Modules, Splitting Sequences, and Direct Sums Maria Ross Department - PowerPoint PPT Presentation

Modules, Splitting Sequences, and Direct Sums Maria Ross Department of Mathematics and Computer Science University of Puget Sound May 1, 2017 Modules, Splitting Sequences, and Direct Sums Maria Ross

Modules Over Rings Definition A left R-module M over a ring R is an abelian additive group together with a map R × M → M , denoted by ( r, m ) �→ rm that satisfies the following properties for r, s ∈ R and m, n ∈ M : ◮ ( r + s ) m = rm + sm ◮ r ( m + n ) = rm + rn ◮ ( rs ) m = r ( sm ) ◮ if 1 ∈ R , then 1 m = m Modules, Splitting Sequences, and Direct Sums Maria Ross

Module Examples ◮ Vector Spaces are Matrices over a ring [1] F -modules ◮ R ring, M n ( R ) set of n × n ◮ Rings are matrices over R R -modules ◮ R acts on M n ( R ) by scalar ◮ Abelian Groups multiplication, r �→ rA for r ∈ R are Z -modules and A ∈ M n ( R ) ◮ Ideals are ◮ M n ( R ) is an R − module R -modules Modules, Splitting Sequences, and Direct Sums Maria Ross

Basic Properties Proposition Consider a ring R and an R -module M . Then, for r ∈ R and m ∈ M , ◮ ( r )0 M = 0 M ◮ (0 R ) m = 0 M ◮ ( − r ) m = − ( rm ) = r ( − m ) ◮ ( nr ) m = n ( rm ) = r ( nm ) for all n ∈ Z . Modules, Splitting Sequences, and Direct Sums Maria Ross

Submodules Definition A non-empty subset N of an R − module M is a submodule if for every r, s ∈ R and n, l ∈ N , we have that rn + sl ∈ N . ◮ R ring of integers with ideal 6 Z ⇒ 6 Z is a Z -module ◮ 12 Z subset of 6 Z ◮ x, y ∈ 12 Z ⇒ x = 12 q and y = 12 r for some q, r ∈ Z ◮ ax + by = a (12 q ) + b (12 r ) = 12( aq ) + 12( br ) = 12( aq + br ) ∈ 12 Z ◮ So, 12 Z is a submodule of 6 Z Modules, Splitting Sequences, and Direct Sums Maria Ross

Module Homomorphisms Definition If M and N are R − modules, a module homomorphism from M to N is a mapping f : M → N so that (i) f ( m + n ) = f ( m ) + f ( n ) (ii) f ( rm ) = rf ( m ) for m, n ∈ M and r ∈ R . Modules, Splitting Sequences, and Direct Sums Maria Ross

Quotient Structures Proposition Suppose R is a ring, M an R − module, and N a submodule of M . Then M/N , the quotient group of cosets of N , is an R − module. Modules, Splitting Sequences, and Direct Sums Maria Ross

Quotient Structures Proof. ◮ M/N is an additive abelian group ◮ Define the action of R on M/N by ( r, m + N ) �→ rm + N ◮ By coset operations, for r, s ∈ R and m + N, l + N ∈ M/N , (i) ( r + s )( m + N ) = r ( m + N )+ s ( m + N ) = ( rm + N )+( sm + N ) (ii) r (( m + N ) + ( l + N )) = r ( m + l + N ) = rm + rl + N = ( rm + N ) + ( rl + N ) (iii) ( rs )( m + N ) = ( rsm + N ) = r ( sm + N ) (iv) if 1 ∈ R , then 1( m + N ) = 1 m + N = m + N. Modules, Splitting Sequences, and Direct Sums Maria Ross

Direct Sum of Modules ◮ I set of indices (finite or infinite) ◮ A family ( x i , i ∈ I ) is a function on I whose value at i is x i Definition [8] The external direct sum of the modules M i for i ∈ I is � i ∈ I M i , all families ( x i , i ∈ I ) with x i ∈ M i such that x i = 0 for all except finitely many i . ◮ Addition defined by ( x i ) + ( y i ) = ( x i + y i ) ◮ Scalar multiplication defined by r ( x i ) = ( rx i ) Modules, Splitting Sequences, and Direct Sums Maria Ross

