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✓ 30 results · all verified · 8 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 22 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Tensor Products of Modules

1 · Prerequisites

2 · Summary

Left and right modules over a ring, module homomorphisms with their kernels, images and cokernels, quotient modules and the universal property of a quotient are published, as are the free module on a set with its universal property, direct sums of families with theirs, Hom groups and the maps induced on them, exact and short exact sequences, the characterisations of a projective module, the fact that free modules are projective together with its choice boundary, and injective modules. Linear algebra supplies the standard basis of $F^{n}$ , dimension, the algebraic dual with its dual basis in finite dimension, and the trace of an endomorphism; ideals, ring homomorphisms and quotient groups supply the rest.

For a right module $M$ and a left module $N$ over an arbitrary ring $R$ , the page constructs $M \otimes_{R} N$ by generators and relations, proves it represents the balanced maps out of $M \times N$ , and derives uniqueness up to unique isomorphism. The universal property then yields functoriality, the criterion for a formula on elementary tensors to descend, bimodule actions, associativity, unit isomorphisms, compatibility with arbitrary direct sums, bases and dimension, internal Hom, the trace as a contraction, algebras and their tensor products, and the Hom–tensor adjunction; the module structure and symmetry over a commutative ring are given separately. Right exactness, the identification of $M \otimes_{R} R / I$ with $M / I M$ , flat and faithfully flat modules with the injection and ideal tests, flatness of projectives, restriction and extension of scalars with their adjunction and change-of-rings results, and the injectivity of character duals of flat modules close the page.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

Balanced maps from a right module and a left module, and bilinear maps over a commutative ring

Definition

Let $R$ be a unital ring, let $M$ be a right $R$ -module, let $N$ be a left $R$ -module, and let $A$ be an abelian group, written additively (Unital left and right modules over a ring; unqualified module means left module, Group and abelian group). A map $b : M \times N \to A$ is $R$ -balanced if, for all $m, m^{'} \in M$ , $n, n^{'} \in N$ , and $r \in R$ ,

$b (m + m^{'}, n) = b (m, n) + b (m^{'}, n), b (m, n + n^{'}) = b (m, n) + b (m, n^{'}),$

and

$b (m r, n) = b (m, r n) .$

Thus a balanced map is additive in each variable and identifies the two ways in which a scalar may cross the pair. Additivity implies $b (0, n) = 0 = b (m, 0)$ .

If $R$ is commutative (Commutative ring) and $M, N, P$ are $R$ -modules, a map $b : M \times N \to P$ is $R$ -bilinear if it is $R$ -linear in each variable. Equivalently, it is additive in each variable and satisfies

$b (r m, n) = r b (m, n) = b (m, r n)$

for all $r, m, n$ . After a commutative-ring module is regarded as a right module by $m r : = r m$ , every bilinear map is balanced.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-16Open item page →

The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums

Definition

Let $R$ be a unital ring, $M$ a right $R$ -module, and $N$ a left $R$ -module. Let

$F : = Z^{(M \times N)}$

be the free $Z$ -module on the set $M \times N$ (The free module on a set and its standard basis, Universal property of the free module on a set), and write $e_{(m, n)}$ for its standard basis elements. The additive group of $F$ is abelian. Let $H$ be the subgroup generated (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups) by all elements

$e_{(m + m^{'}, n)} - e_{(m, n)} - e_{(m^{'}, n)},$

$e_{(m, n + n^{'})} - e_{(m, n)} - e_{(m, n^{'})},$

and

$e_{(m r, n)} - e_{(m, r n)}$

as $m, m^{'}$ range over $M$ , $n, n^{'}$ over $N$ , and $r$ over $R$ . Since every subgroup of an abelian group is normal, the quotient group $F / H$ is defined (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ). The tensor product of $M$ and $N$ over $R$ is

$M \otimes_{R} N : = F / H .$

The coset of $e_{(m, n)}$ is the elementary tensor $m \otimes n$ . Every tensor is a finite sum of elementary tensors, because every element of $F$ is a finite $Z$ -linear combination of basis elements and integer coefficients may be absorbed into either additive variable. The defining relations give

$(m + m^{'}) \otimes n = m \otimes n + m^{'} \otimes n, m \otimes (n + n^{'}) = m \otimes n + m \otimes n^{'},$

and

$(m r) \otimes n = m \otimes (r n) .$

In particular $0 \otimes n = 0 = m \otimes 0$ . No $R$ -module structure on $M \otimes_{R} N$ is part of this arbitrary-ring definition; at this stage it is an abelian group.

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

Universal property of the tensor product for balanced maps into abelian groups

Statement

Let $R$ be a unital ring, $M$ a right $R$ -module, $N$ a left $R$ -module, and

$τ : M \times N ⟶ M \otimes_{R} N, τ (m, n) = m \otimes n .$

The map $τ$ is balanced (Balanced maps from a right module and a left module, and bilinear maps over a commutative ring). For every abelian group $A$ and every balanced map $b : M \times N \to A$ , there is a unique group homomorphism

$\overline{b} : M \otimes_{R} N ⟶ A$

such that $\overline{b} (m \otimes n) = b (m, n)$ for all $m, n$ . Consequently composition with $τ$ is a bijection

${Hom}_{A b} (M \otimes_{R} N, A) ≅ {Bal}_{R} (M, N; A) .$

Facts & Assumptions

Given: A unital ring $R$ , a right $R$ -module $M$ , a left $R$ -module $N$ , an abelian group $A$ , and a balanced map $b : M \times N \to A$ .

[L1]

The tensor product is $F / H$ , where $F = Z^{(M \times N)}$ , $H$ is generated by the additivity and balance relations, and $m \otimes n = e_{(m, n)} + H$ (The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums).

[L2]

A balanced map is additive in each variable and satisfies $b (m r, n) = b (m, r n)$ (Balanced maps from a right module and a left module, and bilinear maps over a commutative ring).

[L3]

Every element of $Z^{(X)}$ has a unique finite expression $\sum_{x \in E} k_{x} e_{x}$ with $k_{x} \in Z$ (The free module on a set and its standard basis).

[L4]

Every set map $u : X \to P$ into a left $S$ -module extends uniquely to an $S$ -module homomorphism $S^{(X)} \to P$ taking $e_{x}$ to $u (x)$ (Universal property of the free module on a set).

[L5]

If a group homomorphism $f : G \to K$ kills a normal subgroup $H$ , then it factors uniquely through a group homomorphism $G / H \to K$ (A homomorphism that kills a normal subgroup factors uniquely through the quotient group).

Proof

technique · direct

1.1

Regard $A$ as a $Z$ -module by integer multiplication in its additive group. A group homomorphism between abelian groups is automatically $Z$ -linear: additivity gives $f (n a) = n f (a)$ for $n \geq 0$ , and $f (- a) = - f (a)$ gives the formula for negative integers. Thus $Z$ -module homomorphisms and group homomorphisms between these underlying additive groups are the same maps.

givenalgebra

1.2

If $h : M \otimes_{R} N \to A$ is a group homomorphism, then $(m, n) \mapsto h (m \otimes n)$ is balanced because the elementary tensors satisfy all three relations in [L1].

L1algebra

2.1

Apply [L4] at $S = Z$ and $X = M \times N$ to extend the set map $b$ uniquely to a $Z$ -linear map $\tilde{b} : F \to A$ satisfying $\tilde{b} (e_{(m, n)}) = b (m, n)$ .

L3L4step 1.1

3.1

By [L2], $\tilde{b}$ sends each generator of $H$ to zero: the two additivity generators map respectively to $b (m + m^{'}, n) - b (m, n) - b (m^{'}, n)$ and $b (m, n + n^{'}) - b (m, n) - b (m, n^{'})$ , while the balance generator maps to $b (m r, n) - b (m, r n)$ . Hence $H \subseteq \ker \tilde{b}$ .

L1L2step 2.1algebra

4.1

By [L5], $\tilde{b}$ factors uniquely through a group homomorphism $\overline{b} : F / H = M \otimes_{R} N \to A$ , and $\overline{b} (m \otimes n) = \tilde{b} (e_{(m, n)}) = b (m, n)$ .

L1L5step 2.1step 3.1

5.1

The operations in steps 4.1 and 1.2 are inverse: starting from $b$ recovers its values on every pair, while starting from $h$ produces a homomorphism agreeing with $h$ on every elementary tensor, and those tensors generate $M \otimes_{R} N$ . This proves both the asserted uniqueness and the displayed bijection.

step 4.1step 1.2L1

6.1

No selection is made in the construction. If $M = 0$ or $N = 0$ , every balanced map out of $M \times N$ is zero and the relations make every elementary tensor zero, so the same proof gives $M \otimes_{R} N = 0$ ; the zero ring is covered as well. Since a module contains its zero element, $M \times N$ is never empty, so there is no separate empty-domain case.

L1L2step 5.1∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

Tensor products are unique up to a unique isomorphism carrying elementary tensors to elementary tensors

Statement

Let $T, T^{'}$ be abelian groups equipped with balanced maps $τ : M \times N \to T$ and $τ^{'} : M \times N \to T^{'}$ , and suppose that each pair has the universal property of Universal property of the tensor product for balanced maps into abelian groups. Then there is a unique group isomorphism $u : T \to T^{'}$ such that $u \circ τ = τ^{'}$ . Its inverse is the unique map $v : T^{'} \to T$ with $v \circ τ^{'} = τ$ .

Facts & Assumptions

Given: Two representing pairs $(T, τ)$ and $(T^{'}, τ^{'})$ for balanced maps out of $M \times N$ .

[L1]

For any balanced map from $M \times N$ into an abelian group, a representing pair supplies a unique group homomorphism through which that map factors (Universal property of the tensor product for balanced maps into abelian groups).

Proof

technique · direct

1.1

Apply [L1] for $(T, τ)$ to the balanced map $τ^{'}$ and for $(T^{'}, τ^{'})$ to $τ$ ; this gives unique homomorphisms $u : T \to T^{'}$ and $v : T^{'} \to T$ with $u τ = τ^{'}$ and $v τ^{'} = τ$ .

givenL1

2.1

Both $v \circ u$ and ${id}_{T}$ compose with $τ$ to give $τ$ , so uniqueness in [L1] gives $v \circ u = {id}_{T}$ ; similarly $u \circ v = {id}_{T^{'}}$ .

step 1.1L1

3.1

Thus $u$ is an isomorphism with inverse $v$ , and the same uniqueness clause shows that no other isomorphism carrying $τ$ to $τ^{'}$ exists.

step 2.1L1∎

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

Module homomorphisms induce tensor-product homomorphisms functorially

Statement

Let $f : M \to M^{'}$ be a homomorphism of right $R$ -modules and $g : N \to N^{'}$ a homomorphism of left $R$ -modules. There is a unique group homomorphism

$f \otimes g : M \otimes_{R} N ⟶ M^{'} \otimes_{R} N^{'}$

such that $(f \otimes g) (m \otimes n) = f (m) \otimes g (n)$ . These maps satisfy

${id}_{M} \otimes {id}_{N} = {id}_{M \otimes_{R} N}$

and

$(f^{'} \circ f) \otimes (g^{'} \circ g) = (f^{'} \otimes g^{'}) \circ (f \otimes g) .$

Facts & Assumptions

Given: Homomorphisms $f : M \to M^{'}$ of right $R$ -modules and $g : N \to N^{'}$ of left $R$ -modules.

[L1]

A balanced map $M \times N \to A$ into an abelian group extends uniquely to a group homomorphism $M \otimes_{R} N \to A$ (Universal property of the tensor product for balanced maps into abelian groups).

[L2]

A module homomorphism preserves addition and the relevant scalar action (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct

1.1

The pairing $(m, n) \mapsto f (m) \otimes g (n)$ is additive in both variables by [L2], and $(f (m r)) \otimes g (n) = (f (m) r) \otimes g (n) = f (m) \otimes (r g (n)) = f (m) \otimes g (r n)$ , so it is balanced.

givenL2algebra

2.1

By [L1] the pairing of step 1.1 induces a unique homomorphism $f \otimes g$ with the stated formula.

step 1.1L1

3.1

The maps ${id}_{M} \otimes {id}_{N}$ and ${id}_{M \otimes_{R} N}$ agree on every elementary tensor, so uniqueness in [L1] makes them equal.

step 2.1L1

3.2

The two sides of the composition formula both send $m \otimes n$ to $f^{'} (f (m)) \otimes g^{'} (g (n))$ , so uniqueness in [L1] makes them equal.

step 2.1L1

4.1

Steps 2.1, 3.1 and 3.2 prove existence, uniqueness, identity preservation, and composition preservation.

step 2.1step 3.1step 3.2∎

PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-16Open item page →

A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced

Statement

Let $A$ be an abelian group and let $q : M \times N \to A$ be a function. A prescription

$Q (m \otimes n) : = q (m, n)$

extends to a group homomorphism $Q : M \otimes_{R} N \to A$ if and only if $q$ is balanced. When it exists, the extension is unique.

