Math 581 Problem Set 1 Solutions
... set of k + 1 elements, say B = {b1 , . . . , bk , bk+1 }. We split the injective functions into k + 1 sets Ai where the functions in Ai are the injective functions that send b1 to bi . The functions in the set A1 send b1 to b1 , so are determined by what the function does on the set of k elements {b ...
... set of k + 1 elements, say B = {b1 , . . . , bk , bk+1 }. We split the injective functions into k + 1 sets Ai where the functions in Ai are the injective functions that send b1 to bi . The functions in the set A1 send b1 to b1 , so are determined by what the function does on the set of k elements {b ...
Rings
... Since a ∈ h−1 (J), we have by definition that h(a) ∈ J. Also, we have h(r) ∈ S. Since J is an ideal in S, it is closed under multiplication by any element of S, so h(a)h(r) ∈ S. Since h(a)h(r) = h(ar), we have h(ar) ∈ S. By definition of h−1 (J), this means that ar ∈ h−1 (J), which is what we wanted ...
... Since a ∈ h−1 (J), we have by definition that h(a) ∈ J. Also, we have h(r) ∈ S. Since J is an ideal in S, it is closed under multiplication by any element of S, so h(a)h(r) ∈ S. Since h(a)h(r) = h(ar), we have h(ar) ∈ S. By definition of h−1 (J), this means that ar ∈ h−1 (J), which is what we wanted ...
Ideals - Columbia Math
... 1. I is an additive subgroup of (R, +); 2. (The “absorbing property”) For all r ∈ R and s ∈ I, rs ∈ I; symbolically, we write this as RI ⊆ I. For example, for all d ∈ Z, the cyclic subgroup hdi generated by d is an ideal in Z. A similar statement holds for the cyclic subgroup hdi generated by d in ...
... 1. I is an additive subgroup of (R, +); 2. (The “absorbing property”) For all r ∈ R and s ∈ I, rs ∈ I; symbolically, we write this as RI ⊆ I. For example, for all d ∈ Z, the cyclic subgroup hdi generated by d is an ideal in Z. A similar statement holds for the cyclic subgroup hdi generated by d in ...
