Abstract Algebra Dummit Foote Chapter 13 Field Theory Solutions 2y30m

Abstract Algebra Chapter 13 - Field Theory David S. Dummit and Richard M. Foote Solution Manual by positrón0802

Contents 13 Field Theory 13.1 Basic Theory and Field Extensions . . . . . . . . . 13.2 Algebraic Extensions . . . . . . . . . . . . . . . . . 13.3 Classical Straightedge and Com Constructions 13.4 Splitting Fields and Algebraic Closures . . . . . . . 13.5 Separable and Inseparable Extension . . . . . . . . 13.6 Cyclotomic Polynomials and Extensions . . . . . .

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1 1 4 12 13 15 19

Field Theory

§13.1

Basic Theory and Field Extensions

Exercise 1. Show that p(x) = x3 + 9x + 6 is irreducible in Q[x]. Let θ be a root of p(x). Find the inverse of 1 + θ ∈ Q(θ) Solution: p(x) = x3 + 9x + 6 is irreducible in Z[x] by Eisenstein Criterion with p = 3. By Gauss Lemma, then it 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 2. Show that x3 − 2x − 2 is irreducible over Q and let θ be a root. Compute 1+θ (1 + θ)(1 + θ + θ2 ) and in Q(θ). 1 + θ + θ2 Solution: Let f (x) = x3 − 2x − 2. f is irreducible over Z by Eisenstein Criterion with p = 2, hence over Q by Gauss 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 computing we obtain

1+θ , first we compute (1 + θ + θ2 )−1 . Applying the Euclidean algorithm, 1 + θ + θ2 x3 − 2x − 2 = (x2 + x + 1)(x − 1) − 2x − 1,

and x3 − 2x − 2 = (2x + 1)( 1

x2 x 7 9 − − )− . 2 4 8 8

13.1 Basic Theory and Field Extensions

Evaluating at θ, from this equalities we obtain (θ2 + θ + 1)(θ − 1) = 2θ + 1

and

8 θ2 θ 7 (2θ + 1)−1 = ( − − ). 9 2 4 8

Combining these two equations we obtain 8 2 θ2 θ 7 (θ + θ + 1)(θ − 1)( − − ) = 1. 9 2 4 8 So, 8 θ2 θ 7 2θ2 θ 5 (θ2 + θ + 1)−1 = (θ − 1)( − − ) = − + + , 9 2 4 8 3 3 3 3 where we used θ = 2θ + 2 again. Therefore, 2θ2 θ 5 θ2 2θ 1 1+θ = (1 + θ)(− + + ) = − + + . 1 + θ + θ2 3 3 3 3 3 3 Exercise 3. Show that x3 + x + 1 is irreducible over F2 and let θ be a root. Compute the powers of θ in F2 (θ). Solution: Since 03 + 0 + 1 = 1 and 11 + 1 + 1 = 1 in F2 , then x3 + x + 1 is irreducible over F2 . Since θ is root of 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 4. Prove directly that the map a + b 2 → a − b 2 is an isomorphism of Q( 2) with itself. Solution: 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), and

√ √ √ ϕ((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 6∈ √ 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 5. Suppose α is a rational root of a monic polynomial in Z[x]. Prove that α is an integer. Solution: Let α = p/q be a root of a monic polynomial p(x) = xn + · · · + a1 x + a0 over Z, with gcd(p, q) = 1. Then p p p ( )n + an−1 ( )n−1 + · · · + a1 + a0 = 0. q q q Multiplying this equation by q n one obtains pn + an−1 pn−1 q + · · · + a1 pq n−1 + a0 q n = 0 ⇒

q(an−1 pn−1 + · · · + a1 pq n−2 + a0 q n−1 ) = −pn . 2

13.1 Basic Theory and Field Extensions Thus, every prime that divides q divides pn as well, so divides p. Since gcd(p, q) = 1, there is no prime dividing q, hence q = ±1. The result follows. Exercise 6. Show that if α is a root of an xn + an−1 xn−1 + · · · + a1 x + a0 then an α is a n−1 root of the monic polynomial xn + an−1 xn−1 + an an−2 xn−2 + · · · + an−2 n a1 x + an a0 . Solution: This is straightforward. If an αn + an−1 αn−1 + · · · + a1 α + a0 = 0, then

(an α)n + an−1 (an α)n−1 + an an−2 (an α)n−2 + · · · + ann−2 a1 (an α) + ann−1 a0 n−1 n−1 = ann αn + an−1 + ann−1 an−2 αn−2 + · · · + an−1 n an−1 α n a1 α + an a0 n n−1 + an−2 αn−2 + · · · + a1 α + a0 ) = 0. = an−1 n (an α + an−1 α

Exercise 7. Prove that x3 − nx + 2 is irreducible for n 6= −1, 3, 5. Solution: If x3 − nx + 2 is reducible it must have a linear factor, hence a root. By 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 is irreducible for n 6= −1, 3, 5. Exercise 8. Prove that x5 − ax − 1 ∈ Z[x] is irreducible unless a = 0, 2 or −1. The first two correspond to linear factors, the third corresponds to the factorization (x2 − x + 1)(x3 + x2 − 1). Solution: 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 irreducible polynomials of degrees 2 and 3. Case 1. If x5 − ax − 1 has a root, then, by Rational Root Theorem, it must be α = ±1. If α = 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 of degrees 2 and 3 respectively, such that x5 − ax − 1 = f (x)g(x). Write f (x) = x2 + bx + c and g(x) = x3 + rx2 + sx + t, where b, c, r, s, t ∈ Z. Then x5 − ax − 1 = (x2 + bx + c)(x3 + rx2 + sx + t) = x5 + (b + r)x4 + (br + c + s)x3 + (bs + cr + t)x2 + (bt + cs) + tc. Equating coefficients leads to b+r =0 br + c + s = 0 bs + 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 =0 br − 1 + s = 0 bs − 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, that is, r2 + 1 − s = 0 and rs + r − 1 = 0. Adding these last two equations we obtain 3

13.2 Algebraic Extensions 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, so then r2 + rs + r − s = 0 becomes rs + r = 1, that is, r(s + 1) = 1. Hence, r = 1 and s = 0, or r = −1 and s = −2. If r = 1 and s = 0, then (r + 1)(r + s) = 2s leads to 2 = 0, a contradiction. If r = −1 and s = −2, it leads to 0 = −4, another contradiction. Therefore, (c, t) = (−1, 1) is impossible. We now to the case (c, t) = (1, −1). Case 2.2. Suppose that (c, t) = (1, −1). The system of equations reduces to b+r =0 br + 1 + s = 0 bs + 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. Then b = −1 or r = −s, so one more time we have two cases. If r = −s, then br + 1 + s = 0 becomes br + 1 − r = 0. Hence, b = −r and br + 1 − r = 0 gives r2 + r − 1 = 0. By Rational Root Theorem, this equation has no roots on Z. Since r ∈ Z, we have a contradiction. Now suppose b = −1. From b = −r we obtain r = 1, so, from br + 1 + s = 0 we obtain s = 0. Finally, from −b + s = −a we obtain a = −1. Therefore, the solution (b, c, r, s, t) = (−1, 1, 1, 0, −1) is consistent and we obtain the factorization x5 − ax − 1 = (x2 + bx + c)(x3 + rx2 + sx + t) = (x2 − x + 1)(x3 + x2 − 1).

