Abstract AlgebraChapter 13 - Field Theory David S. Dummit and Richard M. Foote Solution Manual by positrón0802 Contents 13 Field Theory 1 13.1 Basic Theory and Field Extensions . . . . . . . . . . . . . . . . . . . . . . . . . 1 13.2 Algebraic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 13.3 Classical Straightedge and Compass Constructions . . . . . . . . . . . . . . . . 12 13.4 Splitting Fields and Algebraic Closures . . . . . . . . . . . . . . . . . . . . . . . 13 13.5 Separable and Inseparable Extension . . . . . . . . . . . . . . . . . . . . . . . . 15 13.6 Cyclotomic Polynomials and Extensions . . . . . . . . . . . . . . . . . . . . . . 19 13 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 θ2 − θ + 10 (1 + θ)−1 = . 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 . 1+θ For computing , first we compute (1 + θ + θ2 )−1 . Applying the Euclidean algorithm, 1 + θ + θ2 we obtain x3 − 2x − 2 = (x2 + x + 1)(x − 1) − 2x − 1, and x2 x 7 9 x3 − 2x − 2 = (2x + 1)( − − )− . 2 4 8 8 1 13.1 Basic Theory and Field Extensions Evaluating at θ, from this equalities we obtain 8 θ2 θ 7 (θ2 + θ + 1)(θ − 1) = 2θ + 1 and (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, 1+θ 2θ2 θ 5 θ2 2θ 1 = (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( 2), then Q) a = b and c = d, so ϕ is injective. Also, given a + b 2 √ 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 First suppose (c. then a = 0. 2 or −1. then n = 3. The result follows. From ct = −1 we deduce (c. 3. Case 2. hence q = ±1. Prove that x3 − nx + 2 is irreducible for n 6= −1. Solution: If x3 − nx + 2 is reducible it must have a linear factor. r2 + 1 − s = 0 and rs + r − 1 = 0. Case 1. every prime that divides q divides pn as well. By Rational Root Theorem. and if α = 2. then n = 1. Exercise 8. the third corresponds to the factorization (x2 − x + 1)(x3 + x2 − 1). 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 = ann αn + an−1 n an−1 α n−1 + ann−1 an−2 αn−2 + · · · + an−1 n−1 n a1 α + an a0 = an−1 n n (an α + an−1 α n−1 + an−2 αn−2 + · · · + a1 α + a0 ) = 0. −1). it must be α = ±1. 5. Exercise 6. put b = −r into second and third equations to obtain −r2 − 1 + s = 0 and −rs − r + 1 = 0. If α = −1 is a root. such that x5 − ax − 1 = f (x)g(x). 5. Case 2. Exercise 7. Now. 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. Prove that x5 − ax − 1 ∈ Z[x] is irreducible unless a = 0. Since gcd(p. so divides p. that is. then. The first two correspond to linear factors. then n = 5. hence a root. q) = 1. s. 1). Solution: We subdivide this exercise in cases and subcases. then a = 2. Now. If x5 − ax − 1 has a root. where b. ±2. Show that if α is a root of an xn + an−1 xn−1 + · · · + a1 x + a0 then an α is a root of the monic polynomial xn + an−1 xn−1 + an an−2 xn−2 + · · · + an−2 n−1 n a1 x + an a0 . x3 − nx + 2 is irreducible for n 6= −1. If an αn + an−1 αn−1 + · · · + a1 α + a0 = 0. Solution: This is straightforward. Write f (x) = x2 + bx + c and g(x) = x3 + rx2 + sx + t. If α = −1 or 2. if α is a root of x3 − nx + 2. c. t) = (1. Therefore. r. Adding these last two equations we obtain 3 . 13. so the possibilities are α = ±1.1. suppose that there exists f (x) and g(x) irreducible monic polynomials over Z of degrees 2 and 3 respectively. by Rational Root Theorem. t ∈ Z. 1) of (c. t) = (−1. Then the system of equations reduces to b+r =0 br − 1 + s = 0 bs − r + 1 = 0 b − s = −a. 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. 3. t) = (−1. then α must divide its constant term. which give us two cases.1 Basic Theory and Field Extensions Thus. there is no prime dividing q. if α = −1. If α = 1 is a root. Equating coefficients leads to b+r =0 br + c + s = 0 bs + cr + t = 0 bt + cs = −a ct = −1. so. From b = −r we obtain r = 1. −1) is consistent and we obtain the factorization x5 − ax − 1 = (x2 + bx + c)(x3 + rx2 + sx + t) = (x2 − x + 1)(x3 + x2 − 1). r(s + 1) = 1. Finally. then F2 (θ2 ) ∼ = F2 /(h(x)) has 8 elements and F3 (θ2 ) ∼ = F3 /(h(x)) has 27 elements. t) = (−1. Prove that |F| = pn for some positive integer n. 13. t) = (1. another contradiction. Thus r2 + rs + r + s = 2s. its prime subfield is (isomorphic to) Fp = Z/pZ. it leads to 0 = −4. 9 and 27 elements by adjoining a root of f (x) to the field F where f (x) = g(x) of h(x) and F = F2 of F3 . 1) is impossible. c. then [F : Fp ] = n for some n ∈ Z+ . so then r2 + rs + r − s = 0 becomes rs + r = 1. t) = (1. t) = (−1. that is. this equation has no roots on Z. Therefore. 0. then F2 (θ) ∼ = F2 /(g(x)) has 4 elements and F3 (θ) ∼ = F3 /(g(x)) has 9 elements. The system of equations reduces to b+r =0 br + 1 + s = 0 bs + r − 1 = 0 −b + s = −a. §13. from r2 + 1 − s = 0 we have r2 = s − 1. Solution: Since the characteristic of F is p. then (r + 1)(r + s) = 2s leads to 2 = 0. we have a contradiction. 1. so that (b + 1)(r + s) = 0. from br + 1 + s = 0 we obtain s = 0. Since r ∈ Z. Now suppose b = −1. Then b = −1 or r = −s. If r = −1 and s = −2. s. so one more time we have two cases. We now pass to the case (c. then br + 1 + s = 0 becomes br + 1 − r = 0. If r = 1 and s = 0. Therefore |F| = |Fp |[F:Fp ] = pn . Since F is finite. −1). Now. r. or r = −1 and s = −2.2 Algebraic Extensions r2 + rs + r − s = 0. By Rational Root Theorem. from −b + s = −a we obtain a = −1. 8.2 Algebraic Extensions Exercise 1. Obtain fields of 4. is θ2 is a root of h. Adding the second and third equation we obtain b(r + s) + r + s = 0. −1). Case 2. Let F be a finite field of characteristic p. Hence. Let g(x) = x2 + x − 1 and let h(x) = x3 − x + 1. b = −r and br + 1 − r = 0 gives r2 + r − 1 = 0. We can consider F as a vector space over Fp . r = 1 and s = 0. Exercise 2. The multiplication table for F2 /(g(x)) is · 0 1 x x+1 0 0 0 0 0 1 0 1 x x+1 x 0 x x+1 x x+1 0 x+1 x x The multiplication table for F3 /(g(x)) is 4 . (c. Now. Solution: Note that g and h are irreducible over F2 and F3 . is θ is a root of g. the solution (b. Write down the multiplication tables for the field with 4 and 9 elements and show that the nonzero elements form a cyclic group. a contradiction. If r = −s. Hence. so (r + 1)(r + s) = 2s. Suppose that (c. 1. Therefore.2. Furthermore. . . . hence x3 − 2 is irreducible over F . . . hence α = −β 2 + 3β + 2 ∈ Q(β) = Q. Moreover. so [Q(β) : Q] = 1 or 3. 2 + 3 √ has degree 2 over Q. ζ 3 3 and ζ 2 3 3. hence a root in F . where ζ is the primitive 3’rd root of unity. where √ that every a. none of this elements √ √ √ is in F . . . Therefore. the minimal polynomial is x2 − 2x + 2.2 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.. Therefore. Clearly F (αi ) ⊂ F (α1 . αn ) is the composite of the fields F (α1 ). Similarly.√Note √ element of F is of the form a + bi. Prove that x3 − 2 and x3 − 3 are irreducible over F . Then β 2 = (1 + α + α2 )2 = 1 + 2α + 3α2 + 2α3 + α4 = 5 + 4α + 3α2 . . αn ). so Q ⊂ Q(β) ⊂ Q(α). hence θ is a root of x2 − 4x + 1. let α = 3 2 and β = 1 + α + α2 . Now let K be a field such that F (αi ) ⊂ K for all i. . which is irreducible by Eisenstein with p = 2. The roots of x3 − 2 are 3 2. ζ = exp(2πi/3) = cos(2π/3) + i sin(2π/3) = − 2 + 2 . . √ √ √ Exercise 4. . Prove directly from the definitions that the field F (α1 . Therefore. its minimal polynomial is of degree at least 2. . where we used α3 = 2. an ∈ F . if they were reducible they must have a linear factor. α2 . . F (α2 ). . 1 √ 3 √ i. αn ) ⊂ K. Exercise 5. Let F = Q(i). b ∈ Q. suppose [Q(β) : Q] = 1. . Then θ2 − 4θ = 7 + 4 3 − 8 − 4 3 = −1. 13. √ note that √ (2 + 3)2 = 4 + 4 3 + 3 = 7 + 4 3. If θ is an element of F (α1 . Exercise 3. x2 − 4x √ +1 is irreducible over Q (because θ ∈ 6 Q). ζ 3 2 and ζ 2 3 2. the roots of x3 − 3 are 3 3. . [Q(β) : Q] = 3. so x 2 − 4x + 1 is the minimal polynomial of 2 + 3. F (αn ). Solution: Since the polynomials have degree 3. . . .e. Q(β) = Q so β ∈ Q. . We try conjugation. . then θ is of the form θ = a1 α1 + · · · + an αn . and obtain (x − (1 + i))(x − (1 − i)) = x2 − 2x + 2. . Hence x3 − 3 is irreducible over F . For a contradiction. . αn ) for all 1 ≤ i ≤ n. αn ) is the smallest field containing F (α1 ). Let θ = 2 + 3. . . and by the same argument non of this elements is in F . . Determine the minimal polynomial over Q for the element 1 + i. Solution: We have to prove that F (α1 . . So β 2 − 3β = 5 + 2α + 3α2 − 3(1 + α + α2 ) = 2 − α. that is. a contradiction. Thus F (α1 . F (αn ). . Exercise 6. √ Therefore. 