Projective Modules

Recall that Hom(M, -) is left-exact: for any module M and exact , we get an exact sequence

Definition. A module M is projective if Hom(M, -) is exact, i.e. if for any surjective N→N”, the resulting Hom_R(M, N) → Hom_R(M, N”) is also surjective.

Unwinding the definition, we get: if f:N→N” is surjective and g:M→N” is any map, then there exists h:M→N such that fh = g.

Some basic properties:

Proposition.

• The base ring R is projective over itself.
• If is a collection of projective modules, then so is
• If M⊕M’ is projective, then so are M and M’.

Proof

For the first property, Hom_R(R, N) is identified with N and thus is identified with N→N”.

For the second, we use the fact that Hom(-, N) turns direct sums into direct products.

• Suppose N→N” is surjective.
• Since M_i is projective, each Hom_R(M_i, N)→Hom_R(M_i, N”) is surjective.
• Hence ∏_i Hom_R(M_i, N) → ∏_i Hom_R(M_i, N”) is surjective.
• But this is Hom_R(⊕_iM_i, N) → Hom_R(⊕_iM_i, N”).
• Thus ⊕_iM_i is projective.

Finally, if M⊕M’ is projective, then any surjective N→N” gives a surjective map Hom_R(M⊕M’, N)→Hom_R(M⊕M’, N”), which is the direct sum of Hom_R(M, N)→Hom_R(M, N”) and Hom_R(M’, N)→Hom_R(M’, N”). Hence both maps are surjective too and M, M’ are projective. ♦

The following lemma will be used multiple times.

Lemma. If f : M→P is a surjective map to a projective module P, then P is a direct summand of M, i.e. there exists a module Q such that

Proof

Indeed, given the identity i: P→P there is a map h : P→M such that fh = i. By the splitting lemma, P is a direct summand of M.

Corollary.

M is projective if and only if it is a direct summand of a free module.

Proof

⇐: from the above properties, R is projective, so a direct sum of R‘s (i.e. a free module) is also projective. Hence if M is a direct summand of a free module, M is projective.

⇒: pick a free module and a surjective map f: F→M (e.g. F can be freely generated by elements of M). By the above lemma, M is a direct summand of F. ♦

Here’s a special case.

Lemma. If R is a semisimple ring, then every module is projective.

Proof

Every R-module is a direct sum of simple modules, and each simple module is a direct summand of R (and thus projective). ♦

Example

If we let R be the ring of 2 × 2 upper-triangular matrices with real entries, then from:

we find at least three examples of projective submodules of R. Exercise: find a non-projective R-module.

When the Base Ring is Artinian

Suppose now R is artinian (and hence noetherian). For the rest of this article, we will only look at finitely generated R-modules.

Theorem. There is a bijection between

1. simple R-modules M up to isomorphism, and
2. finitely generated indecomposable projective R-modules P up to isomprhism,

via , where J := J(R) is the Jacobson radical of R.

Proof

Note that all modules are of finite length since they’re artinian and noetherian.

Step 1. If P, Q are projective, then Hom(P, Q) → Hom(P/JP, Q/JQ) is surjective.

If f : P→Q is a map of projective modules, we have f(JP) =J·f(P) ⊆ JQ, so f induces a map P/JP→Q/JQ. Conversely, if g : P/JP → Q/JQ is a map, then composing with the canonical map gives P → Q/JQ; since P is projective, this pulls back to a map P → Q.

Step 2. If P is projective and indecomposable, then P/JP is simple.

By step 1, End(P) → End(P/JP) is a surjective ring homomorphism. Since P is indecomposable, End(P) is local, and a quotient of a local ring is local, so End(P/JP) is local and thus P/JP is also indecomposable. But then P/JP is semisimple, so it is simple.

Step 3. If P, Q are projective and indecomposable such that P/JP ≅ Q/JQ, then P ≅ Q.

Suppose g : P/JP → Q/JQ is an isomorphism. By step 1, g comes from some f : P → Q. Since g is surjective, But then so repeating this gives us for all n. Since R is artinian, J is nilpotent so f is surjective. The above lemma then says Q is a direct summand of P, so P = Q ⊕ Q’ for some Q’. Since P is indecomposable, Q’=0 and P≅Q.

Step 4. It remains to show: for simple M, there is a projective indecomposable P such that P/JP ≅ M.

Now, M ≅ R/I for a maximal left ideal I of M. By Krull-Schmidt, write R as a direct sum of indecomposable modules Each is projective since it is a direct summand of R. Among all one of them is non-zero and is surjective (since M is simple). By the above  is simple; since it surjects onto M, we get an isomorphism. ♦

Recall that every simple module is a quotient of R; here is an analogous result for projective indecomposable P.

Lemma. If P is a finitely-generated indecomposable projective module over R, then it is a direct summand of R.

Proof

Since P is projective and finitely-generated, we have a surjective map for some n>0. By the above lemma, since P is projective. Now apply the Krull-Schmidt theorem: since P is indecomposable, it must be a direct summand of R. ♦