§13.2

Algebraic Extensions

Exercise 1. Let F be a finite field of characteristic p. Prove that |F| = pn for some positive integer n. Solution: Since the characteristic of F is p, its prime subfield is (isomorphic to) Fp = Z/pZ. We can consider F 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 2. Let g(x) = x2 + x − 1 and let h(x) = x3 − x + 1. Obtain fields of 4, 8, 9 and 27 elements by ading a root of f (x) to the field F where f (x) = g(x) of h(x) and F = F2 of F3 . Write down the multiplication tables for the field with 4 and 9 elements and show that the nonzero elements form a cyclic group. Solution: 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, then F2 (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+1

0 0 0 0 0

1 0 1 x x+1

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

4

x 0 x x+1 x

x+1 0 x+1 x x

13.2 Algebraic Extensions · 0 1 2 x x+1 x+2 2x 2x + 1 2x + 2

0 0 0 0 0 0 0 0 0 0

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

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

x 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 + 1

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

2x 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 2x

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

In both cases, x is a generator of the cyclic group of nonzero elements. Exercise 3. Determine the minimal polynomial over Q for the element 1 + i. Solution: Since 1+i 6∈ 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 4. Determine the degree over Q of 2 + 3 and 1 + 3 2 + 3 4. √ √ √ √ Solution: First, note that (2 + 3)2 = 4 + 4 3 + 3 = 7 + 4 3. Let θ = 2 + 3. Then √ √ +1 θ2 − 4θ = 7 + 4 3 − 8 − 4 3 = −1, hence θ is a root of x2 − 4x + 1. Moreover, x2 − 4x √ 2 − 4x + 1 is the minimal polynomial of 2 + is irreducible over Q (because θ ∈ 6 Q), so x 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 polynomial x3 − 2 over Q, so [Q(β) : Q] = 1 or 3. For a contradiction, suppose [Q(β) : Q] = 1, that is, Q(β) = Q so β ∈ Q. Then β 2 = (1 + α + α2 )2 = 1 + 2α + 3α2 + 2α3 + α4 = 5 + 4α + 3α2 , where we used α3 = 2. So β 2 − 3β = 5 + 2α + 3α2 − 3(1 + α + α2 ) = 2 − α, hence α = −β 2 + 3β + 2 ∈ Q(β) = Q, a contradiction. Therefore, [Q(β) : Q] = 3. Exercise 5. Let F = Q(i). Prove that x3 − 2 and x3 − 3 are irreducible over F . Solution: Since the polynomials have degree 3, if they were reducible they must have a linear factor, hence a root in F .√Note √ that every √ element of F is of the form a + bi, where a, b ∈ Q. The roots of x3 − 2 are 3 2, ζ 3 2 and ζ 2 3 2, where ζ is the primitive 3’rd root of unity, √ √ 3 1 i.e., ζ = exp(2πi/3) = cos(2π/3) + i sin(2π/3) = − 2 + 2 . Since 3 6∈ 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 ζ 2 3 3, and by the same argument non of this elements is in F . Hence x3 − 3 is irreducible over F . Exercise 6. Prove directly from the definitions that the field F (α1 , α2 , . . . , αn ) is the composite of the fields F (α1 ), F (α2 ), . . . , F (αn ). Solution: We have to prove that F (α1 , . . . , αn ) is the smallest field containing F (α1 ), . . . , F (αn ). Clearly F (αi ) ⊂ F (α1 , . . . , αn ) for all 1 ≤ i ≤ n. Now let K be a field such that F (αi ) ⊂ K for all i. If θ is an element of F (α1 , . . . , αn ), then θ is of the form θ = a1 α1 + · · · + an αn , where a1 , . . . , an ∈ F . Every ai αi is in K, hence θ ∈ K. Thus F (α1 , . . . , αn ) ⊂ K. Therefore, 5

13.2 Algebraic Extensions

F (α1 , . . . , αn ) contains all F (αi ) and is contained in every field containing all F (αi ), hence F (α1 , . . . , αn ) is the composite of the fields F (α1 ), F (α2 ), . . . , F (αn ). √ √ √ √ 2,√ 3) [one inclusion is obvious, for the other Exercise that Q( 2 + 3) = Q( √ 7. √ Prove √ 2 , etc.]. Conclude that [Q( 2+ 3) : Q] = 4. Find an irreducible polynomial consider ( 2+ 3) √ √ satisfied by 2 + 3. √ √ √ √ √ √ √ √ Q( 2, 3), clearly Q( 2 + 3) ⊂ Q( 2, 3). Solution: Since 2 + 3 is in √ √ √ √ √ √ For the other direction we have to 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

√ √ 1 1 2 = (θ3 − 9θ), 3 = (11θ − θ3 ) and θ4 = 49 + 20 6. 2 2 √ √ √ √ √ √ 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. 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 8. Let F be a field of characteristic 6= 2. Let D1 and D2 be elements of F , neither √ √ of which is a square in F . Prove that F ( D1 , D2 ) is of √ degree √ 4 over F if D1 D2 is not a square in F and is of degree 2 over F otherwise. When F ( D1 , D2 ) is of degree 4 over F the field is called a biquadratic extension of F . √ √ Solution: The elements of F ( D1 , D2 ) can be written in the form p p p a + b D1 + c D2 + d D1 D2 , where a, b, c, d ∈ F. We have p p p p p p [F ( D1 , D2 ) : F ] = [F ( D1 , D2 ) : F ( D1 )][F ( D1 ) : F ]. √ √ √ √ √ Since [F ( D1 ) : √ F ] = 2, then [F ( D , D ) : F ] can be 2 or 4. Now, [F ( D , D2 ) : F ] = 2 1 2 1 √ √ if and√only if [F ( D1 , √D2 ) : F ( √D1 )] = 1, and that occurs exactly when x2 − D2 is reducible in F ( D1 ) (i.e., when D2 ∈ F ( D1 )), that is, if there exists a, b ∈ F such that p p (a + b D1 )2 = D2 , so that a2 + 2ab D1 + b2 D12 = D2 . √ Note that ab = 0 as ab 6= 0 implies D1 ∈ F , contrary to the hypothesis. Then a = 0 or b = 0. If b = 0, then D2 is a square in F , contrary to the hypothesis. If a = 0, then√b2 D1 = D2 , and thus D1 D2 = ( Db2 )2 , so D1 D2 is a square in F . So, x2 − D2 is reducible in F ( D1 ) if and only if D1 D2 is a square in F . The result follows. Exercise 9. Let F be a field of characteristic 6= 2. Let a, b be elements the field F with p of √ √ √ b not a square in F . Prove that a necessary and sufficient condition for a + b = m + n forp some m and n in F is that a2 − b is a square in F . Use this to determine when the field √ Q( a + b)(a, b ∈ Q) is biquadratic over Q. p √ √ √ √ √ Solution: Suppose a + b = m+ n for some m, n ∈ F , then a+ b = m+n+2 mn. p √ √ √ √ √ Since b is not a square in F , this means b = 2 mn. We also have a + b − n = m, so q √ √ √ √ b = 2 n( a + b − n). 6

13.2 Algebraic Extensions

Hence,

q √ b = 2 n(a + b) − 2n √ √ ⇒ ( b + 2n)2 = 4n(a + b) √ √ ⇒ b + 4n b + 4n2 = 4n(a + b) √

⇒

b + 4n2 − 4na = 0 4a ±

√

⇒

n=

⇒

p 2n a2 − b = ± . a

16a2 − 16b 8

√ Therefore, since a and n are in F , a2 − b is in F . √ 2 2 Now suppose that p a −√b is a√square√in F , so that a − b ∈ F . We prove that there exists m, n ∈ F such that a + b = m + n. Let √ √ a − a2 − b a + a2 − b and n = . m= 2 2 p √ √ √ Note that m and n are in F as char(F ) 6= 2. We claim a + b = m + n. Indeed, we have p p √ √ √ √ √ (a + b) + 2 a2 − b + (a − b) a+ b+ a− b 2 m= =( ) , 4 2 and n= Thus √

(a +

√

p p √ √ √ √ b) − 2 a2 − b + (a − b) a+ b− a− b 2 =( ) . 4 2

p p √ √ a+ b+ a− b m= 2

and

√

p p √ √ a+ b− a− b n= . 2

Therefore, √

m+

√

p p p p √ √ √ √ q √ a+ b+ a− b a+ b− a− b n= + = a + b, 2 2

as claimed. p √ Now, we this to determine when the field Q(pa + b), a, b ∈ Q, is biquadratic over Q. If √ √ √ √ √ a2 − b is a square in b is not, we have Q( a + b) = Q( m + n) = Q( m, n), so p Q and √ 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 squares in Q. Since √ √ a + a2 − b a − a2 − b b mn = = , 2 2 4 p √ then mn is never a square when b isn’t. Thus, Q( a + b) is biquadratic over Q exactly when a2 − b is a square in Q and neither b, m nor n is a square in Q. p √ Exercise 10. Determine the degree of the extension Q( 3 + 2 2) over Q. p p √ √ Solution: Note that 3 + 2 2 = 3 + 8. Recalling last exercise with a = 3 and b = 8, we have a2 − b = 9 − in Q and b = 8 ispnot. Hence, we find (m = 2 and n = 1 p8 = 1√is a square √ √ √ from last exercise) p 3 +√ 8 = 2 + 1. Therefore, Q( 3 + 2 2) = Q( 2) and the degree of the extension Q( 3 + 2 2) over Q is 2. 7