5 . . . Then β ∈ Q(α). Solution: Since 1+i 6∈ Q. . hence θ ∈ K. Since 3 6∈ Q. √ √ √ √ Solution: First. 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. Determine the degree over Q of 2 + 3 and 1 + 3 2 + 3 4. Now. Every ai αi is in K. where a1 . x is a generator of the cyclic group of nonzero elements. d ∈ F. If a = 0. Exercise 9. We also have a + b − n = m. 13. clearly Q( √ 2 √+ 3) ⊂ Q( 2. p √ √ √ √ √ Solution: Suppose a + b = m+ n for some m. Exercise 8. So.]. 3) : Q] = 4. Then a = 0 or b = 0. The equality follows.e. The result follows. D2 ) : F ] = [F ( D1 . If b = 0. √ so Q( 2. √ √ √ √ √ Since [F ( D1 ) : √F ] = 2. where a. for the other consider ( 2+ 3)2 .√ 3) [one inclusion is obvious. Let D1 and D2 be elements of F . Let F be a field of characteristic 6= 2. . [Q( 2 + 3) : Q] = [Q( 2. Let θ = 2 + 3. and is satisfied by 2+ 3. √ Prove that Q( 2 + 3) = Q( √ 2. so √ √ √ √ q b = 2 n( a + b − n). n ∈ F . D 2 ) : F ] can be 2 or 4. x2 − D2 is reducible in F ( D1 ) if and only if D1 D2 is a square in F . 3).. so θ4 − 10θ2 + 1 = 0. √ √ √ √ Since [Q( 2+ 3) : Q] = 4. We also have √ √ θ4 − 10θ2 = (49 + 20 6) − 10(5 + 2 6) = −1. [F ( D 1 . . .2 Algebraic Extensions F (α1 . so D1 D2 is a square in F . . then x4 −10x2 +1 is irreducible over Q. So √ 1 √ 1 √ 2 = (θ3 − 9θ). D2 ) : F ] = 2 if and√only if [F ( D1 . then D2 is a square in F . . Find an irreducible polynomial √ √ satisfied by 2 + 3. Prove that F ( D1 . . √ then [F ( √ D 1 . F (αn ). when D2 ∈ F ( D1 )). D2 ) is of √ 4 over F if D1 D2 is not a √ degree square in F and is of degree 2 over F otherwise. αn ) contains all F (αi ) and is contained in every field containing all F (αi ). Conclude that [Q( 2+ 3) : Q] = 4. Prove that a necessary and sufficient condition for a + b = m + n forpsome m and n in F is that a2 − b is a square in F . etc. b ∈ Q) is biquadratic over Q. c. . Hence. and thus D1 D2 = ( Db2 )2 . √ √ Solution: The elements of F ( D1 . √ 3). αn ) is the composite of the fields F (α1 ). Let F be a field of characteristic √ √6= 2. Then √ √ √ √ θ2 = 5 + 2 6. hence F (α1 . When F ( D1 . √ √ p √ √ √ Since b is not a square in F . . F (α2 ). . neither of which is a square in F . b be elements p of √ the field F with √ √ b not a square in F . √ √ √ √ Exercise √ 7. Use this to determine when the field √ Q( a + b)(a. 6 . that is. if there exists a. then√b2 D1 = D2 . 3) ⊂ Q( 2 + 3). then a+ b = m+n+2 mn. . contrary to the hypothesis. 2 2 √ √ √ √ √ √ Therefore √2 ∈ √Q(θ) and 3 √ ∈ Q(θ). √ √ √ √ √ √ √ √ Solution: Since 2 + 3 is in √ Q( 2. and that occurs exactly when x2 − D2 is reducible in F ( D1 ) (i. θ3 = 11 2 + 9 3 and θ4 = 37 + 15 6. b. 3 = (11θ − θ3 ) and θ4 = 49 + 20 6. Now. √D2 ) : F ( √D1 )] = 1. Let a. this means b = 2 mn. √ √ For the other direction we have to prove that 2 and 2 are in Q( 2 + 3). b ∈ F such that p p (a + b D1 )2 = D2 . . so that a2 + 2ab D1 + b2 D12 = D2 . D2 ) is of degree 4 over F the field is called a biquadratic extension of F . D2 ) can be written in the form p p p a + b D1 + c D2 + d D1 D2 . D2 ) : F ( D1 )][F ( D1 ) : F ]. We have p p p p p p [F ( D1 . contrary to the hypothesis. . √ Note that ab = 0 as ab 6= 0 implies D1 ∈ F . 7 . a √ Therefore. 2 2 4 p √ then mn is never a square when b isn’t. √ 2 2 Now suppose that p a −√b is a√square√in F . Recalling last exercise with a = 3 and b = 8. 2 2 Therefore. so √ by last exercise Q( a + b) is biquadratic over Q when a2 − b is a square in Q. Thus. √ 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 ± 16a2 − 16b ⇒ n= 8 p 2n ⇒ a2 − b = ± . so that a − b ∈ F . Q( 3 + 2 2) = Q( 2) and the degree of from last exercise) the extension Q( 3 + 2 2) over Q is 2. Indeed.2 Algebraic Extensions Hence. we this to determine when the field Q(pa + b). Since √ √ a + a2 − b a − a2 − b b mn = = . m nor n is a square in Q. p √ p √ p √ p √ √ √ √ q a+ b+ a− b a+ b− a− b m+ n= + = a + b. Q( a + b) is biquadratic over Q exactly when a2 − b is a square in Q and neither b. We prove that there exists m. we have √ √ √ p √ p √ (a + b) + 2 a2 − b + (a − b) a+ b+ a− b 2 m= =( ) . 2 2 p √ √ √ Note that m and n are in F as char(F ) 6= 2. is biquadratic over Q. 4 2 and √ √ √ p √ p √ (a + b) − 2 a2 − b + (a − b) a+ b− a− b 2 n= =( ) . we have Q( a + b) = Q( m + n) = Q( m. a. we find (m = 2 and n = 1 √ √ √ p 3 +√ 8 = 2 + 1. m. we have a2 − b = 9 −p8 = 1√is a square in Q and b = 8 ispnot. We claim a + b = m + n. If √ √ √ √ √ a2 − b is a square in p Q and b is not. n or mn are squares in Q. n ∈ F such that a + b = m + n. p √ Now. a2 − b is in F . 2 2 as claimed. Hence. b ∈ Q. n). since a and n are in F . Therefore. Determine the degree of the extension Q( 3 + 2 2) over Q. and neither b. Let √ √ a + a2 − b a − a2 − b m= and n = . p √ Exercise 10. 13. 4 2 Thus p √ p √ p √ p √ √ a+ b+ a− b √ a+ b− a− b m= and n= . p √ p √ Solution: Note that 3 + 2 2 = 3 + 8. Then √ √ √ √ √ q q θ = ( 1 + −3 + 1 − −3)2 = (1 + −3) + (2 1 + 3) + (1 − −3) = 6. . consider the polynomial p(x) = x2 − α2 .. 2 Since x2 − 6 is irreducible over Q (Eisenstein p = 2). so that [F (α) : F (α2 )] = 2. Now. Therefore. Suppose F = Q(α1 . say deg f = 2k + 1 for some k ∈ N. p(x) is reducible in F (α2 ) and α ∈ F (α2 ). Note that 3 + 4i = 3 + −16. . Suppose the degree of the extension K/F is a prime p. Since 3 + 4i is the square root of 3 + √ 4i in the first quadrant. a contradiction. For a contradiction. n. . Since p is prime. . Prove that [Q( 3 + 4i + 3 − 4i) : Q] p = 1. (a) Let 3 +√ 4i denote the square root of the complex number 3 + 4i that √ let 3 − lies in the first quadrant and √ 4i denote the square root of 3 − 4i that lies in the fourth quadrant. Suppose 2 ∈ F . 3 + 4i + 3 − 4i ∈ Q.2 Algebraic Extensions √ Exercise 11. so it takes squares roots to √ square roots. 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. . Prove that F (α) is also formally real. Exercise 15. clearly F (α2 ) ⊂ F (α). . Exercise 14. Then [K : F ] = [K : E][E : F ] = p. αk−1 )] = √ 1 or 2. Solution: (a) First. we have a2 −√b = 25 is a square √ in Q and b = −16 is not. Show that some root β of g has odd degree over F and F (β) is not formally real. Note that α ∈ F (α2 ) if and only if p(x) is reducible in F (α2 ). . Solution: Since α2 ∈ F (α). αn ) where αi2 ∈ Q for i = 1. then θ has degree 2 over Q. m 3 Then [F : Q] = 2 for some m ∈ N. . so [F (α) : F ] is even. either [K : E] = 1 or [E : F ] = 1. Solution: Let E be a subfield of K containing F . α2 . for all 1 ≤ k ≤ n. and maps numbers of the first quadrant to the fourth (and reciprocally). αk ) : Q(α √ 1 . [Pick α a counterexample of minimal degree. so is 3 − 4i. Prove that 3 2 6∈ F . Hence. Hence. 13. i. √ Exercise 13. we√ have [Q(α1 . Solution: Note that. Hence √ 3 + 4ip and √3 − 4i are conjugates each other. suppose p(x) is irreducible in F (α2 ). Furthermore. pTherefore. we √ find m = 1 and n = −4 √and thus √3 + 4i = 1 + −4 √ = 1 +√ 2i. . Thus we have to prove α ∈ F (α2 ). Thus [F (α) : F ] = [F (α) : F (α2 )][F (α2 ) : F ] = 2[F (α2 ) : F ]. Suppose there exists a counterexample. Since F (α) is not formally real. Show that any subfield E of K containing F is either K or F . violating the minimality of α. √ its conjugate √ is the square of root of 3 − 4i in the fourth quadrant. . √ p √ (b) Determine the degree of the extension Q( 1 + −3 + 1 − −3). (b) Let θ = 1 + −3 + 1 − −3. .] Solution: We follow the hint. 3 divides 2m . that is. Prove that if [F (α) : F ] is odd then F (α) = F (α2 ). g(x) ∈ F [x] where g(x) has odd degree < deg f . so that p(α) = 0. we find 3 − 4i = 1 − 2i. 3 2 6∈ F . Then Q 3 3 √ ⊂ Q( 2) ⊂ F . The result follows. √ p3 + 4i√+ 3 − 4i = 4. note that the conjugation map a + bi → a − bi is an isomorphism of C. so [Q( 2) : Q] divides [F : Q]. With a = 3 and b = −16. then −1 can be express as 8 . Let α be of minimal degree such that F (α) is not formally real and α having minimal polynomial f of odd degree. Exercise 12. For this purpose.e. . a contradiction. Show that −1 + f (x)g(x) = (p1 (x))2 + · · · + (pm (x))2 for some pi (x). 