An injective module over is an injective object in .
This is the dual notion of a projective module.
Let be a monomorphism in , and let be a map. We must extend to a map . Consider the poset whose elements are pairs where is an intermediate submodule between and and is an extension of , ordered by if contains and extends . By an application of Zorn's lemma, this poset has a maximal element, say . Suppose is not all of , and let be an element not in ; we show that extends to a map , a contradiction.
is well-defined and extends , as desired.
Assume that the axiom of choice holds.
By Baer’s criterion, it suffices to show that for any ideal of , a module homomorphism extends to a map . Since is Noetherian, is finitely generated as an -module, say by elements . Let be the projection, and put . Then for each , is nonzero for only finitely many summands. Taking all of these summands together over all , we see that factors through
for some finite . But a product of injectives is injective, hence extends to a map , which completes the proof.
This is due to Bass and Papp. See (Lam, Theorem 3.46).
Observe that an object is injective precisely if the hom-functor into it sends monomorphisms to epimorphisms, and that preserves monomorphisms by assumption of exactness. With this the statement follows via adjunction isomorphism
We discuss that in the presence of the axiom of choice at least, the category Mod has enough injectives in that every module is a submodule of an injective one. We first consider this for . We do assume prop. 5, which may be proven using Baer's criterion.
Now by the discussion at projective module every abelian group receives an epimorphism from a free abelian group, hence is the quotient group of a direct sum of copies of . Accordingly it embeds into a quotient of a direct sum of copies of .
Here is divisible because the direct sum of divisible groups is again divisible, and also the quotient group of a divisible groups is again divisble. So this exhibits an embedding of any into a divisible abelian group, hence into an injective -module.
Then if has enough injectives, also has enough injectives.
Consider . By the assumption that has enough injectives, there is an injective object and a monomorphism . The adjunct of this is a morphism
and so it is sufficient to show that
is injective in ;
is a monomorphism.
The first point is the statement of lemma 1.
By the assumption that is an exact functor, the image of this sequence under is still exact
Now observe that is a monomorphism: this is because its composite with the adjunction unit is (by the formula for adjuncts) the original morphism , which by construction is a monomorphism. Therefore the exactness of the above sequence means that is the zero morphism; and by the assumption that is a faithful functor this means that already is zero, hence that , hence that is a monomorphism.
In particular if the axiom of choice holds, then has enough injectives.
Observe that the forgetful functor has both a left adjoint (extension of scalars from to ) and a right adjoint (coextension of scalars). Since it has a left adjoint, it is exact. Thus the statement follows via lemma 2 from prop. 2.
Using Baer’s criterion, prop. 1.
The group of rational numbers is injective in Ab, as is the additive group of real numbers and generally that underlying any field. The additive group underlying any vector space is injective. The quotient of any injective group by any other group is injective.
flat object, flat resolution
The notion of injective modules was introduced in
(The dual notion of projective modules was considered explicitly only much later.)
A general discussion can be found in
The general notion of injective objects is in section 9.5, the case of injective complexes in section 14.1.
Baer’s criterion is discussed in many texts, for example
Section 4.2 of