13.2 Algebraic Extensions √ Exercise 11. (a) Let 3 +√ 4i denote the square root of the complex number 3 + 4i that lies in the first quadrant and √ 4i denote the square root of 3 − 4i that lies in the fourth √ let 3 − = 1. quadrant. Prove that [Q( 3 + 4i + 3 − 4i) : Q] p p √ √ (b) Determine the degree of the extension Q( 1 + −3 + 1 − −3). Solution: (a) First, note that the conjugation map a + bi → a − bi is an isomorphism of C, so it takes squares roots to √ square roots, and maps numbers of the first quadrant to the fourth 4i in the first quadrant, its conjugate (and reciprocally). Since 3 + 4i is the square root of 3 + √ √ √ is the square of root of 3 − 4i in the fourth quadrant, so is 3 − 4i. Hence 3 + 4i and 3 − 4i p √ √ 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 2i. Furthermore, we find √ √ √ √ = 1 +√ 3 − 4i = 1 − 2i. pTherefore, p3 + 4i√+ 3 − 4i = 4, i.e., 3 + 4i + 3 − 4i ∈ Q. √ (b) Let θ = 1 + −3 + 1 − −3. Then q q √ √ √ √ √ 2 θ = ( 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. Exercise 12. Suppose the degree of the extension K/F is a prime p. Show that any subfield E of K containing F is either K or F . Solution: 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. Suppose F = Q(α1 , α2 , . . . , αn ) where αi2 ∈ Q for i = 1, 2, . . . , n. Prove that √ 3 2 6∈ F . Solution: Note that, for all 1 ≤ k ≤ n, we√ have [Q(α1 , . . . , αk ) : Q(α 1 or 2. 1 , . . . , αk−1 )] = √ √ 3 3 3 m Then [F : Q] = 2 for some m ∈ N. Suppose 2 ∈ F . Then Q √ ⊂ Q( 2) ⊂ F , so [Q( 2) : Q] divides [F : Q], that is, 3 divides 2m , a contradiction. Hence, 3 2 6∈ F . Exercise 14. Prove that if [F (α) : F ] is odd then F (α) = F (α2 ). Solution: 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 only if 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 15. A field F is said to be formally real if −1 is not expressible as a sum of squares in F . Let F be a formally real field, let f (x) ∈ F [x] be an irreducible polynomial of odd degree and let α be a root of f (x). Prove that F (α) is also formally real. [Pick α a counterexample of minimal degree. Show that −1 + f (x)g(x) = (p1 (x))2 + · · · + (pm (x))2 for some pi (x), g(x) ∈ F [x] where g(x) has odd degree < deg f . Show that some root β of g has odd degree over F and F (β) is not formally real, violating the minimality of α.] Solution: We follow the hint. Suppose there exists a counterexample. Let α be of minimal degree such that F (α) is not formally real and α having minimal polynomial f of odd degree, say deg f = 2k + 1 for some k ∈ N. Since F (α) is not formally real, then −1 can be express as

8

13.2 Algebraic Extensions a sum of squares in F (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 than 4k + 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 of maximal P degree). Now, sincePF is formally real, 0 can’t be 2 expressed as a sum of squares in F . Indeed, if li=1 a2i = 0, then l−1 i=1 (ai /al ) = −1. Therefore 2d 2 2 x 6= 0, so the degree of (p1 (x)) + · · · + (pm (x)) is 2d, as claimed. Hence, the degree of g must be odd by the assertion above. Then g must contain an 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) = (p1 (x))2 + · · · + (pm (x))2 , h(x)

so −1 is a square in F [x]/((h(x)) ∼ = F (β), which means F (β) is not formally real. Therefore, β is a root of an odd degree polynomial h such that F (β) is not formally real. Since deg h < deg f , this contradicts the minimality of α. The result follows. Exercise 16. Let K/F be an algebraic extension and let R be a ring contained in K and containing F . Show that R is a subfield of K containing F . Solution: Let r ∈ R be nonzero. Since r is algebraic over F , there exist an irreducible polynomial p(x) = a0 + a1 x + · · · + xn ∈ F [x] such that p(r) = 0. Note that a0 6= 0 since p is n−1 + · · · + a ). Since a ∈ F ⊂ R and r ∈ R, we have r −1 ∈ R. irreducible. Then r−1 = −a−1 1 i 0 (r Exercise 17. Let f (x) be an irreducible polynomial of degree n over a field F . Let g(x) be any polynomial in F [x]. Prove that every irreducible factor of the composite polynomial f (g(x)) has degree divisible by n. Solution: Let p(x) be an irreducible factor of f (g(x)) of degree m. Let α be a root of p(x). Since p is irreducible, then [F (α) : F ] = deg p(x) = m. Now, since p(x) divides f (g(x)), we have f (g(α)) = 0 and thus g(α) is a root of f (x). Since f is irreducible, this means n = [F (g(α)) : F ]. Note that F (g(α)) ⊂ F (α). Therefore, n = [F (g(α)) : F ] = [F (g(α)) : F (α)][F (α) : F ] = [F (g(α)) : F (α)] · m, so m divides n, that is, deg p divides deg f . Exercise 18. Let k be a field and let k(x) be the field of rational functions in x with P (x) coefficients from k. Let t ∈ k(x) be the rational function with relatively prime polynomials Q(x) P (x), Q(x) ∈ k[x], with Q(x) 6= 0. Then k(x) is an extension of k(t) and to compute its degree it is necessary to compute the minimal polynomial with coefficients in k(t) satisfied by x. (a) Show that the polynomial P (X) − tQ(X) in the variable X and coefficients in k(t) is irreducible over k(t) and has x as a root. [By Gauss Lemma this polynomial is irreducible in (k(t))[X] if and only if it is irreducible in (k[t])[X]. Then note that (k[t])[X] = (k[X])[t].] (b) Show that the degree of P (X) − tQ(X) as a polynomial in X with coefficient in k(t) is the maximum of the degree of P (x) and Q(x). P (x) (c) Show that [k(x) : k(t)] = [k(x) : k( )] = max(degP (x), degQ(x)). Q(x) 9

13.2 Algebraic Extensions

Solution: (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 P (x) in k(t). Now, x is clearly a root of P (X) − tQ(X) since P (x) − tQ(x) = P (x) − Q(x) = Q(x) P (x) − P (x) = 0. (b) Let n = max{degP (x), degQ(x)}. Write P (x) = an xn + · · · + a1 x + a0

and

Q(x) = bn xn + · · · + b1 x + 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 6= 0. Then an , bn ∈ k, but t 6∈ k (as t ∈ k(x)), it cannot be that an = tbn . Thus (an − tbn )X n 6= 0, so the degree 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 19. Let K be an extension of F of degree n. (a) For any α ∈ K prove that α acting by left multiplication on K is an F -linear transformation of K. (b) Prove that K is isomorphic to a subfield of the ring of n × n matrices over F , so the ring of n × n matrices over F contains an isomorphic copy of every extension of F of degree ≤ n. Solution: (a) Fix α in K. Since K is (in particular) a commutative ring, we have α(a+b) = αa + αb and α(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 a F -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) = αkβk = Tα (k)Tβ (k) for every k ∈ K, so T(αβ) = Tα Tβ . Thus ϕ(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 isomorphic copy of every extension of F of degree ≤ n. Exercise 20. Show that if the matrix of the linear transformation "multiplication by α" considered in the previous exercise is A then α is a root of the characteristic polynomial for A. This gives an effective procedure for determining an equation of degree n satisfies by a element α in an extension √ of F of degree √ n. Use √ this procedure to obtain the monic polynomial of degree 3 satisfied by 3 2 and by 1 + 3 2 + 3 4. Solution: The characteristic polynomial of A is p(x) = det(Ix − 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. √ √ 3 3 Now, consider the √ field Q( 3 2) with basis {1, 2, 4} over Q. √ Denote √ √ √ the elements of this basis by 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,     0 0 2 1 2 2 Aα = 1 0 0 and Aβ = 1 1 2 . 0 1 0 1 1 1 10

