the injective envelope of a module

E. L. Lady

Theorem. Let E′ be an injective module and M ⊆ E ⊆ E′. The following conditions are equivalent:

(1) E is a maximal essential extension of M.

(2) E is a minimal injective submodule of E′ containing M.

(3) E is an essential extension of M and is injective.

(4) E is injective and every monomorphism from M into an injective module Q extends to a monomorphism from E into Q.

definition. If the above conditions are satisfied, we say that E is an injective envelope of M.

Lemma. If condition 1) is satisfied, then E is a summand of E′.

proof: By Zorn’s Lemma there exists a submodule F ⊆ E′ maximal with respect to the property F ∩ E = 0. We will show that E′ = E ⊕ F. Let π: E ⊕ F → E be the projection. Since E′ is injective, π extends to a map π′: E′ → E′.

We now claim that E is an essential submodule of π′(E′). In fact, let π′(x) ∈ π′(E′). If x ∈ E ⊕ F then π′(x) = π(x) ∈ E, so we only need consider the case x ∉ E ⊕ F. Then x ∉ F so by the maximality of F, (F ⊕ Rx) ∩ E ≠ 0, so there exists f + rx ≠ 0 ∈ E with f ∈ F, r ∈ R. Then rπ′(x) = π′(f + rx) ∈ π′(E) = π(E) = E and rπ′(x) = π′(f + rx) ≠ 0 since f + rx ∉ F, otherwise F ∩ E ≠ 0. This shows that π′(E′) is an essential extension of E, so by the maximality of E, π′(E′) = E. Thus π′ is an endomorphism of E′ such that π′(e) = π(e) = e for e ∈ E. This shows that E is a summand of E′. ✓

We claim that E′/F is isomorphic to a submodule of E′ which is an essential extension of E. In fact the quotient map E′ → E′/F restricts to a monomorphism ζ: E → E′/F because E ∩ F = 0. Now if x + F ≠ 0 ∈ E′/F then x ∉ F so Rx + F properly contains F, so by the maximality of F we see that (Rx + F) ∩ E ≠ 0, which says that in E′/F the coset x + F has a non-trivial multiple in ζ(E) = (E + F)/F. This means that E′/F is an essential extension of ζ(E). Now since E′ is injective, then inclusion map E ↪ E′ extends to a map θ: E′/F → E′, and since E′/F is an essential extension of ζ(E) it follows that θ is a monomorphism.

0 ⟶ E ⟶^{ζ} E′/F
↓^{⊆}
E′

Thus θ embeds E′/F into a submodule θ(E′/F) of E′ which is an essential extension of E. By the transitivity of essential extensions θ(E′/F) is also an essential extension of M. Since E is a maximal essential extension of M, we conclude that θ(E′/F) = E. From this we in turn conclude that E′/F = ζ(E) = (E + F)/F and so E + F = E′, so that E′ = E ⊕ F.

proof of theorem: (1) ⇒ (2): By the Lemma, E is a summand of E′, hence is injective. Now if M ⊆ E_{1} ⊆ E and E_{1} is injective then E_{1} is a summand of E. Since E is an essential extension of E_{1} , this is only possible if E_{1} = E. Thus E is a minimal injective submodule of E′ containing M.

(2) ⇒ (3): Assume that E is a minimal injective module containing M. Using Zorn’s Lemma, we get a maximal essential extension E_{1} of M with E_{1} ⊆ E. Since E is injective, (1) ⇒ (2) shows that E_{1} is injective. Thus by the minimality of E we conclude E_{1} = E, so that E is an essential extension of M.

(3) ⇒ (4): Immediate from the properties of injective modules and essential extensions.

(4) ⇒ (3): Again by Zorn’s Lemma we get an essential extension E_{1} of M satisfying (1) with E_{1} ⊆ E. So E_{1} is injective, and so by (4) the inclusion map M ↪ E_{1} extends to a monomorphism from E into E_{1} . Then since E_{1} is an essential extension of M, E must also be an essential extension of M. (In fact, of course, we must have E_{1} = E.)

(3) ⇒ (1): Since an injective module is a summand of every module that contains it, it can have no proper essential extensions. It then follows from the transitivity of injective extensions that no module properly containing E can be an essential extension of M.