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Abstract AlgebraChapter 13 - Field Theory

David S. Dummit & Richard M. Foote

Solutions by positró[email protected]

March 31, 2018

Contents

13 Field Theory 113.1 Basic Theory and Field Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Algebraic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.3 Classical Straightedge and Compass Constructions . . . . . . . . . . . . . . . . . . 1013.4 Splitting Fields and Algebraic Closures . . . . . . . . . . . . . . . . . . . . . . . . 1213.5 Separable and Inseparable Extension . . . . . . . . . . . . . . . . . . . . . . . . . . 1313.6 Cyclotomic Polynomials and Extensions . . . . . . . . . . . . . . . . . . . . . . . . 17

13 Field Theory

13.1 Basic Theory and Field Extensions

Exercise 13.1.1.p(x) = x3 + 9x + 6 is irreducible in Z[x] by Eisenstein Criterion with p = 3. By Gauss Lemma, thenit is irreducible in Q[x]. To find (1 + θ)−1, we apply the Euclidean algorithm (long division) to p(x)and 1 + x. We find

x3 + 9x + 6 = (1 + x)(x2 − x + 10) − 4.

Evaluating at θ, we find (1 + θ)(θ2 − θ + 10) = 4. Therefore

(1 + θ)−1 =θ2 − θ + 10

4.

Exercise 13.1.2.Let f (x) = x3 − 2x − 2. f is irreducible over Z by Eisenstein Criterion with p = 2, hence over Q byGauss Lemma. Now, if θ is a root of f , then θ3 = 2θ + 2. Hence

(1 + θ)(1 + θ + θ2) = 1 + 2θ + 2θ2 + θ3 = 3 + 4θ + 2θ2.

For computing1 + θ

1 + θ + θ2 , first we compute (1 + θ + θ2)−1. Applying the Euclidean algorithm, weobtain

x3 − 2x − 2 = (x2 + x + 1)(x − 1) − 2x − 1,

andx3 − 2x − 2 = (2x + 1)(

x2

2−

x4−

78

) −98.

1


Evaluating at θ, from this equalities we obtain

(θ2 + θ + 1)(θ − 1) = 2θ + 1 and (2θ + 1)−1 =89

(θ2

2−θ

4−

78

).

Combining these two equations we obtain

89

(θ2 + θ + 1)(θ − 1)(θ2

2−θ

4−

78

) = 1.

So,

(θ2 + θ + 1)−1 =89

(θ − 1)(θ2

2−θ

4−

78

) = −2θ2

3+θ

3+

53,

where we used θ3 = 2θ + 2 again. Therefore,

1 + θ1 + θ + θ2 = (1 + θ)(−

2θ2

3+θ

3+

53

) = −θ2

3+

2θ3+

13.

Exercise 13.1.3.Since 03 + 0 + 1 = 1 and 11 + 1 + 1 = 1 in F2, then x3 + x + 1 is irreducible over F2. Since θ is rootof x3 + x + 1, then θ3 = −θ − 1 = θ + 1. Hence, the powers of θ in F2(θ) are

θ, θ2, θ3 = θ + 1, θ4 = θ2 + θ, θ5 = θ2 + θ + 1, θ6 = θ2 + 1, and θ7 = 1.

Exercise 13.1.4.Denote this map by ϕ. Then

ϕ(a + b√

2 + c + d√

2) = a + c − b√

2 − d√

2 = ϕ(a + b√

2) + ϕ(c + d√

2),

andϕ((a + b

√2) · (c + d

√2)) = ϕ(ac + 2bd + (ad + bc)

√2)

= ac + 2bd − (ad + bc)√

2

= (a − b√

2)(c − d√

2)

= ϕ(a + b√

2)ϕ(c + d√

2),

hence ϕ is an homomorphism. Moreover, if ϕ(a + b√

2) = ϕ(c + d√

2), then a − b√

2 = c − d√

2,hence (since

√2 < Q) a = b and c = d, so ϕ is injective. Also, given a + b

√2 ∈ Q(

√2), then

ϕ(a − b√

2) = a + b√

2, so ϕ is surjective. Therefore, ϕ is an isomorphism of Q(√

2) with itself.

Exercise 13.1.5.Let α = p/q be a root of a monic polynomial p(x) = xn + · · · + a1x + a0 over Z, with gcd(p, q) = 1.Then

(pq

)n + an−1(pq

)n−1 + · · · + a1pq+ a0 = 0.

Multiplying this equation by qn one obtains

pn + an−1pn−1q + · · · + a1pqn−1 + a0qn = 0⇒ q(an−1pn−1 + · · · + a1pqn−2 + a0qn−1) = −pn.

Thus, every prime that divides q divides pn as well, so divides p. Since gcd(p, q) = 1, there is noprime dividing q, hence q = ±1. The result follows.

2


Exercise 13.1.6.This is straightforward. If

anαn + an−1α

n−1 + · · · + a1α + a0 = 0,

then(anα)n + an−1(anα)n−1 + anan−2(anα)n−2 + · · · + an−2

n a1(anα) + an−1n a0

= annα

n + an−1n an−1α

n−1 + an−1n an−2α

n−2 + · · · + an−1n a1α + an−1

n a0

= an−1n (anα

n + an−1αn−1 + an−2α

n−2 + · · · + a1α + a0) = 0.

Exercise 13.1.7.If x3 − nx + 2 is reducible it must have a linear factor, hence a root. By the Rational Root Theorem,if α is a root of x3 − nx + 2, then α must divide its constant term, so the possibilities are α = ±1,±2.If α = −1 or 2, then n = 3; if α = −1, then n = 1; and if α = 2, then n = 5. Therefore, x3 − nx + 2 isirreducible for n , −1, 3, 5.

Exercise 13.1.8.We subdivide this exercise in cases and subcases.

If x5 − ax − 1 is reducible then it has a root (linear factor) or is a product of two irreduciblepolynomials of degrees 2 and 3.

Case 1. If x5 − ax − 1 has a root, then, by the Rational Root Theorem, it must be α = ±1. Ifα = 1 is a root, then a = 0. If α = −1 is a root, then a = 2.

Case 2. Now, suppose that there exists f (x) and g(x) irreducible monic polynomials over Z ofdegrees 2 and 3 respectively, such that x5 − ax − 1 = f (x)g(x). Write f (x) = x2 + bx + c andg(x) = x3 + r x2 + sx + t, where b, c, r, s, t ∈ Z. Then

x5 − ax − 1 = (x2 + bx + c)(x3 + r x2 + sx + t)

= x5 + (b + r)x4 + (br + c + s)x3 + (bs + cr + t)x2 + (bt + cs) + tc.

Equating coefficients leads tob + r = 0

br + c + s = 0bs + cr + t = 0

bt + cs = −a

ct = −1.From ct = −1 we deduce (c, t) = (−1, 1) of (c, t) = (1,−1), which give us two cases.

Case 2.1. First suppose (c, t) = (−1, 1). Then the system of equations reduces to

b + r = 0br − 1 + s = 0bs − r + 1 = 0

b − s = −a.

Now, put b = −r into second and third equations to obtain −r2 − 1 + s = 0 and −rs − r + 1 = 0, thatis, r2 + 1 − s = 0 and rs + r − 1 = 0. Adding these last two equations we obtain r2 + rs + r − s = 0.Thus r2 + rs + r + s = 2s, so (r + 1)(r + s) = 2s. Now, from r2 + 1 − s = 0 we have r2 = s − 1, sothen r2 + rs + r − s = 0 becomes rs + r = 1, that is, r (s + 1) = 1. Hence, r = 1 and s = 0, or r = −1and s = −2. If r = 1 and s = 0, then (r + 1)(r + s) = 2s leads to 2 = 0, a contradiction. If r = −1and s = −2, it leads to 0 = −4, another contradiction.

3

13.2 Algebraic Extensions

Therefore, (c, t) = (−1, 1) is impossible. We now pass to the case (c, t) = (1,−1).Case 2.2. Suppose that (c, t) = (1,−1). The system of equations reduces to

b + r = 0br + 1 + s = 0bs + r − 1 = 0−b + s = −a.

Adding the second and third equation we obtain b(r + s) + r + s = 0, so that (b+ 1)(r + s) = 0. Thenb = −1 or r = −s, so one more time we have two cases. If r = −s, then br + 1 + s = 0 becomesbr +1− r = 0. Hence, b = −r and br +1− r = 0 gives r2+ r −1 = 0. By the Rational Root Theorem,this equation has no roots on Z. Since r ∈ Z, we have a contradiction. Now suppose b = −1. Fromb = −r we obtain r = 1, so, from br + 1 + s = 0 we obtain s = 0. Finally, from −b + s = −a weobtain a = −1. Therefore, the solution (b, c, r, s, t) = (−1, 1, 1, 0,−1) is consistent and we obtain thefactorization

x5 − ax − 1 = (x2 + bx + c)(x3 + r x2 + sx + t) = (x2 − x + 1)(x3 + x2 − 1).