The balance condition cannot be replaced by a check on tensor symbols alone. In $Z \otimes_{Z} Z$ , the prescription $q (m, n) = m$ is not balanced and does not descend: the relation $2 \otimes 1 = 1 \otimes 2$ would force its value to be both $2$ and $1$ .

Facts & Assumptions

Given: A function $q : M \times N \to A$ into an abelian group.

[L1]

Composition with the elementary-tensor map is a bijection from group homomorphisms $M \otimes_{R} N \to A$ to balanced maps $M \times N \to A$ (Universal property of the tensor product for balanced maps into abelian groups).

[L2]

In the tensor product, $(m r) \otimes n = m \otimes (r n)$ and the elementary-tensor map is additive in each variable (The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums).

[L3]

The integers form a commutative unital ring (The integers form a commutative ring).

Proof

technique · direct

1.1

If $q$ is balanced, [L1] supplies a unique group homomorphism $Q$ satisfying $Q (m \otimes n) = q (m, n)$ .

givenL1

1.2

Conversely, if such a homomorphism $Q$ exists, composing it with the elementary-tensor map gives $q$ ; [L2] and additivity of $Q$ show that $q$ is additive in each variable and satisfies $q (m r, n) = q (m, r n)$ , so $q$ is balanced.

givenL2algebra

1.3

For $R = M = N = Z$ , which is permitted by [L3], balance gives $2 \otimes 1 = 1 \otimes 2$ by [L2]. The function $q (m, n) = m$ assigns $2$ to $(2, 1)$ and $1$ to $(1, 2)$ , so it is not balanced and no homomorphism can have the proposed elementary-tensor values.

L2L3algebra

2.1

Steps 1.1 and 1.2 prove the equivalence and uniqueness, while step 1.3 verifies the asserted failure.

step 1.1step 1.2step 1.3∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

$(S, R)$ -bimodules and commuting left and right scalar actions

Definition

Let $S$ and $R$ be unital rings. An $(S, R)$ -bimodule is an abelian group $N$ that is a left $S$ -module and a right $R$ -module (Unital left and right modules over a ring; unqualified module means left module) such that the two actions commute:

$(s n) r = s (n r)$

for every $s \in S$ , $n \in N$ , and $r \in R$ . It is denoted $_{S} N_{R}$ when the rings need to be displayed.

Every ring $R$ is an $(R, R)$ -bimodule by left and right multiplication. If $R$ is commutative, every left $R$ -module becomes an $(R, R)$ -bimodule by defining $m r : = r m$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

A commuting outer scalar action descends to a tensor product

Statement

Let $M$ be a right $R$ -module and let $_{R} N_{S}$ be an $(R, S)$ -bimodule. There is a unique right $S$ -module structure on $M \otimes_{R} N$ satisfying

$(m \otimes n) s = m \otimes (n s) .$

Dually, if $_{S} M_{R}$ is an $(S, R)$ -bimodule and $N$ is a left $R$ -module, there is a unique left $S$ -module structure satisfying

$s (m \otimes n) = (s m) \otimes n .$

If both outer actions are present, they commute, so the tensor product is an $(S, T)$ -bimodule in the evident handed situation.

Facts & Assumptions

Given: A right $R$ -module $M$ , an $(R, S)$ -bimodule $N$ , and, for the dual assertion, an $(S, R)$ -bimodule $M^{'}$ and a left $R$ -module $N^{'}$ .

[L1]

In a bimodule the left and right scalar actions commute: $(r n) s = r (n s)$ ( $(S, R)$ -bimodules and commuting left and right scalar actions).

[L2]

Every balanced pairing into an abelian group induces a unique homomorphism from the tensor product (Universal property of the tensor product for balanced maps into abelian groups).

[L3]

An elementary-tensor prescription descends exactly when the corresponding pairing is balanced (A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

Proof

technique · direct

1.1

For fixed $s \in S$ , the pairing $(m, n) \mapsto m \otimes (n s)$ is additive in both variables and is balanced because $(m r) \otimes (n s) = m \otimes r (n s) = m \otimes (r n) s$ by [L1].

givenL1algebra

1.2

The left-handed construction is identical: for fixed $s \in S$ , the pairing $(m^{'}, n^{'}) \mapsto (s m^{'}) \otimes n^{'}$ is balanced by the commuting actions, and the induced maps satisfy the left module laws on elementary tensors.

L1L2L3algebra

2.1

By [L2] and [L3], step 1.1 induces an additive endomorphism $x \mapsto x s$ of $M \otimes_{R} N$ with $(m \otimes n) s = m \otimes (n s)$ .

step 1.1L2L3

3.1

On every elementary tensor one has $(x (s + s^{'})) = x s + x s^{'}$ , $(x s) s^{'} = x (s s^{'})$ , and $x 1_{S} = x$ , using the right $S$ -module laws of $N$ . In each identity the two sides are additive maps that induce the same balanced pairing, so uniqueness in [L2] makes them equal on all $x$ . Thus step 2.1 defines a right $S$ -module structure.

step 2.1L2algebra

4.1

When a left $S$ -action and a right $T$ -action are both present, $s ((m \otimes n) t) = (s m) \otimes (n t) = (s (m \otimes n)) t$ on elementary tensors, so the actions commute.

step 3.1step 1.2algebra

5.1

The formulas determine every action map on generators, so uniqueness follows from [L2].

step 2.1step 1.2L2∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

Associativity of tensor products for compatible bimodules

Statement

Let $M$ be a right $R$ -module, let $_{R} N_{S}$ be an $(R, S)$ -bimodule, and let $P$ be a left $S$ -module. There is a canonical group isomorphism

$α_{M, N, P} : (M \otimes_{R} N) \otimes_{S} P ⟶ M \otimes_{R} (N \otimes_{S} P)$

determined by

$α_{M, N, P} ((m \otimes n) \otimes p) = m \otimes (n \otimes p) .$

It respects any compatible outer module actions and is natural in $M, N, P$ .

Facts & Assumptions

Given: A right $R$ -module $M$ , an $(R, S)$ -bimodule $N$ , and a left $S$ -module $P$ .

[L1]

The outer actions make $M \otimes_{R} N$ a right $S$ -module and $N \otimes_{S} P$ a left $R$ -module, with the stated elementary-tensor formulas (A commuting outer scalar action descends to a tensor product).

[L2]

Balanced pairings induce unique homomorphisms from tensor products (Universal property of the tensor product for balanced maps into abelian groups).

[L3]

A formula on elementary tensors descends exactly when its underlying pairing is balanced (A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

Proof

technique · direct

1.1

For fixed $p \in P$ , the pairing $(m, n) \mapsto m \otimes (n \otimes p)$ is $R$ -balanced: $(m r, n)$ and $(m, r n)$ have the same image by the outer left $R$ -action in [L1]. Thus [L2] and [L3] give a homomorphism $a_{p} : M \otimes_{R} N \to M \otimes_{R} (N \otimes_{S} P)$ .

givenL1L2L3

1.2

For fixed $m \in M$ , the pairing $(n, p) \mapsto (m \otimes n) \otimes p$ is $S$ -balanced, so [L2] and [L3] give $b_{m} : N \otimes_{S} P \to (M \otimes_{R} N) \otimes_{S} P$ .

givenL1L2L3

2.1

The pairing $(x, p) \mapsto a_{p} (x)$ from $(M \otimes_{R} N) \times P$ is $S$ -balanced. On generators, $a_{p} ((m \otimes n) s) = m \otimes ((n s) \otimes p) = m \otimes (n \otimes s p) = a_{s p} (m \otimes n)$ by [L1], and additivity extends the equality to every $x$ . Hence [L2] gives $α_{M, N, P}$ with the displayed formula.

step 1.1L1L2L3

2.2

The pairing $(m, y) \mapsto b_{m} (y)$ is $R$ -balanced. It is enough to check generators $y = n \otimes p$ , where $b_{m r} (n \otimes p) = ((m r) \otimes n) \otimes p = (m \otimes r n) \otimes p = b_{m} ((r n) \otimes p) = b_{m} (r (n \otimes p))$ by [L1]. Therefore [L2] gives $β : M \otimes_{R} (N \otimes_{S} P) \to (M \otimes_{R} N) \otimes_{S} P$ .

step 1.2L1L2L3

3.1

The composite $β α$ fixes every tensor $(m \otimes n) \otimes p$ , and $α β$ fixes every tensor $m \otimes (n \otimes p)$ ; successive applications of uniqueness in [L2] show that these composites are the identity maps.

step 2.1step 2.2L2

3.2

The same elementary-tensor calculation shows compatibility with any outer actions, and replacing $m, n, p$ by their images under compatible homomorphisms proves naturality because both candidate composites agree on all elementary tensors.

step 2.1step 2.2L1L2

4.1

Thus $α_{M, N, P}$ is the asserted canonical natural isomorphism with inverse $β$ .

step 3.1step 3.2∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

Over a commutative ring, $M \otimes_{R} N$ is an $R$ -module with $r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$

Statement

Let $R$ be a commutative ring and let $M, N$ be $R$ -modules. The abelian group $M \otimes_{R} N$ has a unique $R$ -module structure for which

$r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$

for every $r \in R$ , $m \in M$ , and $n \in N$ .

Facts & Assumptions

Given: A commutative ring $R$ and $R$ -modules $M, N$ , each regarded on either side by the common scalar action.

[L1]

Balanced pairings induce unique homomorphisms from tensor products (Universal property of the tensor product for balanced maps into abelian groups).

[L2]

In a commutative ring, $r s = s r$ for all $r, s \in R$ (Commutative ring).

[L3]

An elementary-tensor formula descends exactly when its underlying pairing is balanced (A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

Proof

technique · direct

1.1

For fixed $r \in R$ , the pairing $(m, n) \mapsto (r m) \otimes n$ is additive in both variables and balanced: $(r (s m)) \otimes n = ((r s) m) \otimes n = ((s r) m) \otimes n = (r m) \otimes (s n)$ by [L2].

givenL2algebra

2.1

By [L1] and [L3], step 1.1 induces an additive endomorphism $x \mapsto r x$ of $M \otimes_{R} N$ satisfying $r (m \otimes n) = (r m) \otimes n$ . The tensor balance relation also gives $(r m) \otimes n = m \otimes (r n)$ .

step 1.1L1L3

3.1

On elementary tensors, $(r + s) x = r x + s x$ , $r (x + y) = r x + r y$ , $(r s) x = r (s x)$ , and $1_{R} x = x$ follow from the module axioms of $M$ . In each identity the two additive maps induce the same balanced pairing, so uniqueness in [L1] makes the identity hold for every tensor.

step 2.1L1algebra

3.2

Any $R$ -module structure with the displayed formula has, for each $r$ , a scalar-multiplication endomorphism inducing the same balanced pairing as step 2.1. Uniqueness in [L1] therefore makes the structure unique.

step 2.1L1

4.1

Steps 2.1, 3.1 and 3.2 give the asserted unique $R$ -module structure and both elementary-tensor formulas.

step 2.1step 3.1step 3.2∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

Symmetry and associativity isomorphisms for tensor products over a commutative ring

Statement

Let $R$ be a commutative ring and let $L, M, N$ be $R$ -modules. There are natural $R$ -module isomorphisms

$σ_{M, N} : M \otimes_{R} N ⟶ N \otimes_{R} M, σ_{M, N} (m \otimes n) = n \otimes m,$

and

$α_{L, M, N} : (L \otimes_{R} M) \otimes_{R} N ⟶ L \otimes_{R} (M \otimes_{R} N), α ((l \otimes m) \otimes n) = l \otimes (m \otimes n) .$

Moreover $σ_{N, M} σ_{M, N}$ is the identity.

Facts & Assumptions

Given: A commutative ring $R$ and $R$ -modules $L, M, N$ .

[L1]

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

[L2]

Compatible bimodules have the canonical associativity isomorphism carrying $(l \otimes m) \otimes n$ to $l \otimes (m \otimes n)$ (Associativity of tensor products for compatible bimodules).