13.2 Algebraic Extensions The characteristic polynomial of Aα is x3 −2, hence is the monic polynomial of degree 3 satisfied √ 3 by α = 2. Furthermore, the characteristic polynomial of√Aβ is x3 − 3x2 − 3x − 1, hence is the √ 3 monic polynomial of degree 3 satisfied by β = 1 + 2 + 3 4. √ √ Exercise 21. Let K = Q(√ D) for some squarefree integer D. Let α = a + b D be an element of K. Use the basis 1, D for K as a vector space over Q and show that the matrix of the linear transformation "multiplication by α" on K considered in theprevious exercises has √ a bD a bD is an isomorphism of the matrix . Prove directly that the map a + b D 7→ b a b a the field K with a subfield of the ring of 2 × 2 matrices with coefficients in Q. Solution: The matrix√of the linear transformation "multiplication on K is found by √ √ by α" √ acting of α in the basis 1, D. We have α(1) = α = a + b D and α( D) = a D + bD. Hence √ a bD a bD the matrix is . Now let ϕ : K → M2 (Q) be defined by ϕ(a + b D) = . b a b a We have √ √ √ √ a + c (b + d)D a bD c dD ϕ(a+b D+c+d D) = = + = ϕ(a+b D)+ϕ(c+d D), b+d a+c b a d c and √ √ ϕ((a + b D) · (c + d D)) =

c dD a bD ac + bdD (ad + bc)D · = d c b a ad + bc ac + bdD √ √ = 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. Since ϕ(K) is clearly non-zero, then ker(ϕ) 6= K and thus ker(ϕ) = {0}. Hence, ϕ is injective. Therefore, ϕ is an isomorphism of K with a subfield of M2 (Q). Exercise 22. Let K1 and K2 be two finite extensions of a field F contained in the field K. Prove that the F -algebra K1 ⊗F K2 is a field if and only if [K1 K2 : F ] = [K1 : F ][K2 : F ]. Solution: Define ϕ : K1 × K2 → K1 K2 by ϕ(a, b) = ab. We prove that ϕ is F -bilinear. Let a, a1 , a2 ∈ K and b, b1 , b2 ∈ K2 . Then ϕ((a1 , b) + (a2 , b)) = ϕ(a1 + a2 , b) = (a1 + a2 )b = a1 b + a2 b = 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 → K1 K2 . We use Φ to prove both directions. Note that K1 ⊗F K2 have dimension [K1 : F ][K2 : F ] as a vector space over F . First, we suppose [K1 K2 : F ] = [K1 : F ][K2 : F ] and prove K1 ⊗F K2 is a field. In this case K1 ⊗F K2 and K1 K2 have the same dimension over F . Let L = Φ(K1 ⊗F K2 ). We claim L = K1 K2 , i.e. Φ is surjective. Note that L contains K1 and K2 . Since L is a subring of K1 K2 containing K1 (or K2 ), then L is a field (Exercise 16). Hence, L is a field containing both K1 and K2 . Since K1 K2 is the smallest such field (by definition), we have L = K1 K2 . Therefore Φ is surjective, as claimed. So, Φ is an F -algebra surjective homomorphism between F -algebras of the same dimension, hence is an isomorphism. 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, Φ is either injective or trivial. It is clearly nontrivial since Φ(1 ⊗ 1) = 1, so it is injective. Hence, [K1 : F ][K2 : F ] ≤ [K1 K2 : F ]. As we already have [K1 K2 : F ] ≤ [K1 : F ][K2 : F ] (Proposition 21 of the book), the equality follows. 11

13.3 Classical Straightedge and Com Constructions

§13.3

Classical Straightedge and Com Constructions

Exercise 1. Prove that it is impossible to construct the regular 9-gon. Solution: Suppose the 9-gon is constructible. It has angles of 40◦ . Since we can bisect an angle by straightedge and com, the angle of 20◦ would be constructible. But then cos 20◦ and sin 20◦ would be constructible too, a contradiction (see proof of Theorem 24). Exercise 2. Prove that Archimedes’ construction actually trisect the angle θ. [Note the isosceles triangles in figure to prove that β = γ = 2α.]

Solution: Let O, P, Q and R be the points marked in the figure below.

Then α = ∠QP O, β = ∠RQO, γ = ∠QRO, and θ is an exterior angle of 4P RO. Since 4P QO is isosceles, then α = ∠QP O = ∠QOP . Since β is an exterior angle of 4P QO, it equals the sum of the two remote interior angles, i.e., equals ∠QP O + ∠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 4P RO, equals the sum of the two remote interior angles, which are α and γ. Therefore, θ = α + γ = α + β = 3α. Exercise 3. Prove that Conway’s construction indicated in the text actually constructs 2k 1/3 and 2k 2/3 . [One method: let (x, y) be the coordinates of the point C, a the distance from √ 2 y 1−k , (b) B and C and b the distance from A to D; use similar triangles to prove (a) = 1 1+a √ x b+k y 1 − k2 = , (c) = , and also show that (d) (1 − k 2 ) + (b + k)2 = (1 + a)2 ; solve a 1+a x−k 3k these equations for a and b.] Solution: We follow the hint. The distances a, b, x, y and x − k are marked in the figure below.

12

13.4 Splitting Fields and Algebraic Closures

From the figure, using similar triangles for (a), (b) and (c), and Pythagoras Theorem for (d), the 4 relations are clear. Hence, we have √ √ 1 − k2 1 − k2 b+k y y= , x=a , = and (1 − k 2 ) + (b + k)2 = (1 + a)2 . 1+a 1+a x−k 3k √ 3ky implies 3k = (x − k)(1 + a). From the equation for x above, we So, 1 − k 2 = y(1 + a) = x−k find 3k = ( a(b+a) 1+a − k)(1 + a) = a(b + k) − k(1 + a), so b + k = equation and reducing, we get

⇒ ⇒

4k+ka a .

Using this in the last

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

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

a4 + 2a3 − 8k 2 a − 16k 2 = 0.

We let a = 2y to obtain h4 + h3 − k 2 h − k 2 = 0. − k, we find b = 2k 1/3 . Therefore, we can We find h = k 2/3 , hence a = 2k 2/3 . From b = 4k+ka a 1/3 2/3 construct 2k and 2k using Conway’s construction. Exercise 4. The construction of the regular 7-gon amounts to the constructibility of cos(2π/7). We have shall see later (Section 14.5 and Exercise 2 of Section 14.7) that α = 2 cos(2π/7) satisfies the equation x3 + x2 − 2x − 1 = 0. Use this to prove that the regular 7-gon is not constructible by straightedge and com. Solution: Let p(x) = x3 + x2 − 2x − 1 and α = 2 cos(2π/7). By Rational Root Theorem, if p has a root in Q, it must be ±1 since it must divide its constant term. But p(1) = −1 and p(−1) = 1, so p is irreducible 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 com. Exercise 5. Use the fact that α = 2 cos(2π/5) satisfies the equation x2 + x − 1 = 0 to conclude that the regular 5-gon is constructible by straightedge and com. Solution: Let p(x) = x2 + x − 1 = 0 and α = 2 cos(2π/5). By Rational Root Theorem, if p has a root in Q, it must 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 pby straightedge and com, so β = cos(2π/5) is also constructible. 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 com.

§13.4

Splitting Fields and Algebraic Closures

Exercise 1. Determine the splitting field and its degree over Q for x4 − 2. √ √ √ √ 4 − 2. The roots of f are 4 2, − 4 2, i 4 2 and −i 4 2. Hence, the Solution: Let f (x) = x √ √ √ splitting field of f is Q(i, 4 2). So, the splitting field of f has degree [Q(i, 4 2) : Q] = [Q(i, 4 2) : √ √ √ 4 Q( 4 2)][Q( 2) : Q] over Q. Since 4 2 is a root of√the irreducible polynomial x4 − 2 over√Q, √ 2 + 1 is irreducible over Q( 4 2) then [Q( 4 2) : Q] = 4. Furthermore, since i 6∈ Q( 4 2), then x√ √ √ 4 4 having i as a root, so [Q(i, 2) : Q( 2)] = 2. Therefore, [Q(i, 4 2) : Q] = 8. Exercise 2. Determine the splitting field and its degree over Q for x4 + 2. 13

13.4 Splitting Fields and Algebraic Closures Solution: Let f (x) = x4 + 2.√ Let K be the splitting field of f and let L be the splitting 4 field of x4 − 2, that is,√L = Q(i, 2) (last exercise). We claim K = L, so that [K : Q] = 8 by √ 2 2 last exercise. Let ζ = 2 + i 2 . First we prove ζ ∈ L and ζ ∈ K, then we prove K = L. √ √ 4 2 = 2. Since θ ∈ L, then θ 2 ∈ L. We also have We prove ζ ∈ L. This is easy. Let θ = √ 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 β 4 = 1. Then (αβ)4 = α4 β 4 = −2, hence αβ is also a root of x4 + 2. Since the roots of x4 − 1 are ±1, ±i, the roots of x4 + 2 are ±α and ±iα. Since K is generated over Q by there roots, then iα/α = i ∈ K. Now let γ = α2 ∈√K. Since √ 2 γ 2 = α4 = −2, then γ is a root of x2 + 2. Since the roots of √ x + 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 ζ 2 = i, so ζ 4 = −1. Hence, (ζθ)4 = ζ 4 θ4 = −2, so ζθ is a root of x4 + 2. Then, as we proved earlier, the roots of x4 + 2 are ±ζθ and ±iζθ. We also have (ζα)4 = ζ 4 α4 = 2, so ζα is a root of x4 − 2. Then, by last 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 is generated by these roots, we have K ⊂ L. Therefore, K = L, so [K : Q] = 8. Exercise 3. Determine the splitting field and its degree over Q for x4 + x2 + 1. Solution: Let f (x) = x4 +√x2 + 1. Note that f (x) = (x2 + x + 1)(x2 − x + 1), so the roots of √ 3 1 f are ± 2 ± i 2 . Let w = 12 − i 23 , so this roots are w, −w, w, −w, where w denotes the complex √

conjugate of w (i.e., w = 12 + i 23 ). 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 6∈ Q. Therefore, the degree of the splitting field of f is [Q(w) : Q] = 2. Exercise 4. Determine the splitting field and its degree over Q for x6 − 4.