Direct Sum of Modules ◮ For finite I , direct sum corresponds to direct product ◮ M , N are R -modules. Then M ⊕ N = { ( m, n ) | m ∈ M, n ∈ N } ◮ Example: Let M = Z 2 and N = Z 3 be Z -modules, then M ⊕ N = Z 2 ⊕ Z 3 = { (0 , 0) , (0 , 1) , (0 , 2) , (1 , 0) , (1 , 1)(1 , 2) } . Then Z 2 ⊕ Z 3 ∼ = Z 6 . Modules, Splitting Sequences, and Direct Sums Maria Ross

Internal Direct Sum ◮ R -module M with submodules M 1 , M 2 ◮ M is the internal direct sum of M 1 and M 2 if M = M 1 + M 2 and M 1 ∩ M 2 = 0 ◮ Internal direct sum is isomorphic to external direct sum ◮ A direct decomposition of M is M 1 ⊕ M 2 where M ∼ = M 1 ⊕ M 2 ◮ M is indecomposable if M ≇ M 1 ⊕ M 2 for M 1 , M 2 � = 0 Modules, Splitting Sequences, and Direct Sums Maria Ross

Free Modules and Cyclic Modules ◮ Modules with bases are called free ◮ M is a free module, then the rank of M is the number of elements in its basis ◮ An R -module M is cyclic if ∃ a ∈ M so M = aR Proposition A free R -module M is isomorphic to the direct sum of “copies” of R . Modules, Splitting Sequences, and Direct Sums Maria Ross

Free Modules Example Let M and N be free modules over Z , with bases �� 1 � � 0 �� �� 2 � � 0 �� B M = , and B N = , . Then, 0 1 0 2 M = { [ a � : a, b ∈ Z } , and N = { [ 2 a � : a, b ∈ Z } . If M and N b 2 b were vector spaces, they would be isomorphic. However, because our scalar multiples are from a ring that does not have multiplicative inverses, N has no vectors with odd-parity entries. Thus, M and N are not isomorphic. Modules, Splitting Sequences, and Direct Sums Maria Ross

Torsion Let R be an integral domain and M be an R -module: ◮ x ∈ M is a torsion element if rx = 0 for r ∈ R , r � = 0 ◮ T , the set of all torsion elements of M , is a submodule of M ◮ T is called the torsion submodule of M ◮ if T = M , M is a torsion module ◮ if T = { 0 } , M is torsion free Modules, Splitting Sequences, and Direct Sums Maria Ross

Torsion Modules Theorem [5] Let T be a finitely generated torsion module over a PID R , and � a i � ideals of R . Then T is isomorphic to the direct sum of cyclic torsion R -modules; that is, T ∼ = R/ � a 1 � ⊕ · · · ⊕ R/ � a m � for some m and nonzero a i ∈ R . Modules, Splitting Sequences, and Direct Sums Maria Ross

Torsion Modules Theorem [5] If R is a PID, then every finitely generated R -module M is isomorphic to F ⊕ T where F is a finite free R -module and T is a finitely generated torsion R -module, which is of the form T ∼ = � m j = i R/ � a j � . Modules, Splitting Sequences, and Direct Sums Maria Ross

Exact Sequences Suppose R is a ring, M 1 , and M 2 , M 3 are R -modules. A sequence of module homomorphisms f 1 f 2 M 1 M 2 M 3 is exact if im( f 1 ) = ker( f 2 ). Modules, Splitting Sequences, and Direct Sums Maria Ross

Short Exact Sequences An exact sequence of the form f 1 f 2 M 1 M 2 M 3 0 0 is called a short exact sequence . ◮ f 2 is surjective ◮ f 1 is injective Modules, Splitting Sequences, and Direct Sums Maria Ross

Examples of Short Exact Sequences For any R -module M with submodule N , there is a short exact sequence f 1 f 2 M/N 0 N M 0 ◮ f 1 : N → M defined by f 1 ( n ) = n ◮ f 2 : M → M/N defined by f 2 ( m ) = m + N Modules, Splitting Sequences, and Direct Sums Maria Ross

