Abstract Algebra. Chapter 13 - Field Theory David S. Dummit & Richard M. Foote. Solutions by positrón0802

Abstract Algebra Chapter 13 - Field Theory David S. Dummit & Richard M. Foote Solutions by positrón080 positron080@mail.com March 31, 018 Contents 13 Field Theory 1 13.1 Basic Theory and Field Extensions........................... 1 13. Algebraic Extensions.................................. 4 13.3 Classical Straightedge and Compass Constructions.................. 10 13.4 Splitting Fields and Algebraic Closures........................ 1 13.5 Separable and Inseparable Extension.......................... 13 13.6 Cyclotomic Polynomials and Extensions........................ 17 13 Field Theory 13.1 Basic Theory and Field Extensions Exercise 13.1.1. p(x) = x 3 + 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 x 3 + 9x + 6 = (1 + x)(x x + 10) 4. Evaluating at θ, we find (1 + θ)(θ θ + 10) = 4. Therefore Exercise 13.1.. (1 + θ) 1 = θ θ + 10. 4 Let f (x) = x 3 x. f is irreducible over Z by Eisenstein Criterion with p =, hence over Q by Gauss Lemma. Now, if θ is a root of f, then θ 3 = θ +. Hence For computing obtain and (1 + θ)(1 + θ + θ ) = 1 + θ + θ + θ 3 = 3 + 4θ + θ. 1 + θ 1 + θ + θ, first we compute (1 + θ + θ ) 1. Applying the Euclidean algorithm, we x 3 x = (x + x + 1)(x 1) x 1, x 3 x = (x + 1)( x x 4 7 8 ) 9 8. 1

13.1 Basic Theory and Field Extensions Evaluating at θ, from this equalities we obtain (θ + θ + 1)(θ 1) = θ + 1 and (θ + 1) 1 = 8 9 ( θ θ 4 7 8 ). Combining these two equations we obtain So, 8 9 (θ + θ + 1)(θ 1)( θ θ 4 7 8 ) = 1. (θ + θ + 1) 1 = 8 θ (θ 1)( 9 θ 4 7 8 ) = θ 3 + θ 3 + 5 3, where we used θ 3 = θ + again. Therefore, Exercise 13.1.3. 1 + θ = (1 + θ)( θ 1 + θ + θ 3 + θ 3 + 5 3 ) = θ 3 + θ 3 + 1 3. Since 0 3 + 0 + 1 = 1 and 1 1 + 1 + 1 = 1 in F, then x 3 + x + 1 is irreducible over F. Since θ is root of x 3 + x + 1, then θ 3 = θ 1 = θ + 1. Hence, the powers of θ in F (θ) are θ, θ, θ 3 = θ + 1, θ 4 = θ + θ, θ 5 = θ + θ + 1, θ 6 = θ + 1, and θ 7 = 1. Exercise 13.1.4. Denote this map by ϕ. Then ϕ(a + b + c + d ) = a + c b d = ϕ(a + b ) + ϕ(c + d ), and ϕ((a + b ) (c + d )) = ϕ(ac + bd + (ad + bc) ) = ac + bd (ad + bc) = (a b )(c d ) = ϕ(a + b )ϕ(c + d ), hence ϕ is an homomorphism. Moreover, if ϕ(a + b ) = ϕ(c + d ), then a b = c d, hence (since Q) a = b and c = d, so ϕ is injective. Also, given a + b Q( ), then ϕ(a b ) = a + b, so ϕ is surjective. Therefore, ϕ is an isomorphism of Q( ) with itself. Exercise 13.1.5. Let α = p/q be a root of a monic polynomial p(x) = x n + + a 1 x + a 0 over Z, with gcd(p, q) = 1. Then ( p q )n + a n 1 ( p q )n 1 + + a 1 p q + a 0 = 0. Multiplying this equation by q n one obtains p n + a n 1 p n 1 q + + a 1 pq n 1 + a 0 q n = 0 q(a n 1 p n 1 + + a 1 pq n + a 0 q n 1 ) = p n. Thus, every prime that divides q divides p n as well, so divides p. Since gcd(p, q) = 1, there is no prime dividing q, hence q = ±1. The result follows.

13.1 Basic Theory and Field Extensions Exercise 13.1.6. This is straightforward. If a n α n + a n 1 α n 1 + + a 1 α + a 0 = 0, then (a n α) n + a n 1 (a n α) n 1 + a n a n (a n α) n + + an n a 1 (a n α) + an n 1 a 0 = a n nα n + a n 1 n a n 1 α n 1 + a n 1 n a n α n + + an n 1 a 1 α + an n 1 a 0 = a n 1 n (a n α n + a n 1 α n 1 + a n α n + + a 1 α + a 0 ) = 0. Exercise 13.1.7. If x 3 nx + is reducible it must have a linear factor, hence a root. By the Rational Root Theorem, if α is a root of x 3 nx +, then α must divide its constant term, so the possibilities are α = ±1, ±. If α = 1 or, then n = 3; if α = 1, then n = 1; and if α =, then n = 5. Therefore, x 3 nx + is irreducible for n 1, 3, 5. Exercise 13.1.8. We subdivide this exercise in cases and subcases. If x 5 ax 1 is reducible then it has a root (linear factor) or is a product of two irreducible polynomials of degrees and 3. Case 1. If x 5 ax 1 has a root, then, by the 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 =. Case. Now, suppose that there exists f (x) and g(x) irreducible monic polynomials over Z of degrees and 3 respectively, such that x 5 ax 1 = f (x)g(x). Write f (x) = x + bx + c and g(x) = x 3 + r x + sx + t, where b, c, r, s, t Z. Then x 5 ax 1 = (x + bx + c)(x 3 + r x + sx + t) Equating coefficients leads to = x 5 + (b + r)x 4 + (br + c + s)x 3 + (bs + cr + t)x + (bt + cs) + tc. 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.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 r 1 + s = 0 and rs r + 1 = 0, that is, r + 1 s = 0 and rs + r 1 = 0. Adding these last two equations we obtain r + rs + r s = 0. Thus r + rs + r + s = s, so (r + 1)(r + s) = s. Now, from r + 1 s = 0 we have r = s 1, so then r + 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 =. If r = 1 and s = 0, then (r + 1)(r + s) = s leads to = 0, a contradiction. If r = 1 and s =, it leads to 0 = 4, another contradiction. 3

