Throughout is a ring with 1 and all modules are left -modules.
Definition 1. An -module is called projective if it satisfies the following condition: given -modules and -module homomorphisms and where is onto, there exists an -module homomorphism such that
Example. Every free -module is projective. To see this, let be a basis for and suppose that and are -module homomorphisms and is onto. So for every there exists some such that Now define by Since is a basis for the map is well-defined. Clearly is an -module homomorphism and
Thus and so is projective.
Definition 2. Let be -modules and be -module homomorphisms. The sequence is called exact if for all
Remark 1. So, by Definition 2, the sequence is exact if and only if is one-to-one, is onto and
Remark 2. Recall that every -module is a quotient of a free -module. That means there exists a free -module and an onto -module homomorphism Let Then we have an exact sequence where is the inclusion map.
Lemma 2. Let and be -modules. If is free, then and are projective.
Proof. Let We only need to prove that is projective. Define the maps and by and for all and Note that We now prove that is projective. Suppose that we have -module homomorphisms and where is onto. So we have an -module homomorphism and thus, since is projective by the above example, there exists an -module homomorphism such that Let Then
Lemma 3. Let be -modules. Let and be -module homomorphisms such that Then
Proof. First note that the condition implies that is one-to-one because if for some then So we only need to prove that Let Then for some and f_2(b)=0.$ But and so Thus and so Now let be any element of Then because and Hence and so
Definition 3. We say that the exact sequence splits if
Theorem. An -module is projective if and only if every exact sequence splits.
Proof. Suppose first that is projective and is an exact sequence of -modules. So is onto and thus, since is projective, there exists an -module homomorphism such that Therefore, by Lemma 3, we have because is one-to-one. Conversely, suppose that every exact sequence splits. By Remark 2, there exists an exact sequence such that is free. This sequence, by hypothesis, splits and so Thus is free and we’re now done by Lemma 2.