Exercise 13.2.1.Since the characteristic of F is p, its prime subfield is (isomorphic to) Fp = Z/pZ. We can considerF as a vector space over Fp. Since F is finite, then [F : Fp] = n for some n ∈ Z+. Therefore

|F| = |Fp |[F:Fp ] = pn.

Exercise 13.2.2.Note that g and h are irreducible over F2 and F3. Now, is θ is a root of g, then F2(θ) � F2/(g(x))has 4 elements and F3(θ) � F3/(g(x)) has 9 elements. Furthermore, is θ2 is a root of h, thenF2(θ2) � F2/(h(x)) has 8 elements and F3(θ2) � F3/(h(x)) has 27 elements.

The multiplication table for F2/(g(x)) is

· 0 1 x x + 10 0 0 0 01 0 1 x x + 1x 0 x x + 1 x

x + 1 0 x + 1 x x

The multiplication table for F3/(g(x)) is

· 0 1 2 x x + 1 x + 2 2x 2x + 1 2x + 20 0 0 0 0 0 0 0 0 01 0 1 2 x x + 1 x + 2 2x 2x + 1 2x + 22 0 2 1 2x 2x + 2 2x + 1 x x + 2 x + 1x 0 x 2x 2x + 1 1 x + 1 x + 2 2x + 2 2

x + 1 0 x + 1 2x + 2 1 x + 2 2x 2 x 2x + 1x + 2 0 x + 2 2x + 1 x + 1 2x 2 2x + 2 1 x2x 0 2x x x + 2 2 2x + 2 2x + 1 x + 1 1

2x + 1 0 2x + 1 x + 2 2x + 2 x 1 x + 1 2 2x2x + 2 0 2x + 2 x + 1 2 2x + 1 x 1 2x x + 2

4


In both cases, x is a generator of the cyclic group of nonzero elements.

Exercise 13.2.3.Since 1 + i < Q, its minimal polynomial is of degree at least 2. We try conjugation, and obtain

(x − (1 + i))(x − (1 − i)) = x2 − 2x + 2,

which is irreducible by Eisenstein with p = 2. Therefore, the minimal polynomial is x2 − 2x + 2.

Exercise 13.2.4.First, note that (2+

√3)2 = 4+4

√3+3 = 7+4

√3.Let θ = 2+

√3. Then θ2−4θ = 7+4

√3−8−4

√3 = −1,

hence θ is a root of x2 − 4x + 1. Moreover, x2 − 4x + 1 is irreducible over Q (because θ < Q), sox2 − 4x + 1 is the minimal polynomial of 2 +

√3. Therefore, 2 +

√3 has degree 2 over Q.

Now, let α = 3√2 and β = 1 + α + α2. Then β ∈ Q(α), so Q ⊂ Q(β) ⊂ Q(α). We have[Q(α) : Q] = [Q(α) : Q(β)][Q(β) : Q]. Note that [Q(α) : Q] = 3 since α has minimal polynomialx3−2 overQ, so [Q(β) : Q] = 1 or 3. For a contradiction, suppose [Q(β) : Q] = 1, that is,Q(β) = Qso β ∈ Q. Then

β2 = (1 + α + α2)2 = 1 + 2α + 3α2 + 2α3 + α4 = 5 + 4α + 3α2,

where we used α3 = 2. So

β2 − 3β = 5 + 2α + 3α2 − 3(1 + α + α2) = 2 − α,

hence α = −β2 + 3β + 2 ∈ Q(β) = Q, a contradiction. Therefore, [Q(β) : Q] = 3.

Exercise 13.2.5.Since the polynomials have degree 3, if they were reducible they must have a linear factor, hencea root in F. Note that every element of F is of the form a + bi, where a, b ∈ Q. The roots ofx3 − 2 are 3√2, ζ 3√2 and ζ2 3√2, where ζ is the primitive 3’rd root of unity, i.e., ζ = exp(2πi/3) =cos(2π/3) + i sin(2π/3) = − 1

2 +√

32 . Since

√3 < Q, none of this elements is in F, hence x3 − 2 is

irreducible over F. Similarly, the roots of x3 − 3 are 3√3, ζ 3√3 and ζ2 3√3, and by the same argumentnon of this elements is in F. Hence x3 − 3 is irreducible over F.

Exercise 13.2.6.We have to prove that F (α1, . . . , αn) is the smallest field containing F (α1), . . . , F (αn). ClearlyF (αi) ⊂ F (α1, . . . , αn) for all 1 ≤ i ≤ n. Now let K be a field such that F (αi) ⊂ K for all i. If θis an element of F (α1, . . . , αn), then θ is of the form θ = a1α1 + · · · + anαn, where a1, . . . , an ∈ F.Every aiαi is in K , hence θ ∈ K . Thus F (α1, . . . , αn) ⊂ K . Therefore, F (α1, . . . , αn) contains allF (αi) and is contained in every field containing all F (αi), hence F (α1, . . . , αn) is the composite ofthe fields F (α1), F (α2), . . . , F (αn).

Exercise 13.2.7.Since

√2 +√

3 is in Q(√

2,√

3), clearly Q(√

2 +√

3) ⊂ Q(√

2,√

3). For the other direction we haveto prove that

√2 and

√2 are in Q(

√2 +√

3). Let θ =√

2 +√

3. Then

θ2 = 5 + 2√

6, θ3 = 11√

2 + 9√

3 and θ4 = 37 + 15√

6.

So√

2 =12

(θ3 − 9θ),√

3 =12

(11θ − θ3) and θ4 = 49 + 20√

6.

Therefore√

2 ∈ Q(θ) and√

3 ∈ Q(θ), so Q(√

2,√

3) ⊂ Q(√

2 +√

3). The equality follows. Hence,[Q(√

2 +√

3) : Q] = [Q(√

2,√

3) : Q] = 4.

5


We also have

θ4 − 10θ2 = (49 + 20√

6) − 10(5 + 2√

6) = −1, so θ4 − 10θ2 + 1 = 0.

Since [Q(√

2 +√

3) : Q] = 4, then x4 − 10x2 + 1 is irreducible over Q, and is satisfied by√

2 +√

3.

Exercise 13.2.8.The elements of F (

√D1,√

D2) can be written in the form

a + b√

D1 + c√

D2 + d√

D1D2, where a, b, c, d ∈ F .

We have[F (

√D1,

√D2) : F] = [F (

√D1,

√D2) : F (

√D1)][F (

√D1) : F].

Since [F (√

D1) : F] = 2, then [F (√

D1,√

D2) : F] can be 2 or 4. Now, [F (√

D1,√

D2) : F] = 2if and only if [F (

√D1,√

D2) : F (√

D1)] = 1, and that occurs exactly when x2 − D2 is reducible inF (√

D1) (i.e., when√

D2 ∈ F (√

D1)), that is, if there exists a, b ∈ F such that

(a + b√

D1)2 = D2, so that a2 + 2ab√

D1 + b2D21 = D2.

Note that ab = 0 as ab , 0 implies√

D1 ∈ F, contrary to the hypothesis. Then a = 0 or b = 0. Ifb = 0, then D2 is a square in F, contrary to the hypothesis. If a = 0, then b2D1 = D2, and thusD1D2 = (D2

b )2, so D1D2 is a square in F. So, x2 − D2 is reducible in F (√

D1) if and only if D1D2 isa square in F. The result follows.

Exercise 13.2.9.

Suppose√

a +√

b =√

m +√

n for some m, n ∈ F, then a +√

b = m + n + 2√

mn. Since b is not a

square in F, this means√

b = 2√

mn. We also have√

a +√

b −√

n =√

m, so

√b = 2

√n(

√a +√

b −√

n).

Hence,√

b = 2√

n(a +√

b) − 2n

⇒ (√

b + 2n)2 = 4n(a +√

b)

⇒ b + 4n√

b + 4n2 = 4n(a +√

b)

⇒ b + 4n2 − 4na = 0

⇒ n =4a ±

√16a2 − 16b

8

⇒√

a2 − b = ±2na.

Therefore, since a and n are in F,√

a2 − b is in F.Now suppose that a2−b is a square in F, so that

√a2 − b ∈ F. We prove that there exists m, n ∈ F

such that√

a +√

b =√

m +√

n. Let

m =a +√

a2 − b2

and n =a −√

a2 − b2

.

6


Note that m and n are in F as char(F) , 2. We claim√

a +√

b =√

m +√

n. Indeed, we have

m =(a +

√b) + 2

√a2 − b + (a −

√b)

4= (

√a +√

b +√

a −√

b

2)2,

and

n =(a +

√b) − 2

√a2 − b + (a −

√b)

4= (

√a +√

b −√

a −√

b

2)2.

