Module Theory

Module Theory

Module Theory Claudia Menini 2014-2015 Contents Contents 2 1 Modules 4 1.1 Homomorphisms and Quotients ..................... 4 1.2 Quotient Module and Isomorphism Theorems . 10 1.3 Product and Direct Sum of a Family of Modules . 13 1.4 Sum and Direct Sum of Submodules. Cyclic Modules . 21 1.5 Exact sequences and split exact sequences . 27 1.6 HomR (M; N) ............................... 32 2 Free and projective modules 35 3 Injective Modules and Injective Envelopes 41 4 Generators and Cogenerators 59 5 2 × 2 Matrix Ring 65 6 Tensor Product and bimodules 68 6.1 Tensor Product 1 ............................. 68 6.2 Bimodules ................................. 78 6.3 Tensor Product 2 ............................. 82 7 Homology 94 7.1 Categories and Functors ......................... 94 7.2 Snake Lemma ...............................100 7.3 Chain Complexes .............................105 7.4 Homotopies ................................113 7.5 Projective resolutions . 116 7.6 Left Derived functors . 123 7.7 Cochain Complexes and Right Derived Functors . 139 8 Semisimple modules and Jacobson radical 145 9 Chain Conditions. 160 2 CONTENTS 3 10 Progenerators and Morita Equivalence 173 10.1 Progenerators ...............................173 10.2 Frobenius .................................199 Chapter 1 Modules 1.1 Homomorphisms and Quotients R Definition 1.1. Let R be a ring. A left R-module is a pair (M; µM ) where (M+; 0) is an abelian group and R µ = µM : R × M ! M is a map such that, setting a · x = µ ((a; x)) , the following properties are satisfied : M1 a · (x + y) = a · x + a · y; M2 (a + b) · x = a · x + b · x; M3 (a ·R b) x = a · (b · x); M4 1R · x = x for every a; b 2 R and every x; y 2 M. In this case we will say that M is a left R-module. The notation RM will be used to mean that M is a left R-module. Definition 1.2. Let R be a ring and let Rop denote the opposite ring of R.A right R-module is a left Rop-module i.e. it is a pair (M; µ0) where (M+; 0) is an abelian group and 0 R × ! µ = µM : R M M is a map such that, setting a · x = µ0 ((a; x)) , M10 a · (x + y) = a · x + a · y; 4 1.1. HOMOMORPHISMS AND QUOTIENTS 5 M20 (a + b) · x = a · x + b · x; 0 M3 (a ·Rop b) · x = a · (b · x); 0 M4 1R · x = x for every a; b 2 R and every x; y 2 M. In this case we will say that M is a right R-module. The notation MR will be used to mean that M is a right R-module. Note that a ·Rop b = b ·R a so that M3' rewrites as (a ·R b) · x = (b ·Rop a) · x = b · (a · x) For this reason, if M is a right R-module, one usually writes x · a instead of a · x, for every a 2 R, x 2 M. With this notation the conditions M1'), M2'), M3'), M4') may be rephrased as follows: M1" (x + y) · a = x · a + y · a; M2" x · (a + b) = x · a + x · b; M3" x · (a ·R b) = (x · a) · b; M4" x · 1R = x. The abelian group M is called the underlying additive group of the left ( resp. right) R-module M. Given x; y 2 M we will write x − y instead of x + (−y). Remark 1.3. If R is a commutative ring, then every left R-module is, in a natural way, a right R-module, and conversely. In fact, let M be a left R-module, given a; b 2 R, x 2 M, we have a · (b · x) = (a ·R b) x = (b ·R a) x = b · (a · x) . In the same way, if M is a right R-module, given a; b 2 R, x 2 M, we have: (x · a) · b = x (a ·R b) = x · (b ·R a) = (x · b) · a. Therefore, when R is a commutative ring, we will, in general, simply say that M is an R-module. Examples 1.4. 6 CHAPTER 1. MODULES 1. Let G be an abelian group with additive notation. G becomes, in a natural way, a Z- module by defining, for every n 2 Z and x 2 G, n · x = nx where nx denotes the nth power of x in the additive notation. 