√ 6 −4. Note that f (x) = (x3 −2)(x3 + 2). The roots of x3 −2 are 3 2, Solution: Let f (x) = x √ √ ζ 3 2 and ζ 2 3 2, where ζ is the primitive 3’rd root of unity, i.e., ζ = exp(2πi/3) = cos(2π/3) + √ √ √ √ 3 1 i sin(2π/3) = − 2 + 2 . Furthermore, the roots of x3 +2 are − 3 2, −ζ 3 2 and −ζ 2 3 2. Therefore, √ √ √ √ √ 3 the splitting field of f is Q(ζ, 3 2). Then [Q(ζ, 3 2) : Q] = [Q(ζ, 3 2) : Q( 2)][Q( 3 2) : Q]. We √ √ have that 3 2 is a root of the irreducible polynomial x3 − 2√ over Q, so 3 2 has degree 3 over Q. √ 3 3 Furthermore, ζ is a root of x2 + x + 1, irreducible over Q( 2), so ζ has degree 2 over Q( 2). √ Hence, the degree of the splitting field of f is [Q(ζ, 3 2) : Q] = 6. Exercise 5. Let K be a finite extension of F . Prove that K is a splitting field over F if and only if every irreducible polynomial in F [x] that has a root in K splits completely in K[x]. [Use Theorems 8 and 27.] Solution: 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 is an 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 extends to 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, let pi be the minimal polynomial of αi over F , and let f = p1 p2 · · · pn . Since every 14

13.5 Separable and Inseparable Extension

αi is in K, every pi has a root in K, hence splits completely in K. Therefore, f splits completely in K and K is generated over F by its roots, so K is the splitting field of f (x) ∈ F [x]. Exercise 6. Let K1 and K2 be finite extensions of F contained in the field K, and assume both are splitting fields over F . (a) Prove that their composite K1 K2 is a splitting field over F . (b) Prove that K1 ∩ K2 is a splitting field over F . [Use the preceding exercise.] Solution: (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 of f2 . Therefore, f1 f2 splits completely in K1 K2 and K1 K2 is generated over F by its roots, hence is the splitting 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 polynomial in F [x] that has a root, say α, in K1 ∩ K2 . By last exercise, f splits completely in K1 and splits completely in K2 . Since K1 and K2 are contained in K, by the uniqueness of the factorization of f in K, the roots of f in K1 must coincide with its roots in K2 . Hence, f splits completely in (K1 ∩ K2 )[x].

§13.5

Separable and Inseparable Extension

Exercise 1. Prove that the derivative Dx of a polynomial satisfies Dx (f (x) + g(x)) = Dx (f (x))+Dx (g(x)) and Dx (f (x)g(x)) = Dx (f (x))g(x)+Dx (g(x))f (x) for any two polynomials f (x) and g(x). Solution: Let f (x) = an xn + · · · + a1 x + a0 and g(x) = bm xm + · · · + b1 x + b0 be two polynomials. Suppose, without any loss of generality, that n ≥ m. Thus, we can write g(x) = bn xn + · · · + b1 x + b0 , where some of 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)). P Now we prove the formula for the product. Let cn = nk=0 ak bn−k , so that n n 2n X l 2n X X X X f (x)g(x) = ( ak xk )( bk xk ) = ( ak bl−k )xl = cl xl . k=0

k=0

l=0 k=0

l=0

Hence, Dx (f (x)g(x)) = Dx (

2n X

l

cl x ) =

l=0

2n−1 X

(l + 1)cl+1 xl ,

l=0

so the coefficient of xl in Dx (f (x)g(x)) is (l + 1)cl+1 . Now, we have Dx (f (x)) = nan xn−1 +· · ·+2a2 x+a1 , and Dx (g(x)) = nbn xn−1 +· · ·+2b2 x+b1 . So (recall product of polynomials, page 295 of the book) Dx (f (x))g(x) = (

n X

n X kak xk−1 )( bk xk )

k=1 n−1 X

k=0

n 2n−1 l X X X (k + 1)ak+1 xk )( bk xk ) = ( (k + 1)ak+1 bl−k )xl ,

=(

k=0

k=0

15

l=0 k=0

13.5 Separable and Inseparable Extension

and n n X X f (x)Dx (g(x)) = ( ak xk )( kbk xk−1 ) k=0 n X

=(

k

k=1 n−1 X

ak x )(

k=0

k

(k + 1)bk+1 x ) =

k=0

2n−1 X

(

l X

ak (l − k + 1)bl−k+1 )xl .

l=0 k=0

Therefore, the coefficient of xl in Dx (f (x))g(x) + Dx (g(x))f (x) is l l X X ( (k + 1)ak+1 bl−k ) + ( (l − k + 1)ak bl−k+1 ) k=0

k=0

l−1 l X X = (l + 1)al+1 b0 + ( (k + 1)ak+1 bl−k ) + ( (l − k + 1)ak bl−k+1 ) + (l + 1)a0 bl+1

= (l + 1)al+1 b0 + (

= (l + 1)al+1 b0 + (

k=0 l X k=1 l X

k=1

kak bl−k+1 ) + (

l X

(l − k + 1)ak bl−k+1 ) + (l + 1)a0 bl+1

k=1

(l + 1)ak bl−k+1 ) + (l + 1)a0 bl+1

k=1

= (l + 1)(

l+1 X

ak bl−k+1 ) = (l + 1)cl+1 .

k=0

Since all their coefficients are equal, we conclude Dx (f (x)g(x)) = Dx (f (x))g(x)+Dx (g(x))f (x). Exercise 2. Find all irreducible polynomials of degree 1, 2 and 4 over F2 and prove that their product is x16 − x. Solution: The polynomials x and x + 1 are the only (non-constant, i.e. 6= 0, 1) polynomials of degree 1 over F2 ; they are clearly irreducible. A polynomial f (x) ∈ F2 [x] of degree 2 is irreducible over F2 if and only if it does not have a root in F2 , that is, exactly when f (0) = f (1) = 1. Hence, the only irreducible polynomial of degree 2 over F2 is x2 + x + 1. Now, for a polynomial f (x) ∈ F2 [x] of degree 4 to be irreducible, it must have no linear or quadratic factors. We can also apply the condition f (1) = f (0) = 1 to discard the ones with linear factors. Furthermore, f must have an odd number of (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 + 1

x4 + 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 of this four is not irreducible. Hence, the irreducible polynomials of degree 4 over F2 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 polynomials is 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. 16

13.5 Separable and Inseparable Extension Exercise 3. Prove that d divides n if and only if xd − 1 divides xn − 1. [Note that if n = qd + r then xn − 1 = (xqd+r − xr ) + (xr − 1).] Solution: 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), then xd − 1 does not divide xn − 1. Exercise 4. Let a > 1 be an integer. Prove for any positive integers n, d that d divides n if and only if ad − 1 divides an − 1 (cf. the previous exercise). Conclude in particular that Fpd ⊆ Fpn if and only if d divides n. Solution: The first assertion is analogous to the last exercise. d Now, Fpd is defined as the field whose pd elements are the roots of xp − 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 occurs d n exactly when xp −1 − 1 divides xp −1 − 1 (by last exercise). Thus, if d divides n, any root of d d n n xp − x = x(xp −1 − 1) must be a root of xp − x = x(xp −1 − 1), hence Fpd ⊆ Fpn . Conversely, d n if Fpd ⊆ Fpn , then xp −1 − 1 divides xp −1 − 1, so d divides n. Exercise 5. For any prime p and any nonzero a ∈ Fp prove that xp − x + a is irreducible and separable over Fp . [For the irreducibility: One approach − prove first that if α is a root then α + 1 is also a root. Another approach − suppose it’s reducible and compute derivatives.] Solution: 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 fi (x) ∈ Fp [x] is irreducible for all 1 ≤ i ≤ n. Let 1 ≤ i < j ≤ n and let αi be a root of fi and αj be a root of fj , so that fi is the minimal polynomial of αi and fj the minimal polynomial of αj . We prove deg fi = deg fj . Since αi is a root of fi , it is a root of f , hence there exists k1 ∈ Fp such that αi = α + k1 . Similarly, there exists k2 ∈ Fp such that αj = α+k2 . Thus, αi = αj +k1 −k2 , so fi (x+k1 −k2 ) is irreducible having αj as a root, so it must be its minimal polynomial. Hence, fi (x + k1 − k2 ) = fj (x), so deg fi = deg fj , as claimed. Since i and j were arbitrary, then all fi are of the same degree, say q. Then p = deg f = nq, so n = 1 or 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. Q Q n n Exercise 6. Prove that xp −1 − 1 = α∈F×n (x − α). Conclude that α∈F×n α = (−1)p so p p the product of the nonzero elements of a finite field is +1 if p = 2 and −1 is p is odd. For p odd and n = 1 derive Wilson’s Theorem: (p − 1)! ≡ −1 (mod p). n

