Some Extensions in Homological Algebra

Transcription

1 International Journal of Algebra, Vol. 2, 2008, no. 9, Some Extensions in Homological Algebra Ali Asghar Onsori Department of Mathematics Islamic Azad University, Yazd Branch, Yazd, Iran aliasghar Saeid Alikhani Department of Mathematics, Faculty of Science Shiraz University of Technology, Shiraz, Iran Abstract In this article we extend some theorems in Homological Algebra. we show that if 0 X n (k+1) X n k... X n 1 X n 0is an exact sequence of zero sequence, then for every k N, there exist a natural homeomorphism ϕ k : H i (X n ) H i n (X n (k+1) ). Also by using α-sequences, we define CH(A) = Im(A ia i+1 ) Ker(A i+1 A i+2 ), and prove that, if T (A) is a additive exact functor, X is a α complex, and if T is covariant in A, then for every n N, CH n (T (X)) = T (CH n (X)). Mathematics Subject Classification: 16E30, 16E40 Keywords: Homological Algebra, Functor, Exact sequence 1 Introduction We have the following theorem in Homological Algebra Theorem 1.( [1]) Suppose that the above diagram is a commutative diagram of R- modules and R-homomorphisms, such that the rows are exact sequences and columns are 0-sequences. Then, there is a natural R- homomorphism K : H(C 3 C 2 C 1 ) H(A 2 A 1 A 0 ).

2 404 A. A. Onsori and S. Alikhani In the next section we extend Theorem 1 and study α-sequence. In the last section we introduce functor and state some of its properties. 2 Extension of Homomorphisms In this section we represent and prove a main theorem of this article, which extend Theorem 1. Theorem 2. If the following diagram is a commutative diagram of R- modules and R- homomorphisms, such that its row are exact and its column are 0- sequences, then we have the following natural R- homomorphism L : H(D 4 D 3 D 2 ) H(A 2 A 1 A 0 ) Proof. Suppose that λ H(D 4 D 3 D 2 ), so there is an element in D 3 such that denoted by d 3 and d 3 Ker(D 3 D 2 ). Since C 3 D 3 is onto, there exist

3 Some extensions in homological algebra 405 c 3 C 3 such that c 3 (C 3 D 3 )=d 3. Therefore, c 3 (C 3 D 3 D 2 )=c 3 (C 3 C 2 D 2 )=0, and so, c 3 (C 3 C 2 ) Ker(C 2 D 2 )=Im(B 2 C 2 ). Then, there exist b 2 B 2 such that b 2 (B 2 C 2 )=c 3 (C 3 C 2 ) (1.1) but b 2 (B 2 B 1 C 1 )=b 2 (B 2 C 2 C 1 )=c 3 (C 3 C 2 C 1 ) = 0, so, b 2 (B 2 B 1 C 1 ) = 0 and b 2 (B 2 B 1 ) Ker(B 1 C 1 )=Im(A 1 B 1 ). So, there exist a 1 A 1 such that Therefore we have a 1 (A 1 B 1 )=b 2 (B 2 B 1 ) (1.2) a 1 (A 1 A 0 B 0 )=a 1 (A 1 B 1 B 0 )=b 2 (B 2 B 1 B 0 )=0 We ve proved that a 1 (A 1 A 0 ) Ker(A 0 B 0 ), but we know A 0 B 0 is monomorphism, so a 1 (A 1 B 0 ) = 0, then a 1 Ker(A 1 A 0 ) and α H(A 2 A 1 A 0 ). Now we shall prove that L is well defined. If consider (d 3,c 3,b 2,a 1 ) for (d 3,c 3,b 2,a 1 ), since d 3 and d 3 represent λ, sod 3 d 3 = d 4 (D 4 D 3 ),d 4 D 4. Since C 4 D 4 is onto, c 4 (C 4 D 4 )=d 4,sod 3 d 3 = c 4(C 4 D 4 D 3 )=c 4 (C 4 C 3 D 3 ). Now we can write c 3 (C 3 D 3 ) c 3(C 3 D 3 ) c 4 (C 4 C 3 D 3 )=0 c 3 c 3 c 4(C 4 C 3 ) Ker(C 3 D 3 )=Im(B 3 C 3 ) and there exists b 3 B 3 such that b 3 (B 3 C 3 )=c 3 c 3 c 4(C 4 C 3 ) b 3 (B 3 C 3 C 2 )=c 3 (C 3 C 2 ) c 3 (C 3C 2 ) c 4 (C 4 C 3 C 2 ) Now by (1.1) we have: b 3 (B 3 C 3 C 2 )=c 3 (C 3 C 2 ) c 3(C 3 C 2 ) b 3 (B 3 C 3 C 2 )=b 2 (B 2 C 2 ) b 2 (B 2C 2 ) so, b 2 b 2 b 3(B 3 B 2 ) Ker(B 2 C 2 )=Im(A 2 B 2 ), therefore there exist a 2 A 2 such that a 2 (A 2 B 2 ) = b 2 b 2 b 3(B 3 B 2 ) and a 2 (A 2 B 2 B 1 ) = b 2 (B 2 B 1 ) b 2 (B 2B 1 ) b 3 (B 3 B 2 B 1 ), and by (1.2), a 2 (A 2 B 2 B 1 )=a 1 (A 1 B 1 ) a 1(A 1 B 1 )

