PropositionStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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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∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 40 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Stacks Project, Section 10.12: Tensor products (standard reference, not scraped)
C. Dennis, Week 1 recap on tensor products (standard reference, not scraped)