Math 2.0

Commutative algebra Definition

Injective module

Builds on Commutative ringModule over a ringLinear mapExact sequencePrincipal ideal domain

Definition

Let RR be a commutative ring and let EE be an RR-module. The module EE is injective if it satisfies the following extension property. Let MM, NN be RR-modules, i ⁣:M→Ni \colon M \to N an injective RR-linear map, and f ⁣:M→Ef \colon M \to E an RR-linear map. Then there is an RR-linear map g ⁣:N→Eg \colon N \to E with g∘i=fg \circ i = f.

Remarks

The following are equivalent for an RR-module EE:

  1. EE is injective.
  2. Every injective linear map E→NE \to N has a left inverse, that is, it splits.
  3. The functor Hom⁡R(−,E)\Hom_R(-, E) takes exact sequences to exact sequences.
  4. (Baer's criterion) Every linear map from an ideal II of RR to EE extends to RR.

Over a principal ideal domain, a module EE is injective exactly when it is divisible: rE=ErE = E for every nonzero r∈Rr \in R. The same definition works over any ring, commutative or not.

Example: ℚ\Q is an injective ℤ\Z-module. By Baer's criterion it suffices to extend a map f ⁣:nℤ→ℚf \colon n\Z \to \Q with n≠0n \ne 0; with q=f(n)q = f(n), the map g(x)=xq/ng(x) = xq/n on ℤ\Z satisfies g(n)=qg(n) = q.

Non-example: ℤ\Z is not an injective ℤ\Z-module. Take i ⁣:ℤ→ℤi \colon \Z \to \Z, x↦2xx \mapsto 2x, and ff the identity of ℤ\Z. An extension gg would satisfy 1=f(1)=g(2)=2g(1)1 = f(1) = g(2) = 2g(1), which has no solution in ℤ\Z.

Reading path

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  1. Commutative ring concept
  2. Exact sequence concept
  3. Linear map concept
  4. Module over a ring concept
  5. Principal ideal domain concept
  6. Injective module definition

Further reading

  1. Web https://en.wikipedia.org/wiki/Injective_module
About this page
TagRE5T59YZ46
URIm2://RE5T59YZ46
Kinddefinition
WritingW0 stub
MSC13C11, 16D50
DefinesInjective module
UsesCommutative ring, Module over a ring, Linear map, Exact sequence, Principal ideal domain
Changed2 November 2026

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