Thus√

m =

√a +√

b +√

a −√

b

2and

√n =

√a +√

b −√

a −√

b

2.

Therefore,

√m +√

n =

√a +√

b +√

a −√

b

2+

√a +√

b −√

a −√

b

2=

√a +√

b,

as claimed.Now, we this to determine when the field Q(

√a +√

b), a, b ∈ Q, is biquadratic over Q. If a2 − b

is a square in Q and b is not, we have Q(√

a +√

b) = Q(√

m +√

n) = Q(√

m,√

n), so by last exercise

Q(√

a +√

b) is biquadratic over Q when a2 − b is a square in Q, and neither b, m, n or mn are squaresin Q. Since

mn =a +√

a2 − b2

a −√

a2 − b2

=b4,

then mn is never a square when b isn’t. Thus, Q(√

a +√

b) is biquadratic over Q exactly when a2 − bis a square in Q and neither b, m nor n is a square in Q.

Exercise 13.2.10.

Note that√

3 + 2√

2 =√

3 +√

8. Recalling last exercise with a = 3 and b = 8, we have a2 − b =9 − 8 = 1 is a square in Q and b = 8 is not. Hence, we find (m = 2 and n = 1 from last exercise)√

3 +√

8 =√

2+1. Therefore,Q(√

3 + 2√

2) = Q(√

2) and the degree of the extensionQ(√

3 + 2√

2)over Q is 2.

Exercise 13.2.11.(a) First, note that the conjugationmap a+bi → a−bi is an isomorphism ofC, so it takes squares rootsto square roots, and maps numbers of the first quadrant to the fourth (and reciprocally). Since

√3 + 4i

is the square root of 3+4i in the first quadrant, its conjugate is the square of root of 3−4i in the fourthquadrant, so is

√3 − 4i. Hence

√3 + 4i and

√3 − 4i are conjugates each other. Now, we use Exercise

9 again. Note that√

3 + 4i =√

3 +√−16. With a = 3 and b = −16, we have a2 − b = 25 is a square

in Q and b = −16 is not. Hence, we find m = 1 and n = −4 and thus√

3 + 4i = 1 +√−4 = 1 + 2i.

Furthermore, we find√

3 − 4i = 1−2i. Therefore,√

3 + 4i+√

3 − 4i = 4, i.e.,√

3 + 4i+√

3 − 4i ∈ Q.

(b) Let θ =√

1 +√−3 +

√1 −√−3. Then

θ2 = (√

1 +√−3 +

√1 −√−3)2 = (1 +

√−3) + (2

√1 + 3) + (1 −

√−3) = 6.

Since x2 − 6 is irreducible over Q (Eisenstein p = 2), then θ has degree 2 over Q.

7


Exercise 13.2.12.Let E be a subfield of K containing F. Then

[K : F] = [K : E][E : F] = p.

Since p is prime, either [K : E] = 1 or [E : F] = 1. The result follows.

Exercise 13.2.13.Note that, for all 1 ≤ k ≤ n, we have [Q(α1, . . . , αk ) : Q(α1, . . . , αk−1)] = 1 or 2. Then [F : Q] = 2mfor some m ∈ N. Suppose 3√2 ∈ F. Then Q ⊂ Q( 3√2) ⊂ F, so [Q( 3√2) : Q] divides [F : Q], that is, 3divides 2m, a contradiction. Hence, 3√2 < F.

Exercise 13.2.14.Since α2 ∈ F (α), clearly F (α2) ⊂ F (α). Thus we have to prove α ∈ F (α2). For this purpose,consider the polynomial p(x) = x2 − α2, so that p(α) = 0. Note that α ∈ F (α2) if and onlyif p(x) is reducible in F (α2). For a contradiction, suppose p(x) is irreducible in F (α2), so that[F (α) : F (α2)] = 2. Thus

[F (α) : F] = [F (α) : F (α2)][F (α2) : F] = 2[F (α2) : F],

so [F (α) : F] is even, a contradiction. Therefore, p(x) is reducible in F (α2) and α ∈ F (α2).

Exercise 13.2.15.We follow the hint. Suppose there exists a counterexample. Let α be of minimal degree such thatF (α) is not formally real and α having minimal polynomial f of odd degree, say deg f = 2k + 1for some k ∈ N. Since F (α) is not formally real, then −1 can be express as a sum of squares inF (α) � F[x]/(( f (x))). Then, the exists polynomials p1(x), . . . , pm(x), g(x) such that

−1 + f (x)g(x) = (p1(x))2 + · · · + (pm(x))2.

As every element in F[x]/(( f (x)) can be written as a polynomial in α with degree less than deg f ,we have deg pi < 2k + 1 for all i. Thus, the degree in the right hand of the equation is less than4k + 1, so deg g < 2k + 1 as well. We prove that the degree of g is odd by proving that the degree of(p1(x))2+· · ·+(pm(x))2 is even, because then the equation−1+ f (x)g(x) = (p1(x))2+· · ·+(pm(x))2

implies the result. Let d be the maximal degree over all pi, we prove that x2d is the leading term of(p1(x))2 + · · ·+ (pm(x))2. Note that x2d is a sum of squares (of the leading coefficients of the pi’s ofmaximal degree). Now, since F is formally real, 0 can’t be expressed as a sum of squares in F. Indeed,if

∑li=1 a2

i = 0, then∑l−1

i=1(ai/al)2 = −1. Therefore x2d , 0, so the degree of (p1(x))2+ · · ·+ (pm(x))2

is 2d, as claimed. Hence, the degree of g must be odd by the assertion above. Then g must containan irreducible factor of odd degree, say h(x). Since deg g < deg f , we have deg h < deg f as well.Let β be a root of h(x), hence a root of g(x). Then

−1 + h(x)f (x)g(x)

h(x)= (p1(x))2 + · · · + (pm(x))2,

so −1 is a square in F[x]/((h(x)) � F (β), which means F (β) is not formally real. Therefore, β isa root of an odd degree polynomial h such that F (β) is not formally real. Since deg h < deg f , thiscontradicts the minimality of α. The result follows.

Exercise 13.2.16.Let r ∈ R be nonzero. Since r is algebraic over F, there exist an irreducible polynomial p(x) =a0 + a1x + · · · + xn ∈ F[x] such that p(r) = 0. Note that a0 , 0 since p is irreducible. Thenr−1 = −a−1

0 (rn−1 + · · · + a1). Since ai ∈ F ⊂ R and r ∈ R, we have r−1 ∈ R.

8


Exercise 13.2.17.Let p(x) be an irreducible factor of f (g(x)) of degree m. Let α be a root of p(x). Since p isirreducible, then [F (α) : F] = deg p(x) = m. Now, since p(x) divides f (g(x)), we have f (g(α)) = 0and thus g(α) is a root of f (x). Since f is irreducible, this means n = [F (g(α)) : F]. Note thatF (g(α)) ⊂ F (α). Therefore,

m = [F (α) : F] = [F (α) : F (g(α))][F (g(α)) : F] = [F (α) : F (g(α))] · n,

so n divides m, that is, deg f divides deg p.

Exercise 13.2.18.(a) We follow the hint. Since k[t] is an UFD and k (t) is its field of fractions, then, by Gauss Lemma,P(X ) − tQ(X ) is irreducible in k ((t))[X] is and only if it is irreducible in (k[t])[X]. Note that(k[t])[X] = (k[X])[t]. Since P(X ) − tQ(X ) is linear in (k[X])[t], is clearly irreducible in (k[X])[t](i.e., in (k[t])[X]), hence in (k (t))[X]. Thus, P(X ) − tQ(X ) is irreducible in k (t). Now, x is clearlya root of P(X ) − tQ(X ) since P(x) − tQ(x) = P(x) −

P(x)Q(x)

Q(x) = P(x) − P(x) = 0.

(b) Let n = max{degP(x), degQ(x)}. Write

P(x) = anxn + · · · + a1x + a0 and Q(x) = bnxn + · · · + b1x + b0,

where ai, bi ∈ k for all i, so at least one of an or bn is nonzero. The degree of P(X )− tQ(X ) is clearly≤ n, we prove is n. If an or bn is zero then clearly deg (P(X ) − tQ(X )) = n. Suppose an, bn , 0.Then an, bn ∈ k, but t < k (as t ∈ k (x)), it cannot be that an = tbn. Thus (an − tbn)Xn , 0, so thedegree of P(X ) − tQ(X ) is n.(c) Since P(X ) − tQ(X ) is irreducible over k (t) and x is a root by part (a), then [k (x) : k (t)] =degP(X ) − tQ(X ), and this degree equals max{degP(x), degQ(x)} by part (b).

