LINEAR ALGEBRA II: PROJECTIVE MODULES Let R be a ring. By `module' we will mean R-module and by `homomorphism' (respectively isomorphism) we will mean homomorphism (respectively isomorphism) of R-modules, unless explicitly stated otherwise. 1. Definition and motivation Definition 1.1. A module P is projective if, given • any surjective homomorphism " : B C, and • any homomorphism γ : P ! C, there exists a homomorphism β : P ! B such that " ◦ β = γ. Proposition 1.2. Let P be a projective module. Then any surjective homomorphism ∼ " : Q P induces an isomorphism of the form Q = ker(") ⊕ P . Proof. Since P is projective, we may and will define a homomorphism β : P ! Q which satisfies ("◦β)(p) = p for all p 2 P . We may hence define a map α : ker(")⊕P ! Q by the formula α((q; p)) := q+β(p) and proceed to prove that it is an isomorphism. The fact that α is a homomorphism is a straightforward consequence of the fact that so is β. In order to prove injectiveness, we assume that α((q; p)) = q + β(p) = 0 and deduce that β(p) = −q 2 ker(") and hence that p = "(β(p)) = 0. It follows that q = −β(0) = 0 and hence that (q; p) = 0, as required. In order to prove surjectiveness, we let q be any element of Q. We then have that "(q−β("(q))) = "(q)−"(β("(q))) = "(q)−"(q) = 0. So we have that q−β("(q)) belongs to ker(") and finally find that α((q − β("(q));"(q))) = q − β("(q)) + β("(q)) = q. Remark 1.3. Proposition 1.2 shows than when studying the structure of a module, it can be useful to construct a surjective homomorphism to a projective module in order to shift attention to the two components in the direct sum decomposition given by the proposition, whose structure may then turn out to be easier to understand. We proved in Theorem 3.15 b) that free modules have the precise same property that Proposition 1.2 attributes to projective modules. In fact, it is easy to use Theorem 3.15 a) to show that every free module is projective. However, the converse is not true in general, which justifies giving a name to this important class of modules. Theorem 1.4. Every free module is projective. L Proof. Let F = j2J Rfj be a free module and suppose given any surjective homo- morphism " : B C and any homomorphism γ : F ! C. For any j 2 J, let bj be any 1 2 LINEAR ALGEBRA II: PROJECTIVE MODULES element of B such that "(bj) = γ(fj). By Theorem 3.15 a), there exists a homomor- phism β : F ! B such that β(fj) = bj for every j 2 J. But then (" ◦ β)(fj) = γ(fj) for every j 2 J, and since this equality holds for every generator fj of F , it is easy to deduce that " ◦ β = γ, as required. 2. Exact sequences and commutative diagrams Definition 2.1. • Let φ : A ! B and : B ! C be homomorphisms. The φ sequence of homomorphisms A ! B ! C is said to be exact at B if ker( ) = im(φ). • A sequence of homomorphisms M0 ! M1 ! ::: ! Mn ! Mn+1 is called exact if it is exact at M1;:::;Mn. • An exact sequence of the form 0 ! A ! B ! C ! 0 is called a short exact sequence (or s.e.s. for short). Here 0 denotes the trivial module, and both the arrow going from 0 to the A and the arrow going from C to 0 denote the zero homomorphism (i.e., the homomorphism sending every element to zero). Remark 2.2. To understand the significance of short exact sequences, consider the following situations: φ (1) Let 0 ! A ! B ! C ! 0 be a short exact sequence. Exactness at A is equiv- alent to φ being injective, exactness at C is equivalent to being surjective and exactness at B is equivalent to having ker( ) = im(φ). The first isomorphism theorem implies that φ induces an isomorphism of the form A=ker(φ) !∼ im(φ) and that induces an isomorphism of the form B=ker( ) !∼ im( ). So the assumption that the sequence is exact means that φ gives an isomorphism from A to the submodule im(φ) = φ(A) of B and that induces an isomorphism of the form B/φ(A) = B=im(φ) = B=ker( ) !∼ im( ) = C: (2) Conversely, let A be a submodule of a module B. Then we obtain a short exact sequence of the form 0 ! A !ι B !π B=A ! 0 where ι is just the tautological inclusion (i.e., ι(a) = a for all a 2 A) and π is just the natural projection given by π(b) := b + A for any b 2 B. Indeed, ι is clearly injective, π is clearly surjective and ker(π) = A = im(ι), so the sequence is exact. Definition 2.3. We say that a diagram of homomorphisms of the form A −−−!