2. let f (x) ∈ F [x] be an irreducible polynomial of odd degree and let α be a root of f (x). . we use Exercise 9 again. . the degree in the right hand of the equation is less than 4k + 1. . g(x) such that −1 + f (x)g(x) = (p1 (x))2 + · · · + (pm (x))2 . [By Gauss Lemma this polynomial is irreducible in (k(t))[X] if and only if it is irreducible in (k[t])[X]. 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. Let g(x) be any polynomial in F [x]. h(x) so −1 is a square in F [x]/((h(x)) ∼ = F (β). Exercise 18. P (x) (c) Show that [k(x) : k(t)] = [k(x) : k( )] = max(degP (x). Exercise 16. so m divides n. we have deg h < deg f as well. Let f (x) be an irreducible polynomial of degree n over a field F . hence a root of g(x). We prove that the degree of g is odd by proving that the degree of (p1 (x))2 + · · · + (pm (x))2 is even. As every element in F [x]/((f (x)) can be written as a polynomial in α with degree less than deg f . Now. Since p is irreducible. Let K/F be an algebraic extension and let R be a ring contained in K and containing F . 0 can’t be expressed as a sum of squares in F . deg p divides deg f . Then r−1 = −a−1 0 (r n−1 + · · · + a ). Thus. Therefore. . 1 i Exercise 17. which means F (β) is not formally real. Let β be a root of h(x). because then the equation −1 + f (x)g(x) = (p1 (x))2 + · · · + (pm (x))2 implies the result.2 Algebraic Extensions a sum of squares in F (α) ∼ = F [x]/((f (x))). this contradicts the minimality of α. Since f is irreducible. Solution: Let p(x) be an irreducible factor of f (g(x)) of degree m. Q(x) 9 . . Let t ∈ k(x) be the rational function with relatively prime polynomials Q(x) P (x). degQ(x)). as claimed. since p(x) divides f (g(x)). Note that a0 6= 0 since p is irreducible. Then note that (k[t])[X] = (k[X])[t]. Note that F (g(α)) ⊂ F (α). we have r −1 ∈ R. there exist an irreducible polynomial p(x) = a0 + a1 x + · · · + xn ∈ F [x] such that p(r) = 0. Since r is algebraic over F . Prove that every irreducible factor of the composite polynomial f (g(x)) has degree divisible by n. say h(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.] (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). Then. Q(x) ∈ k[x]. The result follows. pm (x). the exists polynomials p1 (x). Since deg h < deg f . Now. then l−1 2 i=1 (ai /al ) = −1. Show that R is a subfield of K containing F . Hence. then [F (α) : F ] = deg p(x) = m. sincePF is formally real. that is. Indeed. we have deg pi < 2k + 1 for all i. if li=1 a2i = 0. Solution: Let r ∈ R be nonzero. n = [F (g(α)) : F ] = [F (g(α)) : F (α)][F (α) : F ] = [F (g(α)) : F (α)] · m. so deg g < 2k + 1 as well. the degree of g must be odd by the assertion above. Let d be the maximal degree over all pi . Since deg g < deg f . Therefore. . Note that x2d is a sum of squares (of the leading coefficients of the pi ’s of maximal P degree). 13. Therefore 2d 2 2 x 6= 0. Let α be a root of p(x). with Q(x) 6= 0. Since a ∈ F ⊂ R and r ∈ R. we prove that x2d is the leading term of (p1 (x))2 + · · · + (pm (x))2 . β is a root of an odd degree polynomial h such that F (β) is not formally real. Let k be a field and let k(x) be the field of rational functions in x with P (x) coefficients from k. we have f (g(α)) = 0 and thus g(α) is a root of f (x). Then f (x)g(x) −1 + h(x) = (p1 (x))2 + · · · + (pm (x))2 . so the degree of (p1 (x)) + · · · + (pm (x)) is 2d. Then g must contain an irreducible factor of odd degree. this means n = [F (g(α)) : F ]. and this degree equals max{degP (x). then [k(x) : k(t)] = degP (X) − tQ(X). if ϕ(α) = ϕ(β). Thus (an − tbn )X n 6= 0. for every α ∈ K we can associate a F -linear transformation Tα .. p(α) = 0. Let K be an extension of F of degree n. so letting k = 1 we find that ϕ is injective. we have the result. e2 = 3 2 and e3 = 3 4. Solution: The characteristic polynomial of A is p(x) = det(Ix − A). then. Then define ϕ : K → Mn (F ) by ϕ(α) = Tα . Since k[t] is an UFD and k(t) is its field of fractions. 13. b. 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. degQ(x)}. 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. Exercise 20. where ai .e. so at least one of an or bn is nonzero. so the degree of P (X) − tQ(X) is n. Use √ this procedure to obtain the monic polynomial of degree 3 satisfied by 3 2 and by 1 + 3 2 + 3 4. By part (a). For every k ∈ K. (c) Since P (X) − tQ(X) is irreducible over k(t) and x is a root by part (a). then T(α+β) (k) = (α + β)(k) = αk + βk = Tα (k) + Tβ (k) for every k ∈ K. so det(Iα √ − A) = 0 in K. respectively. 0 0 2 1 2 2 Aα = 1 0 0 and Aβ = 1 1 2 . If. (a) For any α ∈ K prove that α acting by left multiplication on K is an F -linear transfor- mation of K. Therefore. 3 4} over Q. P (X) − tQ(X) is irreducible in k((t))[X] is and only if it is irreducible in (k[t])[X]. We also have β(e1 ) = e1 + e2 + e3 . ϕ(K) is isomorphic to a subfield of Mn (F ). (b) Let n = max{degP (x). Note that (k[t])[X] = (k[X])[t]. in (k[t])[X]). Indeed. bn ∈ k. Exercise 19. β(e2 ) = 2e1 + e2 + e3 and β(e3 ) = 2e1 + 2e2 + e3 . We claim ϕ is an isomorphism. α(e2 ) = e3 and α(e3 ) = 2e1 .2 Algebraic Extensions Solution: (a) We follow the hint. Therefore. λ ∈ K. λ ∈ F . the associated matrices of the their linear transformations are. Let α = 3 2 and β = 1 + 3 2 + 3 4. bn 6= 0. We also have T(αβ) (k) = (αβ)(k) = αkβk = Tα (k)Tβ (k) for every k ∈ K. we have α(a+b) = αa + αb and α(λa) = λ(αa) for all a. If an or bn is zero then clearly deg (P (X) − tQ(X)) = n. degQ(x)} by part (b). Thus ϕ(α + β) = ϕ(α) + ϕ(β) and ϕ(αβ) = ϕ(α)ϕ(β) (since the basis is fixed). Solution: (a) Fix α in K. Since K is (in particular) a commutative ring. (b) Fix a basis for K as a vector space over F . β ∈ K. Denote by Tα the matrix of Tα with respect to the basis fixed above. hence in (k(t))[X]. Since P (X)−tQ(X) is linear in (k[X])[t]. hence T(α+β) = Tα + Tβ . P (X)−tQ(X) is irreducible P (x) in k(t). bi ∈ k for all i. so T(αβ) = Tα Tβ . Write P (x) = an xn + · · · + a1 x + a0 and Q(x) = bn xn + · · · + b1 x + b0 . but t 6∈ k (as t ∈ k(x)). √ Now. then αk = βk for every k ∈ K. consider the √field Q( 3 2) with √ basis {1. Thus. Then α(e1 ) = e2 . so the ring of n × n matrices over F contains an isomorphic copy of every extension of F of degree ≤ n. √ Denote √ the elements of this basis by e1 = 1. Suppose an . 3 √ 2. by Gauss Lemma. (b) Prove that K is isomorphic to a subfield of the ring of n × n matrices over F . is clearly irreducible in (k[X])[t] (i. Now. Then an . in particular. it cannot be that an = tbn . The degree of P (X) − tQ(X) is clearly ≤ n. Now. so ϕ is an homomorphism. 0 1 0 1 1 1 10 . so the ring Mn (F ) contains an isomorphic copy of every extension of F of degree ≤ n. we prove is n. if α. 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. Thus. we have (Iα − A)k = αk − Ak = √ αk − αk = 0. Φ is either injective or trivial. b2 ∈ K2 . Now suppose that K1 ⊗F K2 is a field. so ker(ϕ) is trivial or K. hence is the monic polynomial of degree 3 satisfied 3 by α = 2. b1 ) + ϕ(a. b)) = ϕ(a1 + a2 . We use Φ to prove both directions. In this case Φ is a field homomorphism. for r ∈ F . Prove directly that the map a + b D 7→ is an isomorphism of b a b a the field K with a subfield of the ring of 2 × 2 matrices with coefficients in Q. Use the basis 1. We prove that ϕ is F -bilinear. ϕ is a F -bilinear map. In this case K1 ⊗F K2 and K1 K2 have the same dimension over F . b1 + b2 ) = a(b1 + b2 ) = ab1 + ab1 = ϕ(a. Hence. b). Let α = a + b D be an element of K. 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).e. Solution: The matrix√of the linear transformation "multiplication √ √ by α" √ on K is found by acting of α in the basis 1. ϕ is an isomorphism of K with a subfield of M2 (Q). b) + (a2 . Let a. Since K is a field. Since ϕ(K) is clearly non-zero. Furthermore. Now let ϕ : K → M2 (Q) be defined by ϕ(a + b D) = . Thus. D. Prove that the F -algebra K1 ⊗F K2 is a field if and only if [K1 K2 : F ] = [K1 : F ][K2 : F ]. K1 ⊗F K2 is a field. as claimed. and ϕ((a. then L is a field (Exercise 16). 