Math 210 A : Algebra , Homework 8 Ian Coley

Math 210 A : Algebra , Homework 7

We claim that monomorphisms in Sets and Groups are the usual injective maps and homomorphisms. First, suppose that f : A → B is a set injection and that fg = gh. Then for all c ∈ C, we have f(g(c)) = f(h(c)) implies g(c) = h(c) by injectivitiy. Hence g = h. Conversely, suppose f : A → B is not an injection. Then let f(a) = f(a′) for some a 6= a′ ∈ A. Then let g(c) = a and h(c) = a′ for all c ∈ ...

Math 210 B : Algebra , Homework 3

so ker f ⊂ imh. Hence ker f = imh so the sequence is exact at the middle term. Further, f is injective since Z[X] is a domain, so left multiplication by any element is injective. Finally, h is surjective since the constant polynomials g = a have g(0) = a for every a ∈ Z. Therefore the sequence is exact. The sequence is split over R. The requisite map q : Z→ R so that h ◦ q = 1Z. It is clear tha...

Math 225 B : Differential Geometry , Homework 4

Math 210 A : Algebra , Homework 6

Solution. Let α : Z/nZ → Z/mnZ be defined by α(1) = m. Since 〈1〉 = Z/nZ, this determines the entire homomorphism. Let x ∈ kerα. Then mx ≡ 0, which implies mn | mx so n | x. But since 0 ≤ x < n, we must have x = 0. Therefore the kernel of α is trivial, so α is an injection. Since m | mn, by the fundamental theorem on cyclic groups Z/mnZ has a subgroup of order m isomorphic to Z/mZ. Let β : Z/mnZ...

Math 210 B : Algebra , Homework 4

t(x · 1 + 0 · s) = t · x = 0. Therefore f ′ is injective. Now we need to show that im f ′ = ker g′. We have g′(f ′(x/s)) = g(f(x)/s) = (g ◦ f)(x)/s = 0, so im f ′ ⊂ ker g′. Conversely, if g′(y/s) = 0, then g(y)/s = 0, so there exists t ∈ S so that t · g(y) = g(t · y) = 0. Therefore t · y ∈ ker g, so t · y ∈ im f . Let x ∈ X so that f(x) = t · y. Then f ′(x/(st)) = f(x)/(st) = (t · y)/(st) = y/s...