Algebra (Unknown 27). - Index of

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6 Algebra 2 Proof. Clearly f is surjective. Its kernel is {(n, k) : n + k = 0}. Then n = −k ∈ N ∩ K. A map {(n, k) : n + k = 0} → N ∩ K, (n, −n) ↦→ n is a bijection. 2.5. The submodule M generated by elements x i is the minimal submodule which contains all of them, it consists of finite A-linear combinations of x i ; elements x i ∈ M are called generators of M . The minimal number of generators (if it exists) of M is called the rank of M . M is said to be of finite type if it has a finite number of generators. Every module of finite type has finite rank. An A-module M is called free if M has generators x i such that ∑ a i x i = 0 implies a i = 0 for all i. The set of x i is called then a basis of M . 2.6. Lemma. (1) The module A n = ⊕ 1in A is free of rank n. (2) Let M be an A-module of finite type and let x 1 , . . . , x n be generators of M . Define a homomorphism f: A n → M, (a 1 , . . . , a n ) ↦→ ∑ a i x i . It is surjective. If N is the kernel of f , then M is isomorphic to the quotient module (A) n /N . Thus, every A-module of finite type is isomorphic to a quotient of a free module. (3) Every free module of finite rank n is isomorphic to A n . Proof. (1), (2) follow from the definitions. If M is free and the number of generators is finite equal to n, then the homomorphism (A) n → M is surjective and injective. Elements of N serve as relations for generators of M . As a corollary we deduce that the direct sum of free modules is free: A n ⊕ A m ≃ A n+m . Examples. 1. From linear algebra it is known that every module of finite rank over a field has a basis and is free. 2. Let A = Z and M = Z/nZ for n > 1. Then M has rank 1 and is not a free A-module, since every element of M is annihilated by multiplication by n. 3. Polylinear constructions 3.1. The set of A-module homomorphisms from an A-module M to N is an A-module: af(m) = f(am), (f + g)(m) = f(m) + g(m). It is denoted Hom A (M, N). Examples. Hom A (0, N) = Hom A (M, 0) = 0. Hom A (A, N) ≃ N , Hom A (M, N 1 ⊕ N 2 ) ≃ Hom A (M, N 1 ) ⊕ Hom A (M, N 2 ).
2005/2006 7 3.2. A map f: M × N → R is called A-bilinear if for all m, m 1 , m 2 , n, n 1 , n 2 , a f(m, n 1 + n 2 ) = f(m, n 1 ) + f(m, n 2 ), f(m, an) = af(m, n) f(m 1 + m 2 , n) = f(m 1 , n) + f(m 2 , n), f(am, n) = af(m, n). So for every m the map N → R, n ↦→ f(m, n) is a homomorphism of A-modules and for every n the map M → R, m ↦→ f(m, n) is a homomorphism of A-modules. Note that an A-bilinear map f does not induce a homomorphism of A-modules M ⊕ N → R, since f(a(m, n)) = f(am, an) = a 2 f(m, n) is not equal to af(m, n) in general. Denote the set of all A-bilinear maps f: M × N → R by Bil A (M, N; R). The latter is an A-module with respect to the sum of maps and multiplication of a map by an element of A. Similarly one can define A- n-linear maps. Example. Let A = F be a field, and let M be an F -vector space of dimension d 1 and N be an F -vector space of dimension d 2 . Fix a basis {m i } in M and a basis {n j } in N . Let C be a matrix of order d 1 × d 2 with entries in F . Define a map f: M × N → F , f(m, n) = mCn ◦ where m is written as a row and n ◦ as a column. The map f is an F -bilinear map. Conversely, every F -bilinear map M × N → F is determined by its values on {(m i , n j ) : 1 i d 1 , 1 j d 2 }: f( ∑ a i m i , ∑ b j n j ) = ∑ a i b j f(m i , n j ). Now form a matrix C whose entries are f(m i , n j ). Thus, there is a one-to-one correspondence between bilinear maps M × N → F and matrices of order d 1 × d 2 with entries in F . 3.3. To study A-bilinear maps from M × N to R it is useful to introduce another A-module T and a bilinear map g: M × N → T such that bilinear maps f: M × N → R are in one-to-one correspondence with homomorphisms of A-modules T → R via g. In other words, we define an isomorphism of A-modules Bil A (M, N; R) ≃ Hom A (T, R); T will be the same for all R. To define T first denote by L the free A-module with a basis consisting of elements ∑ l m,n indexed by elements of M × N . So an arbitrary element of L is a finite sum ai l mi ,n i with a i ∈ A, m i ∈ M and n i ∈ N . Let K be the A-submodule of L generated by elements l m1 +m 2 ,n − l m1 ,n − l m2 ,n, l m,n1 +n 2 − l m,n1 − l m,n2 , l am,n − al m,n , l m,an − al m,n (for all a ∈ A, m ∈ M and n ∈ N ). Denote T = L/K. The image of l m,n in T , i.e. the coset l m,n + K is usually denoted by m ⊗ n. Since L is generated by l m,n , the module T is generated by m ⊗ n, i.e. T = { ∑ ai m i ⊗ n i : a i ∈ A, m i ∈ M, n i ∈ N } .
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Page 3 and 4: 2005/2006 3 Proof. If I is a non-ze
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Page 11 and 12: 2005/2006 11 (ii) ⇒ (iii): if the
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Page 15 and 16: 2005/2006 15 Elements a, b are call
Page 17 and 18: 2005/2006 17 principal). More gener
Page 19 and 20: 2005/2006 19 Note that some of thos
Page 21 and 22: 2005/2006 21 7.5. Definition. A mod
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Page 25 and 26: 2005/2006 25 {f/g : f, g ∈ C [X 1
Page 27: 2005/2006 27 have C times the colum

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