Exercise 13.2.19.(a) Fix α in K . Since K is (in particular) a commutative ring, we have α(a + b) = αa + αb andα(λa) = λ(αa) for all a, b, λ ∈ K . If, in particular, λ ∈ F, we have the result.(b) Fix a basis for K as a vector space over F. By part (a), for every α ∈ K we can associate aF-linear transformation Tα. Denote by

(Tα

)the matrix of Tα with respect to the basis fixed above.

Then define ϕ : K → Mn(F) by ϕ(α) =(Tα

). We claim ϕ is an isomorphism. Indeed, if α, β ∈ K ,

then T(α+β) (k) = (α + β)(k) = αk + βk = Tα(k) + Tβ (k) for every k ∈ K , hence T(α+β) = Tα + Tβ .We also have T(αβ) (k) = (αβ)(k) = α(βk) = TαTβ (k) for every k ∈ K , so T(αβ) = TαTβ . Thusϕ(α+ β) = ϕ(α)+ϕ(β) and ϕ(αβ) = ϕ(α)ϕ(β) (since the basis is fixed), so ϕ is an homomorphism.Now, if ϕ(α) = ϕ(β), then αk = βk for every k ∈ K , so letting k = 1 we find that ϕ is injective.Therefore, ϕ(K ) is isomorphic to a subfield of Mn(F), so the ring Mn(F) contains an isomorphiccopy of every extension of F of degree ≤ n.

Exercise 13.2.20.The characteristic polynomial of A is p(x) = det(I x − A). For every k ∈ K , we have (Iα − A)k =αk − Ak = αk − αk = 0, so det(Iα − A) = 0 in K . Therefore, p(α) = 0.

Now, consider the field Q( 3√2) with basis {1, 3√2, 3√4} over Q. Denote the elements of this basisby e1 = 1, e2 =

3√2 and e3 =3√4. Let α = 3√2 and β = 1+ 3√2+ 3√4. Then α(e1) = e2, α(e2) = e3 and

α(e3) = 2e1. We also have β(e1) = e1 + e2 + e3, β(e2) = 2e1 + e2 + e3 and β(e3) = 2e1 + 2e2 + e3.Thus, the associated matrices of the their linear transformations are, respectively,

Aα =*..,

0 0 21 0 00 1 0

+//-

and Aβ =*..,

1 2 21 1 21 1 1

+//-.

9


The characteristic polynomial of Aα is x3 − 2, hence is the monic polynomial of degree 3 satisfied byα =

3√2. Furthermore, the characteristic polynomial of Aβ is x3 − 3x2 − 3x − 1, hence is the monicpolynomial of degree 3 satisfied by β = 1 + 3√2 + 3√4.

Exercise 13.2.21.The matrix of the linear transformation "multiplication by α" on K is found by acting of α in the basis

1,√

D. We have α(1) = α = a + b√

D and α(√

D) = a√

D + bD. Hence the matrix is(a bDb a

).

Now let ϕ : K → M2(Q) be defined by ϕ(a + b√

D) =(a bDb a

).

We have

ϕ(a+ b√

D+ c+ d√

D) =(a + c (b + d)Db + d a + c

)=

(a bDb a

)+

(c dDd c

)= ϕ(a+ b

√D)+ ϕ(c+ d

√D),

and

ϕ((a + b√

D) · (c + d√

D)) =(ac + bdD (ad + bc)Dad + bc ac + bdD

)=

(a bDb a

)·

(c dDd c

)= ϕ(a + b

√D)ϕ(c + d

√D),

so ϕ is an homomorphism. Since K is a field, its ideals are {0} and K , so ker(ϕ) is trivial or K . Sinceϕ(K ) is clearly non-zero, then ker(ϕ) , K and thus ker(ϕ) = {0}. Hence, ϕ is injective. Therefore,ϕ is an isomorphism of K with a subfield of M2(Q).

Exercise 13.2.22.Define ϕ : K1 × K2 → K1K2 by ϕ(a, b) = ab. We prove that ϕ is F-bilinear. Let a, a1, a2 ∈ K andb, b1, b2 ∈ K2. Then

ϕ((a1, b) + (a2, b)) = ϕ(a1 + a2, b) = (a1 + a2)b = a1b + a2b = ϕ(a1, b) + ϕ(a2, b),

and

ϕ((a,1 b) + (a, b2)) = ϕ(a, b1 + b2) = a(b1 + b2) = ab1 + ab1 = ϕ(a, b1) + ϕ(a, b2).

We also have, for r ∈ F, ϕ(ar, b) = (ar)b = a(rb) = ϕ(rb). Therefore, ϕ is a F-bilinear map.Hence, ϕ induces a F-algebra homomorphism Φ : K1 ⊗F K2 → K1K2. We use Φ to prove bothdirections. Note that K1 ⊗F K2 have dimension [K1 : F][K2 : F] as a vector space over F.

First, we suppose [K1K2 : F] = [K1 : F][K2 : F] and prove K1 ⊗F K2 is a field. In this caseK1 ⊗F K2 and K1K2 have the same dimension over F. Let L = Φ(K1 ⊗F K2). We claim L = K1K2,i.e. Φ is surjective. Note that L contains K1 and K2. Since L is a subring of K1K2 containing K1 (orK2), then L is a field (Exercise 16). Hence, L is a field containing both K1 and K2. Since K1K2 isthe smallest such field (by definition), we have L = K1K2. Therefore Φ is surjective, as claimed. So,Φ is an F-algebra surjective homomorphism between F-algebras of the same dimension, hence is anisomorphism. Thus, K1 ⊗F K2 is a field.

Now suppose that K1 ⊗F K2 is a field. In this case Φ is a field homomorphism. Therefore, Φis either injective or trivial. It is clearly nontrivial since Φ(1 ⊗ 1) = 1, so it is injective. Hence,[K1 : F][K2 : F] ≤ [K1K2 : F]. As we already have [K1K2 : F] ≤ [K1 : F][K2 : F] (Proposition 21of the book), the equality follows.

10

13.3 Classical Straightedge and Compass Constructions

13.3 Classical Straightedge and Compass Constructions

Exercise 13.3.1.Suppose the 9-gon is constructible. It has angles of 40◦. Since we can bisect an angle by straightedgeand compass, the angle of 20◦ would be constructible. But then cos 20◦ and sin 20◦ would beconstructible too, a contradiction (see proof of Theorem 24).

Exercise 13.3.2.Let O, P,Q and R be the points marked in the figure below.

Then α = ∠QPO, β = ∠RQO, γ = ∠QRO, and θ is an exterior angle of 4PRO. Since 4PQO isisosceles, then α = ∠QPO = ∠QOP. Since β is an exterior angle of 4PQO, it equals the sum of thetwo remote interior angles, i.e., equals ∠QPO + ∠QOP. This two angles equals α, hence β = 2α.Now, since 4QRO is isosceles, then β = γ. Finally, since θ is an exterior angle of 4PRO, equals thesum of the two remote interior angles, which are α and γ. Therefore, θ = α + γ = α + β = 3α.

Exercise 13.3.3.We follow the hint. The distances a, b, x, y and x − k are marked in the figure below.

From the figure, using similar triangles for (a), (b) and (c), and Pythagoras Theorem for (d), the4 relations are clear. Hence, we have

y =

√1 − k2

1 + a, x = a

b + k1 + a

,y

x − k=

√1 − k2

3kand (1 − k2) + (b + k)2 = (1 + a)2.

So,√

1 − k2 = y(1 + a) = 3kyx−k implies 3k = (x − k)(1 + a). From the equation for x above, we find

3k = ( a(b+a)1+a − k)(1 + a) = a(b + k) − k (1 + a), so b + k = 4k+ka

a . Using this in the last equationand reducing, we get

(1 − k2) + (b + k)2 = (1 + a)2

⇒ (1 − k2) + (4k + ka

a)2 = (1 + a)2

⇒ a2(1 − k2) + (4k + ka)2 = a2(1 + a)2

⇒ a2 − (ka)2 + (4k)2 + 8k2a + (ka)2 = a2 + 2a3 + a4

⇒ a4 + 2a3 − 8k2a − 16k2 = 0.

11

13.4 Splitting Fields and Algebraic Closures

We let a = 2y to obtainh4 + h3 − k2h − k2 = 0.

We find h = k2/3, hence a = 2k2/3. From b = 4k+kaa − k, we find b = 2k1/3. Therefore, we can

construct 2k1/3 and 2k2/3 using Conway’s construction.

Exercise 13.3.4.Let p(x) = x3 + x2 − 2x − 1 and α = 2 cos(2π/7). By the Rational Root Theorem, if p has a rootin Q, it must be ±1 since it must divide its constant term. But p(1) = −1 and p(−1) = 1, so p isirreducible over Q. Therefore, α is of degree 3 over Q, hence [Q(α) : Q] cannot be a power of 2.Since we can’t construct α, the regular 7-gon is not constructible by straightedge and compass.

