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Quotient by a divisible subgroup

PDF p. 13: Final group-theoretic paragraph

Statement

Let $0 \to D \overset{i}{\to} G \overset{π}{\to} A \to 0$ be a short exact sequence of abelian groups. If $D$ is divisible and $G$ is torsion-free, then $A$ is torsion-free.

Here divisible means that, for every positive integer $n$ and every $d \in D$ , there is an $e \in D$ with $n e = d$ . Torsion-free means that $n x = 0$ for a positive integer $n$ implies $x = 0$ . Short exactness means that $i$ is injective, $π$ is surjective, and $im i = \ker π$ . These are the stated module definitions applied over $Z$ ; positive integers suffice because multiplication by a negative integer differs only by negation.

Statement references

Divisible modules over an integral domain

Annihilators, torsion elements and the torsion subset of a module

Exact sequences and short exact sequences of modules

Facts & Assumptions

Given: A short exact sequence $0 \to D \overset{i}{\to} G \overset{π}{\to} A \to 0$ of abelian groups, with $D$ divisible and $G$ torsion-free.

Goal: Prove that $A$ is torsion-free.

[F1]

Short exactness gives surjectivity of $π$ and $\ker π = im i$ (Exact sequences and short exact sequences of modules).

[F2]

Divisibility of $D$ means that multiplication by each positive integer is surjective on $D$ (Divisible modules over an integral domain).

[F3]

Since $G$ is torsion-free, an element of $G$ killed by a positive integer is zero; the same criterion characterizes torsion-freeness of $A$ (Annihilators, torsion elements and the torsion subset of a module).

Proof

technique · direct lifting and subtraction

1.1givenF1choose

Let $a \in A$ and let $n$ be a positive integer such that $n a = 0$ . By surjectivity choose $g \in G$ with $π (g) = a$ . Then $π (n g) = n π (g) = 0$ , so exactness gives $n g = i (d)$ for some $d \in D$ .

2.1F1F2F3step 1.1choose∎

By divisibility choose $e \in D$ with $n e = d$ . As $i$ is a homomorphism, $n (g - i (e)) = n g - i (n e) = i (d) - i (d) = 0$ . Torsion-freeness of $G$ therefore gives $g = i (e)$ . Consequently $a = π (g) = π (i (e)) = 0$ . Since $a$ and $n$ were arbitrary, $A$ is torsion-free.

Depends on

Used by · 1

Torsion freeness after removing a smooth-locus integral curve

Dependency tree · 3 levels

Arrows run from prerequisites to the items that use them. External source nodes link directly to the cited text. The view shows up to three dependency steps; open a prerequisite to continue.

Sources

Jean Fasel and Maria Yakerson, Vector bundles on singular surfaces and threefolds · PDF p. 13, final group-theoretic paragraph in the proof of Proposition 6.1