Math 202 -assignment 7
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چکیده
a. N1 = {(i1, i2, . . . , in) : ik ∈ Ik for all k ∈ {1, 2, . . . , n}} and b. N2 = {(x1, x2, . . . , xn) : ∑n k=1 xk = 0}. Proof. To prove (a), it suffices to show, by Proposition 1, that N1 is nonempty and x + ry ∈ N1 for all r ∈ R and all x, y ∈ N1. For the first condition, (0, 0, . . . , 0) ∈ N1 since Ik is a subgroup of R containing the additive identity 0 for all k ∈ {1, 2, . . . , n}. That is, N1 is nonempty. For the second condition, let x = (ik)k∈Z+ , y = (yk)k∈Z+ ∈ N1 and let r ∈ R. Then, by definition of addition and scalar multiplication,
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Math 202 - Assignment 7 Solutions
Exercise 10.3.2. Let R be a commutative ring with identity. For all positive integers n and m, R ∼= R if and only if n = m. Proof. Let φ : R → R be an isomorphism of R-modules and let I E R be a maximal ideal. Then the map φ̄ : R → R/IR given by φ̄(α) = φ(α) is a morphism of R-modules. Moreover ker φ̄ = {α ∈ R | φ̄(α) = 0} = {α ∈ R | φ(α) ∈ IR} = φ−1(IRm) = IR. Therefore by the first isomorphism th...
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a. N1 = {(i1, i2, . . . , in) : ik ∈ Ik for all k ∈ {1, 2, . . . , n}} and b. N2 = {(x1, x2, . . . , xn) : ∑n k=1 xk = 0}. Proof. To prove (a), it suffices to show, by Proposition 1, that N1 is nonempty and x + ry ∈ N1 for all r ∈ R and all x, y ∈ N1. For the first condition, (0, 0, . . . , 0) ∈ N1 since Ik is a subgroup of R containing the additive identity 0 for all k ∈ {1, 2, . . . , n}. Tha...
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