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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

Over a PID, injective modules are exactly divisible modules Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 21 results over 6 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)