Math 210 B : Algebra , Homework 4

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  • Ian Coley
چکیده

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, so y/s ∈ im f ′. Therefore the sequence is exact in at the middle term. Finally, we need to show that g′ is surjective. Let z/s ∈ S−1Z. Then there exists y ∈ Y so that g(y) = z. Then g′(y/s) = g(y)/s = z/s, so g′ is surjective. This completes the proof.

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تاریخ انتشار 2014