Analogies between Z and F [ T ] : Homework
نویسنده
چکیده
1. Let f, g ∈ C[T ] be nonconstant and relatively prime. a) If f−g 6= 0, show deg(f−g) ≥ (1/2) deg f+1, or equivalently deg f ≤ 2(deg(f−g)−1). b) Find infinitely many examples where equality occurs in the conclusion of part a. Start with an example where deg f = 3 and deg g = 2. c) When the hypothesis of relative primality is dropped, is part a still true? d) Find a lower bound on deg(f − g) in terms of deg g. 2. Assume the abc conjecture for some ε. Show there is a constant Cε such that, for each integer d 6= 0, any solution to the equation y = x + d in relatively prime integers x and y has |x| ≤ Cε|d| and |y| ≤ Cε|d|. Of course, Cε depends on ε, but it does not depend on d. (The exponents can be written more simply as 2(1 + ε′) and 3(1 + ε′), but ε′ is not the ε for which we are assuming the abc conjecture.) Can you remove the condition that x and y are relatively prime? 3. Show ε = 0 does not work in the abc-conjecture by considering a = 3 n − 1, b = 1, and c = 32n for large n, or by considering a = 2p(p−1) − 1, b = 1, and c = 2p(p−1) for large primes p. To start, show a ≡ 0 mod 2 in the first case and a ≡ 0 mod p in the second case. After showing the need for ε in the abc conjecture, use either of these examples to show κε →∞ as ε→ 0. 4. For relatively prime a, b ≥ 1, set c = a+ b and
منابع مشابه
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...
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