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 the matrix . It is clearly nontrivial since Φ(1 ⊗ 1) = 1. the characteristic polynomial √ of√Aβ is x3 − 3x2 − 3x − 1. [K1 : F ][K2 : F ] ≤ [K1 K2 : F ]. its ideals are {0} and K. We claim L = K1 K2 . 3 √ √ Exercise 21. b2 )) = ϕ(a. Hence √ a bD a bD the matrix is . Since L is a subring of K1 K2 containing K1 (or K2 ). Therefore. b1 . b2 ). b) + ϕ(a2 .1 b) + (a. Hence. L is a field containing both K1 and K2 . We have α(1) = α = a + b D and α( D) = a D + bD. so ϕ is an homomorphism. Therefore. Exercise 22. b) = (ar)b = a(rb) = ϕ(rb). Let L = Φ(K1 ⊗F K2 ). b) = (a1 + a2 )b = a1 b + a2 b = ϕ(a1 . ϕ(ar. Since K1 K2 is the smallest such field (by definition). We also have. Hence. So. Note that L contains K1 and K2 . so it is injective. b) = ab. Let K = Q(√ D) for some squarefree integer D. Let K1 and K2 be two finite extensions of a field F contained in the field K. Note that K1 ⊗F K2 have dimension [K1 : F ][K2 : F ] as a vector space over F . Φ is an F -algebra surjective homomorphism between F -algebras of the same dimension. Solution: Define ϕ : K1 × K2 → K1 K2 by ϕ(a. As we already have [K1 K2 : F ] ≤ [K1 : F ][K2 : F ] (Proposition 21 of the book). Φ is surjective. a2 ∈ K and b. a1 .2 Algebraic Extensions The characteristic √ polynomial of Aα is x3 −2. Then ϕ((a1 . ϕ is injective. 11 . then ker(ϕ) 6= K and thus ker(ϕ) = {0}. First. 13. we have L = K1 K2 . hence is an isomorphism. ϕ induces a F -algebra homomorphism Φ : K1 ⊗F K2 → K1 K2 . we suppose [K1 K2 : F ] = [K1 : F ][K2 : F ] and prove K1 ⊗F K2 is a field. Therefore. Therefore Φ is surjective. the equality follows. Hence. i. hence is the monic polynomial of degree 3 satisfied by β = 1 + 2 + 3 4. b+d a+c b a d c and √ √ ac + bdD (ad + bc)D a bD c dD ϕ((a + b D) · (c + d D)) = = · ad + bc ac + bdD b a d c √ √ = ϕ(a + b D)ϕ(c + d D). since θ is an exterior angle of 4P RO. θ = α + γ = α + β = 3α. the angle of 20◦ would be constructible. a contradiction (see proof of Theorem 24). Finally. It has angles of 40◦ . Exercise 2. Since 4P QO is isosceles. then β = γ. Prove that Archimedes’ construction actually trisect the angle θ. 12 . (c) = . Since we can bisect an angle by straightedge and compass. and also show that (d) (1 − k 2 ) + (b + k)2 = (1 + a)2 . (b) √ 1 1+a x b+k y 1 − k2 = . Exercise 3. Now. hence β = 2α. equals ∠QP O + ∠QOP . Prove that Conway’s construction indicated in the text actually constructs 2k 1/3 and 2k 2/3 . 13.3 Classical Straightedge and Compass Constructions Exercise 1. solve a 1+a x−k 3k these equations for a and b. This two angles equals α. equals the sum of the two remote interior angles. Therefore. b. P. then α = ∠QP O = ∠QOP . a the distance √ from y 1−k 2 B and C and b the distance from A to D. Then α = ∠QP O. But then cos 20◦ and sin 20◦ would be constructible too.] Solution: Let O. Solution: Suppose the 9-gon is constructible. y and x − k are marked in the figure below. since 4QRO is isosceles. which are α and γ. [One method: let (x. i. β = ∠RQO. use similar triangles to prove (a) = . The distances a. Q and R be the points marked in the figure below.. γ = ∠QRO. Since β is an exterior angle of 4P QO. y) be the coordinates of the point C.3 Classical Straightedge and Compass Constructions §13.] Solution: We follow the hint. [Note the isosceles triangles in figure to prove that β = γ = 2α. Prove that it is impossible to construct the regular 9-gon. x. and θ is an exterior angle of 4P RO.e. it equals the sum of the two remote interior angles. then x√2 + 1 is irreducible over Q( 4 2) √ √ having i as a root. the regular 7-gon is not constructible by straightedge and compass. the regular 5-gon is constructible by straightedge and compass. 2) : Q( 2)] = 2. then [Q( 4 2) : Q] = 4. Determine the splitting field and its degree over Q for x4 + 2. the √ splitting field of f has degree [Q(i. Furthermore. α is of degree 2 over Q. it must be ±1. The construction of the regular 7-gon amounts to the constructibility of cos(2π/7). Determine the splitting field and its degree over Q for x4 − 2. i 4 2 and −i 4 2. From the equation for x above. We let a = 2y to obtain h4 + h3 − k 2 h − k 2 = 0. hence [Q(α) : Q] cannot be a power of 2. it must be ±1 since it must divide its constant term.5 and Exercise 2 of Section 14. We have shall see later (Section 14. so it is constructible. §13. Therefore. so b + k = 4k+ka a . Using this in the last equation and reducing. 4 2). as sin(2π/5) = 1 − cos2 (2π/5). 4 2) : Q] = [Q(i. By Rational Root Theorem. We find h = k 2/3 . so p is irreducible over Q.4 Splitting Fields and Algebraic Closures From the figure. we find b = 2k 1/3 . 1+a 1+a x−k 3k √ 3ky So. x=a . Hence. if p has a root in Q. since i 6∈ Q( 4 2). − 4 2. Therefore. the Solution: Let f (x) = √ x √ √ splitting √ field √ of f is Q(i. √ √ √ √ 4 − 2. 1 − k 2 = y(1 + a) = x−k implies 3k = (x − k)(1 + a). hence a = 2k 2/3 . Since we can’t construct α. [Q(i. Use this to prove that the regular 7-gon is not constructible by straightedge and compass. we can construct 2k 1/3 and 2k 2/3 using Conway’s construction. From b = 4k+ka a − k. Solution: Let p(x) = x3 + x2 − 2x − 1 and α = 2 cos(2π/7). p is irreducible over Q. Hence. = and (1 − k 2 ) + (b + k)2 = (1 + a)2 . sin(2π/5) is also constructible. (b) and (c). Since p(1) = 1 and p(−1) = −1. Therefore. so β = cos(2π/5) is also constructible. α is of degree 3 over Q. But p(1) = −1 and p(−1) = 1. we have √ √ 1 − k2 b+k y 1 − k2 y= . Hence. 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 compass. we get (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. So. Since 4 2 is a root of√the irreducible polynomial x4 − 2 over√Q. The roots of f are 4 2. Exercise 4. 4 2) : Q] = 8. We can bisect an angle pby straightedge and compass. Therefore. Exercise 5. Solution: Let p(x) = x2 + x − 1 = 0 and α = 2 cos(2π/5). By Rational Root Theorem. using similar triangles for (a). 4 4 Exercise 2. 13. so [Q(i. 13 .7) that α = 2 cos(2π/7) satisfies the equation x3 + x2 − 2x − 1 = 0. the 4 relations are clear.4 Splitting Fields and Algebraic Closures Exercise 1. Finally. we find 3k = ( a(b+a) 1+a − k)(1 + a) = a(b + k) − k(1 + a). if p has a root in Q. 4 2) : Q( 4 2)][Q( √ 4 2) : Q] over Q. and Pythagoras Theorem for (d). e. . so all roots of x4 − 2 are in K. and K(β) is the splitting field for f ∼ over F (β). Furthermore. We prove β ∈ K. Since every 14 .. w). For every 1 ≤ i ≤ n. By Theorem 8. The roots of x3 −2 are 3 2. irreducible over √ Q( 3 2). ζ ∈ K. Let K be a finite extension of F . We also have i ∈ L. We have to prove i ∈ K and 2 ∈ K. αn ) for some α1 . so [K : Q] = 8. K = L. hence αβ is also a root of x4 + 2. so this roots are w. so ζα is a root of x4 − 2. . Therefore. Therefore. √ 4 2) (last exercise). Then α4 = −2 and θ4 = 2. Let ζ = 2 + i 2 . where w denotes the complex 1 3 √ conjugate of w (i. −w. by last exercise. Then.4 Splitting Fields and Algebraic Closures Solution: Let f (x) = x4 + 2. √ We prove ζ ∈ K. Solution: √ Let f (x) = x4 +√x2 + 1. Since K = K(α). Since L is generated by these roots. so that α4 = −2. ζ and θ are in L. then L ⊂ K. −ζ 3 2 and −ζ 2 3 2. so ζθ is a root of x4 + 2. Now suppose that every irreducible polynomial in F [x] that has a root in K splits completely in K[x]. . Note that ζ 2 = i. Then (αβ)4 = α4 β 4 = −2. Then. so 2. Therefore. then we prove K = L. −w. then K = F (α1 . then γ is a root of x2 + 2. Since K is generated by these roots. ζ = exp(2πi/3) = cos(2π/3) + 1 √ 3 √ √ √ i sin(2π/3) = − 2 + 2 . . Therefore. Let α be a root of x4 + 2 and θ be a root of x4 − 2. i ∈ L implies ζ ∈ L.e. as we proved earlier. In either case γ/i ∈ K. Hence. First suppose that K is a splitting field over F . so that g splits completely ∼ in K[x]. Hence. 3 2). the roots of x4 + 2 are ±α and ±iα. we have ζα ∈ K. Let g(x) be an irreducible polynomial in F [x] with a root α ∈ K. Determine the splitting field and its degree over Q for x6 − 4. 