Exercise 13.3.5.Let p(x) = x2+ x−1 = 0 and α = 2 cos(2π/5). By the Rational Root Theorem, if p has a root inQ, itmust be ±1. Since p(1) = 1 and p(−1) = −1, p is irreducible over Q. Hence, α is of degree 2 over Q,so it is constructible. We can bisect an angle by straightedge and compass, so β = cos(2π/5) is alsoconstructible. Finally, as sin(2π/5) =

√1 − cos2(2π/5), sin(2π/5) is also constructible. Therefore,

the regular 5-gon is constructible by straightedge and compass.

13.4 Splitting Fields and Algebraic Closures

Exercise 13.4.1.Let f (x) = x4 − 2. The roots of f are 4√2, − 4√2, i 4√2 and −i 4√2. Hence, the splitting field of f isQ(i, 4√2). So, the splitting field of f has degree [Q(i, 4√2) : Q] = [Q(i, 4√2) : Q( 4√2)][Q( 4√2) : Q] overQ. Since 4√2 is a root of the irreducible polynomial x4−2 overQ, then [Q( 4√2) : Q] = 4. Furthermore,since i < Q( 4√2), then x2 + 1 is irreducible over Q( 4√2) having i as a root, so [Q(i, 4√2) : Q( 4√2)] = 2.Therefore, [Q(i, 4√2) : Q] = 8.

Exercise 13.4.2.Let f (x) = x4 + 2. Let K be the splitting field of f and let L be the splitting field of x4 − 2, that is,L = Q(i, 4√2) (last exercise). We claim K = L, so that [K : Q] = 8 by last exercise. Let ζ =

√2

2 + i√

22 .

First we prove ζ ∈ L and ζ ∈ K , then we prove K = L.We prove ζ ∈ L. This is easy. Let θ = 4√2. Since θ ∈ L, then θ2 =

√2 ∈ L. We also have i ∈ L,

so√

2, i ∈ L implies ζ ∈ L.We prove ζ ∈ K . We have to prove i ∈ K and

√2 ∈ K . Let α be a root of x4 + 2, so that α4 = −2.

Let β be a root of x4 − 1, so that β4 = 1. Then (αβ)4 = α4 β4 = −2, hence αβ is also a root ofx4 + 2. Since the roots of x4 − 1 are ±1,±i, the roots of x4 + 2 are ±α and ±iα. Since K is generatedover Q by there roots, then iα/α = i ∈ K . Now let γ = α2 ∈ K . Since γ2 = α4 = −2, then γ is aroot of x2 + 2. Since the roots of x2 + 2 are i

√2 and −i

√2, then γ is one of this roots. In either case

γ/i ∈ K , which implies√

2 ∈ K . Therefore, ζ ∈ K .Now we prove L = K proving both inclusions.Let α be a root of x4 + 2 and θ be a root of x4 − 2. Then α4 = −2 and θ4 = 2. Note that ζ2 = i, so

ζ4 = −1. Hence, (ζθ)4 = ζ4θ4 = −2, so ζθ is a root of x4 + 2. Then, as we proved earlier, the rootsof x4 + 2 are ±ζθ and ±iζθ. We also have (ζα)4 = ζ4α4 = 2, so ζα is a root of x4 − 2. Then, bylast exercise, the roots of x4 − 2 are ±ζα and ±iζα. Now, since ζ and α are in K , we have ζα ∈ K .We also have i ∈ K , so all roots of x4 − 2 are in K . Since L is generated by these roots, then L ⊂ K .Similarly, ζ and θ are in L, so ζθ ∈ L; since i ∈ L, then all roots of x4 + 2 are in L. Since K isgenerated by these roots, we have K ⊂ L. Therefore, K = L, so [K : Q] = 8.

12

13.5 Separable and Inseparable Extension

Exercise 13.4.3.Let f (x) = x4 + x2 + 1. Note that f (x) = (x2 + x + 1)(x2 − x + 1), so the roots of f are ±1

2 ± i√

32 .

Let w = 12 − i

√3

2 , so this roots are w,−w,w,−w, where w denotes the complex conjugate of w (i.e.,w = 1

2 + i√

32 ). Hence, the splitting field of f is Q(w,w). Since w + w = 1, then Q(w,w) = Q(w).

Furthermore, w is a root of x2 − x + 1, that is irreducible over Q since w < Q. Therefore, the degreeof the splitting field of f is [Q(w) : Q] = 2.

Exercise 13.4.4.Let f (x) = x6 − 4. Note that f (x) = (x3 − 2)(x3 + 2). The roots of x3 − 2 are 3√2, ζ 3√2 and ζ2 3√2,where ζ is the primitive 3’rd root of unity, i.e., ζ = exp(2πi/3) = cos(2π/3)+ i sin(2π/3) = − 1

2 +√

32 .

Furthermore, the roots of x3 + 2 are − 3√2, −ζ 3√2 and −ζ2 3√2. Therefore, the splitting field of f isQ(ζ, 3√2). Then [Q(ζ, 3√2) : Q] = [Q(ζ, 3√2) : Q( 3√2)][Q( 3√2) : Q]. We have that 3√2 is a root ofthe irreducible polynomial x3 − 2 over Q, so 3√2 has degree 3 over Q. Furthermore, ζ is a root ofx2 + x + 1, irreducible over Q( 3√2), so ζ has degree 2 over Q( 3√2). Hence, the degree of the splittingfield of f is [Q(ζ, 3√2) : Q] = 6.

Exercise 13.4.5.We follow the hint. First suppose that K is a splitting field over F. Hence, there exists f (x) ∈ F[x]such that K is the splitting field of f . Let g(x) be an irreducible polynomial in F[x]with a root α ∈ K .Let β be any root of g. We prove β ∈ K , so that g splits completely in K[x]. By Theorem 8, there isan isomorphism ϕ : F (α)

∼−→ F (β) such that ϕ(α) = β. Furthermore, K (α) is the splitting field for

f over F (a), and K (β) is the splitting field for f over F (β). Therefore, by Theorem 28, ϕ extendsto an isomorphism σ : K (α)

∼−→ K (β). Since K = K (α), then [K : F] = [K (α) : F] = [K (β) : F],

so K = K (β). Thus, β ∈ K .Now suppose that every irreducible polynomial in F[x] that has a root in K splits completely in

K[x]. Since [K : F] is finite, then K = F (α1, . . . , αn) for some α1, . . . , αn. For every 1 ≤ i ≤ n, letpi be the minimal polynomial of αi over F, and let f = p1p2 · · · pn. Since every αi is in K , every pihas a root in K , hence splits completely in K . Therefore, f splits completely in K and K is generatedover F by its roots, so K is the splitting field of f (x) ∈ F[x].

Exercise 13.4.6.(a) Let K1 be the splitting field of f1(x) ∈ F[x] over F and K2 the splitting field of f2(x) ∈ F[x]over F. Thus, K1 is generated over F by the roots of f1, and K2 is generated over F by the roots off2. Therefore, f1 f2 splits completely in K1K2 and K1K2 is generated over F by its roots, hence is thesplitting field of f1 f2(x) ∈ F[x].(b) We follow the hint. By last exercise, we have to prove that every irreducible polynomial in F[x]that has a root in K1∩K2 splits completely in (K1∩K2)[x]. So, let f (x) be an irreducible polynomialin F[x] that has a root, say α, in K1 ∩ K2. By last exercise, f splits completely in K1 and splitscompletely in K2. Since K1 and K2 are contained in K , by the uniqueness of the factorization of f inK , the roots of f in K1 must coincide with its roots in K2. Hence, f splits completely in (K1∩K2)[x].


Exercise 13.5.1.Let f (x) = anxn + · · · + a1x + a0 and g(x) = bmxm + · · · + b1x + b0 be two polynomials. Suppose,without any loss of generality, that n ≥ m. Thus, we canwrite g(x) = bnxn+· · ·+b1x+b0, where someof the last coefficients bi could be zero. We have f (x)+g(x) = (an+bn)xn+· · ·+(a1+b1)x+(a0+b0),so

Dx ( f (x) + g(x)) = n(an + bn)xn−1 + · · · + 2(a2 + b2)x + (a1 + b1) = Dx ( f (x)) + Dx (g(x)).

13


Now we prove the formula for the product. Let cn =∑n

k=0 akbn−k , so that

f (x)g(x) = (n∑

k=0ak xk )(

n∑k=0

bk xk ) =2n∑l=0

(l∑

k=0akbl−k )xl =

2n∑l=0

clxl .

Hence,

Dx ( f (x)g(x)) = Dx (2n∑l=0

clxl) =2n−1∑l=0

(l + 1)cl+1xl,

so the coefficient of xl in Dx ( f (x)g(x)) is (l + 1)cl+1.Now, we have Dx ( f (x)) = nanxn−1+ · · ·+2a2x+a1, and Dx (g(x)) = nbnxn−1+ · · ·+2b2x+b1.

