Injective means divisible
Over a principal ideal domain, Baer's criterion collapses to a single condition: a module is injective exactly when it is divisible, meaning every element can be divided by every non-zero ring element. For abelian groups this classifies the injectives completely as direct sums of copies of ℚ and of Prüfer groups, and it gives every module an injective resolution of length at most one — the mirror of the projective situation.
Learning objectives
- Characterise injectivity over a PID as divisibility.
- Classify the injective abelian groups.
- Construct a length-one injective resolution.
- Explain the symmetry with the projective case and where it breaks.
Section 01Divisibility
A module D over a domain R is divisible when for every d ∈ D and non-zero r ∈ R there is d′ with rd′ = d. Over a PID this is exactly injectivity, by Baer's criterion applied to the ideals (r).
| Group | Divisible? | Injective? |
|---|---|---|
| ℚ | Yes | Yes |
| ℚ/ℤ | Yes | Yes |
| ℤ(p∞) (Prüfer) | Yes | Yes |
| ℝ, ℂ | Yes | Yes |
| ℤ | No | No |
| ℤ/nℤ | No | No — but injective as a ℤ/nℤ-module when n is squarefree |
ℤ/pℤ is not injective over ℤ, but it is injective over ℤ/pℤ. Whenever the phrase ‘injective module’ appears, the ring must be named — the property is not intrinsic to the underlying abelian group.
Section 02Classification over ℤ
Every divisible abelian group is a direct sum of copies of ℚ and of Prüfer groups ℤ(p∞), and the multiset of summands is an invariant. The torsion-free part is a ℚ-vector space; the p-primary part is a sum of Prüfer groups.
ℤ(p∞) is the union of the cyclic groups of order pn, equivalently the p-primary part of ℚ/ℤ. It is the smallest injective containing ℤ/pℤ, so it is the injective hull of ℤ/pℤ — the standard first example of a hull.
Section 03Short injective resolutions
- Embed M in an injective module I0 — over ℤ, embed in a divisible group.
- Set I1 = I0/M.
- I1 is divisible, being a quotient of a divisible module. This is the step that fails over a general ring.
- So 0 → M → I0 → I1 → 0 is an injective resolution of length 1.
- Hence Extn(−, M) = 0 for n ≥ 2, agreeing with the projective computation.
This closure property is what makes the resolution stop. It is the injective mirror of ‘submodules of free modules are free’, and both are special to hereditary rings.
ReferenceFrequently asked questions
Why is ℚ/ℤ so useful?
Because it is injective and cogenerates: a module is zero if and only if its dual into ℚ/ℤ is zero. That makes Hom(−, ℚ/ℤ) an exact faithful functor, which converts statements about a module into dual statements that are often easier to check — flatness criteria being the standard use.
Is a divisible group ever free?
Only the zero group. A non-zero free abelian group has elements not divisible by 2, so divisibility and freeness are incompatible except trivially. This is a compact illustration of how far the projective and injective worlds sit apart.
Does divisible imply injective over any domain?
Over a Dedekind domain, yes. Over a general domain, divisibility is necessary but not sufficient; injectivity is the stronger condition, and Baer's criterion is what must be checked.
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