unit for every $r$ .

step 1.2L2

3.1

Steps 1.1, 2.1, and 1.3 prove the equivalence.

step 1.1step 2.1step 1.3∎

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Determinant trick for Nakayama

Statement

Let $R$ be a commutative ring, let $I ⊴ R$ be an ideal, and let $M$ be a finitely generated left $R$ -module. If $I M = M$ , then there exists $a \in I$ such that

(1 - a) M = 0.

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I ⊴ R$ , and a finitely generated left $R$ -module $M$ with $I M = M$ .

[L1]

A finitely generated module has a finite generating set (Generated submodule, cyclic and finitely generated modules, module basis and free module).

[L2]

The submodule $I M$ consists of finite sums of products $i m$ with $i \in I$ and $m \in M$ (The submodule $I M$ generated by products of elements of an ideal $I$ with elements of a module $M$ ).

[L3]

For a positive-size square matrix $A$ over a commutative ring, $A adj (A) = \det (A) I$ (For every positive-sized square matrix over a commutative ring, $A adj (A) = adj (A) A = \det (A) I$ ).

Proof

technique · direct

1.1

If $M = 0$ , then $a = 0 \in I$ satisfies $(1 - a) M = 0$ . So assume $M \neq 0$ and choose generators $m_{1}, \dots, m_{n}$ with $n \geq 1$ by [L1].

L1givenchoose

2.1

Since $I M = M$ , each generator has the form $m_{i} = \sum_{j} a_{i j} m_{j}$ with $a_{i j} \in I$ . Writing $A = (a_{i j})$ and $m = (m_{1}, \dots, m_{n})^{T}$ , this is $(I_{n} - A) m = 0$ .

L2step 1.1algebra

3.1

Multiply the relation of step 2.1 by $adj (I_{n} - A)$ . By [L3], this gives $\det (I_{n} - A) m = 0$ , so $\det (I_{n} - A)$ annihilates every generator and hence all of $M$ .

L3step 2.1algebra

3.2

Expanding $\det (I_{n} - A)$ , the identity permutation contributes $1$ , and every other term contains at least one entry of $A$ , hence lies in $I$ . Therefore $\det (I_{n} - A) = 1 - a$ for some $a \in I$ .

step 2.1algebra

4.1

Step 3.1 and step 3.2 give $(1 - a) M = 0$ for some $a \in I$ .

step 3.1step 3.2∎

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

Assuming the Axiom of Choice, Nakayama's lemma

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring, let $I ⊴ R$ satisfy $I \subseteq J (R)$ , and let $M$ be a finitely generated left $R$ -module. If $I M = M$ , then $M = 0$ .

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I ⊴ R$ with $I \subseteq J (R)$ , and a finitely generated left $R$ -module $M$ with $I M = M$ .

[L1]

An element $x$ lies in $J (R)$ exactly when $1 - r x$ is a unit for every $r \in R$ (Assuming the Axiom of Choice, an element lies in the Jacobson radical exactly when one minus any multiple is a unit).

[L2]

If $I M = M$ for finite $M$ , then $(1 - a) M = 0$ for some $a \in I$ (Determinant trick for Nakayama).

Proof

technique · direct

1.1

By [L2], choose $a \in I$ with $(1 - a) M = 0$ . Since $a \in I \subseteq J (R)$ , [L1] makes $1 - a$ a unit.

L1L2givenchoose

2.1

Multiplying the equality $(1 - a) m = 0$ by $(1 - a)^{- 1}$ shows $m = 0$ for every $m \in M$ . Therefore $M = 0$ .

step 1.1algebra∎

CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26Open item page →

Assuming the Axiom of Choice, generators modulo an ideal in the Jacobson radical lift to generators

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring, let $I ⊴ R$ satisfy $I \subseteq J (R)$ , and let $M$ be a finitely generated left $R$ -module. If elements $x_{1}, \dots, x_{r} \in M$ generate $M / I M$ , then $x_{1}, \dots, x_{r}$ generate $M$ .

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I ⊴ R$ with $I \subseteq J (R)$ , a finitely generated left $R$ -module $M$ , and elements $x_{1}, \dots, x_{r} \in M$ whose images generate $M / I M$ .

[L1]

If a finite module $Q$ satisfies $I Q = Q$ and $I \subseteq J (R)$ , then $Q = 0$ (Assuming the Axiom of Choice, Nakayama's lemma).

[L2]

The submodule $I M$ consists of finite sums of products $i m$ with $i \in I$ and $m \in M$ (The submodule $I M$ generated by products of elements of an ideal $I$ with elements of a module $M$ ).

[L3]

A finite list of elements generates the submodule it spans (Generated submodule, cyclic and finitely generated modules, module basis and free module).

Proof

technique · direct

1.1

Let $N$ be the submodule of $M$ generated by $x_{1}, \dots, x_{r}$ . The hypothesis on $M / I M$ means every element of $M$ is congruent modulo $I M$ to an element of $N$ , so $M = N + I M$ .