4 406 A. A. Onsori and S. Alikhani Since diagram is commutative,a 2 (A 2 A 1 B 1 )=a 1 (A 1 B 1 ) a 1 (A 1B 1 ), and we know that A 1 B 1 is monomorphism, a 2 (A 2 A 1 )=a 1 a 1, therefore, L is well defined. Also it s obvious that L is a R- homomorphism. Here, we state the definition of α- sequence. Definition 1. The sequence A i A i+1 A i+2 of R-modules and R- homomorphisms is said α- sequence, if Ker(A i+1 A i+2 ) Im(A i A i+1 ) Note. We can state the Theorem 2, with α-sequences of X n, and also substitute the condition exact of sequences in the row, with α- sequences. Now we obtain another main theorem of this article, which is about 0- sequences. Theorem 3. If 0 X n (k+1) X n k... X n 1 X n 0 is a exact sequences of zero sequence, then for every k N, and integer fixed i, there exist a natural homeomorphism ϕ k : H i (X n ) H i n (X n (k+1) ). Proof. By induction on k. The theorem is true by Theorems 1 and 2. Now we suppose that the theorem is true for l k 1, and we prove it for l = k. We consider the following diagram. By induction hypothesis, there is natural homomorphism, θ : H i (X n ) H i k+2 (X n k+1 ) and by Theorem 1, there exist homomorphism δ : H i k+2 (X n k+1 ) H i k (X n (k+1) ). Now by choosing ϕ = δoθ we have the result. 3 Definition and Properties of Functor In this section we introduce functor and then state some of its properties. Definition 2. If A 2 A 1 A 0 is a α-sequence, then we denote functor by CH and define with the following form: CH(A) = Im(A 2A 1 ) Ker(A 1 A 0 )

5 Some extensions in homological algebra 407 Example 1. If f : Ker(A 1 A 0 ) Im(A 2 A 1 ) is inclusion mapping, then CH(A) =coker(f). Example 2. If A 2 A 1 A 0 and B 2 B 1 B 0 are two α- sequences, and f : A B commute the following diagram, then, we can define ψ : CH(A) CH(B) with ψ(a 1 + Ker(A 1 A 0 )) = b 1 + Ker(B 1 B 0 ),b 1 = a 1 (A 1 B 1 ). Since a 1 Im(A 2 A 1 ), there exist a 2 A 2 such that b 1 = a 1 (A 1 B 1 )=a 2 (A 2 A 1 B 1 )=a 2 (A 2 B 2 B 1 )=[a 2 (A 2 B 2 )](B 2 B 1 ) so, b 1 Im(B 2 B 1 ), therefore the above mapping is well defined. Note that if Im(A 2 A 1 ) project into Ker(B 1 B 0 ), then CH(A) CH(B) is zero.