[L3]

Balanced pairings induce unique homomorphisms from tensor products (Universal property of the tensor product for balanced maps into abelian groups).

[L4]

Tensor maps induced by module homomorphisms preserve identities and compositions (Module homomorphisms induce tensor-product homomorphisms functorially).

Proof

technique · direct

1.1

The pairing $(m, n) \mapsto n \otimes m$ is balanced because $(r m, n)$ maps to $n \otimes (r m) = (r n) \otimes m$ , which is also the image of $(m, r n)$ ; hence [L3] induces $σ_{M, N}$ .

givenL1L3

1.2

Regard all three modules as $(R, R)$ -bimodules. Then [L2] supplies the displayed associativity isomorphism, and [L1] shows it is $R$ -linear on elementary tensors.

givenL1L2

2.1

Applying the same construction in the opposite order gives $σ_{N, M}$ , and their composite fixes every $m \otimes n$ ; uniqueness in [L3] makes the composite the identity, so $σ_{M, N}$ is an isomorphism.

step 1.1L3

2.2

The map $σ_{M, N}$ is $R$ -linear because $σ (r (m \otimes n)) = σ ((r m) \otimes n) = n \otimes (r m) = r (n \otimes m)$ .

step 1.1L1

2.3

Naturality of both maps follows by applying [L4]: after replacing the variables by their images under module homomorphisms, the two candidate composites agree on every elementary tensor, hence agree by [L3].

L3L4step 1.1step 1.2

3.1

Steps 1.1 through 2.3 prove the two natural $R$ -linear isomorphisms and the involutivity of symmetry.

step 1.1step 2.1step 2.2step 1.2step 2.3∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$

Statement

Let $R$ be a unital ring. For every left $R$ -module $N$ and every right $R$ -module $M$ , the maps

$λ_{N} : R \otimes_{R} N ⟶ N, r \otimes n ⟼ r n,$

and

$ρ_{M} : M \otimes_{R} R ⟶ M, m \otimes r ⟼ m r,$

are group isomorphisms. Their inverses are $n \mapsto 1_{R} \otimes n$ and $m \mapsto m \otimes 1_{R}$ , respectively. The maps respect every displayed outer module structure.

Facts & Assumptions

Given: A unital ring $R$ , a left $R$ -module $N$ , and a right $R$ -module $M$ .

[L1]

Balanced pairings induce unique homomorphisms from tensor products (Universal property of the tensor product for balanced maps into abelian groups).

[L2]

An elementary-tensor prescription descends exactly when the corresponding pairing is balanced (A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

Proof

technique · direct

1.1

The pairing $(r, n) \mapsto r n$ is additive in each variable and satisfies $(r s) n = r (s n)$ , so it is balanced; [L1] and [L2] give $λ_{N}$ .

givenL1L2

1.2

The pairing $(m, r) \mapsto m r$ is balanced, so [L1] and [L2] give $ρ_{M}$ . Define the additive map $η_{M} (m) = m \otimes 1_{R}$ . Then $ρ_{M} η_{M} (m) = m$ , while $η_{M} ρ_{M} (m \otimes r) = m r \otimes 1_{R} = m \otimes r$ by balance; uniqueness in [L1] makes the second composite the identity. Thus $ρ_{M}$ and $η_{M}$ are inverse.

L1L2algebra

2.1

The additive map $η_{N} : N \to R \otimes_{R} N$ , $n \mapsto 1_{R} \otimes n$ , satisfies $λ_{N} η_{N} (n) = n$ , while $η_{N} λ_{N} (r \otimes n) = 1_{R} \otimes r n = r \otimes n$ by balance. Uniqueness in [L1] makes the second composite the identity, so $λ_{N}$ and $η_{N}$ are inverse.

step 1.1L1algebra

3.1

Each map commutes with any outer scalar action by associativity of that action, checked on elementary tensors. The calculations remain valid for the zero ring and for zero modules, where all displayed maps are the unique maps between zero groups.

step 1.1step 2.1step 1.2algebra

4.1

Therefore the regular module is a left and right tensor unit with the stated natural formulas.

step 2.1step 1.2step 3.1∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

Finite iterated tensor products represent multilinear maps independently of parenthesization

Statement

Let $R$ be a commutative ring and let $M_{1}, \dots, M_{k}$ be a finite list of $R$ -modules. Any parenthesized tensor product

$T = M_{1} \otimes_{R} \dots \otimes_{R} M_{k}$

represents $R$ -multilinear maps from $M_{1} \times \dots \times M_{k}$ : for every $R$ -module $P$ , composition with $(m_{1}, \dots, m_{k}) \mapsto m_{1} \otimes \dots \otimes m_{k}$ is a bijection from ${Hom}_{R} (T, P)$ to the set of multilinear maps into $P$ . Different parenthesizations are connected by the unique isomorphism preserving pure tensors.

For $k = 0$ , take $T = R$ and identify zero-variable multilinear maps with chosen elements of $P$ . For $k = 1$ , take $T = M_{1}$ .

Facts & Assumptions

Given: A commutative ring $R$ , a finite list of $R$ -modules, and an $R$ -module $P$ .

[L1]

The binary tensor product represents balanced, hence over a commutative ring bilinear, maps (Universal property of the tensor product for balanced maps into abelian groups).

[L2]

Tensor products over a commutative ring have canonical symmetry and associativity isomorphisms preserving elementary tensors (Symmetry and associativity isomorphisms for tensor products over a commutative ring).

[L3]

The regular module is a tensor unit (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

Proof

technique · induction

1.1

For $k = 0$ , an $R$ -linear map $R \to P$ is uniquely determined by the image of $1_{R}$ , and every $p \in P$ defines such a map by $r \mapsto r p$ ; this is the required representation of maps from the one-point empty product.

baseL3

1.2

For $k = 1$ , the identity $M_{1} \to M_{1}$ represents linear maps from $M_{1}$ by composition.

base

1.3

Assume a parenthesized product $T_{k}$ represents $k$ -linear maps. A $(k + 1)$ -linear map is equivalently a bilinear map $T_{k} \times M_{k + 1} \to P$ : first use the induction bijection with the last variable fixed, and then use multilinearity to see that the resulting dependence on the last variable is linear.

ihL1

2.1

By [L1], the bilinear maps in step 1.3 correspond uniquely to linear maps $T_{k} \otimes_{R} M_{k + 1} \to P$ , proving the representing property for $k + 1$ .

step 1.3L1

3.1

By [L2], any two parenthesizations are joined by composites of elementary associativity isomorphisms preserving pure tensors. Any two such comparison maps agree on every pure tensor, so the representing uniqueness proved in step 2.1 makes them equal.

step 2.1L2

4.1

The base cases and induction step establish the representation for every finite $k$ , including the empty and singleton cases, and step 3.1 proves independence of parenthesization.

step 1.1step 1.2step 2.1step 3.1discharge-induction∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

Tensor products commute with arbitrary direct sums

Statement

Let $R$ be a commutative ring, let $(M_{i})_{i \in I}$ be any family of $R$ -modules, and let $N$ be an $R$ -module. The homomorphism induced by the coordinate inclusions is a natural $R$ -module isomorphism

$Φ : ⨁_{i \in I} (M_{i} \otimes_{R} N) ⟶ (⨁_{i \in I} M_{i}) \otimes_{R} N,$

$Φ (ȷ_{i} (m \otimes n)) = ȷ_{i} (m) \otimes n .$

The same holds with the direct sum in the second variable: there is a natural $R$ -module isomorphism

$Φ^{'} : ⨁_{i \in I} (N \otimes_{R} M_{i}) ⟶ N \otimes_{R} (⨁_{i \in I} M_{i}), Φ^{'} (ȷ_{i} (n \otimes m)) = n \otimes ȷ_{i} (m) .$

The assertion includes $I = \emptyset$ , when both sides are zero.

Facts & Assumptions

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

[L1]

Balanced pairings induce unique homomorphisms from tensor products (Universal property of the tensor product for balanced maps into abelian groups).

[L2]

Tensor products over a commutative ring carry their canonical $R$ -module structure (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 direct sum consists of finite-support tuples, with coordinate inclusions $ȷ_{i}$ , and the empty direct sum is zero (The direct sum of an indexed family of modules).

[L4]

Every family of homomorphisms $f_{i} : M_{i} \to P$ extends uniquely to a homomorphism $⨁_{i} M_{i} \to P$ whose value is the finite sum of the coordinate values (Universal property of a direct sum of modules).

[L5]

Over a commutative ring there is a natural isomorphism $σ_{M, N} : M \otimes_{R} N \to N \otimes_{R} M$ with $σ_{M, N} (m \otimes n) = n \otimes m$ , and $σ_{N, M} σ_{M, N}$ is the identity (Symmetry and associativity isomorphisms for tensor products over a commutative ring).

Proof

technique · direct

1.1

For each $i$ , the pairing $(m, n) \mapsto ȷ_{i} (m) \otimes n$ is bilinear, so [L1] gives $ϕ_{i} : M_{i} \otimes_{R} N \to (⨁_{i} M_{i}) \otimes_{R} N$ . By [L4], the family $(ϕ_{i})$ induces the displayed map $Φ$ .

givenL1L2L4

1.2

Define $b : (⨁_{i} M_{i}) \times N \to ⨁_{i} (M_{i} \otimes_{R} N)$ by $b ((m_{i}), n) = (m_{i} \otimes n)_{i}$ . This tuple has finite support by [L3], and coordinatewise calculation shows that $b$ is bilinear.

L2L3algebra

2.1

By [L1], the pairing in step 1.2 induces $Ψ : (⨁_{i} M_{i}) \otimes_{R} N \to ⨁_{i} (M_{i} \otimes_{R} N)$ with $Ψ ((m_{i}) \otimes n) = (m_{i} \otimes n)_{i}$ .

step 1.2L1

3.1

The composite $Ψ Φ$ fixes every coordinate generator $ȷ_{i} (m \otimes n)$ , so it is the identity by [L4]; the composite $Φ Ψ$ fixes every elementary tensor $(m_{i}) \otimes n$ , so it is the identity by [L1].

step 1.1step 2.1L1L4

4.1

Thus $Φ$ and $Ψ$ are inverse isomorphisms. They are natural because maps $u_{i} : M_{i} \to M_{i}^{'}$ and $v : N \to N^{'}$ make the two composites send each coordinate generator $ȷ_{i} (m \otimes n)$ to $ȷ_{i} (u_{i} (m) \otimes v (n))$ . When $I = \emptyset$ , [L3] makes both sides zero and the same construction yields the unique isomorphism.

step 3.1L1L4L3

5.1

For the second variable, put $Φ^{'} : = σ_{⨁_{i} M_{i}, N} \circ Φ \circ (⨁_{i} σ_{N, M_{i}})$ , where the middle direct sum of the symmetries is the map induced by [L4] from the family $ȷ_{i} \circ σ_{N, M_{i}}$ . Each $σ$ is an isomorphism by [L5] and $Φ$ is one by step 4.1, so $Φ^{'}$ is an isomorphism; it is natural as a composite of natural isomorphisms. Tracing a coordinate generator gives $Φ^{'} (ȷ_{i} (n \otimes m)) = σ (Φ (ȷ_{i} (m \otimes n))) = σ (ȷ_{i} (m) \otimes n) = n \otimes ȷ_{i} (m)$ , which is the displayed formula. At $I = \emptyset$ both sides are again zero.

step 4.1L3L4L5∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

The elementary tensors of two bases form the product basis of the tensor product

Statement

Let $R$ be a commutative ring. If $M$ is free with basis $(e_{i})_{i \in I}$ and $N$ is free with basis $(f_{j})_{j \in J}$ , then $M \otimes_{R} N$ is free with basis

$(e_{i} \otimes f_{j})_{(i, j) \in I \times J} .$

Equivalently, the canonical map $R^{(I \times J)} \to M \otimes_{R} N$ sending the standard basis vector at $(i, j)$ to $e_{i} \otimes f_{j}$ is an isomorphism. This includes an empty basis in either factor.

Facts & Assumptions

Given: A commutative ring $R$ and free modules $M, N$ with bases indexed by $I, J$ .

[L1]

Tensor products commute with arbitrary direct sums in either variable: $⨁_{i} (M_{i} \otimes_{R} N) ≅ (⨁_{i} M_{i}) \otimes_{R} N$ and $⨁_{i} (N \otimes_{R} M_{i}) ≅ N \otimes_{R} (⨁_{i} M_{i})$ (Tensor products commute with arbitrary direct sums).

