Projective Modules

Recall that Hom(M, -) is left-exact: for any module M and exact $0\to N' \to N\to N''$, we get an exact sequence $0\to \text{Hom}_R(M, N') \to \text{Hom}_R(M, N) \to \text{Hom}_R(M,N'').$

Definition. A module M is projective if Hom(M, -) is exact, i.e. if for any surjective N→N”, the resulting HomR(M, N) → HomR(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 $\{M_i\}$ is a collection of projective modules, then so is $\oplus_i M_i.$
• If M⊕M’ is projective, then so are M and M’.

Proof

For the first property, HomR(R, N) is identified with N and thus $\text{Hom}_R(R, N) \to \text{Hom}_R(R, N'')$ is identified with NN”.

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

• Suppose NN” is surjective.
• Since Mi is projective, each HomR(Mi, N)→HomR(Mi, N”) is surjective.
• Hence ∏i HomR(Mi, N) → ∏i HomR(Mi, N”) is surjective.
• But this is HomR(⊕iMi, N) → HomR(⊕iMi, N”).
• Thus ⊕iMis projective.

Finally, if MM’ is projective, then any surjective NN” gives a surjective map HomR(MM’, N)→HomR(MM’, N”), which is the direct sum of HomR(M, N)→HomR(M, N”) and HomR(M’, N)→HomR(M’, N”). Hence both maps are surjective too and MM’ 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 $M\cong P \oplus Q.$

Proof

Indeed, given the identity iPP there is a map hPM such that fh = i. By the splitting lemmaP 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 fFM (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:

$R = \left\{\begin{pmatrix} * & * \\ 0 & * \end{pmatrix}\right\} = \left\{ \begin{pmatrix} * & 0 \\ 0 & 0 \end{pmatrix}\right\} \oplus \left\{ \begin{pmatrix} 0 & * \\ 0 & *\end{pmatrix}\right\} = \left\{ \begin{pmatrix} a & a \\ 0 & 0\end{pmatrix}\right\} \oplus \left\{ \begin{pmatrix} 0 & * \\ 0 & * \end{pmatrix} \right\}$

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 $P\mapsto M := P/JP$, 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 PQ are projective, then Hom(PQ) → Hom(P/JPQ/JQ) is surjective.

If fPQ is a map of projective modules, we have f(JP) =J·f(P) ⊆ JQ, so f induces a map P/JPQ/JQ. Conversely, if gP/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 PQ are projective and indecomposable such that P/JP ≅ Q/JQ, then P ≅ Q.

Suppose gP/JP → Q/JQ is an isomorphism. By step 1, g comes from some fP → Q. Since g is surjective, $Q = f(P) + JQ.$ But then $JQ = f(JP) + J^2 Q,$ so repeating this gives us $Q = f(P) + J^n Q$ 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 PQ ⊕ Q’ for some Q’. Since P is indecomposable, Q’=0 and PQ.

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 $\oplus_i P_i \cong R \to M.$ Each $P_i$ is projective since it is a direct summand of R. Among all $P_i \to M$ one of them is non-zero and is surjective (since M is simple). By the above $P_i/JP_i$ 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 $R^n \to P$ for some n>0. By the above lemma, $R^n\cong P\oplus Q$ since P is projective. Now apply the Krull-Schmidt theorem: since P is indecomposable, it must be a direct summand of R. ♦

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2 Responses to Projective Modules and Artinian Rings

1. 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.

• limsup says:

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