Examples of Short Exact Sequences For ideals I and J of a ring R such that I + J = R , there is a short exact sequence f 1 f 2 I ⊕ J 0 I ∩ J R 0 ◮ f 1 : I ∩ J → I ⊕ J is the map f 1 ( x ) = ( x, − x ) ◮ f 2 : I ⊕ J → R is addition where ker( f 2 ) = { ( x, − x ) | x ∈ I ∩ J } Modules, Splitting Sequences, and Direct Sums Maria Ross

Examples of Short Exact Sequences L and M are R − modules with direct sum L ⊕ M . There is a short exact sequence f 1 f 2 L ⊕ M 0 L M 0 ◮ f 1 : L → L ⊕ M is the embedding of l ∈ L into L ⊕ M ◮ f 2 : L ⊕ M → M is the projection of x ∈ L ⊕ M onto M , so that f 2 ( f 1 ( l )) = 0 for l ∈ L Modules, Splitting Sequences, and Direct Sums Maria Ross

Splitting Sequences Definition [8] A short exact sequence f 1 f 2 M 1 M 2 M 3 0 0 is said to split on the right if there is a homomorphism g 2 : M 3 → M 2 so that the function composition f 2 ◦ g 2 = 1. The sequence splits on the left if there is a homomorphism g 1 : M 2 → M 1 so that f 1 ◦ g 1 = 1. A sequence that splits on the left and the right splits , and is called a splitting sequence . Modules, Splitting Sequences, and Direct Sums Maria Ross

The Five Lemma Consider the following commutative diagram where both sequences of homomorphisms are exact: f 1 f 2 f 3 f 4 M 1 M 2 M 3 M 4 M 5 h 1 h 2 h 3 h 4 h 5 L 1 L 2 L 3 L 4 L 5 g 1 g 2 g 3 g 4 ◮ h 1 surjective h 3 is an ⇒ ◮ h 2 , h 4 bijective isomorphism. ◮ h 5 injective Modules, Splitting Sequences, and Direct Sums Maria Ross

The Five Lemma Proof Claim 1. If h 2 and h 4 are surjective and h 5 is injective, then h 3 is surjective. f 1 f 2 f 3 f 4 M 1 M 2 M 3 M 4 M 5 h 1 h 2 h 3 h 4 h 5 L 1 L 2 L 3 L 4 L 5 g 1 g 2 g 3 g 4 ◮ x ∈ L 3 ⇒ g 3 ( x ) ∈ L 4 ⇒ g 3 ( x ) = h 4 ( y ) for y ∈ M 4 ◮ g 4 ( g 3 ( x )) = g 4 ( h 4 ( y )) = 0 ◮ g 4 ( h 4 ( y )) = h 5 ( f 4 ( y )) ⇒ g 4 ( g 3 ( x )) = h 5 ( f 4 ( y )) = 0 ◮ h 5 ( f 4 ( y )) = 0 ⇒ f 4 ( y ) ∈ ker( h 5 ) ⇒ f 4 ( y ) = 0. ◮ y ∈ ker( f 4 ) = im( f 3 ) ⇒ y = f 3 ( a ) for some a ∈ M 3 . Modules, Splitting Sequences, and Direct Sums Maria Ross

The Five Lemma Proof f 1 f 2 f 3 f 4 M 1 M 2 M 3 M 4 M 5 h 1 h 2 h 3 h 4 h 5 L 1 L 2 L 3 L 4 L 5 g 1 g 2 g 3 g 4 ◮ g 3 ( x ) = h 4 ( y ) = h 4 ( f 3 ( a )) = g 3 ( h 3 ( a )) ⇒ x − h 3 ( a ) ∈ ker( g 3 ) = im( g 2 ) ◮ x − h 3 ( a ) = g 2 ( b ) for b ∈ L 2 ◮ b = h 2 ( m ) for m ∈ M 2 , and x − h 3 ( a ) = g 2 ( b ) = g 2 ( h 2 ( m )) = h 3 ( f 2 ( m )) ◮ x − h 3 ( a ) = h 3 ( f 2 ( m )) ⇒ x = h 3 ( a + f 2 ( m )) ◮ x ∈ im( h 3 ) ⇒ h 3 is surjective. Modules, Splitting Sequences, and Direct Sums Maria Ross