LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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Every abelian group embeds in a divisible abelian group

Statement

Every abelian group $A$ admits an injective homomorphism into a divisible abelian group. The construction uses no choice principle.

Facts & Assumptions

Given: An abelian group $A$ , viewed as a $Z$ -module.

[L1]

The canonical map $F = Z^{(A)} \to A$ is surjective and identifies $A$ with $F / K$ , where $K$ is its kernel (Every module is a quotient of a free module).

[F1]

An abelian group is divisible if multiplication by every nonzero integer is surjective (Divisible modules over an integral domain).

[F2]

Direct sums consist of finite-support tuples (The direct sum of an indexed family of modules).

[L2]

$Q$ is a field and therefore permits division by every nonzero integer (The rationals form a field).

Proof

technique · constructive

1.1

Let $F = Z^{(A)}$ , let $ε : F \to A$ be the canonical surjection, and put $K = \ker ε$ ; by [L1], $A ≅ F / K$ .

L1construct

1.2

The group $Q$ is divisible: for a finite-support tuple $q$ and a nonzero integer $n$ , divide each of its finitely many nonzero rational coordinates by $n$ using [L2].

F2L2

2.1

Embed $F$ coordinatewise into $Q : = Q^{(A)}$ , and regard $K$ as a subgroup of $Q$ . Define $D : = Q / K$ and $j : F / K \to D$ by $j (f + K) = f + K$ . This is well defined and injective because $F \cap K = K$ .

step 1.1F2L2construct

2.2

A quotient of a divisible group is divisible: if $q + K \in D$ and $n \neq 0$ , choose $y \in Q$ with $n y = q$ by step 1.2; then $n (y + K) = q + K$ . Thus $D$ is divisible by [F1].

step 1.2F1

3.1

Composing the isomorphism $A ≅ F / K$ from step 1.1 with the injection $j$ of step 2.1 embeds $A$ in the divisible group $D$ . Every division was coordinatewise on finite support, so no choice was used.

step 1.1step 2.1step 2.2discharge-construct∎

Depends on

Used by

A category with enough injectives but not enough projectives Counterexample
An injective resolution of an abelian group beginning with a divisible group Example
Module categories have enough injectives Theorem

Dependency tree · two levels

13 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

A. Kleshchev, Lectures on Abstract Algebra for Graduate Students, sections 3.6, 3.14, and 3.15 (standard reference, not scraped)
The Stacks Project, Algebra (standard reference, not scraped)
P. Hekmati, Homological Algebra, section 3.1 (standard reference, not scraped)