[L2]

The regular module is a tensor unit: $R \otimes_{R} R ≅ R$ via $r \otimes s \mapsto r s$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

[L3]

A free module on $X$ is $R^{(X)} = ⨁_{x \in X} R$ , with standard basis and unique finite coordinate expressions, including $X = \emptyset$ (The free module on a set and its standard basis).

[L4]

A map from the basis set of a free module extends uniquely to a module homomorphism (Universal property of the free module on a set).

Proof

technique · direct

1.1

The chosen bases identify $M$ with $⨁_{i \in I} R$ and $N$ with $⨁_{j \in J} R$ , carrying $e_{i}, f_{j}$ to the corresponding standard basis vectors.

givenL3L4

2.1

Apply [L1] in each variable and then [L2] to obtain canonical isomorphisms $M \otimes_{R} N ≅ ⨁_{i \in I} ⨁_{j \in J} (R \otimes_{R} R) ≅ ⨁_{(i, j) \in I \times J} R = R^{(I \times J)}$ .

step 1.1L1L2L3

3.1

Tracing a coordinate generator through step 2.1 sends it to $e_{i} \otimes f_{j}$ ; by [L3], those images therefore form a basis and every tensor has a unique finite expansion in them.

step 2.1L3

3.2

If $I = \emptyset$ or $J = \emptyset$ , then $I \times J = \emptyset$ , the corresponding factor and the target free module are zero by [L3], and step 2.1 is the unique isomorphism between zero modules.

step 2.1L3

4.1

This proves the product-basis assertion in all cases.

step 3.1step 3.2∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

$R^{m} \otimes_{R} R^{n} ≅ R^{m n}$ with the product basis, and $\dim_{F} (V \otimes_{F} W) = \dim_{F} V \dim_{F} W$

Statement

Let $R$ be a commutative ring and $m, n \in N$ . The standard finite free modules satisfy

$R^{m} \otimes_{R} R^{n} ≅ R^{m n},$

with the tensor products of standard basis vectors corresponding to the standard basis indexed by $m \times n$ .

If $F$ is a field and $V, W$ are finite-dimensional $F$ -vector spaces, then

$\dim_{F} (V \otimes_{F} W) = (\dim_{F} V) (\dim_{F} W) .$

Both assertions include a zero rank or zero-dimensional factor.

Facts & Assumptions

Given: Natural numbers $m, n$ , a commutative ring $R$ , and finite-dimensional vector spaces $V, W$ over a field $F$ .

[L1]

Tensor products of free modules with bases indexed by $I, J$ have basis indexed by $I \times J$ (The elementary tensors of two bases form the product basis of the tensor product).

[L2]

For any unital ring $R$ , the free module $R^{(m)}$ has its standard basis indexed by $m$ , including the empty basis at $m = 0$ (The free module on a set and its standard basis). For a field $F$ , this is the usual basis of $F^{m}$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L3]

The dimension of a finite-dimensional vector space is the unique natural number equinumerous with a basis; the zero space has dimension zero (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

Proof

technique · direct

1.1

Apply [L1] to the standard bases indexed by $m$ and $n$ . Their product is indexed by $m \times n$ , which has $m n$ elements, giving the first isomorphism and its product basis.

givenL1algebra

1.2

Choose bases of $V$ and $W$ with respectively $p = \dim_{F} V$ and $q = \dim_{F} W$ elements. By [L1] their elementary tensors form a basis of $V \otimes_{F} W$ indexed by $p \times q$ , hence with $p q$ elements.

L1L3choose

1.3

If $m = 0$ or $n = 0$ , or if $p = 0$ or $q = 0$ , [L2] makes one basis empty and [L1] makes the product basis empty, so both sides are the zero module or have dimension zero as asserted.

L1L2L3

2.1

By [L3], step 1.2 gives $\dim_{F} (V \otimes_{F} W) = p q = (\dim_{F} V) (\dim_{F} W)$ .

step 1.2L3

3.1

Steps 1.1 through 2.1 prove both formulas, including their zero boundaries.

step 1.1step 2.1step 1.3∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

The $R$ -module ${Hom}_{R} (M, N)$ over a commutative ring

Definition

Let $R$ be a commutative ring and let $M, N$ be $R$ -modules. The abelian group ${Hom}_{R} (M, N)$ of module homomorphisms (The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition) becomes an $R$ -module under pointwise scalar multiplication

$(r f) (m) : = r f (m) .$

The function $r f$ is $R$ -linear because, for $s \in R$ ,

$(r f) (s m) = r s f (m) = s r f (m) = s (r f) (m),$

where the middle equality uses commutativity (Commutative ring). The remaining module laws hold pointwise in $N$ . This module is the internal Hom module and is denoted ${Hom}_{R} (M, N)$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

For finite-dimensional $V$ , the canonical map $V^{*} \otimes_{F} W \to {Hom}_{F} (V, W)$ is an isomorphism

Statement

Let $V$ be a finite-dimensional vector space over a field $F$ , and let $W$ be any $F$ -vector space. The bilinear map

$(ϕ, w) ⟼ [v \mapsto ϕ (v) w]$

induces a natural isomorphism

$Φ : V^{*} \otimes_{F} W ⟶ {Hom}_{F} (V, W) .$

For a basis $(v_{1}, \dots, v_{n})$ of $V$ with dual basis $(v_{1}^{*}, \dots, v_{n}^{*})$ , its inverse is

$Ψ (T) = \sum_{i = 1}^{n} v_{i}^{*} \otimes T (v_{i}) .$

The empty sum gives the assertion when $V = 0$ .

Facts & Assumptions

Given: A finite-dimensional $F$ -vector space $V$ , an $F$ -vector space $W$ , and a basis $(v_{i})_{1 \leq i \leq n}$ of $V$ with dual family $(v_{i}^{*})$ .

[L1]

Bilinear maps from $V^{*} \times W$ induce unique homomorphisms from $V^{*} \otimes_{F} W$ (Universal property of the tensor product for balanced maps into abelian groups).

[L2]

Tensor products of bases have product basis (The elementary tensors of two bases form the product basis of the tensor product).

[L3]

Over the commutative field $F$ , ${Hom}_{F} (V, W)$ is an $F$ -module under pointwise scalar multiplication (The $R$ -module ${Hom}_{R} (M, N)$ over a commutative ring).

[L4]

The algebraic dual is $V^{*} = L (V, F)$ (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ).

[L5]

The dual family of a finite basis is a basis of $V^{*}$ (The dual family of a finite basis is a basis of the dual space, with the same dimension).

Proof

technique · direct

1.1

The map $(ϕ, w) \mapsto [v \mapsto ϕ (v) w]$ is bilinear, so [L1] gives an $F$ -linear map $Φ : V^{*} \otimes_{F} W \to {Hom}_{F} (V, W)$ .

givenL1L3L4

1.2

Define $Ψ (T) = \sum_{i} v_{i}^{*} \otimes T (v_{i})$ . This is an $F$ -linear map because evaluation and the finite sum are linear in $T$ .

givenL3L5construct

2.1

For $T \in {Hom}_{F} (V, W)$ and $v = \sum_{i} v_{i}^{*} (v) v_{i}$ , one has $(Φ Ψ (T)) (v) = \sum_{i} v_{i}^{*} (v) T (v_{i}) = T (v)$ , so $Φ Ψ$ is the identity.

step 1.1step 1.2L5algebra

2.2

For an elementary tensor $ϕ \otimes w$ , one has $Ψ Φ (ϕ \otimes w) = \sum_{i} v_{i}^{*} \otimes ϕ (v_{i}) w = (\sum_{i} ϕ (v_{i}) v_{i}^{*}) \otimes w = ϕ \otimes w$ , because $(v_{i}^{*})$ is the dual basis; elementary tensors generate, so $Ψ Φ$ is the identity.

step 1.1step 1.2L2L5algebra

2.3

The map $Φ$ is natural in $W$ : for $h : W \to W^{'}$ , both routes send $ϕ \otimes w$ to the map $v \mapsto ϕ (v) h (w)$ . It is contravariantly natural in $V$ : for $a : V^{'} \to V$ , both routes send $ϕ \otimes w$ to $[v^{'} \mapsto ϕ (a (v^{'})) w]$ . Equality on elementary tensors gives both naturality squares.

step 1.1L1algebra

3.1

Thus $Φ$ is a natural isomorphism with inverse $Ψ$ . If $V = 0$ , its basis and dual basis are empty, both $V^{*} \otimes_{F} W$ and ${Hom}_{F} (V, W)$ are zero, and the same formulas are the unique inverse maps.

step 2.1step 2.2step 2.3L2L5∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

Under ${End}_{F} (V) ≅ V^{*} \otimes_{F} V$ , tensor contraction is the trace

Statement

Let $V$ be a finite-dimensional vector space over $F$ . Under the isomorphism

$V^{*} \otimes_{F} V ⟶ {End}_{F} (V), ϕ \otimes v ⟼ [x \mapsto ϕ (x) v],$

the contraction map $c : V^{*} \otimes_{F} V \to F$ , defined by $c (ϕ \otimes v) = ϕ (v)$ , corresponds to the trace $T \mapsto tr (T)$ .

Facts & Assumptions

Given: A finite-dimensional $F$ -vector space $V$ and the canonical Hom-tensor isomorphism with $W = V$ .

[L1]

The canonical isomorphism sends $ϕ \otimes v$ to the rank-one endomorphism $x \mapsto ϕ (x) v$ (For finite-dimensional $V$ , the canonical map $V^{*} \otimes_{F} W \to {Hom}_{F} (V, W)$ is an isomorphism).

[L2]

The trace of an endomorphism is the sum of the diagonal entries of its matrix in any basis, and is zero in dimension zero (The basis-independent trace of an endomorphism of a finite-dimensional vector space).

[L3]

A bilinear pairing induces a unique linear map from a tensor product (Universal property of the tensor product for balanced maps into abelian groups).

Proof

technique · direct

1.1

Evaluation $(ϕ, v) \mapsto ϕ (v)$ is bilinear, so [L3] induces the linear contraction map $c : V^{*} \otimes_{F} V \to F$ with $c (ϕ \otimes v) = ϕ (v)$ .

givenL3

1.2

Choose a basis $(e_{1}, \dots, e_{n})$ of $V$ and write $v = \sum_{j} v_{j} e_{j}$ . For $T_{ϕ, v} (x) = ϕ (x) v$ , the coefficient of $e_{i}$ in $T_{ϕ, v} (e_{i})$ is $ϕ (e_{i}) v_{i}$ .

givenL1algebra

1.3

If $V = 0$ , the tensor product and endomorphism space are zero and both maps are the zero map by [L2].

L1L2L3

2.1

By [L2], $tr (T_{ϕ, v}) = \sum_{i} ϕ (e_{i}) v_{i} = ϕ (\sum_{i} v_{i} e_{i}) = ϕ (v) = c (ϕ \otimes v)$ .

step 1.1step 1.2L2algebra

3.1

Matrix diagonal sums are linear, so trace is linear by [L2]. Therefore trace after [L1] and contraction are linear maps inducing the same bilinear pairing by step 2.1; uniqueness in [L3] proves that they agree everywhere.

step 1.1step 2.1L1L2L3algebra

4.1

Thus tensor contraction is precisely trace under the canonical isomorphism, including the zero-dimensional case.

step 3.1step 1.3∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

Algebras over a commutative ring, central structure maps, and algebra homomorphisms

Definition

Let $R$ be a commutative ring (Commutative ring). An $R$ -algebra is a unital ring $A$ together with a unital ring homomorphism

$η_{A} : R ⟶ A$

(Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ) whose image is central: $η_{A} (r) a = a η_{A} (r)$ for every $r \in R$ and $a \in A$ . The induced scalar action is $r a : = η_{A} (r) a$ , making $A$ an $R$ -module and multiplication $A \times A \to A$ bilinear.

An $R$ -algebra homomorphism $f : A \to B$ is a unital ring homomorphism satisfying $f \circ η_{A} = η_{B}$ . Such a map is automatically $R$ -linear. An $R$ -algebra is commutative when its underlying ring is commutative.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

The tensor product of $R$ -algebras has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$

Statement

Let $R$ be a commutative ring and let $A, B$ be $R$ -algebras. The $R$ -module $A \otimes_{R} B$ has a unique $R$ -algebra structure satisfying

$(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$

and

$1_{A \otimes_{R} B} = 1_{A} \otimes 1_{B}, r ⟼ r (1_{A} \otimes 1_{B}) .$

If $A$ and $B$ are commutative, then $A \otimes_{R} B$ is commutative.