13. Algebraic Extensions Therefore, (c, t) = ( 1, 1) is impossible. We now pass to the case (c, t) = (1, 1). Case.. 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 r + r 1 = 0. By the Rational Root Theorem, this equation has no roots on Z. Since r Z, we have a contradiction. Now suppose b = 1. 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 x 5 ax 1 = (x + bx + c)(x 3 + r x + sx + t) = (x x + 1)(x 3 + x 1). 13. Algebraic Extensions Exercise 13..1. Since the characteristic of F is p, its prime subfield is (isomorphic to) F p = Z/pZ. We can consider F as a vector space over F p. Since F is finite, then [F : F p ] = n for some n Z +. Therefore Exercise 13... F = F p [F:F p] = p n. Note that g and h are irreducible over F and F 3. Now, is θ is a root of g, then F (θ) F /(g(x)) has 4 elements and F 3 (θ) F 3 /(g(x)) has 9 elements. Furthermore, is θ is a root of h, then F (θ ) F /(h(x)) has 8 elements and F 3 (θ ) F 3 /(h(x)) has 7 elements. The multiplication table for F /(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 F 3 /(g(x)) is 0 1 x x + 1 x + x x + 1 x + 0 0 0 0 0 0 0 0 0 0 1 0 1 x x + 1 x + x x + 1 x + 0 1 x x + x + 1 x x + x + 1 x 0 x x x + 1 1 x + 1 x + x + x + 1 0 x + 1 x + 1 x + x x x + 1 x + 0 x + x + 1 x + 1 x x + 1 x x 0 x x x + x + x + 1 x + 1 1 x + 1 0 x + 1 x + x + x 1 x + 1 x x + 0 x + x + 1 x + 1 x 1 x x + 4

13. Algebraic Extensions In both cases, x is a generator of the cyclic group of nonzero elements. Exercise 13..3. Since 1 + i Q, its minimal polynomial is of degree at least. We try conjugation, and obtain (x (1 + i))(x (1 i)) = x x +, which is irreducible by Eisenstein with p =. Therefore, the minimal polynomial is x x +. Exercise 13..4. First, note that (+ 3) = 4+4 3+3 = 7+4 3. Let θ = + 3. Then θ 4θ = 7+4 3 8 4 3 = 1, hence θ is a root of x 4x + 1. Moreover, x 4x + 1 is irreducible over Q (because θ Q), so x 4x + 1 is the minimal polynomial of + 3. Therefore, + 3 has degree over Q. Now, let α = 3 and β = 1 + α + α. 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 x 3 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 where we used α 3 =. So β = (1 + α + α ) = 1 + α + 3α + α 3 + α 4 = 5 + 4α + 3α, β 3β = 5 + α + 3α 3(1 + α + α ) = α, hence α = β + 3β + Q( β) = Q, a contradiction. Therefore, [Q( β) : Q] = 3. Exercise 13..5. 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 x 3 are 3, ζ 3 and ζ 3, where ζ is the primitive 3 rd root of unity, i.e., ζ = exp(πi/3) = cos(π/3) + i sin(π/3) = 1 + 3. Since 3 Q, none of this elements is in F, hence x 3 is irreducible over F. Similarly, the roots of x 3 3 are 3 3, ζ 3 3 and ζ 3 3, and by the same argument non of this elements is in F. Hence x 3 3 is irreducible over F. Exercise 13..6. We have to prove that F(α 1,..., α n ) is the smallest field containing F(α 1 ),..., F(α n ). Clearly F(α i ) F(α 1,..., α n ) for all 1 i n. Now let K be a field such that F(α i ) K for all i. If θ is an element of F(α 1,..., α n ), then θ is of the form θ = a 1 α 1 + + a n α n, where a 1,..., a n F. Every a i α i is in K, hence θ K. Thus F(α 1,..., α n ) K. Therefore, F(α 1,..., α n ) contains all F(α i ) and is contained in every field containing all F(α i ), hence F(α 1,..., α n ) is the composite of the fields F(α 1 ), F(α ),..., F(α n ). Exercise 13..7. Since + 3 is in Q(, 3), clearly Q( + 3) Q(, 3). For the other direction we have to prove that and are in Q( + 3). Let θ = + 3. Then So θ = 5 + 6, θ 3 = 11 + 9 3 and θ 4 = 37 + 15 6. = 1 (θ3 9θ), 3 = 1 (11θ θ3 ) and θ 4 = 49 + 0 6. Therefore Q(θ) and 3 Q(θ), so Q(, 3) Q( + 3). The equality follows. Hence, [Q( + 3) : Q] = [Q(, 3) : Q] = 4. 5

13. Algebraic Extensions We also have θ 4 10θ = (49 + 0 6) 10(5 + 6) = 1, so θ 4 10θ + 1 = 0. Since [Q( + 3) : Q] = 4, then x 4 10x + 1 is irreducible over Q, and is satisfied by + 3. Exercise 13..8. The elements of F( D 1, D ) can be written in the form We have a + b D 1 + c D + d D 1 D, where a, b, c, d F. [F( D 1, D ) : F] = [F( D 1, D ) : F( D 1 )][F( D 1 ) : F]. Since [F( D 1 ) : F] =, then [F( D 1, D ) : F] can be or 4. Now, [F( D 1, D ) : F] = if and only if [F( D 1, D ) : F( D 1 )] = 1, and that occurs exactly when x D is reducible in F( D 1 ) (i.e., when D F( D 1 )), that is, if there exists a, b F such that (a + b D 1 ) = D, so that a + ab D 1 + b D 1 = D. Note that ab = 0 as ab 0 implies D 1 F, contrary to the hypothesis. Then a = 0 or b = 0. If b = 0, then D is a square in F, contrary to the hypothesis. If a = 0, then b D 1 = D, and thus D 1 D = ( D b ), so D 1 D is a square in F. So, x D is reducible in F( D 1 ) if and only if D 1 D is a square in F. The result follows. Exercise 13..9. Suppose a + b = m + n for some m, n F, then a + b = m + n + mn. Since b is not a square in F, this means b = mn. We also have a + b n = m, so Hence, b = n( a + b n). b = n(a + b) n ( b + n) = 4n(a + b) b + 4n b + 4n = 4n(a + b) b + 4n 4na = 0 n = 4a ± 16a 16b 8 a b = ± n a. Therefore, since a and n are in F, a b is in F. Now suppose that 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 m = a + a b and n = a a b. 6

