False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-16

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False: tensoring preserves injections

Statement

False claim: tensoring an injective module homomorphism with a fixed module always gives an injective homomorphism.

The injection $Z \overset{\cdot 2}{\to} Z$ becomes the zero map after tensoring with $Z / 2$ over $Z$ .

Facts & Assumptions

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

[L1]

$Z$ is a commutative ring (The integers form a commutative ring), and multiplication by a nonzero integer can be cancelled (The integers have no zero divisors; multiplicative cancellation).

[L2]

In $Z / 2$ , modular multiplication by $2$ is zero (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 tensor-unit isomorphism $Z \otimes_{Z} N ≅ N$ sends $a \otimes n$ to $a n$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

[L4]

Tensor products preserve right-exact sequences, but this statement does not assert preservation of injections (Tensoring is right exact).

Refutation

technique · direct

1.1givenL1

The map $u : Z \to Z$ , $u (a) = 2 a$ , is injective: if $2 a = 2 b$ , cancellation in [L1] gives $a = b$ .

1.2givenL2L3

Under the unit identifications [L3], the map $u \otimes 1_{Z / 2}$ is multiplication by $2$ on $Z / 2$ , hence is zero by [L2].

2.1step 1.1step 1.2L2L4∎

The zero map on $Z / 2$ is not injective because $\overline{1} \neq \overline{0}$ by [L2]. Thus step 1.1 is an injection whose tensor map is not injective, refuting the claim. This is consistent with [L4], which guarantees right exactness only.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Stacks Project, Section 10.12: Tensor products (standard reference, not scraped)
Christopher Dennis, Week 4 (standard reference, not scraped)