Solution: By definition, Fpn is the field whose pn elements are the roots of xp − x over n n Fp . Since xp − 1 = x(xp −1 − 1), clearly Y n xp −1 − 1 = (x − α). α∈F× pn

17

13.5 Separable and Inseparable Extension

Set x = 0. Then −1 =

Y

(−α) = (−1)p

n −1

n −1

(−1)p

α

α∈F× pn

α∈F× pn

⇒

Y

(−1) = (−1)p

n −1

(−1)p

n −1

Y

α

α∈F× pn n

Y

(−1)p =

⇒

α.

α∈F× pn

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 Y α, −1 = α∈F× p

so taking module p we find [1][2] · · · [p − 1] = [−1], i.e., (p − 1)! ≡ −1 (mod p). Exercise 7. Suppose K is a field of characteristic p which is not a perfect field: K 6= K p . Prove there exist irreducible inseparable polynomials over K. Conclude that there exist inseparable finite extensions of K. Solution: Let a ∈ K such that a 6= bp for every b ∈ K. Let f (x) = xp − a. We prove that f is irreducible and inseparable. If α is a root of xp − a, then xp − a = (x − α)p , so α is a multiple root of f (with multiplicity p), hence f is inseparable. Now, let g(x) be an irreducible factor of f (x). Note that α 6∈ K, otherwise a = αp , contrary to the hypothesis. Then g(x) = (x − α)k for some k ≤ p. Using the binomial theorem, we have g(x) = (x − α)k = xk − kαxk−1 + · · · + (−α)k . Therefore, kα ∈ K. Since α 6∈ K, then k = p, so g = f . Hence, f is irreducible. We conclude that K(α) is an inseparable finite extension of K. Exercise 8. Prove that f (x)p = f (xp ) for any polynomial f (x) ∈ Fp [x]. Solution: Let f (x) = an xn + . . . + a1 x + a0 ∈ Fp [x]. Since Fp has characteristic p, then (a + b)p = ap + bp for any a, b ∈ Fp . Easily we can generalize this to a finite number of , so that (x1 + · · · + xn )p = xp1 + · · · + 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 = (an xn + . . . + a1 x + a0 )p = apn xnp + . . . + ap1 xp + ap0 = an xnp + . . . + a1 xp + a0 = f (xp ). pi Exercise 9. Show that the binomial coefficient pn pi is the coefficient of x in the expansion p i p n of (1 + x)pn . Working over Fp show that this is the coefficient of (x ) in (1 + x ) and hence pn n prove that pi ≡ i (mod p). Solution: By the binomial theorem, we have pn X pn i pn (1 + x) = x, i i=0

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

(1 + x ) =

pn X pn i (x ) = x = (1 + x)pn i k

n X n i=0

p i

i=0

18

13.6 Cyclotomic Polynomials and Extensions

over Fp , hence

pn pi



≡

n i



(mod p).

Exercise 10. Let f (x1 , x2 , . . . , xn ) ∈ Z[x1 , x2 , . . . , xn ] be a polynomial in the variables x1 , x2 , . . . , xn with integer coefficients. For any prime p prove that the polynomial f (x1 , x2 , . . . , xn )p − f (xp1 , xp2 , . . . , xpn ) ∈ Z[x1 , x2 , . . . , xn ] has all its coefficients divisible by p. Solution: This is equivalent to prove that for any prime number p, we have f (x1 , x2 , . . . , xn )p = f (xp1 , xp2 , . . . , xpn ) in Fp [x1 , x2 , . . . , xn ]. Let X aγ1 ,...,γn xγ11 . . . xγnn f (x1 , x2 , . . . , xn ) = γ1 ,...,γn =0

be an element of Fp [x1 , x2 , . . . , xn ]. Since Fp has characteristic p, then (x1 + · · · + xn )p = xp1 + · · · + xpn for any x1 , · · · , xn ∈ Fp . Furthermore, by Fermat’s Little Theorem, ap = a for every a ∈ Fp . Hence, over Fp we have X X f (x1 , x2 , . . . , xn )p = ( aγ1 ,...,γn xγ11 . . . xγnn )p = (aγ1 ,...,γn xγ11 . . . xγnn )p X X p p pγn p 1 = apγ1 ,...,γn (xγ11 . . . xγnn )p = aγ1 ,...,γn (xpγ 1 . . . xn ) = f (x1 , x2 , . . . , xn ). Exercise 11. Suppose K[x] is a polynomial ring over the field K and F is a subfield of K. If F is a perfect field and f (X) ∈ F [x] has no repeated irreducible factors in F [x], prove that f (x) has no repeated irreducible factors in K[x]. Solution: Let f (x) ∈ F [x] with no repeated irreducible factors in F [x]. We can suppose f is monic. Then f = f1 f2 · · · fn for some monic irreducible polynomials fi (x) ∈ F [x]. Since F is perfect, f is separable, hence all fi has distinct roots. Thus, f splits in linear factors in the closure of F , hence splits in linear factors in the closure of K. Therefore, f (x) has no repeated irreducible factors in K[x].

§13.6

Cyclotomic Polynomials and Extensions

Exercise 1. Suppose m and n are relatively prime positive integers. Let ζm be a primitive mth root of unity and let ζn be a primitive nth root of unity. Prove that ζm ζn is a primitive mnth root of unity. Solution: Since (ζm ζn )mn = 1, then ζm ζn is an mnth root of unity. Now, let 1 ≤ k < mn. k ζ k . For a contradiction, suppose ζ k ζ k = 1. Then, ζ k = 1 and ζ k = 1, Then (ζm ζn )k = ζm n m n m n hence m divides k and n divides k, so mn divides k (as m 6= n). Since 1 ≤ k < mn, this is impossible. So, (ζm ζn )k 6= 1 for all 1 ≤ k < mn and thus the order of ζm ζn is mn. Therefore, ζm ζn generates the cyclic group of all mnth roots of unity, that is, ζm ζn is a primitive mnth root of unity. Exercise 2. Let ζn be a primitive nth root of unity and let d be a divisor of n. Prove that is a primitive (n/d)th root of unity. Solution: Since (ζnd )(n/d) = ζnn = 1, then ζnd is an (n/d)th root of unity. Now let 1 ≤ k < (n/d). Then d (ζn )k = ζnkd . Since 1 ≤ kd < n, then ζnkd 6= 1, so (ζnd )k 6= 1. Hence, the order of ζnd is (n/d), so it generates the cyclic group of all (n/d)th roots of unity, that is, ζnd is a primitive (n/d)th root of unity. ζnd

Exercise 3. Prove that if a field contains the nth roots of unity for n odd then it also contains the 2nth roots of unity. 19

13.6 Cyclotomic Polynomials and Extensions Solution: Let F be a field that contains the nth roots of unity for n odd and let ζ be a root of unity. If ζ n = 1, then ζ ∈ F , so suppose ζ n 6= 1. Since ζ 2n = 1, then ζ n is a root 2 of x − 1. Since the roots of this polynomial are 1 and −1, and ζ n 6= 1, then ζ n = −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 . 2nth

Exercise 4. Prove that is n = pk m where p is a prime and m is relatively prime to p then there are precisely m distinct nth roots of unity over a field of characteristic p. Solution: Let F be a field with char F = p. The roots of unity over F are the roots of k k xn − 1 = xp m − 1 = (xm − 1)p , so are the roots of xm − 1. Now, since m is relatively prime to p, so is xm − 1 and its derivative mxm−1 , so xm − 1 has no multiple roots. Hence, the m different roots of xm − 1 are precisely the m distinct nth roots of unity over F . Exercise 5. Prove there are only a finite number of roots of unity in any finite extension K of Q. √ Solution: 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 > 4N 2 √ and there exits some nth root of unity 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 there are only a finite number of roots of unity. Exercise 6. Prove that for n odd, n > 1, Φ2n (x) = Φn (−x). Solution: 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 a primitive nth root of unity. Note that ζ2 = −1, so that ζ2 ζn = −ζn . Since n is odd, 2 and n are relatively prime. So, by Exercise 1, ζ2 ζn 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), hence Φ2n (x) = Φn (−x). Exercise 7. Use the Möbius Inversion Formula indicated in Section 14.3 to prove Φn (x) =