α B ? ? γ? β? y y C −−−!δ D LINEAR ALGEBRA II: PROJECTIVE MODULES 3 is commutative if β ◦ α = δ ◦ γ. This notion generalizes in an obvious way to more complicated diagrams of homomorphisms. Lemma 2.4. Let µ " A1 −−−! A2 −−−! A3 −−−! 0 ? ? ? α1? α2? α3? y y y µ0 "0 0 −−−! B1 −−−! B2 −−−! B3 be a commutative diagram with exact rows and assume that both α1 and α3 are iso- morphisms. Then α2 is an isomorphism. Proof. We first show that α2 is injective: suppose that a 2 A2 is such that α2(a) = 0. 0 Then (α3 ◦ ")(a) = (" ◦ α2)(a) = 0. Since α3 is injective, this implies that "(a) = 0. 0 0 0 0 Since ker(") = im(µ), there exists a 2 A1 such that µ(a ) = a. Then (µ ◦ α1)(a ) = 0 0 (α2 ◦ µ)(a ) = α2(a) = 0. But, since both µ and α1 are injective, this implies that a0 = 0, so that a = µ(a0) = µ(0) = 0, as required. We now show that α2 is surjective: let b 2 B2. Since α3 is surjective, there exists 0 0 a 2 A3 such that α3(a) = " (b), and since " is surjective, there exists a 2 A2 such that "(a0) = a. But then 0 0 0 0 0 0 0 0 0 " (b − α2(a )) = " (b) − (" ◦ α2)(a ) = " (b) − (α3 ◦ ")(a ) = " (b) − " (b) = 0: 0 0 0 0 0 0 Since ker(" ) = im(µ ), there exists b 2 B1 such that µ (b ) = b − α2(a ). Since α1 is 00 00 0 surjective, there exists a 2 A1 such that α1(a ) = b . But then 00 0 0 00 0 0 0 0 α2(µ(a ) + a ) = (µ ◦ α1)(a ) + α2(a ) = µ (b ) + α2(a ) = b; and we have proved that α2 is surjective. µ Definition 2.5. A short exact sequence 0 ! A ! B !" C ! 0 splits if there exists a homomorphism σ : C ! B with the property that " ◦ σ = idC . The homomorphism σ is then called a splitting. Remark 2.6. For any modules A and C, the short exact sequence 0 ! A !ιA A⊕C !πC C ! 0, where ιA(a) = (a; 0) and πC ((a; c)) = c, splits. Conversely, the following also holds: µ " Lemma 2.7. If the short exact sequence 0 ! A ! B ! C ! 0 splits, then B ∼= A ⊕ B. Proof. We fix a splitting σ : C ! B and define a homomorphism : A ⊕ C ! B by setting ((a; c)) := µ(a) + σ(c). We then have that (ιA(a)) = ((a; 0)) = µ(a) and that "( ((a; c))) = "(µ(a)) + "(σ(c)) = 0 + idC (c) = c = πC ((a; c)), where we have used the fact that ker(") = im(µ) so " ◦ µ = 0. We deduce the existence of a 4 LINEAR ALGEBRA II: PROJECTIVE MODULES commutative diagram with exact rows 0 −−−! A −−−!ιA A ⊕ C −−−!πC C −−−! 0 ? ? ? ? ? ? idAy y idC y µ 0 −−−! A −−−! B −−−!" C −−−! 0: Since idA and idC are clearly bijective, Lemma 2.4 implies that is an isomorphism. 3. The group of homomorphisms For any modules A and B, we write HomR(A; B) for the set of homomorphisms from A to B. We define a binary operation on HomR(A; B) as follows: given φ, 2 HomR(A; B), we define an element φ + of HomR(A; B) by the formula (φ + )(a) := φ(a) + (a); a 2 A: Lemma 3.1. The map φ + : A ! B is indeed an element of HomR(A; B). Fur- thermore, this binary operation makes HomR(A; B) into an abelian group (or, equiv- alently, a Z-module). Proof. The first assertion of the lemma is straightforward to prove, as is associativity of the binary operation. The identity element of HomR(A; B) with respect to this binary operation is the zero homomorphism given by sending any element of A to the identity element 0 of B, and for any φ 2 HomR(A; B), we obtain an inverse by setting (−φ)(a) := −φ(a) for every a 2 A. Let β : B1 ! B2 be a homomorphism (of R-modules). We define a map β∗ = HomR(A; β) : HomR(A; B1) ! HomR(A; B2) by setting β∗(φ) := β ◦ φ. Lemma 3.2. (1) β∗ is a Z-module homomorphism. 0 0 0 (2) Given β : B2 ! B3, we have that (β ◦ β)∗ = β∗ ◦ β∗. (3) (idB)∗ = idHomR(A;B) where, for any set S, idS denotes the identity function S ! S (i.e., idS(s) = s for all s 2 S). Proof. We prove claim (1) by obtaining identities β∗(φ + )(a) = β((φ + )(a)) = β(φ(a) + (a)) = β(φ(a)) + β( (a)) = β∗(φ)(a) + β∗( )(a) = (β∗(φ) + β∗( ))(a): Claim (2) is clear since 0 0 0 0 0 (β ◦ β)∗(φ) = (β ◦ β) ◦ φ = β ◦ (β ◦ φ) = β∗(β∗(φ)) = (β∗ ◦ β∗)(φ): Claim (3) is clear since (idB)∗(φ) = idB ◦ φ = φ, so that (idB)∗ is the identity map on HomR(A; B), as required.