## Projective Modules

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

Definition. A module M isprojectiveif 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*is projective, each Hom_{i}(_{R}*M*,_{i}*N*)→Hom(_{R}*M*,_{i}*N”*) is surjective. - Hence ∏
Hom_{i}(_{R}*M*,_{i}*N*) → ∏Hom_{i}(_{R}*M*,_{i}*N”*) is surjective. - But this is Hom
(⊕_{R}_{i}*M*,_{i}*N*) → Hom(⊕_{R}_{i}*M*,_{i}*N”*). - Thus ⊕
_{i}*M*is projective._{i }

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

- simple R-modules M up to isomorphism, and
- 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*. ♦

In your example of 2 x 2 upper-triangular matrices, I only see two isomorphism classes of projective modules: the [[a,a],[0,0]] guy is isomorphic to the [[a,0],[0,0]] guy by the obvious map.

Yes, I said there are three examples of projective submodules, but two of them are isomorphic.