Facts & Assumptions

Given: A commutative ring $R$ and central unital $R$ -algebras $A, B$ .

[L1]

In an $R$ -algebra, the structure map is central and multiplication is $R$ -bilinear (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

[L2]

Finite tensor products represent multilinear maps (Finite iterated tensor products represent multilinear maps independently of parenthesization).

[L3]

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

[L4]

An elementary-tensor formula descends exactly when its underlying pairing is balanced (A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

Proof

technique · direct

1.1

The map $(a, b, a^{'}, b^{'}) \mapsto a a^{'} \otimes b b^{'}$ is $R$ -multilinear: centrality in [L1] lets a scalar move among the four variables without changing the value.

givenL1L3algebra

2.1

By [L2], step 1.1 induces an $R$ -bilinear multiplication $(A \otimes_{R} B) \times (A \otimes_{R} B) \to A \otimes_{R} B$ with the displayed pure-tensor formula; [L4] ensures that the formula has descended before any ring laws are used.

step 1.1L2L4

3.1

Both sides of associativity are trilinear in the three tensor arguments and agree on pure tensors by associativity in $A$ and $B$ ; uniqueness in [L2] makes them equal everywhere. Both distributive laws hold because the multiplication from step 2.1 is bilinear.

step 2.1L2algebra

3.2

Left and right multiplication by $1_{A} \otimes 1_{B}$ are linear maps that agree with the identity on every pure tensor, so uniqueness in the two-factor case of [L2] makes them the identity maps.

step 2.1L2algebra

3.3

If $A$ and $B$ are commutative, then $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'} = a^{'} a \otimes b^{'} b = (a^{'} \otimes b^{'}) (a \otimes b)$ . The two bilinear multiplication maps therefore induce the same four-variable multilinear map, so [L2] gives commutativity for arbitrary tensors.

step 2.1L2algebra

4.1

The map $R \to A \otimes_{R} B$ , $r \mapsto r (1_{A} \otimes 1_{B})$ , is a unital ring homomorphism and its image is central, checked on pure tensors using [L1] and [L3]. Thus the resulting ring is an $R$ -algebra.

step 2.1step 3.2L1L3algebra

5.1

Uniqueness in [L2] forces the multiplication from its displayed pure-tensor formula, while the identity and structure map are then forced by the displayed elements. Steps 2.1 through 4.1 prove existence and all asserted properties.

L2step 2.1step 3.1step 3.2step 4.1step 3.3∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

Universal mapping property of the tensor product of commutative algebras

Statement

Let $A, B, C$ be commutative $R$ -algebras. For every pair of $R$ -algebra homomorphisms $f : A \to C$ and $g : B \to C$ , there is a unique $R$ -algebra homomorphism

$h : A \otimes_{R} B ⟶ C$

such that $h (a \otimes 1) = f (a)$ and $h (1 \otimes b) = g (b)$ . It is given by

$h (a \otimes b) = f (a) g (b) .$

Thus $A \otimes_{R} B$ , with its two canonical maps, is the coproduct of $A$ and $B$ among commutative $R$ -algebras.

Facts & Assumptions

Given: Commutative $R$ -algebras $A, B, C$ and $R$ -algebra maps $f : A \to C$ , $g : B \to C$ .

[L1]

The tensor product algebra has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ and identity $1 \otimes 1$ (The tensor product of $R$ -algebras has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ ).

[L2]

Balanced pairings induce unique homomorphisms from tensor products (Universal property of the tensor product for balanced maps into abelian groups).

[L3]

An elementary-tensor formula descends exactly when the corresponding pairing is balanced (A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

Proof

technique · direct

1.1

The canonical maps $j_{A} (a) = a \otimes 1$ and $j_{B} (b) = 1 \otimes b$ are $R$ -algebra homomorphisms: [L1] gives their multiplication and identity laws, and $j_{A} (r a) = r j_{A} (a)$ and $j_{B} (r b) = r j_{B} (b)$ show compatibility with the structure maps.

givenL1algebra

1.2

The pairing $(a, b) \mapsto f (a) g (b)$ is $R$ -bilinear: additivity is distributivity in $C$ , and $f (r a) g (b) = r f (a) g (b) = f (a) r g (b) = f (a) g (r b)$ because $C$ is commutative and both maps respect $R$ .

givenalgebra

2.1

By [L2] and [L3], step 1.2 induces a unique $R$ -linear map $h : A \otimes_{R} B \to C$ satisfying $h (a \otimes b) = f (a) g (b)$ .

step 1.2L2L3

3.1

On pure tensors, [L1] gives $h ((a \otimes b) (a^{'} \otimes b^{'})) = f (a a^{'}) g (b b^{'}) = f (a) g (b) f (a^{'}) g (b^{'})$ , where commutativity of $C$ permits the middle factors to switch; hence $h$ is multiplicative.

step 2.1L1algebra

3.2

One has $h (1 \otimes 1) = 1_{C}$ , $h (a \otimes 1) = f (a)$ , and $h (1 \otimes b) = g (b)$ , so $h$ is an $R$ -algebra homomorphism with the required restrictions.

step 2.1algebra

4.1

If $h^{'}$ has the same restrictions, then $a \otimes b = (a \otimes 1) (1 \otimes b)$ by [L1], so $h^{'} (a \otimes b) = f (a) g (b) = h (a \otimes b)$ . The underlying group homomorphisms consequently induce the same balanced pairing, and uniqueness in [L2] gives $h^{'} = h$ .

step 3.2L1L2

5.1

Step 1.1 supplies the two coproduct maps, and steps 2.1 through 4.1 prove the asserted universal mapping property.

step 1.1step 2.1step 3.1step 3.2step 4.1∎

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-16Open item page →

Hom-tensor adjunction: ${Hom}_{R} (M \otimes_{R} N, P) ≅ {Hom}_{R} (M, {Hom}_{R} (N, P))$

Statement

Let $R$ be a commutative ring and let $M, N, P$ be $R$ -modules. There is a natural $R$ -module isomorphism

${Hom}_{R} (M \otimes_{R} N, P) ≅ {Hom}_{R} (M, {Hom}_{R} (N, P)) .$

It sends $F$ to the map $m \mapsto [n \mapsto F (m \otimes n)]$ and sends $u : M \to {Hom}_{R} (N, P)$ to the homomorphism determined by $m \otimes n \mapsto u (m) (n)$ .

Facts & Assumptions

Given: A commutative ring $R$ and $R$ -modules $M, N, P$ .

[L1]

The internal Hom is an $R$ -module under $(r f) (n) = r f (n)$ (The $R$ -module ${Hom}_{R} (M, N)$ over a commutative ring).

[L2]

Bilinear maps from $M \times N$ into $P$ correspond uniquely to homomorphisms $M \otimes_{R} N \to P$ (Universal property of the tensor product for balanced maps into abelian groups).

[L3]

Scalar multiplication on the tensor product satisfies $r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$ (Over a commutative ring, $M \otimes_{R} N$ is an $R$ -module with $r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$ ).

Proof

technique · direct

1.1

Given $F : M \otimes_{R} N \to P$ , define $cur (F) (m) (n) = F (m \otimes n)$ . For fixed $m$ this is $R$ -linear in $n$ by [L3], and the dependence on $m$ is $R$ -linear by the same formula, so $cur (F) : M \to {Hom}_{R} (N, P)$ is an $R$ -module homomorphism.

givenL1L3algebra

1.2

Given $u : M \to {Hom}_{R} (N, P)$ , the pairing $(m, n) \mapsto u (m) (n)$ is bilinear by $R$ -linearity of $u$ and each $u (m)$ ; [L2] therefore induces a unique $uncur (u) : M \otimes_{R} N \to P$ .

givenL1L2

2.1

For every $F, m, n$ , $uncur (cur (F)) (m \otimes n) = F (m \otimes n)$ , so uniqueness in [L2] makes $uncur cur$ the identity.

step 1.1step 1.2L2

2.2

For every $u, m, n$ , $cur (uncur (u)) (m) (n) = u (m) (n)$ , so the two Hom-valued maps are equal and $cur uncur$ is the identity.

step 1.1step 1.2

2.3

Both assignments are $R$ -linear pointwise, and precomposition or postcomposition with homomorphisms commutes with evaluation; hence the bijection is an $R$ -module isomorphism natural in all three variables, contravariantly in the Hom source variables and covariantly in $P$ .

step 1.1step 1.2L1algebra

3.1

Steps 2.1 through 2.3 prove the natural Hom-tensor adjunction.

step 2.1step 2.2step 2.3∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

Tensoring is right exact

Statement

Let $R$ be a commutative ring, let

$A \overset{f}{\to} B \overset{g}{\to} C ⟶ 0$

be an exact sequence of $R$ -modules, and let $N$ be an $R$ -module. Then

$A \otimes_{R} N \overset{f \otimes 1}{\to} B \otimes_{R} N \overset{g \otimes 1}{\to} C \otimes_{R} N ⟶ 0$

is exact. Thus tensoring preserves cokernels and surjections, but no injectivity at the left is asserted.

Facts & Assumptions

Given: An exact sequence $A \overset{f}{\to} B \overset{g}{\to} C \to 0$ and an $R$ -module $N$ over a commutative ring $R$ .

[L1]

Exactness means that $g$ is surjective and $im f = \ker g$ (Exact sequences and short exact sequences of modules).

[L2]

Module homomorphisms induce tensor homomorphisms with $(f \otimes 1) (a \otimes n) = f (a) \otimes n$ and functorial composition (Module homomorphisms induce tensor-product homomorphisms functorially).

[L3]

Balanced pairings induce unique homomorphisms from tensor products (Universal property of the tensor product for balanced maps into abelian groups).

[L4]

An elementary-tensor formula descends exactly when the corresponding pairing is balanced (A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

[L5]

For a submodule $K \leq X$ , the quotient module $X / K$ has cosets $x + K$ and its induced scalar action (Quotient module $M / N$ with scalar multiplication on additive cosets).

[L6]

A module homomorphism that kills $K$ factors uniquely through $X / K$ (A module homomorphism vanishing on $N$ factors uniquely through $M / N$ ).

[L7]

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

[L8]

Every tensor is a finite sum of elementary tensors (The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums).

Proof

technique · direct

1.1

The map $g \otimes 1$ is surjective: by [L8], every tensor is a finite sum of elementary tensors $c \otimes n$ , and [L1] supplies $b \in B$ with $g (b) = c$ , so $c \otimes n = (g \otimes 1) (b \otimes n)$ .

givenL1L2L8choose

1.2

Functoriality gives $(g \otimes 1) (f \otimes 1) = (g f) \otimes 1 = 0$ , so $im (f \otimes 1) \subseteq \ker (g \otimes 1)$ .

L1L2

2.1

Let $Q = (B \otimes_{R} N) / im (f \otimes 1)$ . The image is a submodule because [L2] and [L7] make $f \otimes 1$ $R$ -linear. Since $g \otimes 1$ kills it, [L6] gives $\overline{g} : Q \to C \otimes_{R} N$ .

step 1.2L2L5L6L7

3.1

For $c \in C$ and $n \in N$ , choose any $b \in B$ with $g (b) = c$ and set $q (c, n) = [b \otimes n] \in Q$ . If $b^{'}$ is another lift, then $b - b^{'} \in \ker g = im f$ by [L1], so $[b \otimes n] = [b^{'} \otimes n]$ ; hence $q$ is well defined.

