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prime in R[x, y, z]. If u ∈ R[x, y, z], let ū ∈ R be the coset containing u. Show R is not a UFD. In particular x · x =¯ · z are two distinct irreducible factorizations of x2. Show R/(¯ is isomorphic to R[y, z]/(yz), which is not a domain. Exercise In Group Theory If G is an additive abelian group, a subgroup H of G is said to be maximal if H = G and there are no subgroups properly between H and G. Show that H is maximal iff G/H ≈ Zp for some prime p. For simplicity, consider the case G = Q. Which one of the following is true? 1) If a ∈ Q, then there is a maximal subgroup H of Q which contains a. 2) Q contains no maximal subgroups. Splitting Short Exact Sequences Suppose B is an R-module and K is a submodule of B. As defined in the chapter on linear algebra, K is a summand of B provided ∃ a submodule L of B with K + L = B and K ∩ L =0. In this case we write K ⊕ L = B. When is K a summand of B? It turns out that¯ K is a summand of B iff there is a splitting map from B/K to B. In particular, if B/K is free, K must be a summand of B. This is used below to show that if R is a PID, then every submodule of Rn is free. Suppose R is a ring, B and C are R-modules, and g : B → C is a 1) K is a summand of B. 2) g has a right inverse, i.e., ∃ a homomorphism h : C → B with g ◦ h = I : C → C. (h is called a splitting map.) Proof Suppose 1) is true, i.e., suppose ∃ a submodule L of B with K ⊕ L = B. Then (g|L) : L → C is an isomorphism. If i : L → B is inclusion, then h defined by h = i ◦ (g|L)-1 is a right inverse of g. Now suppose 2) is true and h : C → B is a right inverse of g. Then h is injective, K + h(C) = B and K ∩ h(C) = 0. Thus K ⊕ h(C) =B. Theorem 1 surjective homomorphism with kernel K. Then the following are equivalent. ¯ ¯ y ¯ ¯ x) ¯ Chapter 6 Appendix 115 Definition Suppose f : A → B and g : B → C are R-module homomorphisms. The statement that 0 → A → B → C → 0 is a short exact sequence (s.e.s) means f is injective, g is surjective and f(A) = ker(g). The canonical split s.e.s. is A → A ⊕ C → C where f = i1 and g = π2. A short exact sequence is said to split if ∃ an isomorphism B → A ⊕ C such that the following diagram commutes. g B ≈ A ⊕ C f A i1 C π2 We now restate the previous theorem in this terminology. Theorem 1.1 A short exact sequence 0 → A → B → C → 0 splits iff f(A) is a summand of B, iff B → C has a splitting map. If C is a free R-module, there is a splitting map and thus the sequence splits. Proof has a splitting map. Showing these properties are equivalent to the splitting of the sequence is a good exercise in the art of diagram chasing. Now suppose C has a free basis T ⊂ C, and g : B → C is surjective. There exists a function h : T → B such that g ◦ h(c) =c for each c ∈ T . The function h extends to a homomorphism from C to B which is a right inverse of g. Theorem 2 If R is a commutative ring, then the following are equivalent. 1) R is a PID. 2) Every submodule of RR is a free R-module of dimension ≤ 1. This theorem restates the ring property of PID as a module property. Although this theorem is transparent, it is a precursor to the following classical result. Theorem 3 If R is a PID and A ⊂ Rn is a submodule, then A is a free R-module of dimension ≤ n. Thus subgroups of Zn are free Z-modules of dimension ≤ n. Proof From the previous theorem we know this is true for n = 1. Suppose n1 and the theorem is true for submodules of Rn-1. Suppose A ⊂ Rn is a submodule. We know from the previous theorem f(A) is a summand of B iff B → C f g ≈ 116 Appendix Chapter 6 Consider the following short exact sequences, where f : Rn-1 → Rn-1 ⊕R is inclusion and g = π : Rn-1 ⊕ R → R is the projection. 0 -→ Rn-1 -→ Rn-1 ⊕ R -→ R -→ 0 0 -→ A ∩ Rn-1 -→ A -→ π(A) -→ 0 By induction, A ∩ Rn-1 is free of dimension ≤ n - 1. If π(A) =0, then A ⊂ Rn-1. If π(A) =0, it is free of dimension 1 and the sequence splits by Theorem 1.1. In either case, A is a free submodule of dimension ≤ n. Exercise is a free Z-module of dimension 1. Euclidean Domains The ring Z possesses the Euclidean algorithm and the polynomial ring F [x] has the division algorithm. The concept of Euclidean domain is an abstraction of these properties. The axioms are so miniscule that it is surprising you get this much juice out of them. However they are exactly what you need, and it is possible to just play around with matrices and get some deep results. If R is a Euclidean domain and M is a finitely generated R-module, then M is the sum of cyclic modules. This is one of the great classical theorems of abstract algebra, and you don’t have to worry about it becoming obsolete. Here N will denote the set of all non-negative integers, not just the set of positive integers. Definition A domain R is a Euclidean domain provided ∃ φ : (R-0) -→ N such that if a, b ∈ (R - 0), then 1) φ(a) ≤ φ(ab). 2) ∃ q, r ∈ R such that a = bq + r with r =0 or φ(r) φ(b). Examples of Euclidean Domains Z with φ(n) =|n|. A field F with φ(a) =1 ∀ a =0 or with φ(a) =0 ∀ a =0. F [x] where F is a field with φ(f¯= a0 + a1x + · · · + anxn) = deg(f). Z[i] ={a + bi : a, b ∈ Z} = Gaussian integers with φ(a + bi) =a2 + b2. Let A ⊂ Z2 be the subgroup generated by {(6, 24), (16, 64)}. Show A f π ¯ ¯ ¯ ¯ ¯ [ Pobierz całość w formacie PDF ] |
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