Note that m and n are in F as char(f). We claim and Thus Therefore, m = (a + b) + a b + (a b) 4 n = (a + b) a b + (a b) 4 a + b + a b m = a + b + a b m + n = and + 13. Algebraic Extensions a + b = m + n. Indeed, we have a + b + a b = ( ), a + b a b = ( ). a + b a b n =. a + b a b = a + b, as claimed. Now, we this to determine when the field Q( a + b), a, b Q, is biquadratic over Q. If a b is a square in Q and b is not, we have Q( a + b) = Q( m + n) = Q( m, n), so by last exercise Q( a + b) is biquadratic over Q when a b is a square in Q, and neither b, m, n or mn are squares in Q. Since mn = a + a b a a b = b 4, then mn is never a square when b isn t. Thus, Q( a + b) is biquadratic over Q exactly when a b is a square in Q and neither b, m nor n is a square in Q. Exercise 13..10. Note that 3 + = 3 + 8. Recalling last exercise with a = 3 and b = 8, we have a b = 9 8 = 1 is a square in Q and b = 8 is not. Hence, we find (m = and n = 1 from last exercise) 3 + 8 = +1. Therefore, Q( 3 + ) = Q( ) and the degree of the extension Q( 3 + ) over Q is. Exercise 13..11. (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 (and reciprocally). Since 3 + 4i is the square root of 3 + 4i in the first quadrant, its conjugate is the square of root of 3 4i in the fourth quadrant, so is 3 4i. Hence 3 + 4i and 3 4i are conjugates each other. Now, we use Exercise 9 again. Note that 3 + 4i = 3 + 16. With a = 3 and b = 16, we have a b = 5 is a square in Q and b = 16 is not. Hence, we find m = 1 and n = 4 and thus 3 + 4i = 1 + 4 = 1 + i. Furthermore, we find 3 4i = 1 i. Therefore, 3 + 4i + 3 4i = 4, i.e., 3 + 4i + 3 4i Q. (b) Let θ = 1 + 3 + 1 3. Then θ = ( 1 + 3 + 1 3) = (1 + 3) + ( 1 + 3) + (1 3) = 6. Since x 6 is irreducible over Q (Eisenstein p = ), then θ has degree over Q. 7

13. Algebraic Extensions Exercise 13..1. 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..13. Note that, for all 1 k n, we have [Q(α 1,..., α k ) : Q(α 1,..., α k 1 )] = 1 or. Then [F : Q] = m for some m N. Suppose 3 F. Then Q Q( 3 ) F, so [Q( 3 ) : Q] divides [F : Q], that is, 3 3 divides m, a contradiction. Hence, F. Exercise 13..14. Since α F(α), clearly F(α ) F(α). Thus we have to prove α F(α ). For this purpose, consider the polynomial p(x) = x α, so that p(α) = 0. Note that α F(α ) if and only if p(x) is reducible in F(α ). For a contradiction, suppose p(x) is irreducible in F(α ), so that [F(α) : F(α )] =. Thus [F(α) : F] = [F(α) : F(α )][F(α ) : F] = [F(α ) : F], so [F(α) : F] is even, a contradiction. Therefore, p(x) is reducible in F(α ) and α F(α ). Exercise 13..15. 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 = k + 1 for some k N. Since F(α) is not formally real, then 1 can be express as a sum of squares in F(α) F[x]/(( f (x))). Then, the exists polynomials p 1 (x),..., p m (x), g(x) such that 1 + f (x)g(x) = (p 1 (x)) + + (p m (x)). 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 p i < k + 1 for all i. Thus, the degree in the right hand of the equation is less than 4k + 1, so deg g < k + 1 as well. We prove that the degree of g is odd by proving that the degree of (p 1 (x)) + +(p m (x)) is even, because then the equation 1+ f (x)g(x) = (p 1 (x)) + +(p m (x)) implies the result. Let d be the maximal degree over all p i, we prove that x d is the leading term of (p 1 (x)) + + (p m (x)). Note that x d is a sum of squares (of the leading coefficients of the p i s of maximal degree). Now, since F is formally real, 0 can t be expressed as a sum of squares in F. Indeed, if l i=1 ai = 0, then l 1 i=1 (a i/a l ) = 1. Therefore x d 0, so the degree of (p 1 (x)) + +(p m (x)) is d, 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) h(x) = (p 1 (x)) + + (p m (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 13..16. Let r R be nonzero. Since r is algebraic over F, there exist an irreducible polynomial p(x) = a 0 + a 1 x + + x n F[x] such that p(r) = 0. Note that a 0 0 since p is irreducible. Then r 1 = a 1 0 (rn 1 + + a 1 ). Since a i F R and r R, we have r 1 R. 8

13. Algebraic Extensions Exercise 13..17. Let p(x) be an irreducible factor of f (g(x)) of degree m. Let α be a root of p(x). Since p 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, m = [F(α) : F] = [F(α) : F(g(α))][F(g(α)) : F] = [F(α) : F(g(α))] n, so n divides m, that is, deg f divides deg p. Exercise 13..18. (a) We follow the hint. Since k[t] is an UFD and k(t) is its field of fractions, then, by Gauss Lemma, P(X) tq(x) is irreducible in k((t))[x] is and only if it is irreducible in (k[t])[x]. Note that (k[t])[x] = (k[x])[t]. Since P(X) tq(x) is linear in (k[x])[t], is clearly irreducible in (k[x])[t] (i.e., in (k[t])[x]), hence in (k(t))[x]. Thus, P(X) tq(x) is irreducible in k(t). Now, x is clearly a root of P(X) tq(x) since P(x) tq(x) = P(x) P(x) Q(x) = P(x) P(x) = 0. Q(x) (b) Let n = max{degp(x), degq(x)}. Write P(x) = a n x n + + a 1 x + a 0 and Q(x) = b n x n + + b 1 x + b 0, where a i, b i k for all i, so at least one of a n or b n is nonzero. The degree of P(X) tq(x) is clearly n, we prove is n. If a n or b n is zero then clearly deg (P(X) tq(x)) = n. Suppose a n, b n 0. Then a n, b n k, but t k (as t k(x)), it cannot be that a n = tb n. Thus (a n tb n )X n 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 13..19. (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 T α. Denote by ( T α ) the matrix of Tα with respect to the basis fixed above. Then define ϕ : K M n (F) by ϕ(α) = ( T α ). We claim ϕ is an isomorphism. Indeed, if α, β K, then T (α+β) (k) = (α + β)(k) = αk + βk = T α (k) + T β (k) for every k K, hence T (α+β) = T α + T β. We also have T (αβ) (k) = (α β)(k) = α( βk) = T α T β (k) for every k K, so T (αβ) = T α T β. Thus ϕ(α + β) = ϕ(α) + ϕ( β) and ϕ(α β) = 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 M n (F), so the ring M n (F) contains an isomorphic copy of every extension of F of degree n. Exercise 13..0. The characteristic polynomial of A is p(x) = det(i x A). For every k K, we have (Iα A)k = αk Ak = αk αk = 0, so det(iα A) = 0 in K. Therefore, p(α) = 0. Now, consider the field Q( 3 ) with basis {1, 3, 3 4} over Q. Denote the elements of this basis by e 1 = 1, e = 3 and e 3 = 3 4. Let α = 3 and β = 1 + 3 + 3 4. Then α(e 1 ) = e, α(e ) = e 3 and α(e 3 ) = e 1. We also have β(e 1 ) = e 1 + e + e 3, β(e ) = e 1 + e + e 3 and β(e 3 ) = e 1 + e + e 3. Thus, the associated matrices of the their linear transformations are, respectively, A α = 0 0 1 0 0 0 1 0 and 9 A β = 1 1 1. 1 1 1