L1step 2.1choose

4.1

The function $q : C \times N \to Q$ is bilinear: lifts of sums may be taken as sums of lifts, scalar multiples as scalar multiples, and the tensor relations give the required equalities. By [L3] and [L4] it induces $h : C \otimes_{R} N \to Q$ .

step 3.1L3L4L7

5.1

For $b \otimes n$ , one has $h \overline{g} ([b \otimes n]) = h (g (b) \otimes n) = [b \otimes n]$ , while for $c \otimes n$ and a lift $b$ one has $\overline{g} h (c \otimes n) = \overline{g} ([b \otimes n]) = c \otimes n$ ; generators and uniqueness make $h$ and $\overline{g}$ inverse.

step 2.1step 3.1step 4.1L3

6.1

Since $\overline{g}$ is an isomorphism, the kernel of $g \otimes 1$ is exactly the submodule quotiented out in step 2.1, namely $im (f \otimes 1)$ . Together with step 1.1, this is right exactness.

step 1.1step 2.1step 5.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

The submodule $I M$ generated by products of elements of an ideal $I$ with elements of a module $M$

Definition

Let $R$ be a commutative ring, let $I ⊴ R$ be an ideal (Left, right and two-sided ideals), and let $M$ be an $R$ -module (Unital left and right modules over a ring; unqualified module means left module). The product of $I$ and $M$ is

$I M : = {\sum_{k = 1}^{t} i_{k} m_{k} : t \in N, i_{k} \in I, m_{k} \in M},$

where the case $t = 0$ is the empty sum $0$ . It is the submodule of $M$ generated by the products $i m$ : sums and negatives remain of the displayed form, and $r (i m) = (r i) m$ with $r i \in I$ .

Thus $0 M = 0$ and $R M = M$ ; the latter follows from $m = 1_{R} m$ .

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

$M \otimes_{R} R / I ≅ M / I M$ naturally

Statement

Let $R$ be a commutative ring, $I ⊴ R$ an ideal, and $M$ an $R$ -module. There is a natural $R$ -module isomorphism

$M \otimes_{R} (R / I) ≅ M / I M, m \otimes (r + I) ⟼ r m + I M .$

Both sides also carry the induced $R / I$ -module structure, and the isomorphism is $R / I$ -linear. For $I = 0$ it is the tensor-unit isomorphism, while for $I = R$ both sides are zero.

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I$ , and an $R$ -module $M$ .

[L1]

Tensoring an exact sequence ending in zero preserves exactness at the two rightmost terms (Tensoring is right exact).

[L2]

The tensor-unit isomorphism sends $m \otimes r$ to $r m$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

[L3]

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

[L4]

The quotient module $M / I M$ consists of cosets with the induced scalar action (Quotient module $M / N$ with scalar multiplication on additive cosets).

[L5]

A homomorphism that kills a submodule factors uniquely through the quotient module (A module homomorphism vanishing on $N$ factors uniquely through $M / N$ ).

[L6]

Over a commutative ring the natural symmetry $σ_{A, B} : A \otimes_{R} B \to B \otimes_{R} A$ , $a \otimes b \mapsto b \otimes a$ , is an isomorphism (Symmetry and associativity isomorphisms for tensor products over a commutative ring).

Proof

technique · direct

1.1

The sequence $I \to R \to R / I \to 0$ is exact. Tensoring on the right by $M$ and applying [L1] gives the exact sequence $I \otimes_{R} M \to R \otimes_{R} M \to (R / I) \otimes_{R} M \to 0$ . The symmetry isomorphisms of [L6] carry it termwise to $M \otimes_{R} I \to M \otimes_{R} R \to M \otimes_{R} (R / I) \to 0$ , and since $σ$ commutes with the induced maps on elementary tensors, that sequence is exact too.

givenL1L6

2.1

Under [L2], the image of $M \otimes_{R} I \to M \otimes_{R} R ≅ M$ consists exactly of finite sums $i m$ , hence is $I M$ by [L3].

step 1.1L2L3

3.1

Exactness in step 1.1 identifies $M \otimes_{R} (R / I)$ with the cokernel of the first map, which by step 2.1 is $M / I M$ ; [L5] gives the resulting isomorphism.

step 1.1step 2.1L4L5

4.1

Tracing $m \otimes (r + I)$ through the quotient gives $r m + I M$ . Multiplication by an element of $I$ acts as zero on both sides, so the map and its inverse are $R / I$ -linear.

step 3.1L3L4algebra

5.1

If $I = 0$ , step 4.1 is [L2]. If $I = R$ , then [L3] gives $I M = M$ and $R / I = 0$ , so both sides are zero.

step 4.1L2L3L4

6.1

This proves the natural $R$ -linear and $R / I$ -linear isomorphism in every boundary case.

step 3.1step 4.1step 5.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

Flat and faithfully flat modules and ring homomorphisms

Definition

Let $R$ be a commutative ring and let $M$ be an $R$ -module. The module $M$ is flat if the functor $- \otimes_{R} M$ preserves exact sequences: whenever $A \to B \to C$ is exact, so is

$A \otimes_{R} M ⟶ B \otimes_{R} M ⟶ C \otimes_{R} M .$

Since tensoring is always right exact (Tensoring is right exact, Exact sequences and short exact sequences of modules), the definition asks for the remaining left-hand exactness. Its equivalent formulation as preservation of injections is proved separately rather than built into the definition.

The module $M$ is faithfully flat if a sequence of $R$ -modules is exact exactly when its tensor with $M$ is exact.

For a unital ring homomorphism $f : R \to S$ (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ) between commutative rings, $S$ is an $R$ -module by $r \cdot s = f (r) s$ . The map $f$ is flat, respectively faithfully flat, when this $R$ -module is flat, respectively faithfully flat.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

Flatness is equivalent to preserving injections and to the ideal and finitely generated ideal tests

Statement

Let $R$ be a commutative ring and let $M$ be an $R$ -module. The following are equivalent:

$M$ is flat.
For every injection $K ↪ N$ , the induced map $K \otimes_{R} M \to N \otimes_{R} M$ is injective.
For every ideal $I \subseteq R$ , the multiplication map $I \otimes_{R} M \to M$ , $a \otimes m \mapsto a m$ , is injective.
The map in claim 3 is injective for every finitely generated ideal $I$ .

Facts & Assumptions

Given: A commutative ring $R$ and an $R$ -module $M$ .

[L1]

Flatness means that $- \otimes_{R} M$ preserves exact sequences (Flat and faithfully flat modules and ring homomorphisms).

[L2]

Tensoring is right exact (Tensoring is right exact).

[L3]

Tensor products commute with direct sums and $R \otimes_{R} M ≅ M$ ; consequently $R^{n} \otimes_{R} M ≅ M^{n}$ (Tensor products commute with arbitrary direct sums, The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

[L4]

A finite list $x_{1}, \dots, x_{n}$ in a module determines a homomorphism $R^{n} \to N$ taking $e_{j}$ to $x_{j}$ ; if the list generates $N$ , this map is surjective. Also $R^{0} = 0$ (The free module on a set and its standard basis, Universal property of the free module on a set).

[L5]

In a commutative ring, every submodule of the regular module $R$ is an ideal (Left, right and two-sided ideals).

[L6]

A tensor product is the quotient of the free $Z$ -module on pairs by the subgroup generated by the additive and balance relations (The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums). Consequently, a tensor expression and a derivation that it is zero involve only finitely many generators and defining relations.

Proof

technique · direct

1.1

Claim 1 implies claim 2 by applying [L1] to $0 \to K \to N$ ; claim 2 implies claim 3 by taking the inclusion $I ↪ R$ and using $R \otimes_{R} M ≅ M$ ; claim 3 implies claim 4 by restriction to finitely generated ideals.

L1L2L3L5

1.2

Assume claim 4. Claim 3 follows: an element $x \in I \otimes_{R} M$ is represented by finitely many coefficients from $I$ , hence comes from $I_{0} \otimes_{R} M$ for the finitely generated ideal $I_{0}$ they generate; if its image in $M$ is zero, injectivity for $I_{0}$ makes its representative zero, so $x = 0$ .

givenL5

1.3

Let $K ↪ N$ be injective and let $x \in K \otimes_{R} M$ map to zero. By [L6], the tensor $x$ uses finitely many elements of $K$ , and a derivation of its zero image uses only finitely many generators and defining relations in $N \otimes_{R} M$ . Hence there are finitely generated submodules $K_{0} \subseteq K$ and $N_{0} \subseteq N$ with $K_{0} \subseteq N_{0}$ for which $x$ comes from an element killed in $N_{0} \otimes_{R} M$ .

givenL6

2.1

Under claim 3, for every submodule $L \subseteq R^{n}$ , the map $L \otimes_{R} M \to M^{n}$ is injective, by induction on $n$ . For $n = 0$ it is the unique map $0 \to 0$ ; for $n = 1$ it is claim 3 by [L5].

step 1.2L3L4

2.2

Present $N_{0} = R^{n} / L$ using [L4], and let $L^{'} \subseteq R^{n}$ be the inverse image of $K_{0}$ . Right exactness identifies $N_{0} \otimes_{R} M$ with $M^{n} / im (L \otimes_{R} M)$ and $K_{0} \otimes_{R} M$ with the quotient of $L^{'} \otimes_{R} M$ by $im (L \otimes_{R} M)$ .

step 1.3L2L3L4

3.1

For the induction step, let $Q \subseteq R^{n}$ , put $Q^{'} : = Q \cap (R \oplus 0^{n - 1})$ , and let $Q^{''}$ be the image of $Q$ in $R^{n - 1}$ . Tensoring the exact rows $0 \to Q^{'} \to Q \to Q^{''} \to 0$ and $0 \to R \to R^{n} \to R^{n - 1} \to 0$ gives right-exact rows by [L2]; the left and right vertical maps are injective by the $n = 1$ case and the induction hypothesis from step 2.1.

step 2.1L2L3

4.1

If $z \in Q \otimes_{R} M$ maps to zero in $M^{n}$ , its image in $Q^{''} \otimes_{R} M$ maps to zero in $M^{n - 1}$ and hence is zero by the right vertical injection in step 3.1. Right exactness of the top row lifts $z$ from some $y \in Q^{'} \otimes_{R} M$ . The image of $y$ in $R \otimes_{R} M$ maps to the zero image of $z$ in $M^{n}$ ; the left vertical injection in step 3.1 makes $y = 0$ , and therefore $z = 0$ . This completes the induction of step 2.1.

step 3.1L2L3

5.1

By step 4.1, both $L \otimes_{R} M$ and $L^{'} \otimes_{R} M$ inject into $M^{n}$ . Therefore the induced map of the quotients in step 2.2 is injective, so $x = 0$ . This proves claim 2 from claim 4.

step 4.1step 2.2

6.1

Finally claim 2 and right exactness [L2] imply claim 1: for an exact sequence, replace the left map by the injection of its image into the middle term; tensoring preserves that injection by claim 2 and preserves the remaining image and cokernel statements by right exactness.

step 5.1L1L2

7.1

Steps 1.1 through 6.1 prove the cycle of equivalences. The zero ideal, $n = 0$ , and zero module cases occur explicitly in steps 1.2 and 2.1; no choice is made, and both directions of every equivalence have been supplied.

step 1.1step 1.2step 2.1step 3.1step 4.1step 1.3step 2.2step 5.1step 6.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

Every projective module over a commutative ring is flat

Statement

Every projective module over a commutative ring is flat. This implication requires no form of the Axiom of Choice.

Facts & Assumptions

Given: A commutative ring $R$ and a projective $R$ -module $P$ .

[L1]

A module is flat exactly when tensoring with it preserves injections (Flatness is equivalent to preserving injections and to the ideal and finitely generated ideal tests).

[L2]

Tensor products commute with arbitrary direct sums in either variable; in particular $A \otimes_{R} ⨁_{x \in X} R ≅ ⨁_{x \in X} (A \otimes_{R} R)$ (Tensor products commute with arbitrary direct sums).

[L3]

The regular module is a tensor unit (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

[L4]

Every projective module is, without choice, a direct summand of its canonical free cover (Equivalent characterizations of projective modules).

[L5]

Tensor maps preserve identities and compositions (Module homomorphisms induce tensor-product homomorphisms functorially).

Proof

technique · direct

1.1

Every free module $F = ⨁_{x \in X} R$ is flat: for an injection $u : A \to B$ , [L2] and [L3] identify $u \otimes 1_{F}$ with the direct sum of copies of $u$ , which is injective coordinatewise; now apply [L1].

L1L2L3

1.2

By [L4], there are a free module $F$ and homomorphisms $i : P \to F$ , $p : F \to P$ with $p i = {id}_{P}$ .

givenL4

2.1

Let $u : A \to B$ be injective and suppose $x \in A \otimes_{R} P$ satisfies $(u \otimes 1_{P}) (x) = 0$ . Functoriality [L5] gives $(u \otimes 1_{F}) ((1_{A} \otimes i) (x)) = (1_{B} \otimes i) ((u \otimes 1_{P}) (x)) = 0$ .

step 1.2L5

3.1

The map $u \otimes 1_{F}$ is injective by step 1.1, so $(1_{A} \otimes i) (x) = 0$ ; applying $1_{A} \otimes p$ and using $(1_{A} \otimes p) (1_{A} \otimes i) = 1_{A} \otimes (p i) = id$ gives $x = 0$ .

step 1.1step 1.2step 2.1L5

4.1

Thus $- \otimes_{R} P$ preserves every injection, and [L1] makes $P$ flat. The proof used the canonical splitting supplied by projectivity and made no family of choices.

step 3.1L1L4∎

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

Under the stated choice boundary, free modules are projective and hence flat

Statement

Let $R$ be a commutative ring and let $F$ be a free $R$ -module with basis indexed by $X$ .

