Everything reverses, but the proofs do not
Injectivity is projectivity with every arrow reversed: maps into an injective module extend along injections. The formal duality is exact, but the two notions behave quite differently in practice. Free modules make projectives easy to produce, and there is no dual construction; instead, Baer's criterion reduces injectivity to a test on ideals, and a separate argument shows every module embeds in an injective one — which is what injective resolutions need.
Learning objectives
- State the extension property defining injectivity.
- Apply Baer's criterion to test a module.
- Characterise injective abelian groups as the divisible ones.
- Explain why enough injectives requires an argument where projectives do not.
- Describe the injective hull and its uniqueness.
Section 01The extension property
I is injective when for every injection A ↪ B and every map A → I, an extension B → I exists.
Every 0 → A → B → P → 0 splits. Hom(P, −) is exact. Produced from free modules.
Every 0 → I → B → C → 0 splits. Hom(−, I) is exact. Requires a construction with no free analogue.
Reversing arrows turns every statement about projectives into a true statement about injectives. It does not turn the constructions into each other: there is no dual of a basis. Producing injectives is genuinely harder, which is why Baer's criterion matters.
Section 02Baer's criterion
- To test whether I is injective, it suffices to consider injections of ideals into the ring itself.
- For every left ideal J ⊆ Λ and every homomorphism f: J → I:
- check that f extends to a homomorphism Λ → I. Equivalently, f is given by right multiplication by some element of I.
- If every such f extends, then I is injective.
- The proof extends a partial map maximally by Zorn's lemma and uses the criterion to show the maximal extension is total.
Over ℤ the ideals are (n), and the criterion says a map (n) → I extends to ℤ exactly when every element of I is divisible by n. Hence:
ℚ, ℚ/ℤ and the Prüfer groups ℤ(p∞) are the injective abelian groups, and every injective abelian group is a direct sum of copies of them. ℤ itself is not injective, which is why Ext1(−, ℤ) is interesting.
Section 03Enough injectives
That every module is a quotient of a projective is immediate. The dual statement — that every module embeds in an injective — requires work.
- Stage 01Abelian groups firstEmbed M in a divisible group by presenting it as a quotient of a free group and enlarging ℤ to ℚ.
- Stage 02Change ringsFor a Λ-module M, use the adjunction Homℤ(Λ, −) to carry a divisible abelian group to an injective Λ-module.
- Stage 03EmbedCompose to embed M in an injective Λ-module.
- Stage 04IterateRepeating on the cokernel produces an injective resolution of arbitrary length.
The functor Homℤ(Λ, −) is right adjoint to restriction of scalars, and right adjoints of exact functors preserve injectives. This one observation converts the abelian-group case into the general case, and it is the first place adjointness earns its keep in the subject.
Section 04Injective hulls
Every module M has an injective hull: an injective E(M) containing M as an essential submodule, minimal among injectives containing M, and unique up to isomorphism over M.
Projective covers do not always exist — rings over which they do are called perfect. This asymmetry is one of the few places where the projective and injective theories genuinely diverge rather than merely mirror each other.
ReferenceFrequently asked questions
Why is Baer's criterion enough?
Because a maximal partial extension, produced by Zorn's lemma, can be shown total: if some element were outside its domain, the criterion applied to the ideal of ring elements carrying that element into the domain would extend it further, contradicting maximality.
Is ℚ/ℤ injective?
Yes — it is divisible, hence injective as an abelian group. It also serves as a dualizing object: Hom(−, ℚ/ℤ) is exact and faithful, which makes it the standard tool for character-module arguments and for proving flatness criteria.
Do I need injectives if I already have projectives?
For contravariant left exact functors and for right derived functors of covariant left exact functors, yes. Ext can be computed from either a projective resolution in the first variable or an injective resolution in the second, and the agreement of the two is a theorem worth knowing.
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