Topics in algebra Chapter IV: Commutative rings and modules I - 1

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Definition 4.24 A pair of modules homomorphisms, L f −→ M g −→ N is said to be exact at M if kerg = Imf. A finite sequence of modules homomorphisms, M0 f1 −→ M1 f2 −→ · · · fn−1 −→ Mn−1 fn −→ Mn is said to be exact (resp. a complex) if kerfi+1 = Imfi (resp. Imfi ⊆ kerfi+1) for i = 1, 2, . . . , n − 1. An infinite sequence of modules homomorphisms, · · · fi−1 −→ Mi−1 fi −→ Mi fi+1 −→ Mi+1 fi+2 −→ · · · is said to be exact (resp. a complex) if kerfi+1 = Imfi (resp. Imfi ⊆ kerfi+1) for every i. Definition 4.25 A finite sequence of modules homomorphisms, 0 −→ L f −→ M g −→ N −→ 0 is said to be a short exact sequence (s.e.s. for short ) if it is exact. Remark 4.26 If 0 −→ L f −→ M g −→ N −→ 0 is a s.e.s. then f is 1-1 and g is onto. Example 4.27 Let R be a ring and I be an ideal of R; then 0 −→ I i −→ R π −→ R/I −→ 0 is a s.e.s., where i is the inclusion map and π is the canonical surjective map. Example 4.28 Let f : M −→ N be an R-homomorphism; then f induces an exact sequence: 0 −→ kerf i −→ M f −→ N π −→ N/Imf −→ 0, where i is the inclusion map and π is the canonical surjective map. Example 4.29 Let M and N be R-modules; then 0 −→ M f −→ M ⊕ N g −→ N −→ 0 is a s.e.s., where f is the R-homomorphism given by f(x) = (x, 0) and g is the R-homomorphism given by g(x, y) = y, where x ∈ M and y ∈ N. There are several important results of short exact sequences. Lemma 4.30 Let 0 −→ L f −→ M g −→ N −→ 0 be a s.e.s.; then the following hold: 126
(a) If M is f.g. then so is N. (b) If L and N are f.g. then so is M. Proof. (a) If M is generated by {x1, . . . , xn}, then it is clear that N is generated by {g(x1), . . . , g(xn)}. (b) Suppose that L is generated by {x1, . . . , xn} and N is generated by {y1, . . . , yk}. Let {z1, . . . , zk} ⊆ M such that g(zi) = yi for every i; then we shall see that M is generated by {z1, . . . , zk, f(x1), . . . , f(xn)}. To show this, let x ∈ M; then g(x) ∈ N, so that there are r1 . . . , rk ∈ R such that k k g(x) = riyi, it follows that x − rizi ∈ kerg = Imf. Therefore, there i=1 are s1, . . . , sn ∈ R such that x − i=1 k rizi = i=1 n sif(xi). Thus, we conclude that M is generated by {z1, . . . , zk, f(x1), . . . , f(xn)}. There is a well-known result about s.e.s.: called The snake’s lemma. Lemma 4.31 (The snake’s lemma) Let i=1 0 −→ L f −→ M g −→ N −→ 0 ↓ α ↓ β ↓ γ 0 −→ L ′ f ′ −→ M ′ g ′ −→ N ′ −→ 0 be a commutative diagram of R-modules and R-homomorphisms such that each row is a s.e.s.. Then there is a long exact sequence 0 −→ kerα f1 −→ kerβ g1 −→ kerγ d −→ L ′ /Imα f ′ 1 ′ g −→ M /Imβ ′ 1 ′ −→ N /Imγ −→ 0, where f1 (resp. g1) is an R-homomorphism induced by f (resp. g) and f ′ 1 (resp. g ′ 1) is an R-homomorphism induced by f ′ (resp. g ′ ). Proof. (Sketch) 0 −→ kerα f1 −→ kerβ is exact: Let x ∈ kerα such that f1(x) = 0; then x = 0 as f is 1-1, so that kerf1 = 0. kerα f1 −→ kerβ g1 −→ kerγ is exact: Let x ∈ kerg1; then g(x) = 0. Since kerg = Imf, there is an element y ∈ L such that x = f(y), it follows that (f ′ ◦ α)(y) = (β ◦ f)(y) = β(x) = 0. Since f ′ is 1-1, α(y) = 0 and then y ∈ kerα. Therefore x = f1(y). 127
Page 1 and 2: Topics in algebra Chapter IV: Commu
Page 3 and 4: Example 1.7 Let R be the field of r
Page 5 and 6: Example 1.19 Let f : R −→ S be
Page 7 and 8: no left ideal other than R and 0. 1
Page 9 and 10: Corollary 2.6 Every maximal ideal i
Page 11 and 12: In a domain R, two elements a, b ar
Page 13 and 14: Example 2.25 Let R = Z[i] = {a + bi
Page 15 and 16: Proof. By assumption, there are a,
Page 17 and 18: 6. Find all solutions of the equati
Page 19 and 20: does not exist. 30. In Z[i], find g
Page 21 and 22: Definition 3.5 Let R be a ring and
Page 23 and 24: (a) Q is a P -primary ideal. (b)
Page 25 and 26: Example 3.20 Let F be a field; then
Page 27 and 28: 4 Modules, homomorphisms and exact
Page 29: As in ring theory, we have some fun
Page 33 and 34: (b) If α and γ are epimorphism th
Page 35 and 36: 5. Find a s.e.s. 0 −→ L f −
Page 37 and 38: y assumption, it follows that F ∼
Page 39 and 40: f. (⇐) Let D be an injective Z-mo
Page 41 and 42: g(a) = ax for all a ∈ I. 4. Q is
Page 43 and 44: Proof. ˜ ψ is 1-1: Let f ∈ HomR
Page 45 and 46: f((x, y)) = x ⊗ y = f ′ (x ⊗
Page 47 and 48: Proof. Let ˜ f : M × RS −→ MS

Topics in algebra Chapter IV: Commutative rings and modules I - 1

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