So (recall product of polynomials, page 295 of the book)

Dx ( f (x))g(x) = (n∑

k=1kak xk−1)(

n∑k=0

bk xk )

= (n−1∑k=0

(k + 1)ak+1xk )(n∑

k=0bk xk ) =

2n−1∑l=0

(l∑

k=0(k + 1)ak+1bl−k )xl,

and

f (x)Dx (g(x)) = (n∑

k=0ak xk )(

n∑k=1

kbk xk−1)

= (n∑

k=0ak xk )(

n−1∑k=0

(k + 1)bk+1xk ) =2n−1∑l=0

(l∑

k=0ak (l − k + 1)bl−k+1)xl .

Therefore, the coefficient of xl in Dx ( f (x))g(x) + Dx (g(x)) f (x) is

(l∑

k=0(k + 1)ak+1bl−k ) + (

l∑k=0

(l − k + 1)akbl−k+1)

= (l + 1)al+1b0 + (l−1∑k=0

(k + 1)ak+1bl−k ) + (l∑

k=1(l − k + 1)akbl−k+1) + (l + 1)a0bl+1

= (l + 1)al+1b0 + (l∑

k=1kakbl−k+1) + (

l∑k=1

(l − k + 1)akbl−k+1) + (l + 1)a0bl+1

= (l + 1)al+1b0 + (l∑

k=1(l + 1)akbl−k+1) + (l + 1)a0bl+1

= (l + 1)(l+1∑k=0

akbl−k+1) = (l + 1)cl+1.

Since all their coefficients are equal, we conclude Dx ( f (x)g(x)) = Dx ( f (x))g(x) + Dx (g(x)) f (x).

Exercise 13.5.2.The polynomials x and x + 1 are the only (non-constant, i.e. , 0, 1) polynomials of degree 1 overF2; they are clearly irreducible. A polynomial f (x) ∈ F2[x] of degree 2 is irreducible over F2 ifand only if it does not have a root in F2, that is, exactly when f (0) = f (1) = 1. Hence, the onlyirreducible polynomial of degree 2 over F2 is x2 + x + 1. Now, for a polynomial f (x) ∈ F2[x] of

14


degree 4 to be irreducible, it must have no linear or quadratic factors. We can also apply the conditionf (1) = f (0) = 1 to discard the ones with linear factors. Furthermore, f must have an odd number ofterms (or it will be 0), and must have constant term 1 (or x will be a factor). We are left with

x4 + x3 + x2 + x + 1 x4 + x3 + 1x4 + x2 + 1 x4 + x + 1.

For any of this polynomials to be irreducible, it can’t be factorized as two quadratic irreducible factors.Since x2+x+1 is the only irreducible polynomial of degree 2 over F2, only (x2+x+1)2 = x4+x2+1 ofthis four is not irreducible. Hence, the irreducible polynomials of degree 4 overF2 are x4+x3+x2+x+1,x4 + x3 + 1 and x4 + x + 1.

Now, since x + 1 = x − 1 in F2, we have (x + 1)(x4 + x3 + x2 + x + 1) = x5 − 1. We also calculate(x2+ x+1)(x4+ x+1)(x4+ x3+1) = x10+ x5+1. So, the product of all this irreducible polynomialsis

x(x + 1)(x2 + x + 1)(x4 + x + 1)(x4 + x3 + 1)(x4 + x3 + x2 + x + 1)

= x(x5 − 1)(x10 + x5 + 1) = x16 − x.

Exercise 13.5.3.We follow the hint. Suppose d divides n, so that n = qd for some q ∈ Z. Then xn − 1 = xqd − 1 =(xd − 1)(xqd−d + xqd−2d + . . . + xd + 1). So xd − 1 divides xn − 1.

Conversely, suppose d does not divide n. Then n = qd + r for some q, r ∈ Z with 0 < r < d.Thus xn − 1 = (xqd+r − xr ) + (xr − 1) = xr (xqd − 1) + (xr − 1) = xr (xd − 1)(xqd−d + xqd−2d +

. . .+ xd + 1) + (xr − 1). Since xd − 1 divides the first term, but doesn’t divide xr − 1 (as r < d), thenxd − 1 does not divide xn − 1.

Exercise 13.5.4.The first assertion is analogous to the last exercise.

Now, Fpd is defined as the field whose pd elements are the roots of xpd− x over Fp. Similarly

is defined Fpn . Take a = p. So, d divides n if and only if pd − 1 divides pn − 1, and that occursexactly when xpd−1 − 1 divides xpn−1 − 1 (by last exercise). Thus, if d divides n, any root ofxpd− x = x(xpd−1 − 1) must be a root of xpn

− x = x(xpn−1 − 1), hence Fpd ⊆ Fpn . Conversely, ifFpd ⊆ Fpn , then xpd−1 − 1 divides xpn−1 − 1, so d divides n.

Exercise 13.5.5.Let f (x) = xp − x + a. Let α be a root of f (x). First we prove f is separable. Since (α + 1)p − (α +1) + a = αp + 1− α − 1+ a = 0, then α + 1 is also a root of f (x). This gives p distinct roots of f (x)given by α + k with k ∈ Fp, so f is separable.

Now we prove f is irreducible. Let f = f1 f2 · · · fn where f i (x) ∈ Fp[x] is irreducible for all1 ≤ i ≤ n. Let 1 ≤ i < j ≤ n and let αi be a root of f i and α j be a root of f j , so that f i is the minimalpolynomial of αi and f j the minimal polynomial of α j . We prove deg f i = deg f j . Since αi is a rootof f i, it is a root of f , hence there exists k1 ∈ Fp such that αi = α+ k1. Similarly, there exists k2 ∈ Fpsuch that α j = α + k2. Thus, αi = α j + k1 − k2, so f i (x + k1 − k2) is irreducible having α j as a root,so it must be its minimal polynomial. Hence, f i (x + k1− k2) = f j (x), so deg f i = deg f j , as claimed.Since i and j were arbitrary, then all f i are of the same degree, say q. Then p = deg f = nq, so n = 1or n = p (as p is prime). If n = p, then all roots of f are in Fp, so α ∈ Fp and thus 0 = αp −α+a = a,contrary to the hypothesis. Therefore, n = 1, so f is irreducible.

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Exercise 13.5.6.By definition, Fpn is the field whose pn elements are the roots of xpn

− x over Fp. Since xpn− 1 =

x(xpn−1 − 1), clearlyxpn−1 − 1 =

∏α∈F×

pn

(x − α).

Set x = 0. Then−1 =

∏α∈F×

pn

(−α) = (−1)pn−1

∏α∈F×

pn

α

⇒ (−1)pn−1(−1) = (−1)p

n−1(−1)pn−1

∏α∈F×

pn

α

⇒ (−1)pn

=∏α∈F×

pn

α.

Hence, the product of the nonzero elements is +1 if p = 2 and −1 is p is odd. For p odd and n = 1,we have

−1 =∏α∈F×p

α,

so taking module p we find [1][2] · · · [p − 1] = [−1], i.e., (p − 1)! ≡ −1 (mod p).

Exercise 13.5.7.Let a ∈ K such that a , bp for every b ∈ K . Let f (x) = xp − a. We prove that f is irreducibleand inseparable. If α is a root of xp − a, then xp − a = (x − α)p, so α is a multiple root of f (withmultiplicity p), hence f is inseparable. Now, let g(x) be an irreducible factor of f (x). Note thatα < K , otherwise a = αp, contrary to the hypothesis. Then g(x) = (x − α)k for some k ≤ p. Usingthe binomial theorem, we have

g(x) = (x − α)k = xk − kαxk−1 + · · · + (−α)k .

Therefore, kα ∈ K . Since α < K , then k = p, so g = f . Hence, f is irreducible. We conclude thatK (α) is an inseparable finite extension of K .

Exercise 13.5.8.Let f (x) = anxn + . . .+ a1x + a0 ∈ Fp[x]. Since Fp has characteristic p, then (a+ b)p = ap + bp forany a, b ∈ Fp. Easily we can generalize this to a finite number of terms, so that (x1 + · · · + xn)p =xp

1 + · · · + xpn for any x1, · · · , xn ∈ Fp. Furthermore, by Fermat’s Little Theorem, ap = a for every

a ∈ Fp. So, over Fp, we have

f (x)p = (anxn + . . . + a1x + a0)p = apn xnp + . . . + ap

1 xp + ap0 = anxnp + . . . + a1xp + a0 = f (xp).

Exercise 13.5.9.By the binomial theorem, we have

(1 + x)pn =pn∑i=0

(pni

)xi,

so the coefficient of xpi in the expansion of (1 + x)pn is(pnpi

).