13. Algebraic Extensions The characteristic polynomial of A α is x 3, hence is the monic polynomial of degree 3 satisfied by α = 3. Furthermore, the characteristic polynomial of A β is x 3 3x 3x 1, hence is the monic polynomial of degree 3 satisfied by β = 1 + 3 + 3 4. Exercise 13..1. The matrix of the linear transformation "multiplication by α" on K is found by acting of α in the basis 1, D. We have α(1) = α = a + b D and α( D) = a ( ) a bd D + bd. Hence the matrix is. b a Now let ϕ : K M (Q) be defined by ϕ(a + b ( ) a bd D) =. b a We have ϕ(a + b D + c + d ( ) ( ) ( ) a + c (b + d)d a bd c dd D) = = + = ϕ(a + b D) + ϕ(c + d D), b + d a + c b a d c and ϕ((a + b D) (c + d ( ) ( ) ( ) ac + bdd (ad + bc)d a bd c dd D)) = = ad + bc ac + bdd b a d c = ϕ(a + b D)ϕ(c + d D), so ϕ is an homomorphism. Since K is a field, its ideals are {0} and K, so ker(ϕ) is trivial or K. Since ϕ(k) is clearly non-zero, then ker(ϕ) K and thus ker(ϕ) = {0}. Hence, ϕ is injective. Therefore, ϕ is an isomorphism of K with a subfield of M (Q). Exercise 13... Define ϕ : K 1 K K 1 K by ϕ(a, b) = ab. We prove that ϕ is F-bilinear. Let a, a 1, a K and b, b 1, b K. Then and ϕ((a 1, b) + (a, b)) = ϕ(a 1 + a, b) = (a 1 + a )b = a 1 b + a b = ϕ(a 1, b) + ϕ(a, b), ϕ((a, 1 b) + (a, b )) = ϕ(a, b 1 + b ) = a(b 1 + b ) = ab 1 + ab 1 = ϕ(a, b 1 ) + ϕ(a, b ). 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 : K 1 F K K 1 K. We use Φ to prove both directions. Note that K 1 F K have dimension [K 1 : F][K : F] as a vector space over F. First, we suppose [K 1 K : F] = [K 1 : F][K : F] and prove K 1 F K is a field. In this case K 1 F K and K 1 K have the same dimension over F. Let L = Φ(K 1 F K ). We claim L = K 1 K, i.e. Φ is surjective. Note that L contains K 1 and K. Since L is a subring of K 1 K containing K 1 (or K ), then L is a field (Exercise 16). Hence, L is a field containing both K 1 and K. Since K 1 K is the smallest such field (by definition), we have L = K 1 K. 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, K 1 F K is a field. Now suppose that K 1 F K 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, [K 1 : F][K : F] [K 1 K : F]. As we already have [K 1 K : F] [K 1 : F][K : F] (Proposition 1 of the book), the equality follows. 10

13.3 Classical Straightedge and Compass Constructions 13.3 Classical Straightedge and Compass Constructions Exercise 13.3.1. Suppose the 9-gon is constructible. It has angles of 40. Since we can bisect an angle by straightedge and compass, the angle of 0 would be constructible. But then cos 0 and sin 0 would be constructible too, a contradiction (see proof of Theorem 4). Exercise 13.3.. Let O, P, Q and R be the points marked in the figure below. Then α = QPO, β = RQO, γ = QRO, and θ is an exterior angle of PRO. Since PQO is isosceles, then α = QPO = QOP. Since β is an exterior angle of PQO, it equals the sum of the two remote interior angles, i.e., equals QPO + QOP. This two angles equals α, hence β = α. Now, since QRO is isosceles, then β = γ. Finally, since θ is an exterior angle of PRO, equals the sum of the two remote interior angles, which are α and γ. Therefore, θ = α + γ = α + β = 3α. Exercise 13.3.3. We follow the hint. The distances a, b, x, y and x k are marked in the figure below. From the figure, using similar triangles for (a), (b) and (c), and Pythagoras Theorem for (d), the 4 relations are clear. Hence, we have 1 k y = 1 + a, x = a b + k 1 + a, y 1 k x k = and (1 k ) + (b + k) = (1 + a). 3k So, 1 k = y(1 + a) = 3ky x k implies 3k = (x k)(1 + a). From the equation for x above, we find 3k = ( a(b+a) 1+a k)(1 + a) = a(b + k) k(1 + a), so b + k = 4k+ka a. Using this in the last equation and reducing, we get (1 k ) + (b + k) = (1 + a) (1 k 4k + ka ) + ( ) = (1 + a) a a (1 k ) + (4k + ka) = a (1 + a) a (ka) + (4k) + 8k a + (ka) = a + a 3 + a 4 a 4 + a 3 8k a 16k = 0. 11

13.4 Splitting Fields and Algebraic Closures We let a = y to obtain h 4 + h 3 k h k = 0. We find h = k /3, hence a = k /3. From b = 4k+ka a k, we find b = k 1/3. Therefore, we can construct k 1/3 and k /3 using Conway s construction. Exercise 13.3.4. Let p(x) = x 3 + x x 1 and α = cos(π/7). By the 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. Since we can t construct α, the regular 7-gon is not constructible by straightedge and compass. Exercise 13.3.5. Let p(x) = x + x 1 = 0 and α = cos(π/5). By the 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 over Q, so it is constructible. We can bisect an angle by straightedge and compass, so β = cos(π/5) is also constructible. Finally, as sin(π/5) = 1 cos (π/5), sin(π/5) is also constructible. Therefore, the regular 5-gon is constructible by straightedge and compass. 13.4 Splitting Fields and Algebraic Closures Exercise 13.4.1. Let f (x) = x 4. The roots of f are 4, 4, i 4 and i 4. Hence, the splitting field of f is Q(i, 4 ). So, the splitting field of f has degree [Q(i, 4 ) : Q] = [Q(i, 4 ) : Q( 4 )][Q( 4 ) : Q] over Q. Since 4 is a root of the irreducible polynomial x 4 over Q, then [Q( 4 ) : Q] = 4. Furthermore, since i Q( 4 ), then x + 1 is irreducible over Q( 4 ) having i as a root, so [Q(i, 4 ) : Q( 4 )] =. Therefore, [Q(i, 4 ) : Q] = 8. Exercise 13.4.. Let f (x) = x 4 +. Let K be the splitting field of f and let L be the splitting field of x 4, that is, L = Q(i, 4 ) (last exercise). We claim K = L, so that [K : Q] = 8 by last exercise. Let ζ = + i. First we prove ζ L and ζ K, then we prove K = L. We prove ζ L. This is easy. Let θ = 4. Since θ L, then θ = L. We also have i L, so, i L implies ζ L. We prove ζ K. We have to prove i K and K. Let α be a root of x 4 +, so that α 4 =. Let β be a root of x 4 1, so that β 4 = 1. Then (α β) 4 = α 4 β 4 =, hence α β is also a root of x 4 +. Since the roots of x 4 1 are ±1, ±i, the roots of x 4 + are ±α and ±iα. Since K is generated over Q by there roots, then iα/α = i K. Now let γ = α K. Since γ = α 4 =, then γ is a root of x +. Since the roots of x + are i and i, then γ is one of this roots. In either case γ/i K, which implies K. Therefore, ζ K. Now we prove L = K proving both inclusions. Let α be a root of x 4 + and θ be a root of x 4. Then α 4 = and θ 4 =. Note that ζ = i, so ζ 4 = 1. Hence, (ζθ) 4 = ζ 4 θ 4 =, so ζθ is a root of x 4 +. Then, as we proved earlier, the roots of x 4 + are ±ζθ and ±iζθ. We also have (ζα) 4 = ζ 4 α 4 =, so ζα is a root of x 4. Then, by last exercise, the roots of x 4 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 x 4 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 x 4 + are in L. Since K is generated by these roots, we have K L. Therefore, K = L, so [K : Q] = 8. 1

