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Baer's criterion for injective modules

Statement

Assume the Axiom of Choice. A left $R$ -module $I$ is injective if and only if every homomorphism $f : J \to I$ from a left ideal $J \leq R$ extends to a homomorphism $R \to I$ .

The forward implication is choice-free. The converse uses AC through Zorn's lemma.

Facts & Assumptions

Given: A unital ring $R$ and a left $R$ -module $I$ .

[F1]

Injectivity is extension of homomorphisms along every module monomorphism (Injective modules and the extension property).

[L1]

Under AC, every nonempty poset in which every chain has an upper bound has a maximal element (Zorn's lemma, The Axiom of Choice).

Proof

technique · direct

1.1

If $I$ is injective, apply [F1] to the inclusion of any left ideal $J ↪ R$ to extend each $J \to I$ to $R \to I$ .

assume-hypF1

1.2

Conversely, assume the ideal-extension condition. Given a submodule $N \leq M$ and a homomorphism $f : N \to I$ , let $P$ be the poset of extensions $(N^{'}, f^{'})$ with $N \leq N^{'} \leq M$ , ordered by further extension. It is nonempty because $(N, f) \in P$ .

assume-hypconstruct

2.1

The union of a chain of compatible extensions is a submodule and carries the unique map agreeing with every map in the chain, so it is an upper bound. By [L1], choose a maximal extension $(N_{0}, f_{0})$ .

step 1.2L1choose

3.1

If $N_{0} \neq M$ , choose $x \in M ∖ N_{0}$ and put $J = {r \in R : r x \in N_{0}}$ , a left ideal. The map $h : J \to I$ , $h (r) = f_{0} (r x)$ , is $R$ -linear and by hypothesis extends to $H : R \to I$ . Put $y = H (1)$ .

step 2.1chooseconstruct

4.1

Define $f_{1} : N_{0} + R x \to I$ by $f_{1} (n + r x) = f_{0} (n) + r y$ . If $n + r x = n^{'} + r^{'} x$ , then $(r - r^{'}) x = n^{'} - n \in N_{0}$ , so $r - r^{'} \in J$ and $f_{0} (n^{'} - n) = H (r - r^{'}) = (r - r^{'}) y$ ; hence the formula is well defined. It is linear and extends $f_{0}$ .

step 3.1algebraconstruct

5.1

Since $x \notin N_{0}$ , the domain $N_{0} + R x$ strictly contains $N_{0}$ , contradicting maximality. Thus $N_{0} = M$ , so $f$ extends to $M$ and $I$ is injective by [F1].

step 2.1step 3.1step 4.1F1

6.1

Steps 1.1 and 5.1 prove both directions, with Zorn used only in the converse.

step 1.1step 5.1∎

Depends on

Used by

Dependency tree · two levels

11 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)