Assuming the Axiom of Choice, $F$ is projective and therefore flat.
If $X$ is finite, only finite choice is needed for projectivity; if $X = \emptyset$ , no choice is needed.
Regardless of choice, $F$ is flat, because tensoring with $F$ is a direct sum of copies of the identity tensor functor.

Facts & Assumptions

Given: A commutative ring $R$ and a free $R$ -module $F$ with basis indexed by $X$ .

[L1]

Under AC every free module is projective; a finite basis requires only finite choice, and an empty basis requires none (Free modules are projective, with the exact choice boundary).

[L2]

Every projective module over a commutative ring is flat without choice (Every projective module over a commutative ring is flat).

[L3]

[L4]

The regular module is a tensor unit (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

Proof

technique · direct

1.1

Under AC, [L1] makes $F$ projective and [L2] then makes it flat. The refined finite and empty-basis choice bounds are exactly those stated in [L1].

givenL1L2

1.2

Independently of AC, write $F = ⨁_{x \in X} R$ . By [L3] and [L4], tensoring an exact sequence with $F$ gives the direct sum, over $X$ , of the original exact sequence; kernels and images are computed coordinatewise, so the result remains exact. Thus $F$ is flat without any choice principle.

L3L4algebra

2.1

Step 1.1 establishes the projective route with its precise choice boundary, while step 1.2 establishes flatness unconditionally; the two routes are logically distinct.

step 1.1step 1.2∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

For flat $M$ , one has $I M \cap J M = (I \cap J) M$

Statement

Let $R$ be a commutative ring, let $I, J \subseteq R$ be ideals, and let $M$ be a flat $R$ -module. Then

$I M \cap J M = (I \cap J) M .$

Facts & Assumptions

Given: Ideals $I, J$ of a commutative ring $R$ and a flat $R$ -module $M$ .

[L1]

Tensoring an exact sequence with a flat module preserves exactness (Flat and faithfully flat modules and ring homomorphisms).

[L2]

There is a natural isomorphism $M \otimes_{R} (R / I) ≅ M / I M$ , and similarly for $J$ ( $M \otimes_{R} R / I ≅ M / I M$ naturally).

[L3]

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

[L4]

Exactness at a module is equality of the incoming image and outgoing kernel (Exact sequences and short exact sequences of modules).

[L5]

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

[L6]

Proof

technique · direct

1.1

The sequence $0 \to I \cap J \to R \to R / I \oplus R / J$ , whose last displayed map sends $r$ to $(r + I, r + J)$ , is exact because its kernel is exactly $I \cap J$ .

givenL4

2.1

Tensor step 1.1 with the flat module $M$ . By [L1], the resulting sequence is exact and begins $0 \to (I \cap J) \otimes_{R} M \to M \to (R / I \otimes_{R} M) \oplus (R / J \otimes_{R} M)$ , using [L5].

step 1.1L1L5

3.1

The symmetry of [L6] identifies $R / I \otimes_{R} M$ with $M \otimes_{R} (R / I)$ and likewise for $J$ , so [L2] applies and the last map in step 2.1 is $m \mapsto (m + I M, m + J M)$ , whose kernel is $I M \cap J M$ .

step 2.1L2L6

3.2

The image of $(I \cap J) \otimes_{R} M \to M$ is the set of finite sums of products $a m$ with $a \in I \cap J$ , namely $(I \cap J) M$ by [L3].

step 2.1L3

4.1

Exactness in step 2.1 identifies the image in step 3.2 with the kernel in step 3.1, proving $(I \cap J) M = I M \cap J M$ . The calculation also covers $I = 0$ , $J = 0$ , $I = R$ , or $J = R$ .

step 3.1step 3.2L4∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

A short exact sequence with flat quotient remains short exact after tensoring

Statement

Let

$0 ⟶ A \overset{i}{\to} B \overset{p}{\to} C ⟶ 0$

be a short exact sequence of modules over a commutative ring $R$ . If $C$ is flat, then for every $R$ -module $N$ the sequence

$0 ⟶ A \otimes_{R} N \overset{i \otimes 1}{\to} B \otimes_{R} N \overset{p \otimes 1}{\to} C \otimes_{R} N ⟶ 0$

is short exact.

Facts & Assumptions

Given: A short exact sequence $0 \to A \overset{i}{\to} B \overset{p}{\to} C \to 0$ with $C$ flat, and an $R$ -module $N$ .

[L1]

Flatness makes tensoring preserve injections (Flat and faithfully flat modules and ring homomorphisms).

[L2]

Tensoring is right exact (Tensoring is right exact).

[L3]

Free modules are flat, and flatness means that tensoring preserves exact sequences; hence tensoring a short exact sequence with a free module preserves short exactness (Under the stated choice boundary, free modules are projective and hence flat, Flat and faithfully flat modules and ring homomorphisms).

[L4]

Every module admits a canonical surjection from a free module (Every module is a quotient of a free module).

[L5]

A short exact sequence has an injective first map, a surjective second map, and image equal to kernel (Exact sequences and short exact sequences of modules).

Proof

technique · direct

1.1

By [L4], choose a surjection $ε : F \to N$ from a free module and let $K = \ker ε$ , so $0 \to K \to F \to N \to 0$ is short exact.

L4L5choose

2.1

Tensor the sequence in step 1.1 with each of $A, B, C$ . By [L2], the three resulting columns $X \otimes_{R} K \to X \otimes_{R} F \to X \otimes_{R} N \to 0$ are right exact. The map $C \otimes_{R} K \to C \otimes_{R} F$ is injective by flatness of $C$ and [L1].

step 1.1L1L2

2.2

Tensor the given short exact sequence with $F$ . Since $F$ is flat by [L3], the middle row $0 \to A \otimes_{R} F \to B \otimes_{R} F \to C \otimes_{R} F \to 0$ is short exact. Tensoring it with $K$ and $N$ gives right-exact bottom and top rows by [L2].

givenstep 1.1L2L3

3.1

Let $x \in A \otimes_{R} N$ map to zero in $B \otimes_{R} N$ . By right exactness of the $A$ -column, lift $x$ to $y \in A \otimes_{R} F$ . Its image $y_{B} \in B \otimes_{R} F$ maps to zero in $B \otimes_{R} N$ , so right exactness of the $B$ -column gives $z \in B \otimes_{R} K$ mapping to $y_{B}$ .

step 2.1step 2.2choose

4.1

The image of $z$ in $C \otimes_{R} K$ maps in $C \otimes_{R} F$ to the image of $y_{B}$ , which is zero because $y$ came from $A \otimes_{R} F$ . The injectivity in step 2.1 therefore makes the image of $z$ in $C \otimes_{R} K$ zero.

step 2.1step 3.1L5

5.1

By right exactness of the bottom row in step 2.2, choose $w \in A \otimes_{R} K$ mapping to $z$ . In $B \otimes_{R} F$ , the images of $y$ and of $w$ are both $y_{B}$ ; injectivity of $A \otimes_{R} F \to B \otimes_{R} F$ from step 2.2 makes $y$ the image of $w$ .

step 2.2step 3.1step 4.1choose

6.1

The composite $A \otimes_{R} K \to A \otimes_{R} F \to A \otimes_{R} N$ is zero because $K \to F \to N$ is zero. Hence step 5.1 gives $x = 0$ , proving $i \otimes 1$ injective.

step 3.1step 5.1algebra

7.1

Right exactness [L2] already gives exactness at $B \otimes_{R} N$ , surjectivity onto $C \otimes_{R} N$ , and the terminal zero. Together with step 6.1, the tensored sequence is short exact.

step 6.1L2L5∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-16Open item page →

Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$

Definition

Let $f : R \to S$ be a unital homomorphism of commutative rings (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

For an $S$ -module $N$ , its restriction of scalars along $f$ , denoted ${Res}_{R}^{S} N$ , is the same abelian group with $R$ -action

$r \cdot n : = f (r) n .$

For an $R$ -module $M$ , its extension of scalars along $f$ is

$S \otimes_{R} M .$

Here $S$ is an $(S, R)$ -bimodule with left action by multiplication and right action $s \cdot r : = s f (r)$ ( $(S, R)$ -bimodules and commuting left and right scalar actions). The induced outer action

$s^{'} (s \otimes m) : = (s^{'} s) \otimes m$

makes $S \otimes_{R} M$ an $S$ -module (A commuting outer scalar action descends to a tensor product).

Restriction acts on a homomorphism by leaving its underlying function unchanged. Extension sends $u : M \to M^{'}$ to $1_{S} \otimes u : S \otimes_{R} M \to S \otimes_{R} M^{'}$ .

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-16Open item page →

Extension of scalars is left adjoint to restriction of scalars

Statement

Let $f : R \to S$ be a homomorphism of commutative rings, let $M$ be an $R$ -module, and let $N$ be an $S$ -module. There is a natural bijection

${Hom}_{S} (S \otimes_{R} M, N) ≅ {Hom}_{R} (M, {Res}_{R}^{S} N) .$

It sends $F$ to $m \mapsto F (1_{S} \otimes m)$ . Its inverse sends $u : M \to {Res}_{R}^{S} N$ to the $S$ -linear map determined by

$s \otimes m ⟼ s u (m) .$

Facts & Assumptions

Given: A ring map $f : R \to S$ , an $R$ -module $M$ , and an $S$ -module $N$ .

[L1]

Restriction uses $r n = f (r) n$ , while extension is the $S$ -module $S \otimes_{R} M$ with $s^{'} (s \otimes m) = s^{'} s \otimes m$ (Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$ ).

[L2]

Balanced pairings induce unique homomorphisms from tensor products (Universal property of the tensor product for balanced maps into abelian groups).

[L3]

An elementary-tensor prescription descends exactly when its underlying pairing is balanced (A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

Proof

technique · direct

1.1

If $F : S \otimes_{R} M \to N$ is $S$ -linear, define $Θ (F) (m) = F (1 \otimes m)$ . Then $Θ (F) (r m) = F (1 \otimes r m) = F (f (r) \otimes m) = f (r) F (1 \otimes m)$ , so $Θ (F)$ is $R$ -linear into the restriction of $N$ .

givenL1algebra

1.2

If $u : M \to {Res}_{R}^{S} N$ is $R$ -linear, the pairing $(s, m) \mapsto s u (m)$ is $R$ -balanced because $(s f (r)) u (m) = s u (r m)$ . By [L2] and [L3] it induces a homomorphism $Λ (u) : S \otimes_{R} M \to N$ .

givenL1L2L3

2.1

The map $Λ (u)$ is $S$ -linear because $Λ (u) (s^{'} (s \otimes m)) = Λ (u) (s^{'} s \otimes m) = s^{'} s u (m) = s^{'} Λ (u) (s \otimes m)$ .

step 1.2L1algebra

2.2

For $u$ , one has $Θ (Λ (u)) (m) = Λ (u) (1 \otimes m) = u (m)$ .

step 1.1step 1.2

2.3

Precomposition in $M$ and postcomposition in $N$ commute with both displayed formulas, so the inverse bijections are natural in both modules.

step 1.1step 1.2algebra

3.1

For $F$ , one has $Λ (Θ (F)) (s \otimes m) = s F (1 \otimes m) = F (s \otimes m)$ by $S$ -linearity, so uniqueness on elementary tensors gives $Λ Θ (F) = F$ .

step 1.1step 1.2step 2.1L2

4.1

Steps 2.2, 2.3 and 3.1 prove that extension of scalars is left adjoint to restriction of scalars.

step 2.2step 3.1step 2.3∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

Change of rings: $N \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$

Statement

Let $R \to S$ be a homomorphism of commutative rings, let $N$ be a right $S$ -module, and let $M$ be a left $R$ -module. There is a natural group isomorphism

$N \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$

given by

$n \otimes m ⟼ n \otimes (1_{S} \otimes m),$

with inverse $n \otimes (s \otimes m) \mapsto n s \otimes m$ .

