MAT 150C, Spring 2017Solutions to HW 5

Page 472-473: 1.2. Let F be a field, not of characteristic 2, and letx2 + bx+ c = 0 be a quadratic equation with coefficients in F . Prove that ifδ is an element in F such that δ2 = b2 − 4c then x = (−b + δ)/2 solves thequadratic equation in F . Prove also that if b2 − 4c is not a square in F , thepolynomial has no roots in F .

Solution: Since F is not of characteristic 2, 2 = 1 + 1 is invertible in F .Let us complete the square:

x2 + bx+ c = (x+ b/2)2 − 1

4(b2 − 4c) = 0

if and only if

(x+ b/2)2 =1

4(b2 − 4c), 4(x+ b/2)2 = (2(x+ b/2))2 = b2 − 4cc.

If b2 − 4c is not a square in F , there are no solutions. If b2 − 4c = δ2, then

(2(x+ b/2))2 = δ2, (2(x+ b/2))2− δ2 = (2(x+ b/2)− δ)(2(x+ b/2) + δ) = 0,

2(x+ b/2) = ±δ, x+ β/2 = ±δ/2, x = (−b± δ)/2.

2.1. Let α be a complex root of the polynomial x3 − 3x + 4. Find theinverse of α2 + α + 1 in the form a+ bα + cα2.

Solution: Let m(x) = x3 − 3x + 4. Then 1(x2 + x + 1) = x2 + x + 1mod m(x),

x(x2 + x+ 1) = x3 + x2 + x = m(x) + x2 + 4x− 4 = x2 + 4x− 4 mod m(x),

x2(x2 + x+ 1) = x4 + x3 + x2 = xm(x) + x3 + 4x2 − 4x =

(x+ 1)m(x) + 4x2 − x− 4 = 4x2 − x− 4 mod m(x).

Then

(x− 1)(x2 + x+ 1) = 3x− 5 mod m(x), (x2 − 4)(x2 + x+ 1) = −5x− 8,

1

so

(5(x−1)+3(x2−4))(x2+x+1) = 5(3x−5)+3(−5x−8) = −25−24 = −49 mod m(x).

Therefore the inverse of x2 + x+ 1 modulo m(x) equals

−1

49(5(x− 1) + 3(x2 − 4)) =

−1

49(3x2 + 5x− 17).

3.3. Let ζn = e2πi/n. Prove that ζ5 /∈ Q(ζ7).

Solution: For each prime p ζp is a root of the polynomial

xp − 1

x− 1= xp−1 + xp−2 + . . .+ 1,

which is irreducible over Q1. Therefore [Q(ζp) : Q] = p − 1. Suppose thatζ5 ∈ Q(ζ7), then Q(ζ5) ⊂ Q(ζ7) and

[Q(ζ7) : Q] = [Q(ζ7) : Q(ζ5)] · [Q(ζ5) : Q].

But [Q(ζ7) : Q] = 6, [Q(ζ5) : Q] = 4 and 4 does not divide 6. Contradiction,therefore ζ5 /∈ Q(ζ7).

3.6. Let a be a positive rational number that is not a square in Q. Provethat 4

√a has degree 4 over Q.

Solution: The number 4√a is a root of polynomial f(x) = x4 − a of

degree 4, so it is sufficient to prove that f(x) is irreducible over Q. We canfactor f(x) over C:

f(x) = (x2 −√a)(x2 +

√a) = (x− 4

√a)(x+ 4

√a)(x− i 4

√a)(x+ i 4

√a).

Suppose that f(x) is reducible, then it has either linear or quadratic factor.Since a is not a square, 4

√a is not rational and f(x) cannot have a linear

factor. A quadratic factor must combine two of the linear factors above.Since it should have real coefficients, it combines either the first two or thelast two factors. But the corresponding products are equal to (x2−

√a) and

(x2 +√a) which do not have rational coefficients since

√a is not rational.

Contradiction, hence f(x) is irreducible.

1We use it without proof

2