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]. Thus. 13. that is. the roots of x4 − 2 are ±ζα and ±iζα. w) = Q(w). . Furthermore. w. since ζ and α are in K. ζ is a root of x2 + x + 1. Exercise 3. Determine the splitting field and its degree over Q for x4 + x2 + 1. Since the roots of x4 − 1 are ±1. Hence. Note that f (x) = (x3 −2)(x3 + 2). 3 2) : Q(√ 3 2)][Q( 3 2) : Q]. i. [Use Theorems 8 and 27. 3 2) : Q] = [Q(ζ. so the roots of f are ± 2 ± i 2 . Now. then γ is one of this roots. which implies 2 ∈ K. Let θ = 2. Since [K : F ] is finite. Then [Q(ζ.] Solution: We follow the hint. so ζ 4 = −1. First we prove ζ ∈ L and ζ ∈ K. Since γ 2 = α4 = −2. w is a root of x2 − x + 1. w = 12 + i 23 ). We also have (ζα)4 = ζ 4 α4 = 2. (ζθ)4 = ζ 4 θ4 = −2. Hence. 3 2) : Q] = 6.√ Let K be the splitting field of f and let L be the splitting field of x4 − 2. so 3 2 has degree 3 over √ Q. so ζθ ∈ L. so that [K : Q] = 8 by 2 2 last exercise. Note that f (x) = (x2 + x + 1)(x2 − x + 1). We have that 3 2 is a root of the irreducible polynomial x3 − 2√ over Q. then Q(w. we have K ⊂ L. 6 −4. Exercise 4. β ∈ K. αn . √ √ √ √ √ the splitting √ field of f is Q(ζ. ±i. where ζ is the primitive 3’rd root of unity. Let w = 12 − i 23 .√L = Q(i. the roots of x4 + 2 are ±ζθ and ±iζθ. Since the roots of 2 √ x + 2 are i 2 and −i 2. . Since θ ∈ L. the degree of the splitting field of f is [Q(w) : Q] = 2. This is easy. ϕ extends to an isomorphism σ : K(α) − → K(β). by Theorem 28. Furthermore. Now we prove L = K proving both inclusions. Now let √ γ = α2 ∈√K. that is irreducible over Q since w 6∈ Q. Therefore. then θ 2 ∈ L. Exercise 5. the splitting field of f is Q(w. . there is an isomorphism ϕ : F (α) − → F (β) such that ϕ(α) = β. K(α) is the splitting field for f over F (a). . and let f = p1 p2 · · · pn .. let pi be the minimal polynomial of αi over F . so that β 4 = 1. Let β be any root of g. the roots of x3 +2 are − 3 2. there exists f (x) ∈ F [x] such that K is the splitting field of f . since i ∈ L. We claim K = L. √ √ Solution: √ Let f (x) = x ζ 3 2 and ζ 2 3 2. so K = K(β). Furthermore. then [K : F ] = [K(α) : F ] = [K(β) : F ]. then all roots of x4 + 2 are in L. the degree of the splitting field of f is [Q(ζ. Since K is generated over Q by there roots. Since w + w = 1. Let α be a root of x4 + 2. then iα/α = i ∈ K. We also have i ∈ K. Let β be a root of x4 − 1. so ζ has degree 2 over Q( 3 2). Similarly. √4 2 = √ We prove √ ζ ∈ L. We have f (x) + g(x) = (an + bn )xn + · · · + (a1 + b1 )x + (a0 + b0 ). we can write g(x) = bn xn + · · · + b1 x + b0 . Therefore. Thus. 2n X 2n−1 X l Dx (f (x)g(x)) = Dx ( cl x ) = (l + 1)cl+1 xl . the roots of f in K1 must coincide with its roots in K2 . we have to prove that every irreducible polynomial in F [x] that has a root in K1 ∩ K2 splits completely in (K1 ∩ K2 )[x]. in K1 ∩ K2 . (a) Prove that their composite K1 K2 is a splitting field over F . Since K1 and K2 are contained in K. so K is the splitting field of f (x) ∈ F [x]. (b) Prove that K1 ∩ K2 is a splitting field over F . l=0 l=0 so the coefficient of xl in Dx (f (x)g(x)) is (l + 1)cl+1 . So (recall product of polynomials. By last exercise. Let K1 and K2 be finite extensions of F contained in the field K. Exercise 6. f splits completely in (K1 ∩ K2 )[x]. page 295 of the book) n X Xn Dx (f (x))g(x) = ( kak xk−1 )( bk xk ) k=1 k=0 n−1 X Xn 2n−1 X X l =( (k + 1)ak+1 xk )( bk xk ) = ( (k + 1)ak+1 bl−k )xl . f splits completely in K1 and splits completely in K2 . By last exercise. Suppose.5 Separable and Inseparable Extension Exercise 1. say α. that n ≥ m. we have Dx (f (x)) = nan xn−1 +· · ·+2a2 x+a1 . and assume both are splitting fields over F . So. f splits completely in K and K is generated over F by its roots.5 Separable and Inseparable Extension αi is in K. f1 f2 splits completely in K1 K2 and K1 K2 is generated over F by its roots. [Use the preceding exercise. 13. Thus. Hence. and Dx (g(x)) = nbn xn−1 +· · ·+2b2 x+b1 . §13. let f (x) be an irreducible polynomial in F [x] that has a root. Let cn = nk=0 ak bn−k . 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). hence splits completely in K. Solution: Let f (x) = an xn + · · · + a1 x + a0 and g(x) = bm xm + · · · + b1 x + b0 be two polynomials. so that P Xn Xn 2n X X l 2n X f (x)g(x) = ( ak xk )( bk xk ) = ( ak bl−k )xl = cl xl . Now we prove the formula for the product. (b) We follow the hint. k=0 k=0 l=0 k=0 15 . Therefore. where some of the last coefficients bi could be zero. without any loss of generality. hence is the splitting field of f1 f2 (x) ∈ F [x].] 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 . so Dx (f (x) + g(x)) = n(an + bn )xn−1 + · · · + 2(a2 + b2 )x + (a1 + b1 ) = Dx (f (x)) + Dx (g(x)). Now. by the uniqueness of the factorization of f in K. k=0 k=0 l=0 k=0 l=0 Hence. and K2 is generated over F by the roots of f2 . every pi has a root in K. K1 is generated over F by the roots of f1 . the only irreducible polynomial of degree 2 over F2 is x2 + x + 1. 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. Now. Since x2 +x+1 is the only irreducible polynomial of degree 2 over F2 . Find all irreducible polynomials of degree 1. they are clearly irreducible. For any of this polynomials to be irreducible. We can also apply the condition f (1) = f (0) = 1 to discard the ones with linear factors. it can’t be factorized as two quadratic irreducible factors. the irreducible polynomials of degree 4 over F2 are x4 + x3 + x2 + x + 1. since x + 1 = x − 1 in F2 . Hence. 13. f must have an odd number of terms (or it will be 0). only (x2 +x+1)2 = x4 + x2 + 1 of this four is not irreducible. 6= 0. 16 . Now. k=0 k=0 l=0 k=0 Therefore. k=0 Since all their coefficients are equal. Solution: The polynomials x and x + 1 are the only (non-constant. We also calculate (x2 + x + 1)(x4 + x + 1)(x4 + x3 + 1) = x10 + x5 + 1. Furthermore. 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. i. Hence. the coefficient of xl in Dx (f (x))g(x) + Dx (g(x))f (x) is l X l X ( (k + 1)ak+1 bl−k ) + ( (l − k + 1)ak bl−k+1 ) k=0 k=0 l−1 X l X = (l + 1)al+1 b0 + ( (k + 1)ak+1 bl−k ) + ( (l − k + 1)ak bl−k+1 ) + (l + 1)a0 bl+1 k=0 k=1 Xl l X = (l + 1)al+1 b0 + ( kak bl−k+1 ) + ( (l − k + 1)ak bl−k+1 ) + (l + 1)a0 bl+1 k=1 k=1 Xl = (l + 1)al+1 b0 + ( (l + 1)ak bl−k+1 ) + (l + 1)a0 bl+1 k=1 l+1 X = (l + 1)( ak bl−k+1 ) = (l + 1)cl+1 . and must have constant term 1 (or x will be a factor). 1) polynomials of degree 1 over F2 .e. for a polynomial f (x) ∈ F2 [x] of degree 4 to be irreducible. We are left with x4 + x3 + x2 + x + 1 x4 + x3 + 1 x4 + x2 + 1 x4 + x + 1. exactly when f (0) = f (1) = 1. it must have no linear or quadratic factors. we have (x + 1)(x4 + x3 + x2 + x + 1) = x5 − 1. we conclude Dx (f (x)g(x)) = Dx (f (x))g(x)+Dx (g(x))f (x). x4 + x3 + 1 and x4 + x + 1.5 Separable and Inseparable Extension and Xn Xn f (x)Dx (g(x)) = ( ak xk )( kbk xk−1 ) k=0 k=1 Xn n−1 X 2n−1 X l X k k =( ak x )( (k + 1)bk+1 x ) = ( ak (l − k + 1)bl−k+1 )xl . So. Exercise 2. 