False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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False: $m \otimes n = 0$ implies $m = 0$ or $n = 0$

Statement

False claim: if an elementary tensor $m \otimes n$ is zero, then $m = 0$ or $n = 0$ .

In $Z \otimes_{Z} Z / 2$ , the nonzero factors $2$ and $\overline{1}$ satisfy

$2 \otimes \overline{1} = 1 \otimes 2 \overline{1} = 0.$

Facts & Assumptions

Given: The regular $Z$ -module $Z$ and the quotient module $Z / 2$ .

[L1]

$Z$ is a commutative ring, so multiplication by integers supplies its regular module structure (The integers form a commutative ring).

[L2]

Modular arithmetic gives $2 \overline{1} = \overline{0}$ in $Z / 2$ (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold), while the unique representatives $0, 1$ are distinct classes (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ ).

[L3]

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

Refutation

technique · direct

1.1

Balance in the tensor product and [L2] give $2 \otimes \overline{1} = 1 \otimes 2 \overline{1} = 1 \otimes \overline{0} = 0$ .

givenL2algebra

2.1

The integer $2$ is nonzero, and $\overline{1}$ is nonzero by [L2]. Thus neither factor in step 1.1 is zero.

L1L2algebra

3.1

Moreover, [L3] sends $1 \otimes \overline{1}$ to the nonzero class $\overline{1}$ , so the ambient tensor-product group is itself nonzero. Steps 1.1 and 2.1 therefore refute the claim.

step 1.1step 2.1L2L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 62 results over 13 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

Christopher Dennis, Week 1 recap (standard reference, not scraped)

False: m⊗n=0m⊗n=0 implies m=0m=0 or n=0n=0