6 408 A. A. Onsori and S. Alikhani Now we represent and prove the following theorem: Theorem 4. If A f B g C is a translation, such that (i) A 2 B 2 is a homomorphism onto, and (ii) A 1 B 1 C 1 is a zero sequence, then CH(A) CH(f) CH(B) CH(g) CH(C) is exact sequence. Proof. Consider the following commutative diagram. By properties of functors, (CH(g))(CH(f)) = CH(gf) since gf : A C and A 1 C 1 are zero mapping, then CH(gf) is zero, therefore CH(g)CH(f) = 0 and Im(CH(f)) Ker(CH(g)). Suppose that β Ker(CH(g)) such that there is b 1 Im(B 2 B 1 ) which β denoted by it, so, b 2 (B 2 B 1 )=b 1 and a 2 (A 2 B 2 )=b 2. Now since a 2 (A 2 B 2 B 1 )=a 2 (A 2 A 1 B 1 ), then b 1 = a 2 (A 2 A 1 )A 1 B 1 a 2 (A 2 A 1 ) Im(A 2 A 1 ), so Im(CH(f)) = Ker(CH(g)), and we have the result. Theorem 5. Suppose that the following diagram is commutative on R, and its rows are 0- sequence. If A B is onto, then Im(f) Im(g) Im(h) (3.1)

7 Some extensions in homological algebra 409 and are exact. coim(f) coim(g) coim(h) (3.2) Proof. We consider the diagram in Figure 1. By Theorem 4 and diagram in the Figure 1, we see that (3.2) is exact, and also by Theorem 4 and diagram in Figure 2, (3.1) is exact. So we have the result.. Figure1. Figure2 Definition 3. Let for every n N the sequence X 2 X 1 X 0 X n X n+1 X n+2 be a sequence of R- modules and R- homomorphisms, this sequence is said that α-complex, if for every n Z, Ker(X n+1 X n+2 ) Im(X n X n+1 ).

8 410 A. A. Onsori and S. Alikhani Theorem 6. Let T (A) be a additive exact sequence functor with one variable and X be α-complex. Then (i) If T is covariant in A, then for every n Z, there exist the following canonical isomorphism: CH n (T (X)) = T (CH n (X)) (ii) If T is contravariant in A, then for every n Z, we have the following canonical isomorphism: CH n (T (X)) = T (CH n (X)) Proof. We consider the case that T is covariant in A, put A n = Im(X n 1 X n ),B n = Ker(X n X n+1 ) so the homomorphism X n 1 X n, can be obtain from composition with X n 1 A n X n,so0 B n A n CH n (X) 0 is exact sequence and T is a exact functor. Therefore 0 T (CH n (X)) T (A n ) T (B n ) 0 (3.3) is exact sequence. Since T (X n ) T (A n ) 0 is epimorphism (note that 0 A n X n is exact), then is epimorphism, therefore ϕ : Ker(T (X n ) T (B n )) Ker(T (A n ) T (B n )) Ker(T (X n ) T (B n )) Ker(T (X n ) T (A n )) = Ker(T (A n ) T (B n )) so, we have the below exact sequence, 0 Ker(T (Xn ) T (B n )) Ker(T (X n ) T (A n )) T (An ) T (B n ) 0 (3.4) Since X n 1 A n 0 is exact, so 0 T (A n ) T (X n 1 ) is exact. But T (X n ) T (X n 1 ) obtain from composition with T (X n ) T (A n ) T (X n 1 ), we see that, Ker(T (X n ) T (X n 1 )) = Ker(T (X n ) T (A n )) (3.5)

9 Some extensions in homological algebra 411 so T (X n+1 ) T (X n ) T (B n ) is exact (note that B n X n X n+1 is exact), which shows Im[T (X n+1 ) T (X n )] = Ker[T (X n ) T (B n )] (3.6) By substitution (3.5) and (3.6) in (3.4), we ll have the below exact sequence 0 CH n (T (X)) T (A n ) T (B n ) 0 and with comparing (3.3), we have the following isomorphism: T (CH n (X)) = CH n (T (X)). References [1] D. G. Northcott, An introduction to Homological Algebra, Camridge University Press, Received: October 20, 2007