2. Let A be a ring, R be a subring of A. A becomes a left (resp. right) R-module by setting, for every r 2 R, a 2 A, ra ( resp. ar) to be the product of the element r 2 R ⊆ A with the element a 2 A ( resp. of the element a 2 A with the element r 2 R ⊆ A) in the ring A. In particular the rings R; R[X];R[[X]] may be considered as left (resp. right) R-modules. If D is a commutative domain, Q (D) is a D-module. 3. (More generally,) let f : R ! A be a ring homomorphism. Any left A-module A M; µM inherits the structure of a left R-module by setting R A µM ((r; x)) = µM ((f (r) ; x)) for every r 2 R and x 2 M i.e. r · x = f (r) · x for every r 2 A and x 2 M: This module is often denoted by f∗ (M) and called the R-module obtained by restriction of the ring of scalars from A to R. 1.5. If R is a division ring and M is a left (resp. right) R-module we say that M is a left (resp. right) vector space over R. If R is a field, we simply say that M is a vector space over R. Proposition 1.6. Let R be a ring, M a left R-module. Then, for every a; b 2 R and for every x; y 2 M we have : 1. a · 0M = 0M ; 2. 0R · x = 0M ; 3. (−a) · x = −a · x = a · (−x) ; (−a) · (−x) = a · x; 4. a · (x − y) = a · x − a · y; 5. (a − b) · x = a · x − b · x . 6. n (a · x) = (na) · x = a · (nx) for every n 2 Z; a 2 R; x 2 M: 1.1. HOMOMORPHISMS AND QUOTIENTS 7 Proof. 1) Let us start from : a·0M = a (0M + 0M ) = a·0M +a·0M : Adding − (a · 0M ) to both sides we find : 0M = a · 0M : 2) First we look at the obvious : 0R · x = (0R + 0R) x = 0R · x + 0R · x: Adding − (x · 0R) to both sides we find : 0M = 0R · x: 3) From (−a) x + ax = ((−a) + a)x = 0R · x = 0M we obtain that (−a) x = −ax. In a similar way it follows from ax + a (−x) = a(x + (−x)) = a · 0M = 0M that a (−x) = −ax. Moreover : (−a)(−x) = −(a(−x)) = −(−(ax)) = ax: 4) We calculate: a(x − y) = a(x + (−y)) = ax + a(−y) = ax + (−(ay)) = ax − ay. 5) We calculate: (a − b) x = (a + (−b))x = ax + (−b) x = ax + (−bx) = ax − bx. 6) It is easily proved by Induction. 1.7. Let M be an abelian group and let A = End(M) denote the ring of endomor- phisms of M. Then M becomes a left A-module by setting f · x = f (x) every f 2 A and x 2 M: In fact,note that (f ·A g) x = (f ◦ g)·x = (f ◦ g)(x) = f (g (x)) = f·(g · x) for every f; g 2 A and x 2 M: Now let ' : R ! End(M) be a ring morphism. Then, in view of Example 3 in 1.4, we can consider the left R-module '∗ (M) i.e. M becomes a left R-module by setting r · m = '(r)(m) for all r 2 R and for all m 2 M. Conversely let M be a left R-module and let End(M) denote the ring of endomor- phisms of the abelian group underlying the R-module structure of M. For every r 2 R consider the map t : M ! M r . m 7! r · m Clearly tr 2 End(M) and the map : R ! End(M) r 7! tr is a ring morphism. In this way we get: Theorem 1.8. Let R be a ring and let M be an abelian group. The ring morphisms ' : R ! End(M) correspond bijectively to the left R-module structures on M. 8 CHAPTER 1. MODULES Proof. Using notation as above, given a ring morphisms ' : R ! End(M) we have: (r)(m) = r · m = '(r)(m). Conversely, if M is a left R-module we have: r · m = (r)(m). To get an analogous result for right R-modules we have to consider the ring End(M)op which has the same addition as End(M) but where multiplication is defined by f · g = g ◦ f. Rephrasing the foregoing theorem we obtain : Theorem 1.9. Let R be a ring and let M be an abelian group. The ring morphisms ' : R ! End(M)op correspond bijectively to the right R-module structures on M. Definitions 1.10. Let R be a ring and let M be a left R-module. A subset N of M is said to be an R-submodule ( or simply submodule) of M if : 1. N is a subgroup of M; 2. a 2 R and x 2 N implies that a · x 2 N, for every a 2 R and x 2 N. We write N ≤R M to mean that N is a submodule of M. We denote by L(RM) the set of all the submodules of RM.

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    201 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us