13.5 Separable and Inseparable Extension Exercise 13.4.3. Let f (x) = x 4 + x + 1. Note that f (x) = (x + x + 1)(x x + 1), so the roots of f are ± 1 ± i 3 Let w = 1 i 3, so this roots are w, w, w, w, where w denotes the complex conjugate of w (i.e., w = 1 + i 3 ). 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 x x + 1, that is irreducible over Q since w Q. Therefore, the degree of the splitting field of f is [Q(w) : Q] =. Exercise 13.4.4. Let f (x) = x 6 4. Note that f (x) = (x 3 )(x 3 + ). The roots of x 3 are 3, ζ 3 and ζ 3, where ζ is the primitive 3 rd root of unity, i.e., ζ = exp(πi/3) = cos(π/3) +i sin(π/3) = 1 + 3. Furthermore, the roots of x 3 + are 3, ζ 3 and ζ 3. Therefore, the splitting field of f is Q(ζ, 3 ). Then [Q(ζ, 3 ) : Q] = [Q(ζ, 3 ) : Q( 3 )][Q( 3 3 ) : Q]. We have that is a root of the irreducible polynomial x 3 over Q, so 3 has degree 3 over Q. Furthermore, ζ is a root of x + x + 1, irreducible over Q( 3 ), so ζ has degree over Q( 3 ). Hence, the degree of the splitting field of f is [Q(ζ, 3 ) : Q] = 6. Exercise 13.4.5. We follow the hint. First suppose that K is a splitting field over F. Hence, there exists f (x) F[x] such that K is the splitting field of f. Let g(x) be an irreducible polynomial in F[x] with a root α K. Let β be any root of g. We prove β K, so that g splits completely in K[x]. By Theorem 8, there 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 8, 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(α 1,..., α n ) for some α 1,..., α n. For every 1 i n, let p i be the minimal polynomial of α i over F, and let f = p 1 p p n. Since every α i is in K, every p i 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 13.4.6. (a) Let K 1 be the splitting field of f 1 (x) F[x] over F and K the splitting field of f (x) F[x] over F. Thus, K 1 is generated over F by the roots of f 1, and K is generated over F by the roots of f. Therefore, f 1 f splits completely in K 1 K and K 1 K is generated over F by its roots, hence is the splitting field of f 1 f (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 K 1 K splits completely in (K 1 K )[x]. So, let f (x) be an irreducible polynomial in F[x] that has a root, say α, in K 1 K. By last exercise, f splits completely in K 1 and splits completely in K. Since K 1 and K are contained in K, by the uniqueness of the factorization of f in K, the roots of f in K 1 must coincide with its roots in K. Hence, f splits completely in (K 1 K )[x]. 13.5 Separable and Inseparable Extension Exercise 13.5.1. Let f (x) = a n x n + + a 1 x + a 0 and g(x) = b m x m + + b 1 x + b 0 be two polynomials. Suppose, without any loss of generality, that n m. Thus, we can write g(x) = b n x n + +b 1 x+b 0, where some of the last coefficients b i could be zero. We have f (x)+g(x) = (a n +b n )x n + +(a 1 +b 1 )x+(a 0 +b 0 ), so D x ( f (x) + g(x)) = n(a n + b n )x n 1 + + (a + b )x + (a 1 + b 1 ) = D x ( f (x)) + D x (g(x)).. 13

13.5 Separable and Inseparable Extension Now we prove the formula for the product. Let c n = n a kb n k, so that n n f (x)g(x) = ( a k x k )( b k x k ) = n ( l=0 l a k b l k )x l = n l=0 c l x l. Hence, D x ( f (x)g(x)) = D x ( n l=0 n 1 c l x l ) = (l + 1)c l+1 x l, so the coefficient of x l in D x ( f (x)g(x)) is (l + 1)c l+1. Now, we have D x ( f (x)) = na n x n 1 + +a x + a 1, and D x (g(x)) = nb n x n 1 + +b x + b 1. So (recall product of polynomials, page 95 of the book) n n D x ( f (x))g(x) = ( ka k x k 1 )( b k x k ) k=1 n 1 n n 1 l = ( (k + 1)a k+1 x k )( b k x k ) = ( (k + 1)a k+1 b l k )x l, l=0 l=0 and n n f (x)d x (g(x)) = ( a k x k )( kb k x k 1 ) = ( k=1 n n 1 n 1 l a k x k )( (k + 1)b k+1 x k ) = ( a k (l k + 1)b l k+1 )x l. l=0 Therefore, the coefficient of x l in D x ( f (x))g(x) + D x (g(x)) f (x) is l l ( (k + 1)a k+1 b l k ) + ( (l k + 1)a k b l k+1 ) l 1 = (l + 1)a l+1 b 0 + ( (k + 1)a k+1 b l k ) + ( k=1 l (l k + 1)a k b l k+1 ) + (l + 1)a 0 b l+1 l l = (l + 1)a l+1 b 0 + ( ka k b l k+1 ) + ( (l k + 1)a k b l k+1 ) + (l + 1)a 0 b l+1 = (l + 1)a l+1 b 0 + ( k=1 k=1 l (l + 1)a k b l k+1 ) + (l + 1)a 0 b l+1 k=1 l+1 = (l + 1)( a k b l k+1 ) = (l + 1)c l+1. Since all their coefficients are equal, we conclude D x ( f (x)g(x)) = D x ( f (x))g(x) + D x (g(x)) f (x). Exercise 13.5.. The polynomials x and x + 1 are the only (non-constant, i.e. 0, 1) polynomials of degree 1 over F ; they are clearly irreducible. A polynomial f (x) F [x] of degree is irreducible over F if and only if it does not have a root in F, that is, exactly when f (0) = f (1) = 1. Hence, the only irreducible polynomial of degree over F is x + x + 1. Now, for a polynomial f (x) F [x] of 14