Facts & Assumptions

Given: A ring map $R \to S$ , a right $S$ -module $N$ , and a left $R$ -module $M$ .

[L1]

Restriction makes $N$ a right $R$ -module and makes $S$ an $(S, R)$ -bimodule used in $S \otimes_{R} M$ (Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$ ).

[L2]

Compatible bimodules have the associativity isomorphism $(N \otimes_{S} S) \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$ (Associativity of tensor products for compatible bimodules).

[L3]

The tensor-unit isomorphism identifies $N \otimes_{S} S$ with $N$ by $n \otimes s \mapsto n s$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

Proof

technique · direct

1.1

Apply [L2] to $N$ , the $(S, R)$ -bimodule $S$ , and $M$ , then use [L3] on the left factor to obtain $(N \otimes_{S} S) \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$ and hence $N \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$ .

givenL1L2L3

2.1

Tracing elementary tensors through step 1.1 gives $n \otimes m \mapsto n \otimes (1 \otimes m)$ ; tracing the inverse gives $n \otimes (s \otimes m) \mapsto n s \otimes m$ .

step 1.1L2L3

3.1

The two formulas are mutually inverse on elementary tensors. The inverse after the forward map sends $n \otimes m$ to $n 1_{S} \otimes m = n \otimes m$ . In the other direction, balance over $S$ gives $n s \otimes (1_{S} \otimes m) = n \otimes s (1_{S} \otimes m) = n \otimes (s \otimes m)$ . The universal properties therefore make the composites identities.

step 2.1L1L2L3

3.2

Each construction commutes with homomorphisms in $N$ and $M$ because its elementary-tensor formula does, so the isomorphism is natural.

step 2.1algebra

4.1

This proves the change-of-rings isomorphism and its stated inverse.

step 3.1step 3.2∎

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

Extension of scalars carries flat modules to flat modules

Statement

Let $R \to S$ be a homomorphism of commutative rings. If $M$ is a flat $R$ -module, then its extension of scalars $S \otimes_{R} M$ is a flat $S$ -module.

Facts & Assumptions

Given: A ring map $R \to S$ and a flat $R$ -module $M$ .

[L1]

For every right $S$ -module $N$ , there is a natural isomorphism $N \otimes_{S} (S \otimes_{R} M) ≅ N \otimes_{R} M$ after restriction of scalars (Change of rings: $N \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$ ).

[L2]

A module is flat exactly when tensoring with it preserves injections (Flatness is equivalent to preserving injections and to the ideal and finitely generated ideal tests).

[L3]

Restriction of scalars leaves the underlying abelian group and function of a module map unchanged (Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$ ).

Proof

technique · direct

1.1

Let $u : N \to N^{'}$ be an injection of right $S$ -modules. By [L3], it is still injective after restriction to $R$ -modules.

givenL3

2.1

Since $M$ is flat, [L2] makes $u \otimes_{R} 1_{M} : N \otimes_{R} M \to N^{'} \otimes_{R} M$ injective.

step 1.1L2

3.1

Under the natural isomorphisms [L1], the map in step 2.1 is precisely $u \otimes_{S} 1_{S \otimes_{R} M}$ . Hence tensoring over $S$ with $S \otimes_{R} M$ preserves every injection.

step 2.1L1

4.1

By [L2] over the ring $S$ , the extended module $S \otimes_{R} M$ is flat.

step 3.1L2∎

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16Open item page →

Flatness is transitive under a flat change of rings

Statement

Let $R \to S$ be a flat homomorphism of commutative rings. If $N$ is a flat $S$ -module, then $N$ , restricted to an $R$ -module, is flat over $R$ . Consequently a composite of flat ring homomorphisms is flat.

The same assertions hold with "faithfully flat" throughout.

Facts & Assumptions

Given: A flat ring map $R \to S$ and a flat $S$ -module $N$ ; for the faithful assertion, assume both are faithfully flat.

[L1]

A ring map is flat or faithfully flat exactly when its target has that property as a module over its source (Flat and faithfully flat modules and ring homomorphisms).

[L2]

For every right $R$ -module $X$ , change of rings gives $X \otimes_{R} N ≅ (X \otimes_{R} S) \otimes_{S} N$ (Change of rings: $N \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$ ).

[L3]

Restriction of scalars leaves the underlying groups and maps unchanged (Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$ ).

Proof

technique · direct

1.1

Let $A \to B \to C$ be an exact sequence of $R$ -modules. Flatness of $S$ over $R$ makes $A \otimes_{R} S \to B \otimes_{R} S \to C \otimes_{R} S$ exact as a sequence of $S$ -modules.

givenL1

2.1

Flatness of $N$ over $S$ preserves the exactness of step 1.1 after tensoring over $S$ .

givenstep 1.1L1

3.1

By [L2], the sequence in step 2.1 is naturally isomorphic to $A \otimes_{R} N \to B \otimes_{R} N \to C \otimes_{R} N$ , so the restricted $R$ -module $N$ is flat.

step 2.1L2L3

3.2

If both functors are faithful on exactness, the implications in steps 1.1 and 2.1 may be read backwards as well; [L2] then shows that tensoring with $N$ over $R$ reflects exactness, so the restricted module is faithfully flat.

step 1.1step 2.1L1L2

4.1

Taking $N$ to be the target ring of a second flat, respectively faithfully flat, ring map and using [L1] proves the corresponding composition statement.

step 3.1step 3.2L1∎

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-16Open item page →

The character dual of a flat module is injective

Statement

Let $R$ be a commutative ring, let $P$ be a flat $R$ -module, and let $D$ be an injective $Z$ -module. Define the character dual

$P^{+} : = {Hom}_{Z} (P, D)$

with $R$ -action

$(r ϕ) (p) : = ϕ (r p) .$

Then $P^{+}$ is an injective $R$ -module.

Facts & Assumptions

Given: A commutative ring $R$ , a flat $R$ -module $P$ , and an injective $Z$ -module $D$ .

[L1]

Flatness makes $u \otimes 1_{P}$ injective whenever $u$ is an injection of $R$ -modules (Flat and faithfully flat modules and ring homomorphisms).

[L2]

Balanced maps into an abelian group correspond uniquely to group homomorphisms from a tensor product (Universal property of the tensor product for balanced maps into abelian groups).

[L3]

An injective module has the extension property along every injective module homomorphism (Injective modules and the extension property).

[L4]

${Hom}_{Z} (P, D)$ is an abelian group under pointwise addition (The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition).

Proof

technique · direct

1.1

The displayed formula makes $P^{+}$ an $R$ -module: $(r s) ϕ$ and $r (s ϕ)$ agree at every $p$ , and all other module laws hold pointwise in the abelian group [L4].

givenL4algebra

1.2

Let $u : A \to B$ be an injective $R$ -module homomorphism and let $v : A \to P^{+}$ be $R$ -linear. By [L1], $u \otimes 1_{P} : A \otimes_{R} P \to B \otimes_{R} P$ is injective.

givenL1

2.1

For every $R$ -module $M$ , an $R$ -linear map $v : M \to P^{+}$ determines the balanced map $(m, p) \mapsto v (m) (p)$ , and [L2] gives a group homomorphism $\hat{v} : M \otimes_{R} P \to D$ . Conversely a group homomorphism $h : M \otimes_{R} P \to D$ gives $m \mapsto [p \mapsto h (m \otimes p)]$ ; the balance relation makes this map $R$ -linear. These constructions are inverse.

step 1.1L2

3.1

A group homomorphism between abelian groups is automatically $Z$ -linear because additivity gives compatibility with positive integer multiples and with negatives. Thus the group homomorphisms in step 2.1 are precisely the $Z$ -module homomorphisms to which injectivity of $D$ applies.

algebra

4.1

Transpose $v$ by step 2.1 to $\hat{v} : A \otimes_{R} P \to D$ . By step 3.1 and injectivity [L3] at the ring $Z$ , extend it along $u \otimes 1_{P}$ to a homomorphism $\hat{w} : B \otimes_{R} P \to D$ .

step 2.1step 3.1step 1.2L3choose

5.1

Transpose $\hat{w}$ back by step 2.1 to an $R$ -linear map $w : B \to P^{+}$ . Naturality of the evaluation formulas gives $w \circ u = v$ .

step 2.1step 4.1

6.1

Every $R$ -linear map into $P^{+}$ therefore extends along every injection, so [L3] makes $P^{+}$ injective.

step 5.1L3∎

5 · Examples, counterexamples and false statements

None yet.

Sources

Standard references

Recommended treatments; not extraction sources.

Tensor Products of Modules

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Balanced maps from a right module and a left module, and bilinear maps over a commutative ring

Definition

The tensor product M⊗RNM⊗R​N from the additive group underlying the free ZZ-module on M×NM×N, elementary tensors, and finite tensor sums

Definition

Universal property of the tensor product for balanced maps into abelian groups

Statement

Facts & Assumptions

Proof

Tensor products are unique up to a unique isomorphism carrying elementary tensors to elementary tensors

Statement

Facts & Assumptions

Proof

Module homomorphisms induce tensor-product homomorphisms functorially

Statement

Facts & Assumptions

Proof

A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced

Statement

Facts & Assumptions

Proof

(S,R)(S,R)-bimodules and commuting left and right scalar actions

Definition

A commuting outer scalar action descends to a tensor product

Statement

Facts & Assumptions

Proof

Associativity of tensor products for compatible bimodules

Statement

Facts & Assumptions

Proof

Over a commutative ring, M⊗RNM⊗R​N is an RR-module with r(m⊗n)=(rm)⊗n=m⊗(rn)r(m⊗n)=(rm)⊗n=m⊗(rn)

Statement

Facts & Assumptions

Proof

Symmetry and associativity isomorphisms for tensor products over a commutative ring

Statement

Facts & Assumptions

Proof

The regular module is a tensor unit: R⊗RN≅NR⊗R​N≅N and M⊗RR≅MM⊗R​R≅M

Statement

Facts & Assumptions

Proof

Finite iterated tensor products represent multilinear maps independently of parenthesization

Statement

Facts & Assumptions

Proof

Tensor products commute with arbitrary direct sums

Statement

Facts & Assumptions

Proof

The elementary tensors of two bases form the product basis of the tensor product

Statement

Facts & Assumptions

Proof

Rm⊗RRn≅RmnRm⊗R​Rn≅Rmn with the product basis, and dim⁡F(V⊗FW)=dim⁡FV dim⁡FWdimF​(V⊗F​W)=dimF​VdimF​W

Statement

Facts & Assumptions

Proof

The RR-module Hom⁡R(M,N)HomR​(M,N) over a commutative ring

Definition

For finite-dimensional VV, the canonical map V∗⊗FW→Hom⁡F(V,W)V∗⊗F​W→HomF​(V,W) is an isomorphism

Statement

Facts & Assumptions

Proof

Under End⁡F(V)≅V∗⊗FVEndF​(V)≅V∗⊗F​V, tensor contraction is the trace

Statement

Facts & Assumptions

Proof

Algebras over a commutative ring, central structure maps, and algebra homomorphisms

Definition

The tensor product of RR-algebras has multiplication (a⊗b)(a′⊗b′)=aa′⊗bb′(a⊗b)(a′⊗b′)=aa′⊗bb′

Statement

Facts & Assumptions

Proof

Universal mapping property of the tensor product of commutative algebras

The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums

$(S, R)$ -bimodules and commuting left and right scalar actions

Over a commutative ring, $M \otimes_{R} N$ is an $R$ -module with $r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$

The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$

$R^{m} \otimes_{R} R^{n} ≅ R^{m n}$ with the product basis, and $\dim_{F} (V \otimes_{F} W) = \dim_{F} V \dim_{F} W$

The $R$ -module ${Hom}_{R} (M, N)$ over a commutative ring

For finite-dimensional $V$ , the canonical map $V^{*} \otimes_{F} W \to {Hom}_{F} (V, W)$ is an isomorphism

Under ${End}_{F} (V) ≅ V^{*} \otimes_{F} V$ , tensor contraction is the trace

The tensor product of $R$ -algebras has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$

Hom-tensor adjunction: ${Hom}_{R} (M \otimes_{R} N, P) ≅ {Hom}_{R} (M, {Hom}_{R} (N, P))$

The submodule $I M$ generated by products of elements of an ideal $I$ with elements of a module $M$

$M \otimes_{R} R / I ≅ M / I M$ naturally

For flat $M$ , one has $I M \cap J M = (I \cap J) M$

Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$

Change of rings: $N \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$