MTH 461 University of Miami Polynomial Questions Please help me with my math question. Need to show steps. I will give tips for quality works. Math 461 Hom

MTH 461 University of Miami Polynomial Questions Please help me with my math question. Need to show steps. I will give tips for quality works. Math 461
Homework 5
Spring 2020
Drew Armstrong
Problem 1. Use the Rational Root Test to split the following polynomial:
f (x) = 8×3 + 4×2 − 2x − 1 ∈ Q[x].
Problem 2. Symmetric Polynomials. Suppose that the polynomial x3 + x2 + 2x + 3 has
the roots r, s, t in some field. Find some integer coefficients a, b, c ∈ Z such that the polynomial
x3 + ax2 + bx + c has the roots rs, rt, st.
Problem 3. Some Specific Cyclotomic Polynomials. Recall the definition of the nth
cyclotomic polynomial:
Y
Φn (x) =
(x − e2πik/n ).
1≤k≤n
gcd(k,n)=1
(a) If p is prime, show that Φp (x) = 1 + x + x2 + · · · + xp−1 . [Hint: In this case we have
gcd(k, p) = 1 for all 1 ≤ k < p and hence Φp (x) = (x − ω)(x − ω 2 ) · · · (x − ω p−1 ) for ω = e2πi/p . On the other hand, we know that xp − 1 = (x − 1)(x − ω) · · · (x − ω p−1 ).] (b) If n = 2m for some m ≥ 1, show that Φn (x) = 1 + xn/2 . [Hint: Show that the roots of Φn (x) are precisely the (n/2)th roots of −1. First, observe that gcd(k, n) = 1 if and only if k is odd, hence the roots of Φn (x) are (e2πi/n )odd . Second, observe that αn/2 = −1 = ei(π+2πk) implies α = ei(π+2πk)/(n/2) = (e2πi/n )1+2k for all k ∈ Z.] Problem 4. Uniqueness of Quotient and Remainder. Let F be a field and consider polynomials f (x), g(x) ∈ F[x] with g(x) 6= 0(x). (a) Suppose that we have q1 (x), r1 (x), q2 (x), r2 (x) ∈ F[x] satisfying f (x) = q1 (x)g(x) + r1 (x), f (x) = q2 (x)g(x) + r2 (x), deg(r1 ) < deg(g), deg(r2 ) < deg(g). In this case, prove that q1 (x) = q2 (x) and r1 (x) = r2 (x). [Hint: First note that (q1 − q2 )g = (r2 − r1 ). If q1 6= q2 then this implies that deg(r2 − r1 ) ≥ deg(g). On the other hand, we have deg(r2 − r1 ) ≤ max{deg(r1 ), deg(r2 )}.] (b) Now let R ⊆ F be a subring. Suppose that we have f (x), g(x) ∈ R[x] where g(x) has leading coefficient 1, and suppose that f (x) = g(x)q(x) for some q(x) ∈ F[x]. In this case, use part (a) to show that we must actually have q(x) ∈ R[x]. [Hint: Since g(x) ∈ R[x] has leading coefficient 1, we may apply long division to obtain f (x) = g(x)q 0 (x) + r0 (x) for some q 0 (x), r0 (x) ∈ R[x] with deg(r0 ) < deg(g 0 ). On the other hand, we have assumed that f (x) = g(x)q(x) + 0 for some q(x) ∈ F[x]. Apply (a) to show that q(x) = q 0 (x), and hence q(x) ∈ R[x].] Problem 5. Cyclotomic Polynomials Have Integer Coefficients. We will prove in class that cyclotomic polynomials satisfy the following identity: Y xn − 1 = Φd (x). 1≤d≤n d|n Use this identity and Problem 4(b) to prove by induction that Φn (x) ∈ Z[x] for all n ≥ 1. [Hint: Suppose that we have xn − 1 = Φn (x)q(x) for some polynomial q(x) ∈ Z[x]. Then since Φn (x) ∈ C[x] has leading coefficient 1, we can apply Problem 4(b) with R = Z and F = C.] Problem 6. A Property of Quadratic Field Extensions. The construction of C from R can be generalized as follows. Let E ⊇ F be fields and let ι ∈ E be some element satisfying ι 6∈ F and ι2 ∈ F. Then I claim that the following set is a subfield of E: F(ι) := {a + bι : a, b ∈ F}. Furthermore, the conjugation operator (a + bι)∗ = (a − bι) behaves exactly like complex conjugation. Jargon: We say that F(ι) ⊇ F is a quadratic field extension. The following Lemma will be useful in our discussion of impossible constructions: Consider a polynomial f (x) ∈ F[x] of degree 3. If f (x) has some root α ∈ F(ι) in a quadratic field extension then I claim that f (x) also has a root in F. Prove the Lemma. [Hint: Let α ∈ F(ι) be a root of f (x). If α ∈ F then we are done. Otherwise, the conjugate α∗ ∈ F(ι) is another root of f (x), hence by Descartes’ Factor Theorem we have f (x) = (x − α)(x − α∗ )g(x) for some g(x) ∈ F(ι)[x] of degree 1. Use Problem 4(b) to show that g(x) ∈ F[x], hence g(x) has a root in F.] Purchase answer to see full attachment