13.5 Separable and Inseparable Extension 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 terms (or it will be 0), and must have constant term 1 (or x will be a factor). We are left with x 4 + x 3 + x + x + 1 x 4 + x 3 + 1 x 4 + x + 1 x 4 + x + 1. For any of this polynomials to be irreducible, it can t be factorized as two quadratic irreducible factors. Since x +x+1 is the only irreducible polynomial of degree over F, only (x +x+1) = x 4 +x +1 of this four is not irreducible. Hence, the irreducible polynomials of degree 4 over F are x 4 +x 3 +x +x+1, x 4 + x 3 + 1 and x 4 + x + 1. Now, since x + 1 = x 1 in F, we have (x + 1)(x 4 + x 3 + x + x + 1) = x 5 1. We also calculate (x + x +1)(x 4 + x +1)(x 4 + x 3 +1) = x 10 + x 5 +1. So, the product of all this irreducible polynomials is x(x + 1)(x + x + 1)(x 4 + x + 1)(x 4 + x 3 + 1)(x 4 + x 3 + x + x + 1) Exercise 13.5.3. = x(x 5 1)(x 10 + x 5 + 1) = x 16 x. We follow the hint. Suppose d divides n, so that n = qd for some q Z. Then x n 1 = x qd 1 = (x d 1)(x qd d + x qd d +... + x d + 1). So x d 1 divides x n 1. Conversely, suppose d does not divide n. Then n = qd + r for some q, r Z with 0 < r < d. Thus x n 1 = (x qd+r x r ) + (x r 1) = x r (x qd 1) + (x r 1) = x r (x d 1)(x qd d + x qd d +... + x d + 1) + (x r 1). Since x d 1 divides the first term, but doesn t divide x r 1 (as r < d), then x d 1 does not divide x n 1. Exercise 13.5.4. The first assertion is analogous to the last exercise. Now, F p d is defined as the field whose p d elements are the roots of x pd x over F p. Similarly is defined F p n. Take a = p. So, d divides n if and only if p d 1 divides p n 1, and that occurs exactly when x pd 1 1 divides x pn 1 1 (by last exercise). Thus, if d divides n, any root of x pd x = x(x pd 1 1) must be a root of x pn x = x(x pn 1 1), hence F p d F p n. Conversely, if F p d F p n, then x pd 1 1 divides x pn 1 1, so d divides n. Exercise 13.5.5. Let f (x) = x p x + a. Let α be a root of f (x). First we prove f is separable. Since (α + 1) p (α + 1) + a = α p + 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 F p, so f is separable. Now we prove f is irreducible. Let f = f 1 f f n where f i (x) F p [x] is irreducible for all 1 i n. Let 1 i < j n and let α i be a root of f i and α j be a root of f j, so that f i is the minimal polynomial of α i and f j the minimal polynomial of α j. We prove deg f i = deg f j. Since α i is a root of f i, it is a root of f, hence there exists k 1 F p such that α i = α + k 1. Similarly, there exists k F p such that α j = α + k. Thus, α i = α j + k 1 k, so f i (x + k 1 k ) is irreducible having α j as a root, so it must be its minimal polynomial. Hence, f i (x + k 1 k ) = f j (x), so deg f i = deg f j, as claimed. Since i and j were arbitrary, then all f i are of the same degree, say q. Then p = deg f = nq, so n = 1 or n = p (as p is prime). If n = p, then all roots of f are in F p, so α F p and thus 0 = α p α + a = a, contrary to the hypothesis. Therefore, n = 1, so f is irreducible. 15

13.5 Separable and Inseparable Extension Exercise 13.5.6. By definition, F p n is the field whose p n elements are the roots of x pn x over F p. Since x pn 1 = x(x pn 1 1), clearly x pn 1 1 = (x α). α F p n Set x = 0. Then 1 = ( α) = ( 1) pn 1 α α F p n α F p n ( 1) pn 1 ( 1) = ( 1) pn 1 ( 1) pn 1 ( 1) pn = α F p n α. α F p n α Hence, the product of the nonzero elements is +1 if p = and 1 is p is odd. For p odd and n = 1, we have 1 = α, α F p so taking module p we find [1][] [p 1] = [ 1], i.e., (p 1)! 1 (mod p). Exercise 13.5.7. Let a K such that a b p for every b K. Let f (x) = x p a. We prove that f is irreducible and inseparable. If α is a root of x p a, then x p 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 K, otherwise a = α p, 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 = x k kαx k 1 + + ( α) k. Therefore, kα K. Since α K, then k = p, so g = f. Hence, f is irreducible. We conclude that K (α) is an inseparable finite extension of K. Exercise 13.5.8. Let f (x) = a n x n +... + a 1 x + a 0 F p [x]. Since F p has characteristic p, then (a + b) p = a p + b p for any a, b F p. Easily we can generalize this to a finite number of terms, so that (x 1 + + x n ) p = x p 1 + + xp n for any x 1,, x n F p. Furthermore, by Fermat s Little Theorem, a p = a for every a F p. So, over F p, we have f (x) p = (a n x n +... + a 1 x + a 0 ) p = a p n x np +... + a p 1 xp + a p 0 = a nx np +... + a 1 x p + a 0 = f (x p ). Exercise 13.5.9. By the binomial theorem, we have (1 + x) pn = pn i=0 ( pn i ) x i, so the coefficient of x pi in the expansion of (1 + x) pn is ( pn pi ). Since F p has characteristic p, we have (1 + x) pn = 1 + x pn = (1 + x p ) n, so over F p this is the coefficient of (x p ) i in (1 + x p ) n. Furthermore, (1 + x) pn = (1 + x p ) n implies 16