2 and 4 over F2 and prove that their product is x16 − x. Then xn − 1 = xqd − 1 = (xd − 1)(xqd−d + xqd−2d + .] Solution: We follow the hint. Exercise 5. Let 1 ≤ i < j ≤ n and let αi be a root of fi and αj be a root of fj . Since xp − 1 = x(xp −1 − 1). Since (α + 1)p − (α + 1) + a = αp + 1 − α − 1 + a = 0. . n n Exercise 6. d Now. d n if Fpd ⊆ Fpn . Prove for any positive integers n. so f is separable. Similarly. n = 1. r ∈ Z with 0 < r < d. [For the irreducibility: One approach − prove first that if α is a root then α + 1 is also a root. if d divides n. Prove that d divides n if and only if xd − 1 divides xn − 1. Let a > 1 be an integer. then α + 1 is also a root of f (x). Another approach − suppose it’s reducible and compute derivatives. Conclude that α∈F×n α = (−1)p so Q Q p p the product of the nonzero elements of a finite field is +1 if p = 2 and −1 is p is odd. Thus. Take a = p. + xd + 1) + (xr − 1). as claimed. Since i and j were arbitrary. but doesn’t divide xr − 1 (as r < d). n Solution: By definition. + xd + 1). so that fi is the minimal polynomial of αi and fj the minimal polynomial of αj . say q. Conversely. so d divides n. then xd − 1 does not divide xn − 1. Similarly is defined Fpn . Then p = deg f = nq. then all fi are of the same degree. so f is irreducible. Thus. . Thus xn − 1 = (xqd+r − xr ) + (xr − 1) = xr (xqd − 1) + (xr − 1) = xr (xd − 1)(xqd−d + xqd−2d + . α∈F× pn 17 . Exercise 4. Therefore. and that occurs d n exactly when xp −1 − 1 divides xp −1 − 1 (by last exercise). the previous exercise). d that d divides n if and only if ad − 1 divides an − 1 (cf. [Note that if n = qd + r then xn − 1 = (xqd+r − xr ) + (xr − 1). Suppose d divides n. Conclude in particular that Fpd ⊆ Fpn if and only if d divides n. Solution: The first assertion is analogous to the last exercise. contrary to the hypothesis. there exists k2 ∈ Fp such that αj = α+k2 . We prove deg fi = deg fj . so that n = qd for some q ∈ Z. First we prove f is separable. Let α be a root of f (x). Conversely. so α ∈ Fp and thus 0 = αp − α + a = a. If n = p. This gives p distinct roots of f (x) given by α + k with k ∈ Fp . Fpd is defined as the field whose pd elements are the roots of xp − x over Fp .5 Separable and Inseparable Extension Exercise 3. Hence. So xd − 1 divides xn − 1. Let f = f1 f2 · · · fn where fi (x) ∈ Fp [x] is irreducible for all 1 ≤ i ≤ n. For p odd and n = 1 derive Wilson’s Theorem: (p − 1)! ≡ −1 (mod p). then xp −1 − 1 divides xp −1 − 1. So. Fpn is the field whose pn elements are the roots of xp − x over n n Fp . clearly n Y xp −1 − 1 = (x − α). Since αi is a root of fi . d divides n if and only if pd − 1 divides pn − 1. 13. so n = 1 or n = p (as p is prime). Prove that xp −1 − 1 = α∈F×n (x − α). . . it is a root of f . Now we prove f is irreducible. any root of d d n n xp − x = x(xp −1 − 1) must be a root of xp − x = x(xp −1 − 1). so it must be its minimal polynomial. Since xd − 1 divides the first term. αi = αj +k1 −k2 . For any prime p and any nonzero a ∈ Fp prove that xp − x + a is irreducible and separable over Fp . suppose d does not divide n. so deg fi = deg fj . fi (x + k1 − k2 ) = fj (x). hence there exists k1 ∈ Fp such that αi = α + k1 . then all roots of f are in Fp . Then n = qd + r for some q. so fi (x+k1 −k2 ) is irreducible having αj as a root.] Solution: Let f (x) = xp − x + a. hence Fpd ⊆ Fpn . (p − 1)! ≡ −1 (mod p). So. Then Y n −1 Y −1 = (−α) = (−1)p α α∈F× pn α∈F× pn n −1 n −1 n −1 Y ⇒ (−1)p (−1) = (−1)p (−1)p α α∈F× pn n Y ⇒ (−1)p = α. so over Fp this is the coefficient of (xp )i in (1 + xp )n . then k = p. Exercise 8.e. we have pn pn X pn i (1 + x) = x. so g = f . then xp − a = (x − α)p . . Working p i p n over Fp show that this is the coefficient of (x ) in (1 + x ) and hence pn n prove that pi ≡ i (mod p). . by Fermat’s Little Theorem. let g(x) be an irreducible factor of f (x). 13. Note that α 6∈ K. Hence. i. hence f is inseparable. Therefore. so α is a multiple root of f (with multiplicity p). Easily we can generalize this to a finite number of terms. we have g(x) = (x − α)k = xk − kαxk−1 + · · · + (−α)k . . + a1 x + a0 )p = apn xnp + . Suppose K is a field of characteristic p which is not a perfect field: K 6= K p .5 Separable and Inseparable Extension Set x = 0. Let f (x) = xp − a. + a1 xp + a0 = f (xp ). + a1 x + a0 ∈ Fp [x]. Since Fp has characteristic p. Solution: Let a ∈ K such that a 6= bp for every b ∈ K. then (a + b)p = ap + bp for any a. . we have Y −1 = α. α∈F× pn Hence. otherwise a = αp . we have f (x)p = (an xn + . . the product of the nonzero elements is +1 if p = 2 and −1 is p is odd. Prove there exist irreducible inseparable polynomials over K. Prove that f (x)p = f (xp ) for any polynomial f (x) ∈ Fp [x]. over Fp . i i=0 so the coefficient of xpi in the expansion of (1 + x)pn is pn pi . Conclude that there exist inseparable finite extensions of K. xn ∈ Fp . α∈F× p so taking module p we find [1][2] · · · [p − 1] = [−1]. kα ∈ K. ap = a for every a ∈ Fp . We conclude that K(α) is an inseparable finite extension of K. Then g(x) = (x − α)k for some k ≤ p. Solution: By the binomial theorem. Show that the binomial coefficient pn pi pi is the coefficient of x in the expansion of (1 + x)pn . Solution: Let f (x) = an xn + . Exercise 9. Furthermore. (1 + x)pn = (1 + xp )n implies n pn p n X n X p i pn i (1 + x ) = (x ) = x = (1 + x)pn i k i=0 i=0 18 . Since α 6∈ K. For p odd and n = 1. Now. we have (1 + x)pn = 1 + xpn = (1 + xp )n . + ap1 xp + ap0 = an xnp + . f is irreducible. contrary to the hypothesis. Since Fp has characteristic p. . Furthermore. . b ∈ Fp .. Using the binomial theorem. · · · . If α is a root of xp − a. . We prove that f is irreducible and inseparable. Exercise 7. so that (x1 + · · · + xn )p = xp1 + · · · + xpn for any x1 . . Exercise 10... xn ).6 Cyclotomic Polynomials and Extensions pn n over Fp . hence pi ≡ i (mod p). . .. Thus. . . . . . xn ) = γ1 . x2 . . this is impossible. If F is a perfect field and f (X) ∈ F [x] has no repeated irreducible factors in F [x]. we have f (x1 . .. . xn ) ∈ Z[x1 .6 Cyclotomic Polynomials and Extensions Exercise 1. Hence. .. Prove that if a field contains the nth roots of unity for n odd then it also contains the 2nth roots of unity. .. x2 . Then (ζm ζn )k = ζmk ζ k . . We can suppose f is monic. . Then f = f1 f2 · · · fn for some monic irreducible polynomials fi (x) ∈ F [x]. prove that f (x) has no repeated irreducible factors in K[x]. . then (x1 + · · · + xn )p = xp1 + · · · + xpn for any x1 .γn xγ11 . . . . Exercise 11.γn xγ11 . then ζnd is an (n/d)th root of unity. . x2 . . . . ..γn xγ11 . 13. hence all fi has distinct roots. Let ζm be a primitive mth root of unity and let ζn be a primitive nth root of unity. . Exercise 3. ap = a for every a ∈ Fp .. ... . Then (ζn )k = ζnkd .. Prove that ζnd is a primitive (n/d)th root of unity. . . . xn ] be a polynomial in the variables x1 . Suppose K[x] is a polynomial ring over the field K and F is a subfield of K. Then. .. ζm ζn generates the cyclic group of all mnth roots of unity. xn ∈ Fp . Therefore. xn )p = ( apγ1 . . . · · · . Furthermore. Exercise 2. n m n m n hence m divides k and n divides k. . x2 . For any prime p prove that the polynomial f (x1 . xγnn X f (x1 . . xn ] has all its coefficients divisible by p. . x2 . 19 . xp2 . then ζnkd 6= 1. Hence. Let ζn be a primitive nth root of unity and let d be a divisor of n. xpn ) ∈ Z[x1 . Therefore. xn )p = f (xp1 . xn ]. .γn (xpγ p p X X 1 pγn p = 1 . . xn ]. so mn divides k (as m 6= n). . For a contradiction. (ζm ζn )k 6= 1 for all 1 ≤ k < mn and thus the order of ζm ζn is mn. . Since 1 ≤ kd < n.. xγnn )p = (aγ1 . . . so (ζnd )k 6= 1. .. Solution: This is equivalent to prove that for any prime number p.. Now. Solution: Since (ζnd )(n/d) = ζnn = 1. f splits in linear factors in the closure of F .. xp2 . x2 . xγnn )p = aγ1 . ζnd is a primitive (n/d)th root of unity. xn ) = f (x1 . Since F is perfect.. Solution: Since (ζm ζn )mn = 1. ..γn (xγ11 . Let aγ1 . Now let 1 ≤ k < (n/d). So. . Since Fp has characteristic p. x2 . the order of ζnd is (n/d). . ζ k = 1 and ζ k = 1. Suppose m and n are relatively prime positive integers. f is separable. . that is. . Since 1 ≤ k < mn. . xn )p − f (xp1 . . over Fp we have aγ1 . then ζm ζn is an mnth root of unity. suppose ζ k ζ k = 1.. x2 . x2 . . by Fermat’s Little Theorem. . xpn ) in Fp [x1 . xγnn )p X X f (x1 . .. . xn with integer coefficients. .. . hence splits in linear factors in the closure of K. Solution: Let f (x) ∈ F [x] with no repeated irreducible factors in F [x]. Prove that ζm ζn is a primitive mnth root of unity. . ζm ζn is a primitive mnth root of unity. f (x) has no repeated irreducible factors in K[x].. . .. that is.γn =0 be an element of Fp [x1 . let 1 ≤ k < mn. Let f (x1 . §13.. . so d it generates the cyclic group of all (n/d)th roots of unity. . . x2 . x2 . then ζ ∈ F . then f (n) = d|n µ(d)F ( nd ). Therefore. Exercise 5. they are the minimal polynomial of any of its roots. −ζn is clearly a root of Φn (−x). d|n d|n d|n 20 . so are the roots of xm − 1. Φ2n (x) = Φn (−x). Then. Then [K : Q] ≥ [Q(ζ) : Q] = ϕ(n) ≥ n/2 > N . Exercise 7. Let N ∈ N. So lets start with the formula P Y xn − 1 = Φd (x). Note that ζ2 = −1. 