False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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False: $Z / m \otimes_{Z} Z / n$ is nonzero for all positive $m, n$

Statement

False claim: for all positive integers $m, n$ , the tensor product $Z / m \otimes_{Z} Z / n$ is nonzero.

In fact, with the convention that $Z / 1$ is the zero group,

$Z / m \otimes_{Z} Z / n ≅ Z / \gcd (m, n) .$

Thus $m = 2$ and $n = 3$ give a tensor product of two nonzero cyclic groups that is zero.

Facts & Assumptions

Given: Positive integers $m, n$ , and $d : = \gcd (m, n)$ .

[L1]

For a right module $M$ and an ideal $I$ of a commutative ring $R$ , $M \otimes_{R} R / I ≅ M / I M$ ( $M \otimes_{R} R / I ≅ M / I M$ naturally).

[L2]

The subgroup $m Z + n Z$ of $Z$ is $d Z$ ( $a Z + b Z = \gcd (a, b) Z$ and $a Z \cap b Z = lcm (a, b) Z$ ; equivalently, in $(Z, +)$ the subgroup generated by ${a, b}$ is $⟨ \gcd (a, b) ⟩$ and $⟨ a ⟩ \cap ⟨ b ⟩ = ⟨ lcm (a, b) ⟩$ ).

[L3]

Modular addition and multiplication give $Z / q$ its usual quotient-ring operations (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold). For positive $q$ , its classes have the unique representatives $0, \dots, q - 1$ , so $∣ Z / q ∣ = q$ ; in particular, $Z / 1$ is zero while $Z / 2$ and $Z / 3$ are nonzero (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 $Z$ ).

Refutation

technique · direct

1.1L1L3

Apply [L1] to $M = Z / m$ and $I = n Z$ to obtain $Z / m \otimes_{Z} Z / n ≅ (Z / m) / n (Z / m)$ .

1.2L2L3

Define $ϕ : Z / d \to (Z / m) / n (Z / m)$ by $ϕ ([a]_{d}) = [a]_{m} + n (Z / m)$ . If $a - b \in d Z = m Z + n Z$ by [L2], then $[a - b]_{m}$ lies in $n (Z / m)$ , so $ϕ$ is well-defined.

2.1step 1.2L2L3

The map $ϕ$ is surjective because every class in the target is represented by some $[a]_{m}$ . If $ϕ ([a]_{d}) = 0$ , then $[a]_{m} = n [b]_{m}$ for some integer $b$ , so $a - n b \in m Z$ and hence $a \in m Z + n Z = d Z$ by [L2]; therefore $[a]_{d} = 0$ , and $ϕ$ is injective.

3.1step 1.1step 2.1L3∎

Steps 1.1 and 2.1 give the displayed isomorphism. For $(m, n) = (2, 3)$ one has $d = 1$ , so the tensor product is $Z / 1 = 0$ although both $Z / 2$ and $Z / 3$ are nonzero. This refutes the claim.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

32 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

Wenqi Li, Commutative Algebra, Lecture 10 (standard reference, not scraped)

False: Z/m⊗ZZ/nZ/m⊗Z​Z/n is nonzero for all positive m,nm,n