Since Fp has characteristic p, we have (1 + x)pn = 1 + xpn = (1 + xp)n, so over Fp this is thecoefficient of (xp)i in (1 + xp)n. Furthermore, (1 + x)pn = (1 + xp)n implies

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13.6 Cyclotomic Polynomials and Extensions

(1 + xp)n =n∑i=0

(ni

)(xp)i =

pn∑i=0

(pnk

)xi = (1 + x)pn

over Fp, hence(pnpi

)≡

(ni

)(mod p).

Exercise 13.5.10.This is equivalent to prove that for any primenumber p, we have f (x1, x2, . . . , xn)p = f (xp

1 , xp2 , . . . , xp

n )in Fp[x1, x2, . . . , xn]. Let

f (x1, x2, . . . , xn) =∑

γ1,...,γn=0aγ1,...,γn xγ1

1 . . . xγnn

be an element of Fp[x1, x2, . . . , xn].Since Fp has characteristic p, then (x1 + · · · + xn)p = xp

1 + · · · + xpn for any x1, · · · , xn ∈ Fp.

Furthermore, by Fermat’s Little Theorem, ap = a for every a ∈ Fp. Hence, over Fp we have

f (x1, x2, . . . , xn)p = (∑

aγ1,...,γn xγ11 . . . xγnn )p =

∑(aγ1,...,γn xγ1

1 . . . xγnn )p

=∑

apγ1,...,γn (xγ1

1 . . . xγnn )p =∑

aγ1,...,γn (xpγ11 . . . xpγn

n ) = f (xp1 , xp

2 , . . . , xpn ).

Exercise 13.5.11.Let f (x) ∈ F[x] with no repeated irreducible factors in F[x]. We can suppose f is monic. Thenf = f1 f2 · · · fn for some monic irreducible polynomials f i (x) ∈ F[x]. Since F is perfect, f isseparable, hence all f i has distinct roots. Thus, f splits in linear factors in the closure of F, hencesplits in linear factors in the closure of K . Therefore, f (x) has no repeated irreducible factors inK[x].


Exercise 13.6.1.Since (ζmζn)mn = 1, then ζmζn is anmnth root of unity. Now, let 1 ≤ k < mn. Then (ζmζn)k = ζkmζ

kn .

For a contradiction, suppose ζkmζkn = 1. Then, ζkm = 1 and ζkn = 1, hence m divides k and n dividesk, so mn divides k (as m , n). Since 1 ≤ k < mn, this is impossible. So, (ζmζn)k , 1 for all1 ≤ k < mn and thus the order of ζmζn is mn. Therefore, ζmζn generates the cyclic group of all mnth

roots of unity, that is, ζmζn is a primitive mnth root of unity.

Exercise 13.6.2.Since (ζdn )(n/d) = ζnn = 1, then ζdn is an (n/d)th root of unity. Now let 1 ≤ k < (n/d). Then(ζdn )k = ζkdn . Since 1 ≤ kd < n, then ζkdn , 1, so (ζdn )k , 1. Hence, the order of ζdn is (n/d), so itgenerates the cyclic group of all (n/d)th roots of unity, that is, ζdn is a primitive (n/d)th root of unity.

Exercise 13.6.3.Let F be a field that contains the nth roots of unity for n odd and let ζ be a 2nth root of unity. Ifζn = 1, then ζ ∈ F, so suppose ζn , 1. Since ζ2n = 1, then ζn is a root of x2 − 1. Since the roots ofthis polynomial are 1 and −1, and ζn , 1, then ζn = −1. Hence, (−ζ )n = (−1)n(ζ )n = (−1)n+1 = 1(since n is odd), so −ζ ∈ F. Since F is a field, then ζ ∈ F.

Exercise 13.6.4.Let F be a field with char F = p. The roots of unity over F are the roots of xn − 1 = xpkm − 1 =(xm − 1)p

k , so are the roots of xm − 1. Now, since m is relatively prime to p, so is xm − 1 and

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its derivative mxm−1, so xm − 1 has no multiple roots. Hence, the m different roots of xm − 1 areprecisely the m distinct nth roots of unity over F.

Exercise 13.6.5.We use the inequality ϕ(n) ≥

√n/2 for all n ≥ 1. Let K be an extension of Q with infinitely many

roots of unity. Let N ∈ N. Then, there exits n ∈ N such that n > 4N2 and there exits some nth root ofunity ζ ∈ K . Then [K : Q] ≥ [Q(ζ ) : Q] = ϕ(n) ≥

√n/2 > N . Since N was arbitrary, we conclude

that [K : Q] > N for all N ∈ N, so [K : Q] is infinite. Therefore, in any finite extension of Q thereare only a finite number of roots of unity.

Exercise 13.6.6.Since Φ2n(x) and Φn(−x) are irreducible, they are the minimal polynomial of any of its roots. Thus,it is sufficient to find a common root for both. Let ζ2 be the primitive 2th root of unity and let ζn be aprimitive nth root of unity. Note that ζ2 = −1, so that ζ2ζn = −ζn. Since n is odd, 2 and n are relativelyprime. So, by Exercise 1, ζ2ζn is a primitive 2nth root of unity, i.e, a root of Φ2n(x). Furthermore,−ζn is clearly a root of Φn(−x). Thus, −ζn is a common root for both Φ2n(x) and Φn(−x), henceΦ2n(x) = Φn(−x).

Exercise 13.6.7.The Möbius Inversion Formula sates that if f (n) is defined for all nonnegative integers and F (n) =∑

d |n f (d), then f (n) =∑

d |n µ(d)F ( nd ). So lets start with the formula

xn − 1 =∏d |n

Φd (x).

We take natural logarithm in both sides and obtain

ln(xn − 1) = ln(∏d |n

Φd (x)) =∑d |n

lnΦd (x).

So, we use the Möbius Inversion Formula for f (n) = lnΦn(x) and F (n) = ln(xn − 1) to obtain

lnΦn(x) =∑d |n

µ(d) ln(xn/d − 1) =∑d |n

ln(xn/d − 1)µ(d) .

Hence, taking exponentials we obtain

Φn(x) = exp(∑d |n

ln(xn/d − 1)µ(d)) =∏d |n

(xn/d − 1)µ(d) =∏d |n

(xd − 1)µ(n/d) .

Exercise 13.6.8.(a) Since p is prime, in Fp[x] we have (x − 1)p = xp − 1, so Φ` (x) = x`−1

x−1 =(x−1)`x−1 = (x − 1)l−1.

(b) Note that ζ has order ` as being a primitive `th root of unity. Since p f ≡ 1 mod `, then p f −1 = q`for some integer q, hence ζ p f −1 = ζq` = 1, so ζ ∈ Fp f . Now we prove that f is the smallest integerwith that property. Suppose ζ ∈ Fpn for some n. Then ζ is a root of xpn−1 −1, hence ` divides pn −1(see Exercise 13.5.3). Since f is the smallest power of p such that p f ≡ 1 mod `, is the smallestinteger such that ` divides p f − 1, so n ≥ l, as desired. This in fact proves that Fp (ζ ) = Fp f , so theminimal polynomial of ζ over Fp has degree f .(c) Since ζa ∈ Fp (ζ ), clearly Fp (ζa) ⊂ Fp (ζ ). For the other direction we follow the hint. Let b thethe multiplicative inverse of a mod `, i.e ab ≡ 1 mod `. Then (ζa)b = ζ , so ζ ∈ Fp (ζa) and thusFp (ζ ) ⊂ Fp (ζa). The equality follows.

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Now, consider Φ` (x) as a polynomial over Fp[x]. Let ζi for 1 ≤ i ≤ ` be ` distinct primitive `throots of unity. The minimal polynomial of each ζi has degree f by part (b). Hence, the irreduciblefactors of Φ` (x) have degree f . Since Φ` have degree ` − 1, then there must be `−1

f factors, and allof them are different since Φ` (x) is separable.(d) If p = 7, then Φ7(x) = (x − 1)6 by part (a). If p ≡ 1 mod 7, then f = 1 in (b) and all roots havedegree 1, so Φ7(x) splits in distinct linear factors. If p ≡ 6 mod 7, then f = 2 is the smallest integersuch that p f = p2 ≡ 36 ≡ 1 mod 7, so we have 3 irreducible quadratics. If p ≡ 2, 4 mod 7, thenf = 3 is the smallest integer such that p3 ≡ 23, 43 ≡ 8, 64 ≡ 1 mod 7, so we have 2 irreducible cubics.Finally, if p ≡ 3, 5 mod 7, then f = 6 is the smallest integer such that p6 ≡ 36, 56 ≡ 729, 15626 ≡ 1mod 7, hence we have an irreducible factor of degree 6.

Exercise 13.6.9.Let A be an n by n matrix over C for which Ak = I for some integer k ≥ 1. Then the minimalpolynomial of A divides xk − 1. Since we are working over C, there are k distinct roots of thispolynomial, so the minimal polynomial of A can be split in linear factors. Hence, A is diagonalizable.