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) . Let ζ2 be the primitive 2th root of unity and let ζn be a primitive nth root of unity. 2 (−ζ)n = (−1)n (ζ)n = (−1)n+1 = 1 (since n is odd). Thus. so [K : Q] is infinite. then ζ n = −1. If ζ n = 1. √ Solution: We use the inequality ϕ(n) ≥ n/2 for all n ≥ 1. Prove that for n odd. d|n We take natural logarithm in both sides and obtain Y X ln(xn − 1) = ln( Φd (x)) = ln Φd (x). a root of Φ2n (x). and ζ n 6= 1. so is xm − 1 and its derivative mxm−1 . i. So. Solution: Since Φ2n (x) and Φn (−x) are irreducible. Exercise 4. then ζ ∈ F .e. Since ζ 2n = 1. The roots of unity over F are the roots of k k xn − 1 = xp m − 1 = (xm − 1)p . Since N was arbitrary. 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. so that ζ2 ζn = −ζn . so suppose ζ n 6= 1. −ζn is a common root for both Φ2n (x) and Φn (−x). it is sufficient to find a common root for both. d|n d|n So.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 2nth root of unity. by Exercise 1. then ζ n is a root of x − 1. 2 and n are relatively prime. hence Φ2n (x) = Φn (−x). Let K be an extension of Q with infinitely many roots of unity. in any finite extension of Q there are only a finite number of roots of unity. Solution: Let F be a field with char F = p. Since n is odd. Now. 13. d|n Solution: The Möbius Inversion FormulaPsates that if f (n) is defined for all nonnegative integers and F (n) = d|n f (d). Exercise 6. Use the Möbius Inversion Formula indicated in Section 14.3 to prove Y Φn (x) = (xd − 1)µ(n/d) . ζ2 ζn is a primitive 2nth root of unity. Hence. n > 1. Thus. since m is relatively prime to p. 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) . we conclude that [K : Q] > N for all N ∈ N. Since the roots of this polynomial are 1 and −1. so xm − 1 has no multiple roots. the m different roots of xm − 1 are precisely the m distinct nth roots of unity over F . Furthermore. there exits n ∈ N such that n > 4N 2 √ and there exits some nth root of unity ζ ∈ K. Since F is a field. Prove there are only a finite number of roots of unity in any finite extension K of Q. d|n d|n Hence. so −ζ ∈ F . a product of distinct linear factor for p ≡ 1 mod 7. Finally.3). in Fp [x] we have (x−1)p = xp −1. Since we are working over C. Φ` (x) is the product of `−1 f distinct irreducible polynomials of degree f. so Φ7 (x) splits in distinct linear factors.6 Cyclotomic Polynomials and Extensions `−1 Exercise 8. and is irreducible for p ≡ 3. Show that A can be diagonalized. Φ7 (x) = x6 + x5 + · · · + x + 1 is (x − 1)6 for p = 7. This in fact proves that Fp (ζ) = Fpf . in Fp [x]. then f = 3 is the smallest integer such that p3 ≡ 23 . Let ζ denote any fixed primitive `th root of unity. A is diagonalizable. Hence.. Now we prove that f is the smallest integer with that property. so we have 3 irreducible quadratics. If p ≡ 2. Now. Then the minimal polynomial of A divides xk − 1.] Conclude using (b) that. (d) In particular. Suppose ζ ∈ Fpn for some n. (d) If p = 7. so Φ` (x) = xx−1 = (x−1) x−1 = (x−1) l−1 . then f = 1 in (b) and all roots have degree 1. clearly Fp (ζ a ) ⊂ Fp (ζ). then f = 6 is the smallest integer such that p6 ≡ 36 . the irreducible factors of Φ` (x) have degree f . The equality follows. If p ≡ 6 mod 7. so ζ ∈ Fp (ζ a ) and thus Fp (ζ) ⊂ Fp (ζ a ). which is irreducible over Z by Theorem 41. Conclude that the minimal polynomial of ζ over Fp has degree f . 43 ≡ 8. then f = 2 is the smallest integer such that pf = p2 ≡ 36 ≡ 1 mod 7. Then ζ is a root of n xp −1 − 1. so the minimal polynomial of A can be split in linear factors. 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. hence we have an irreducible factor of degree 6. we can prove (by induction) that A = for every positive 0 1 21 . prove that. For the other. For the other direction we follow the hint. Solution: ` `−1 (a) Since p is prime. Since Φ` have degree ` − 1. then there must be `−1 f factors. (b) Note that ζ has order ` as being a primitive `th root of unity. Use the fact that F× pn is a cyclic group to show that n = f is the smallest n power p of p with ζ ∈ Fp . consider Φ` (x) as a polynomial over Fp [x]. Solution: Let A be an n by n matrix over C for which Ak = I for some integer k ≥ 1. Since f is the smallest power of p such that pf ≡ 1 mod `. i. so the minimal polynomial of ζ over Fp has degree f . so ζ ∈ Fpf . 0 1 n 1 nα Computing powers of A. 5 mod 7. there are k distinct roots of this polynomial. viewed in Fp [x]. if p ≡ 3. hence ζ p −1 = ζ q` = 1. hence ` divides pn − 1 (see Exercise 13. If p ≡ 1 mod 7.e. [One inclusion is obvious. 15626 ≡ 1 mod 7. Then (ζ a )b = ζ.5. 5 mod 7. (c) Since ζ a ∈ Fp (ζ). Exercise 9. so n ≥ l. 56 ≡ 729. note that ζ = (ζ a )b where b is the multiplicative inverse of a mod `. 13. Let ζi for 1 ≤ i ≤ ` be ` distinct primitive `th roots of unity. This exercise determines the factorization of Φ` (x) modulo p for any prime p. 64 ≡ 1 mod 7. f is the smallest power of p with pf ≡ 1 mod `.e ab ≡ 1 mod `. Hence. 1 α Now consider A = where α is an element of a field of characteristic p. n (c) Show that Fp (ζ) = Fp (ζ a ) for any integer a not divisible by `. Suppose A is an n by n matrix over C for which Ak= I for some integer 1 α k ≥ 1. a product of 2 irreducible cubics for p ≡ 2. The minimal polynomial of each ζi has degree f by part (b). f then pf − 1 = q` for some integer q. and all of them are different since Φ` (x) is separable. 4 mod 7. then Φ7 (x) = (x − 1)6 by part (a). (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. Since pf ≡ 1 mod `. 4 mod 7. Let b the the multiplicative inverse of a mod `. so we have 2 irreducible cubics. as desired. Let ` be a prime and let Φ` (x) = xx−1 = x`−1 + x`−2 + · · · + x + 1 ∈ Z[x] be the th ` cyclotomic polynomial. is the smallest integer such that ` divides pf − 1. a product of 3 irreducible quadratics for p ≡ 6 mod 7. 6 Cyclotomic Polynomials and Extensions integer n. . which is impossible. Exercise 10. . then x 1 0 for all x ∈ Fpn . it must be α = 0. so that xn − 1 = (xm − 1)p and we get exactly m distinct nth roots of unity. Then. Hence. Since all the eigenvalues are zeros of both the minimal and characteristic polynomial of multiplicity pk . Moreover. . We use Exercise 4 and write k n = pk m for some prime p and some m relatively prime to p. Solution: We’ll work over the algebraic closure of Fpn . . hence is the only invariant factor. Since Fpn has degree n as a vector space over Fp . Exercise 11.5. and ϕ(ab) = (ab)p = ap bp = ϕ(a)ϕ(b). . Prove that ϕ gives an isomorphism of Fpn to itself (such an isomorphism is called an automorphism). Moreover. . ϕ is injective. there exists some non-singular matrix P such that A = P DP −1 . In last exercise we proved that the minimal and characteristic polynomial of ϕ is xn − 1. then xn − 1 is also the characteristic polynomial of ϕ.. Let ϕ denote the Frobenius map x 7→ xp on the finite field Fpn . . for if ϕ satisfies some polynomial xn−1 + · · · + a pn−1 + · · · + a xp + a 1 x + a0 of degree n − 1 (or less) with coefficients in Fp . Let ϕ denote the Frobenius map x 7→ xp on the field Fpn as in the previous exercise. . . Now. Solution: Note that the minimal polynomial of ϕ is xn −1. Fpn ⊂ Fpm and thus n divides m (Exercise 13. if ϕ(a) = 0. then n ϕn (a) = ap = a for all a ∈ Fpn . Solution: Let a. . Since Fpn is finite. Therefore.. Prove that ϕn is the identity map an that no lower power of ϕ is the identity. if A is diagonalizable. . . . since every element of Fpn is a root of xp − x. 13. then ϕ is also surjective n so it is an isomorphism. . then Ap = I. so ϕn is the identity map. b ∈ Fpn . . fix a primitive mth root of unity. the rational canonical form of ϕ over Fp is the companion matrix of xn − 1. . Hence. Then ϕ(a + b) = (a + b)p = ap + bp = ϕ(a) + ϕ(b). so ϕ is and homomorphism. so n ≤ m.. . So. where D is a diagonal matrix whose diagonal entries are the eigenvalues of A. Furthermore. the eigenvalues of ϕ are the nth roots of unity. then D = I.4). . . 