L2L3given

2.1

Passing to the quotient $Q = M / N$ , step 1.1 gives $I Q = Q$ . Since $Q$ is a quotient of the finite module $M$ , it is finite, so [L1] gives $Q = 0$ .

step 1.1L1

3.1

The equality $Q = 0$ means $M = N$ , so $x_{1}, \dots, x_{r}$ generate $M$ .

step 2.1L3∎

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Assuming the Axiom of Choice, minimal generators over a local ring are exactly residue-field bases

Statement

Assume the Axiom of Choice.

Let $(R, m)$ be a local ring with residue field $k = R / m$ , and let $M$ be a finitely generated left $R$ -module. A finite generating set $x_{1}, \dots, x_{r}$ of $M$ is minimal if and only if the images of $x_{1}, \dots, x_{r}$ in $M / m M$ form a $k$ -basis. In particular every minimal generating set of $M$ has the same cardinality.

Facts & Assumptions

Given: A local ring $(R, m)$ , its residue field $k = R / m$ , a finitely generated left $R$ -module $M$ , and elements $x_{1}, \dots, x_{r} \in M$ .

[L1]

If elements generate $M / m M$ , then they generate $M$ (Assuming the Axiom of Choice, generators modulo an ideal in the Jacobson radical lift to generators).

[L2]

A local ring is a nonzero commutative ring with a unique maximal ideal, and its residue field is the quotient by that maximal ideal (A local ring is a nonzero commutative ring with a unique maximal ideal).

[L3]

The submodule $m M$ consists of finite sums of products $a m$ with $a \in m$ and $m \in M$ (The submodule $I M$ generated by products of elements of an ideal $I$ with elements of a module $M$ ).

Proof

technique · direct

1.1

If the images of $x_{1}, \dots, x_{r}$ span $M / m M$ , then [L1] gives that $x_{1}, \dots, x_{r}$ generate $M$ .

L1L2L3

1.2

Suppose the images of $x_{1}, \dots, x_{r}$ are linearly dependent over $k$ . Then there are $a_{1}, \dots, a_{r} \in R$ , not all in $m$ , with $\sum_{i} a_{i} x_{i} \in m M$ . Choose $j$ with $a_{j} \notin m$ . Since $R$ is local, the ideal $m + (a_{j})$ properly contains $m$ , so by maximality it is all of $R$ ; choose $b \in R$ and $c \in m$ with $1 = b a_{j} + c$ . Multiplying the relation by $b$ shows $x_{j}$ lies in the submodule generated by the other $x_{i}$ together with $m M$ . Therefore the other $r - 1$ elements generate $M / m M$ , so [L1] makes them generate $M$ . Thus the original generating set was not minimal.

L1L2L3choose

1.3

Conversely, if $x_{1}, \dots, x_{r}$ generate $M$ but are not minimal, then some $x_{j}$ lies in the submodule generated by the other $x_{i}$ . Passing to $M / m M$ shows that the image of $x_{j}$ lies in the $k$ -span of the other images, so the images are linearly dependent.

L2L3algebra

2.1

Therefore $x_{1}, \dots, x_{r}$ is a minimal generating set of $M$ if and only if its images form a $k$ -basis of $M / m M$ . Any two minimal generating sets give two bases of the same $k$ -vector space, so they have the same cardinality.

step 1.1step 1.2step 1.3∎

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A finite module that vanishes at a prime vanishes on some principal neighbourhood of that prime

Statement

Let $M$ be a finitely generated left $R$ -module and let $p$ be a prime ideal of $R$ . If $M_{p} = 0$ , then there exists $s \notin p$ such that the localisation of $M$ at the multiplicative set

S_{s} : = {1, s, s^{2}, \dots}

is zero.

Facts & Assumptions

Given: A commutative ring $R$ , a finitely generated left $R$ -module $M$ , and a prime ideal $p$ with $M_{p} = 0$ .

[L1]

For a finite module, the support is the set of primes containing its annihilator (For a finite module, support is the set of primes containing the annihilator).

[L2]

The support condition $p \in {Supp}_{R} (M)$ means $M_{p} \neq 0$ (Support of a module).

[L3]

In a localisation, each denominator becomes a unit (The localisation relation is an equivalence relation and fraction arithmetic is well defined).

Proof

technique · direct

1.1

Since $M_{p} = 0$ , [L2] says $p \notin {Supp}_{R} (M)$ . By [L1], ${Ann}_{R} (M) ⊈ p$ , so choose $s \in {Ann}_{R} (M) ∖ p$ .

L1L2choose

2.1

Let $S_{s} = {1, s, s^{2}, \dots}$ . In the localisation $S_{s}^{- 1} M$ , the element $s / 1 \in S_{s}^{- 1} R$ is a unit by [L3], and it annihilates every element because $s$ annihilates all of $M$ . Therefore $S_{s}^{- 1} M = 0$ .

step 1.1L3algebra∎

5 · Examples, counterexamples and false statements

None yet.