13.6 Cyclotomic Polynomials and Extensions (1 + x p ) n = over F p, hence ( pn pi ) ( n i ) (mod p). Exercise 13.5.10. n i=0 ( ) n (x p ) i = i pn i=0 ( ) pn x i = (1 + x) pn k This is equivalent to prove that for any prime number p, we have f (x 1, x,..., x n ) p = f (x p 1, xp,..., xp n) in F p [x 1, x,..., x n ]. Let f (x 1, x,..., x n ) = a γ1,...,γ n x γ 1 1... xγ n n γ 1,...,γ n =0 be an element of F p [x 1, x,..., x n ]. Since F p has characteristic p, then (x 1 + + x n ) p = x p 1 + + xp n for any x 1,, x n F p. Furthermore, by Fermat s Little Theorem, a p = a for every a F p. Hence, over F p we have f (x 1, x,..., x n ) p = ( = Exercise 13.5.11. a γ1,...,γ n x γ 1 1... xγ n n ) p = a p γ 1,...,γ n (x γ 1 1... xγ n n ) p = (a γ1,...,γ n x γ 1 1... xγ n a γ1,...,γ n (x pγ 1 n ) p 1... x pγ n n ) = f (x p 1, xp,..., xp n). Let f (x) F[x] with no repeated irreducible factors in F[x]. We can suppose f is monic. Then f = f 1 f f n for some monic irreducible polynomials f i (x) F[x]. Since F is perfect, f is separable, hence all f i 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 13.6.1. Since (ζ m ζ n ) mn = 1, then ζ m ζ n is an mn th root of unity. Now, let 1 k < mn. Then (ζ m ζ n ) k = ζ k mζ k n. For a contradiction, suppose ζ k mζ k n = 1. Then, ζ k m = 1 and ζ k n = 1, hence m divides k and n divides k, so mn divides k (as m n). Since 1 k < mn, this is impossible. So, (ζ m ζ n ) k 1 for all 1 k < mn and thus the order of ζ m ζ n is mn. Therefore, ζ m ζ n generates the cyclic group of all mn th roots of unity, that is, ζ m ζ n is a primitive mn th root of unity. Exercise 13.6.. Since (ζn d ) (n/d) = ζn n = 1, then ζn d is an (n/d) th root of unity. Now let 1 k < (n/d). Then (ζn d ) k = ζn kd. Since 1 kd < n, then ζn kd 1, so (ζn d ) k 1. Hence, the order of ζn d is (n/d), so it generates the cyclic group of all (n/d) th roots of unity, that is, ζn d is a primitive (n/d) th root of unity. Exercise 13.6.3. Let F be a field that contains the n th roots of unity for n odd and let ζ be a n th root of unity. If ζ n = 1, then ζ F, so suppose ζ n 1. Since ζ n = 1, then ζ n is a root of x 1. Since the roots of this polynomial are 1 and 1, and ζ n 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. Exercise 13.6.4. Let F be a field with char F = p. The roots of unity over F are the roots of x n 1 = x pk m 1 = (x m 1) pk, so are the roots of x m 1. Now, since m is relatively prime to p, so is x m 1 and 17

13.6 Cyclotomic Polynomials and Extensions its derivative mx m 1, so x m 1 has no multiple roots. Hence, the m different roots of x m 1 are precisely the m distinct n th roots of unity over F. Exercise 13.6.5. We use the inequality ϕ(n) n/ 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 and there exits some nth root of unity ζ K. Then [K : Q] [Q(ζ) : Q] = ϕ(n) n/ > 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 13.6.6. Since Φ n (x) and Φ n ( 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 ζ be the primitive th root of unity and let ζ n be a primitive n th root of unity. Note that ζ = 1, so that ζ ζ n = ζ n. Since n is odd, and n are relatively prime. So, by Exercise 1, ζ ζ n is a primitive n th root of unity, i.e, a root of Φ n (x). Furthermore, ζ n is clearly a root of Φ n ( x). Thus, ζ n is a common root for both Φ n (x) and Φ n ( x), hence Φ n (x) = Φ n ( x). Exercise 13.6.7. The Möbius Inversion Formula sates that if f (n) is defined for all nonnegative integers and F(n) = d n f (d), then f (n) = d n µ(d)f( n d ). So lets start with the formula x n 1 = Φ d (x). We take natural logarithm in both sides and obtain ln(x n 1) = ln( Φ d (x)) = ln Φ d (x). d n So, we use the Möbius Inversion Formula for f (n) = ln Φ n (x) and F(n) = ln(x n 1) to obtain ln Φ n (x) = µ(d) ln(x n/d 1) = ln(x n/d 1) µ(d). d n Hence, taking exponentials we obtain Φ n (x) = exp( ln(x n/d 1) µ(d) ) = (x n/d 1) µ(d) = (x d 1) µ(n/d). Exercise 13.6.8. d n d n d n (a) Since p is prime, in F p [x] we have (x 1) p = x p 1, so Φ l (x) = xl 1 x 1 = (x 1)l x 1 = (x 1) l 1. (b) Note that ζ has order l as being a primitive l th root of unity. Since p f 1 mod l, then p f 1 = ql for some integer q, hence ζ p f 1 = ζ ql = 1, so ζ F p f. Now we prove that f is the smallest integer with that property. Suppose ζ F p n for some n. Then ζ is a root of x pn 1 1, hence l divides p n 1 (see Exercise 13.5.3). Since f is the smallest power of p such that p f 1 mod l, is the smallest integer such that l divides p f 1, so n l, as desired. This in fact proves that F p (ζ) = F p f, so the minimal polynomial of ζ over F p has degree f. (c) Since ζ a F p (ζ), clearly F p (ζ a ) F p (ζ). For the other direction we follow the hint. Let b the the multiplicative inverse of a mod l, i.e ab 1 mod l. Then (ζ a ) b = ζ, so ζ F p (ζ a ) and thus F p (ζ) F p (ζ a ). The equality follows. d n d n d n 18