0 0 ··· 1 0 Exercise 12. Let ϕ denote the Frobenius map x 7→ xp on the finite field Fpn as in the previous exercise. we get m Jordan blocks of size pk . if A is diagonalizable. to ensure the field to contain all eigenvalues. each Jordan block each of the form i ζ 1 0 ··· 0 0 0 ζi 1 · · · 0 0 0 0 ζi · · · 0 0 Ji = . Determine the rational canonical form over Fp for ϕ considered as an Fp -linear transformation of the n-dimensional Fp -vector space Fpn . Since A has only one eigenvalue 1. let m be an integer such that m ϕm is the identity map. . . so every element of Fpn must be a root m of xp − x. Now. .. 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 . each one with multiplicity pk . . Now. which is 0 0 ··· 0 1 1 0 · · · 0 0 0 1 · · · 0 0 . . . and thus A = P IP −1 = I. 0 0 0 · · · ζi 1 0 0 0 · · · 0 ζi 22 . then ap = 0 implies a = 0.. say ζ.. . Since pα = 0. Then ap = a for all a ∈ Fpn . 2. . Clearly Z ⊂ CD (x). Let q = pn . . |CD× (xi )| = q mi − 1 for some m1 < n. . (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. x2 . so xa−1 = a−1 x and a−1 ∈ CD (x). is commutative). . Since a ∈ CD (x). . Conclude from (b) that for each i. (b) The nonzero elements D× of D form a multiplicative group. 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. (a) Let Z denote the center of D (i. 13. We already know the Jordan canonical form is given by J0 0 · · · 0 0 J1 · · · 0 0 0 ··· 0 . then CD (x) is a proper subset of D and hence m < n. |CD× (xi )| q i −1 23 . . xr are representatives of the distinct conjugacy classes in D× not contained in the center of D× . 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]. .7) for the group D× is r n X qn − 1 q − 1 = (q − 1) + |CD× (xi )| i=1 where x1 . then CD (x) is a Z-vector space. 0 0 · · · Jm−1 Exercise 13. i. since Z ⊂ CD (x).. (e) Prove that (c) and (d) imply that Φn (q) = ζ primitive (q − ζ) divides q − 1. If Z = Fq prove that D has order q n for some integer n [D is a vector space over Z]. Hence. then |D| = q n for some integer n. so |CD (x)| = q m for some integer m. (b) Let x ∈ D× and let CD (x) be the set of the elements in D that commutes with x. By (a) we have |D× | = q n −1. If x 6∈ Z. and it is commutative by definition of the center. Prove that Z is a field containing Fp for some prime p. . that D = Z is a field. Then |D× : CD× (xi )| = = m . so that Z is isomorphic to Fq . the index |D× : CD× (xi )|) then mi and q i −1 divides n (cf. CD (x) is a division ring.. qn − 1 qn − 1 |Z(D× )| = q − 1 and |CD× (xi )| = q mi − 1. Since D is a vector space over Z. r.. then ax = xa. . Moreover. so it is isomorphic to Fpm for some integer m.e. Since it is finite.e. then a−1 ∈ D. (c) Show that the class equation (Theorem 4. . . Since D is a division ring. (c) The class equation for the group D× is r X |D× | = |Z(D× )| + |D× : CD× (xi )|. so Z is a field. the elements of D which commute with every element of D). 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. Conclude that n = 1. For any x ∈ D× show that the elements of D which commute with x form a division ring which contains Z. .. We prove that every element a ∈ CD (x) has an inverse in CD (x)..e. we have a−1 ax = a−1 xa and thus x = a−1 xa.. . its prime subfield is Fp for some prime p. i=1 where the xi are the representatives of the distinct conjugacy class.6 Cyclotomic Polynomials and Extensions for some 0 ≤ i ≤ m − 1. Now. . . qn − 1 (d) Prove that since m is an integer (namely. . Solution: (a) Z is a division subring of D. Exercise 4 of Section 5). . Hence. r. . . . . . . let ζ 6= 1 be a nth root of unity. . . 2. Then. . Hence.. |CD× (xi )| q mi − 1 i=1 i=1 qn − 1 (d) Since |D× : CD× (xi )| is an integer. n = 1. Exercise 15. P (3). no mth th root of unity. pk x)n + · · · + a1 (N + ap1 p2 . Therefore. . . . . pk . . pk M ) has a prime factor different from p1 . . . q mi − 1 q mi − 1 divides q n − 1. . . . p2 . 13. pk x) is an element of Z[x] and that Q(n) ≡ 1 (mod p1 p2 . pk ). . p2 . . 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). Q(x) ∈ Z[x]. pk . . Conclude that there is some integer M such that Q(M ) has a prime factor different from p1 . 2. then |D× : CD× (xi )| = is an integer. Φn (q) divides (q n − 1)/(q mi − 1) for i = 1. pk n) ≡ a−1 a = 1 (mod p1 . Moreover. contradicting the fact that only the primes p1 . Let N be an integer with P (N ) = a 6= 0. . . . . we have Q(x) = a−1 P (N + ap1 p2 . . none of the pi ’s divide Q(m). . . Let N be an integer such that P (N ) = a 6= 0. Then q divides aQ(m) = P (N + ap1 p2 . for all n ∈ Z+ . p2 . pk . p2 . 2. pk x) = a−1 ((N + ap1 p2 . . . . so (Exercise 13. . Show that Q(x) = a−1 P (N + ap1 p2 . . say p1 . . using the binomial theorem. For a contradiction. . . . . . Therefore. P (2). . . .] Solution: We follow the hint. . . so that Q(m) ≡ 1 (mod pi ) for all i. .4) mi divides n.5. . pk ). so the class equation in (c) implies Φn (q) divides q − 1. . 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 . . .. . P (n). P (2). so Q(n) = a−1 P (N +ap1 p2 . . Exercise 14. pk ) for n = 1. 2. pk are the only primes dividing the values P (n). n = 1. . . . . Hence. Since |Q(m)| > 1. as Φ (x) divides xn − 1.. . . . it must divide i root of unity is a n n n m (x − 1)/(x − 1) for i = 1. p2 . pk and hence that P (N + ap1 p2 . Now let m be a positive integer such that |Q(m)| > 1. . (e) From (d). pk n) ≡ a (mod p1 .. . pk divides the numbers P (1). .6 Cyclotomic Polynomials and Extensions Replacing in the class equation we obtain r r X qn − 1 X qn − 1 q n − 1 = (q − 1) + = (q − 1) + . . p2 . . . . . Since mi < n (no xi is in Z). . Suppose a ∈ Z satisfies Φm (a) ≡ 0 (mod p).. so |q − ζ| > |q − 1| = q − 1. . pk m). Let Q(x) = a−1 P (N + ap1 p2 . 2. . Letting x = q we have Φn (q) divides (q n − 1)/(q mi − 1) i for i = 1. . . . Since Φn (q) = ζ primitive (q − ζ) divides q − 1. Now. D = Z and D is a field. pk x). Q In the complex plane q is closer to 1 that ζ is. Let p be an odd prime not dividing m and let Φm (x) be the mth cyclotomic polynomial. . r. . this is impossible unless n = 1. . P (N +ap1 p2 . . . Prove that a is relatively prime to p and that the order of a in (Z/pZ)× is precisely m. p2 . . [Since Y Y xm − 1 = Φd (x) = Φm (x) Φd (x) d|n d|n d<n 24 . r. suppose there are only finitely many primes dividing the values P (n). [Suppose p1 . . there exists a prime q 6= pi for all i such that q divides Q(m). 2. Let P (x) = xn + · · · + a1 x + a0 be a monic polynomial of degree ≥ 1 over Z. then. Show that if p is an odd prime dividing Φm (a) then either p divides m of p ≡ 1 (mod m). Therefore. Solution: By Exercise 14. then. there are infinitely many primes dividing Φm (1).6 Cyclotomic Polynomials and Extensions we see first that am − 1 ≡ 0 (mod p) i. then am ≡ 1 (mod p). Φm (3).. Let a ∈ Z. Exercise 17. Thus. so then Φd (a) ≡ 0 (mod p) for some d < m. mam−1 ≡ 0 mod p. there must exists infinitely many primes p with p ≡ 1 (mod m). .e. so a is relatively prime to p. .com positron0802@mail. that is. so that Φd (a) ≡ 0 (mod p) for some d < m.com 25 . a ≡ 1 (mod p). this implies m divides p − 1. a contradiction. Hence. but then xm − 1 would have a as a multiple root mod p. by Exercise 16. Since Φm (a) ≡ 0 (mod p). If the order of a mod p were less that m. Hence. positron0802. a is a multiple root of xm − 1. If p does not divide m. Prove there are infinitely many primes p with p ≡ 1 (mod m). For a contradiction. impossible since p does not divide m nor a.wordpress. . suppose ad ≡ 1 (mod p) for some d dividing m. the order of a in (Z/pZ)× is precisely m Exercise 16. 13. We prove that the order of a is m. Since only finitely of them can divide m. b = am−1 ). Any comments or suggestions about the solutions are welcome. a is relatively prime to p and the order of a in F× × p is m. p ≡ 1 (mod m). there exist b such that ba ≡ 1 mod p (indeed.. then ad ≡ 1 (mod p) for some d dividing m. Solution: Let p be an odd prime dividing Φm (a). so is also a root of its derivative mam−1 .] Solution: We follow the hint. by (c). Since |Fp | = p − 1. Φm (2).
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