Y (xd − 1)µ(n/d) . d|n

Solution: The Möbius Inversion FormulaPsates that if f (n) is defined for all nonnegative P 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 =

Y

Φd (x).

d|n

We take natural logarithm in both sides and obtain Y X ln(xn − 1) = ln( Φd (x)) = ln Φd (x). d|n

d|n

So, we use the Möbius Inversion Formula for f (n) = ln Φn (x) and F (n) = ln(xn − 1) to obtain X X ln Φn (x) = µ(d) ln(xn/d − 1) = ln(xn/d − 1)µ(d) . d|n

d|n

Hence, taking exponentials we obtain X Y Y Φn (x) = exp( ln(xn/d − 1)µ(d) ) = (xn/d − 1)µ(d) = (xd − 1)µ(n/d) . d|n

d|n

20

d|n

13.6 Cyclotomic Polynomials and Extensions `−1

Exercise 8. Let ` be a prime and let Φ` (x) = xx−1 = x`−1 + x`−2 + · · · + x + 1 ∈ Z[x] be the th ` cyclotomic polynomial, which is irreducible over Z by Theorem 41. This exercise determines the factorization of Φ` (x) modulo p for any prime p. Let ζ denote any fixed primitive `th root of unity. (a) Show that if p = ` then Φ` (x) = (x − 1)`−1 ∈ F` [x] (b) Suppose p 6= ` and let f denote the order of p mod ` , i.e., f is the smallest power of p with pf ≡ 1 mod `. Use the fact that F× pn is a cyclic group to show that n = f is the smallest n n power p of p with ζ ∈ Fp . Conclude that the minimal polynomial of ζ over Fp has degree f . (c) Show that Fp (ζ) = Fp (ζ a ) for any integer a not divisible by `. [One inclusion is obvious. For the other, note that ζ = (ζ a )b where b is the multiplicative inverse of a mod `.] Conclude using (b) that, in Fp [x], Φ` (x) is the product of `−1 f distinct irreducible polynomials of degree f. (d) In particular, prove that, viewed in Fp [x], Φ7 (x) = x6 + x5 + · · · + x + 1 is (x − 1)6 for p = 7, a product of distinct linear factor for p ≡ 1 mod 7, a product of 3 irreducible quadratics for p ≡ 6 mod 7, a product of 2 irreducible cubics for p ≡ 2, 4 mod 7, and is irreducible for p ≡ 3, 5 mod 7. Solution: ` `−1 l−1 . (a) Since p is prime, in Fp [x] we have (x−1)p = xp −1, so Φ` (x) = xx−1 = (x−1) x−1 = (x−1) (b) Note that ζ has order ` as being a primitive `th root of unity. Since pf ≡ 1 mod `, f then pf − 1 = q` for some integer q, hence ζ p −1 = ζ q` = 1, so ζ ∈ Fpf . Now we prove that f is the smallest integer with that property. Suppose ζ ∈ Fpn for some n. Then ζ is a root of n xp −1 − 1, hence ` divides pn − 1 (see Exercise 13.5.3). Since f is the smallest power of p such that pf ≡ 1 mod `, is the smallest integer such that ` divides pf − 1, so n ≥ l, as desired. This in fact proves that Fp (ζ) = Fpf , so the minimal polynomial of ζ over Fp has degree f . (c) Since ζ a ∈ Fp (ζ), clearly Fp (ζ a ) ⊂ Fp (ζ). For the other direction we follow the hint. Let b the the multiplicative inverse of a mod `, i.e ab ≡ 1 mod `. Then (ζ a )b = ζ, so ζ ∈ Fp (ζ a ) and thus Fp (ζ) ⊂ Fp (ζ a ). The equality follows. Now, consider Φ` (x) as a polynomial over Fp [x]. Let ζi for 1 ≤ i ≤ ` be ` distinct primitive `th roots of unity. The minimal polynomial of each ζi has degree f by part (b). Hence, the irreducible factors of Φ` (x) have degree f . Since Φ` have degree ` − 1, then there must be `−1 f factors, and all of 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 have degree 1, so Φ7 (x) splits in distinct linear factors. If p ≡ 6 mod 7, then f = 2 is the smallest integer such that pf = p2 ≡ 36 ≡ 1 mod 7, so we have 3 irreducible quadratics. If p ≡ 2, 4 mod 7, then f = 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 ≡ 1 mod 7, hence we have an irreducible factor of degree 6. Exercise 9. Suppose A is an n by n matrix over C for which Ak= I for some integer 1 α k ≥ 1. Show that A can be diagonalized. Show that the matrix A = where α is an 0 1 element of a field of characteristic p satisfies Ap = I and cannot be diagonalized if α 6= 0. Solution: Let A be an n by n matrix over C for which Ak = I for some integer k ≥ 1. Then the minimal polynomial of A divides xk − 1. Since we are working over C, there are k distinct roots of this polynomial, so the minimal polynomial of A can be split in linear factors. Hence, A is diagonalizable. 1 α Now consider A = where α is an element of a field of characteristic p. 0 1 1 nα n Computing powers of A, we can prove (by induction) that A = for every positive 0 1 21

13.6 Cyclotomic Polynomials and Extensions integer n. Since pα = 0, then Ap = I. Now, if A is diagonalizable, there exists some non-singular matrix P such that A = P DP −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 = P IP −1 = I. So, if A is diagonalizable, it must be α = 0. Exercise 10. Let ϕ denote the Frobenius map x 7→ xp on the finite field Fpn . Prove that ϕ gives an isomorphism of Fpn to itself (such an isomorphism is called an automorphism). Prove that ϕn is the identity map an that no lower power of ϕ is the identity. Solution: Let a, b ∈ Fpn . Then ϕ(a + b) = (a + b)p = ap + bp = ϕ(a) + ϕ(b), and ϕ(ab) = (ab)p = ap bp = ϕ(a)ϕ(b), so ϕ is and homomorphism. Moreover, if ϕ(a) = 0, then ap = 0 implies a = 0. Hence, ϕ is injective. Since Fpn is finite, then ϕ is also surjective n so it is an isomorphism. Furthermore, since every element of Fpn is a root of xp − x, then n ϕn (a) = ap = a for all a ∈ Fpn , so ϕn is the identity map. Now, let m be an integer such that m ϕm is the identity map. Then ap = a for all a ∈ Fpn , so every element of Fpn must be a root m of xp − x. Hence, Fpn ⊂ Fpm and thus n divides m (Exercise 13.5.4), so n ≤ m. Exercise 11. Let ϕ denote the Frobenius map x 7→ xp on the finite field Fpn as in the previous exercise. Determine the rational canonical form over Fp for ϕ considered as an Fp -linear transformation of the n-dimensional Fp -vector space Fpn . Solution: Note that the minimal polynomial of ϕ is xn −1, for if ϕ satisfies some polynomial pn−1 + · · · + a xp + a 1 x + a0 of degree n − 1 (or less) with coefficients in Fp , then x 1 0 for all x ∈ Fpn , which is impossible. Since Fpn has degree n as a vector space over Fp , then xn − 1 is also the characteristic polynomial of ϕ, hence is the only invariant factor. Therefore, the rational canonical form of ϕ over Fp is the companion matrix of xn − 1, which is   0 0 ··· 0 1 1 0 · · · 0 0    0 1 · · · 0 0   .  .. .. . . .. ..  . . . . . 0 0 ··· 1 0 xn−1 + · · · + a

Exercise 12. Let ϕ denote the Frobenius map x 7→ xp on the field Fpn as in the previous exercise. Determine the Jordan canonical form (over a field containing all the eigenvalues) for ϕ considered as an Fp -linear transformation of the n-dimensional Fp -vector space Fpn . Solution: We’ll work over the algebraic closure of Fpn , to ensure the field to contain all eigenvalues. In last exercise we proved that the minimal and characteristic polynomial of ϕ is xn − 1. Moreover, the eigenvalues of ϕ are the nth roots of unity. We use Exercise 4 and write k n = pk m for some prime p and some m relatively prime to p, so that xn − 1 = (xm − 1)p 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 and characteristic polynomial of multiplicity pk , we get m Jordan blocks of size pk . Now, fix a primitive mth root of unity, say ζ. Then, each Jordan block each of the form  i  ζ 1 0 ··· 0 0  0 ζi 1 · · · 0 0     0 0 ζi · · · 0 0    Ji =  . . . . .. ..  . . . . . . . . . .    0 0 0 · · · ζi 1  0 0 0 · · · 0 ζi