Now consider A =(1 α

0 1

)where α is an element of a field of characteristic p.

Computing powers of A, we can prove (by induction) that An =

(1 nα0 1

)for every positive integer

n. Since pα = 0, then Ap = I. Now, if A is diagonalizable, there exists some non-singular matrix Psuch that A = PDP−1, where D is a diagonal matrix whose diagonal entries are the eigenvalues of A.Since A has only one eigenvalue 1, then D = I, and thus A = PIP−1 = I. So, if A is diagonalizable,it must be α = 0.

Exercise 13.6.10.Let a, b ∈ Fpn . Then ϕ(a + b) = (a + b)p = ap + bp = ϕ(a) + ϕ(b), and ϕ(ab) = (ab)p = apbp =

ϕ(a)ϕ(b), so ϕ is and homomorphism. Moreover, if ϕ(a) = 0, then ap = 0 implies a = 0. Hence, ϕis injective. Since Fpn is finite, then ϕ is also surjective so it is an isomorphism. Furthermore, sinceevery element of Fpn is a root of xpn

− x, then ϕn(a) = apn= a for all a ∈ Fpn , so ϕn is the identity

map. Now, let m be an integer such that ϕm is the identity map. Then apm= a for all a ∈ Fpn , so

every element of Fpn must be a root of xpm− x. Hence, Fpn ⊂ Fpm and thus n divides m (Exercise

13.5.4), so n ≤ m.

Exercise 13.6.11.Note that the minimal polynomial of ϕ is xn−1, for if ϕ satisfies some polynomial xn−1+ · · ·+a1x+a0of degree n − 1 (or less) with coefficients in Fp, then xpn−1

+ · · · + a1xp + a0 for all x ∈ Fpn , whichis impossible. Since Fpn has degree n as a vector space over Fp, then xn − 1 is also the characteristicpolynomial of ϕ, hence is the only invariant factor. Therefore, the rational canonical form of ϕ overFp is the companion matrix of xn − 1, which is

*........,

0 0 · · · 0 11 0 · · · 0 00 1 · · · 0 0...

.... . .

......

0 0 · · · 1 0

+////////-

.

Exercise 13.6.12.We’ll work over the algebraic closure of Fpn , to ensure the field to contain all eigenvalues. In lastexercise we proved that the minimal and characteristic polynomial of ϕ is xn − 1. Moreover, theeigenvalues of ϕ are the nth roots of unity. We use Exercise 4 and write n = pkm for some prime p

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and some m relatively prime to p, so that xn − 1 = (xm − 1)pk and we get exactly m distinct nth roots

of unity, each one with multiplicity pk . Since all the eigenvalues are zeros of both the minimal andcharacteristic polynomial of multiplicity pk , we get m Jordan blocks of size pk . Now, fix a primitivemth root of unity, say ζ . Then, each Jordan block each of the form

Ji =

*..........,

ζ i 1 0 · · · 0 00 ζ i 1 · · · 0 00 0 ζ i · · · 0 0...

......

. . ....

...

0 0 0 · · · ζ i 10 0 0 · · · 0 ζ i

+//////////-

for some 0 ≤ i ≤ m − 1. We already know the Jordan canonical form is given by

*........,

J0 0 · · · 00 J1 · · · 00 0 · · · 0...

.... . .

...

0 0 · · · Jm−1

+////////-

.

Exercise 13.6.13.(a) Z is a division subring of D, and it is commutative by definition of the center, so Z is a field.Since it is finite, its prime subfield is Fp for some prime p, so it is isomorphic to Fpm for some integerm. Let q = pn, so that Z is isomorphic to Fq. Since D is a vector space over Z , then |D | = qn forsome integer n.(b) Let x ∈ D× and let CD (x) be the set of the elements in D that commutes with x. ClearlyZ ⊂ CD (x). We prove that every element a ∈ CD (x) has an inverse in CD (x). Since a ∈ CD (x),then ax = xa. Since D is a division ring, then a−1 ∈ D. Moreover, we have a−1ax = a−1xa andthus x = a−1xa, so xa−1 = a−1x and a−1 ∈ CD (x). Hence, CD (x) is a division ring. Now, sinceZ ⊂ CD (x), then CD (x) is a Z-vector space, so |CD (x) | = qm for some integer m. If x < Z , thenCD (x) is a proper subset of D and hence m < n.(c) The class equation for the group D× is

|D× | = |Z (D×) | +r∑i=1|D× : CD× (xi) |,

where the xi are the representatives of the distinct conjugacy class. By (a) we have |D× | = qn − 1,

|Z (D×) | = q − 1 and |CD× (xi) | = qmi − 1. Then |D× : CD× (xi) | =qn − 1|CD× (xi) |

=qn − 1qmi − 1

. Replacingin the class equation we obtain

qn − 1 = (q − 1) +r∑i=1

qn − 1|CD× (xi) |

= (q − 1) +r∑i=1

qn − 1qmi − 1

.

(d) Since |D× : CD× (xi) | is an integer, then |D× : CD× (xi) | =qn − 1qmi − 1

is an integer. Hence, qmi − 1

divides qn − 1, so (Exercise 13.5.4) mi divides n. Since mi < n (no xi is in Z), no mthi root of unity

is a nth root of unity. Therefore, as Φn(x) divides xn − 1, it must divide (xn − 1)/(xmi − 1) fori = 1, 2, . . . , r .. Letting x = q we have Φn(q) divides (qn − 1)/(qmi − 1) for i = 1, 2, . . . , r .

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(e) From (d), Φn(q) divides (qn − 1)/(qmi − 1) for i = 1, 2, . . . , r , so the class equation in (c) impliesΦn(q) divides q − 1. Now, let ζ , 1 be a nth root of unity. In the complex plane q is closer to 1 thatζ is, so |q − ζ | > |q − 1| = q − 1. Since Φn(q) =

∏ζ primitive(q − ζ ) divides q − 1, this is impossible

unless n = 1. Hence, D = Z and D is a field.

Exercise 13.6.14.We follow the hint. Let P(x) = xn+ · · ·+a1x+a0 be a monic polynomial of degree ≥ 1 over Z. For acontradiction, suppose there are only finitely many primes dividing the values P(n), n = 1, 2, . . ., sayp1, p2, . . . , pk . Let N be an integer such that P(N ) = a , 0. Let Q(x) = a−1P(N + ap1p2 . . . pk x).Then, using the binomial theorem, we have

Q(x) = a−1P(N + ap1p2 . . . pk x)

= a−1((N + ap1p2 . . . pk x)n + · · · + a1(N + ap1p2 . . . pk x) + a0)

= a−1(Nn + an−1Nn−1 + · · · + a1N + a0 + R(x))

= a−1(P(N ) + R(x))

= 1 + a−1R(x)

for some polynomial R(x) ∈ Z[x] divisible by ap1p2 . . . pk . Hence, Q(x) ∈ Z[x]. Moreover, forall n ∈ Z+, P(N + ap1p2 . . . pkn) ≡ a (mod p1, p2, . . . , pk), so Q(n) = a−1P(N + ap1p2 . . . pkn) ≡a−1a = 1 (mod p1, p2, . . . , pk). Now let m be a positive integer such that |Q(m) | > 1, so thatQ(m) ≡ 1 (mod pi) for all i. Therefore, none of the pi’s divide Q(m). Since |Q(m) | > 1, there existsa prime q , pi for all i such that q divides Q(m). Then q divides aQ(m) = P(N + ap1p2 . . . pkm),contradicting the fact that only the primes p1, p2, . . . , pk divides the numbers P(1), P(2), . . ..

Exercise 13.6.15.We follow the hint. Since Φm(a) ≡ 0 (mod p), then am ≡ 1 (mod p). Hence, there exist b such thatba ≡ 1 mod p (indeed, b = am−1), so a is relatively prime to p. We prove that the order of a is m.For a contradiction, suppose ad ≡ 1 (mod p) for some d dividing m, so that Φd (a) ≡ 0 (mod p) forsome d < m. Thus, a is a multiple root of xm − 1, so is also a root of its derivative mam−1. Hence,mam−1 ≡ 0 mod p, impossible since p does not divide m nor a. Therefore, the order of a in (Z/pZ)×

is precisely m

Exercise 13.6.16.Let p be an odd prime dividing Φm(a). If p does not divide m, then, by (c), a is relatively prime to pand the order of a in F×p is m. Since |F×p | = p− 1, this implies m divides p− 1, that is, p ≡ 1 (mod m).

Exercise 13.6.17.By Exercise 14, there are infinitely many primes dividing Φm(1),Φm(2),Φm(3), . . .. Since onlyfinitely of them can divide m, then, by Exercise 16, there must exists infinitely many primes p withp ≡ 1 (mod m).

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