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13.6 Cyclotomic Polynomials and Extensions

for some 0 ≤ i ≤ m − 1. We already know the Jordan canonical form is given by   J0 0 · · · 0  0 J1 · · · 0     0 0 ··· 0   .  .. .. . . ..  . . . .  0

0

· · · Jm−1

Exercise 13. (Wedderburn’s Theorem on Finite Division Rings) This exercise outlines a proof (following Witt) of Wedderburn’s Theorem that a finite division ring D is a field (i.e., is commutative). (a) Let Z denote the center of D (i.e., the elements of D which commute with every element of D). Prove that Z is a field containing Fp for some prime p. If Z = Fq prove that D has order q n for some integer n [D is a vector space over Z]. (b) The nonzero elements D× of D form a multiplicative group. For any x ∈ D× show that the elements of D which commute with x form a division ring which contains Z. Show that this division ring if of order q m for some integer m and that m < n if x is not an element of Z. (c) Show that the class equation (Theorem 4.7) for the group D× is n

q − 1 = (q − 1) +

r X i=1

qn − 1 |CD× (xi )|

where x1 , x2 , . . . , xr are representatives of the distinct conjugacy classes in D× not contained in the center of D× . Conclude from (b) that for each i, |CD× (xi )| = q mi − 1 for some m1 < n. qn − 1 is an integer (namely, the index |D× : CD× (xi )|) then mi and (d) Prove that since m q i −1 divides n (cf. Exercise 4 of Section 5). Conclude that Φn (x) divides (xn − 1)/(xmi − 1) and hence that the integer Φn (q) divides (q n − 1)/(q mi − Q 1) for i = 1, 2, . . . , r. (e) Prove that (c) and (d) imply that Φn (q) = ζ primitive (q − ζ) divides q − 1. Prove that |q − ζ| > q − 1 (complex absolute value) for any root of unity ζ 6= 1 [note that 1 is the closest point on the unit circle in C to the point q on the real line]. Conclude that n = 1, i.e., that D = Z is a field. Solution: (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 integer m. Let q = pn , so that Z is isomorphic to Fq . Since D is a vector space over Z, then |D| = q n for some integer n. (b) Let x ∈ D× and let CD (x) be the set of the elements in D that commutes with x. Clearly Z ⊂ 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−1 ax = a−1 xa and thus x = a−1 xa, so xa−1 = a−1 x and a−1 ∈ CD (x). Hence, CD (x) is a division ring. Now, since Z ⊂ CD (x), then CD (x) is a Z-vector space, so |CD (x)| = q m for some integer m. If x 6∈ Z, then CD (x) is a proper subset of D and hence m < n. (c) The class equation for the group D× is |D× | = |Z(D× )| +

r X

|D× : CD× (xi )|,

i=1

where the xi are the representatives of the distinct conjugacy class. By (a) we have |D× | = q n −1, qn − 1 qn − 1 |Z(D× )| = q − 1 and |CD× (xi )| = q mi − 1. Then |D× : CD× (xi )| = = m . |CD× (xi )| q i −1 23

13.6 Cyclotomic Polynomials and Extensions

Replacing in the class equation we obtain q n − 1 = (q − 1) +

r X i=1

r

X qn − 1 qn − 1 = (q − 1) + . |CD× (xi )| q mi − 1 i=1

qn − 1 is an integer. Hence, q mi − 1 q mi − 1 divides q n − 1, so (Exercise 13.5.4) mi divides n. Since mi < n (no xi is in Z), no th root of unity. Therefore, as Φ (x) divides xn − 1, it must divide mth n i root of unity is a n n m i (x − 1)/(x − 1) for i = 1, 2, . . . , r.. Letting x = q we have Φn (q) divides (q n − 1)/(q mi − 1) for i = 1, 2, . . . , r. (e) From (d), Φn (q) divides (q n − 1)/(q mi − 1) for i = 1, 2, . . . , r, so the class equation in (c) implies Φn (q) divides q − 1. Now, let ζ 6= 1 be a nth root of unity. In the complex plane q is Q closer to 1 that ζ 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. (d) Since |D× : CD× (xi )| is an integer, then |D× : CD× (xi )| =

Exercise 14. Given any monic polynomial P (x) ∈ Z[x] of degree at leat one show that there are infinitely many distinct prime divisors of the integers P (1), P (2), P (3), . . . , P (n), . . . [Suppose p1 , p2 , . . . , pk are the only primes dividing the values P (n), n = 1, 2, . . .. Let N be an integer with P (N ) = a 6= 0. Show that Q(x) = a−1 P (N + ap1 p2 . . . pk x) is an element of Z[x] and that Q(n) ≡ 1 (mod p1 p2 . . . pk ) for n = 1, 2, . . .. Conclude that there is some integer M such that Q(M ) has a prime factor different from p1 , p2 , . . . , pk and hence that P (N + ap1 p2 . . . pk M ) has a prime factor different from p1 , p2 , . . . pk .] Solution: We follow the hint. Let P (x) = xn + · · · + a1 x + a0 be a monic polynomial of degree ≥ 1 over Z. For a contradiction, suppose there are only finitely many primes dividing the values P (n), n = 1, 2, . . ., say p1 , p2 , . . . , pk . Let N be an integer such that P (N ) = a 6= 0. Let Q(x) = a−1 P (N + ap1 p2 . . . pk x). Then, using the binomial theorem, we have Q(x) = a−1 P (N + ap1 p2 . . . pk x) = a−1 ((N + ap1 p2 . . . pk x)n + · · · + a1 (N + ap1 p2 . . . pk x) + a0 ) = a−1 (N n + an−1 N n−1 + · · · + a1 N + a0 + R(x)) = a−1 (P (N ) + R(x)) = 1 + a−1 R(x) for some polynomial R(x) ∈ Z[x] divisible by ap1 p2 . . . pk . Hence, Q(x) ∈ Z[x]. Moreover, for all n ∈ Z+ , P (N +ap1 p2 . . . pk n) ≡ a (mod p1 , p2 , . . . , pk ), so Q(n) = a−1 P (N +ap1 p2 . . . pk n) ≡ a−1 a = 1 (mod p1 , p2 , . . . , pk ). Now let m be a positive integer such that |Q(m)| > 1, so that Q(m) ≡ 1 (mod pi ) for all i. Therefore, none of the pi ’s divide Q(m). Since |Q(m)| > 1, there exists a prime q 6= pi for all i such that q divides Q(m). Then q divides aQ(m) = P (N + ap1 p2 . . . pk m), contradicting the fact that only the primes p1 , p2 , . . . , pk divides the numbers P (1), P (2), . . .. Exercise 15. Let p be an odd prime not dividing m and let Φm (x) be the mth cyclotomic polynomial. Suppose a ∈ Z satisfies Φm (a) ≡ 0 (mod p). Prove that a is relatively prime to p and that the order of a in (Z/pZ)× is precisely m. [Since Y Y xm − 1 = Φd (x) = Φm (x) Φd (x) d|n

d|n d

24

13.6 Cyclotomic Polynomials and Extensions we see first that am − 1 ≡ 0 (mod p) i.e., a ≡ 1 (mod p). If the order of a mod p were less that m, then ad ≡ 1 (mod p) for some d dividing m, so then Φd (a) ≡ 0 (mod p) for some d < m. but then xm − 1 would have a as a multiple root mod p, a contradiction.] Solution: We follow the hint. Since Φm (a) ≡ 0 (mod p), then am ≡ 1 (mod p). Hence, there exist b such that ba ≡ 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) for some 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 16. Let a ∈ Z. Show that if p is an odd prime dividing Φm (a) then either p divides m of p ≡ 1 (mod m). Solution: Let p be an odd prime dividing Φm (a). If p does not divide m, then, by (c), a is × relatively prime to p and the order of a in F× p is m. Since |Fp | = p − 1, this implies m divides p − 1, that is, p ≡ 1 (mod m). Exercise 17. Prove there are infinitely many primes p with p ≡ 1 (mod m). Solution: By Exercise 14, there are infinitely many primes dividing Φm (1), Φm (2), Φm (3), . . .. Since only finitely of them can divide m, then, by Exercise 16, there must exists infinitely many primes p with p ≡ 1 (mod m).

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