13.6 Cyclotomic Polynomials and Extensions Now, consider Φ l (x) as a polynomial over F p [x]. Let ζ i for 1 i l be l distinct primitive l th roots of unity. The minimal polynomial of each ζ i has degree f by part (b). Hence, the irreducible factors of Φ l (x) have degree f. Since Φ l have degree l 1, then there must be l 1 f factors, and all of them are different since Φ l (x) is separable. (d) If p = 7, then Φ 7 (x) = (x 1) 6 by part (a). If p 1 mod 7, then f = 1 in (b) and all roots have degree 1, so Φ 7 (x) splits in distinct linear factors. If p 6 mod 7, then f = is the smallest integer such that p f = p 36 1 mod 7, so we have 3 irreducible quadratics. If p, 4 mod 7, then f = 3 is the smallest integer such that p 3 3, 4 3 8, 64 1 mod 7, so we have irreducible cubics. Finally, if p 3, 5 mod 7, then f = 6 is the smallest integer such that p 6 3 6, 5 6 79, 1566 1 mod 7, hence we have an irreducible factor of degree 6. Exercise 13.6.9. Let A be an n by n matrix over C for which A k = I for some integer k 1. Then the minimal polynomial of A divides x k 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 Computing powers of A, we can prove (by induction) that A n = for every positive integer 0 1 n. Since pα = 0, then A p = I. Now, if A is diagonalizable, there exists some non-singular matrix P such that A = PDP 1, where D is a diagonal matrix whose diagonal entries are the eigenvalues of A. Since A has only one eigenvalue 1, then D = I, and thus A = PIP 1 = I. So, if A is diagonalizable, it must be α = 0. Exercise 13.6.10. Let a, b F p n. Then ϕ(a + b) = (a + b) p = a p + b p = ϕ(a) + ϕ(b), and ϕ(ab) = (ab) p = a p b p = ϕ(a)ϕ(b), so ϕ is and homomorphism. Moreover, if ϕ(a) = 0, then a p = 0 implies a = 0. Hence, ϕ is injective. Since F p n is finite, then ϕ is also surjective so it is an isomorphism. Furthermore, since every element of F p n is a root of x pn x, then ϕ n (a) = a pn = a for all a F p n, so ϕ n is the identity map. Now, let m be an integer such that ϕ m is the identity map. Then a pm = a for all a F p n, so every element of F p n must be a root of x pm x. Hence, F p n F p m and thus n divides m (Exercise 13.5.4), so n m. Exercise 13.6.11. Note that the minimal polynomial of ϕ is x n 1, for if ϕ satisfies some polynomial x n 1 + +a 1 x +a 0 of degree n 1 (or less) with coefficients in F p, then x pn 1 + + a 1 x p + a 0 for all x F p n, which is impossible. Since F p n has degree n as a vector space over F p, then x n 1 is also the characteristic polynomial of ϕ, hence is the only invariant factor. Therefore, the rational canonical form of ϕ over F p is the companion matrix of x n 1, which is Exercise 13.6.1. 0 0 0 1 1 0 0 0 0 1 0 0........ 0 0 1 0 We ll work over the algebraic closure of F p n, to ensure the field to contain all eigenvalues. In last exercise we proved that the minimal and characteristic polynomial of ϕ is x n 1. Moreover, the eigenvalues of ϕ are the n th roots of unity. We use Exercise 4 and write n = p k m for some prime p 19

13.6 Cyclotomic Polynomials and Extensions and some m relatively prime to p, so that x n 1 = (x m 1) pk and we get exactly m distinct n th roots of unity, each one with multiplicity p k. Since all the eigenvalues are zeros of both the minimal and characteristic polynomial of multiplicity p k, we get m Jordan blocks of size p k. Now, fix a primitive m th root of unity, say ζ. Then, each Jordan block each of the form ζ i 1 0 0 0 0 ζ i 1 0 0 0 0 ζ J i = i 0 0........ 0 0 0 ζ i 1 0 0 0 0 ζ i for some 0 i m 1. We already know the Jordan canonical form is given by Exercise 13.6.13. J 0 0 0 0 J 1 0 0 0 0....... 0 0 J m 1 (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 F p for some prime p, so it is isomorphic to F p m for some integer m. Let q = p n, so that Z is isomorphic to F q. Since D is a vector space over Z, then D = q n for some integer n. (b) Let x D and let C D (x) be the set of the elements in D that commutes with x. Clearly Z C D (x). We prove that every element a C D (x) has an inverse in C D (x). Since a C D (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 C D (x). Hence, C D (x) is a division ring. Now, since Z C D (x), then C D (x) is a Z-vector space, so C D (x) = q m for some integer m. If x Z, then C D (x) is a proper subset of D and hence m < n. (c) The class equation for the group D is D = Z(D ) + r D : C D (x i ), i=1 where the x i are the representatives of the distinct conjugacy class. By (a) we have D = q n 1, Z(D ) = q 1 and C D (x i ) = q m i 1. Then D : C D (x i ) = qn 1 = qn 1 C D (x i ) q m. Replacing i 1 in the class equation we obtain q n 1 = (q 1) + r i=1 q n 1 C D (x i ) = (q 1) + r i=1 q n 1 q m i 1. (d) Since D : C D (x i ) is an integer, then D : C D (x i ) = qn 1 q m is an integer. Hence, q m i 1 i 1 divides q n 1, so (Exercise 13.5.4) m i divides n. Since m i < n (no x i is in Z), no mi th root of unity is a n th root of unity. Therefore, as Φ n (x) divides x n 1, it must divide (x n 1)/(x m i 1) for i = 1,,..., r.. Letting x = q we have Φ n (q) divides (q n 1)/(q m i 1) for i = 1,,..., r. 0

13.6 Cyclotomic Polynomials and Extensions (e) From (d), Φ n (q) divides (q n 1)/(q m i 1) for i = 1,,..., r, so the class equation in (c) implies Φ n (q) divides q 1. Now, let ζ 1 be a n th root of unity. In the complex plane q is closer to 1 that ζ is, so q ζ > q 1 = q 1. Since Φ n (q) = ζ primitive(q ζ) divides q 1, this is impossible unless n = 1. Hence, D = Z and D is a field. Exercise 13.6.14. We follow the hint. Let P(x) = x n + + a 1 x + a 0 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,,..., say p 1, p,..., p k. Let N be an integer such that P(N) = a 0. Let Q(x) = a 1 P(N + ap 1 p... p k x). Then, using the binomial theorem, we have Q(x) = a 1 P(N + ap 1 p... p k x) = a 1 ((N + ap 1 p... p k x) n + + a 1 (N + ap 1 p... p k x) + a 0 ) = a 1 (N n + a n 1 N n 1 + + a 1 N + a 0 + R(x)) = a 1 (P(N) + R(x)) = 1 + a 1 R(x) for some polynomial R(x) Z[x] divisible by ap 1 p... p k. Hence, Q(x) Z[x]. Moreover, for all n Z +, P(N + ap 1 p... p k n) a (mod p 1, p,..., p k ), so Q(n) = a 1 P(N + ap 1 p... p k n) a 1 a = 1 (mod p 1, p,..., p k ). Now let m be a positive integer such that Q(m) > 1, so that Q(m) 1 (mod p i ) for all i. Therefore, none of the p i s divide Q(m). Since Q(m) > 1, there exists a prime q p i for all i such that q divides Q(m). Then q divides aq(m) = P(N + ap 1 p... p k m), contradicting the fact that only the primes p 1, p,..., p k divides the numbers P(1), P(),.... Exercise 13.6.15. We follow the hint. Since Φ m (a) 0 (mod p), then a m 1 (mod p). Hence, there exist b such that ba 1 mod p (indeed, b = a m 1 ), so a is relatively prime to p. We prove that the order of a is m. For a contradiction, suppose a d 1 (mod p) for some d dividing m, so that Φ d (a) 0 (mod p) for some d < m. Thus, a is a multiple root of x m 1, so is also a root of its derivative ma m 1. Hence, ma m 1 0 mod p, impossible since p does not divide m nor a. Therefore, the order of a in (Z/pZ) is precisely m Exercise 13.6.16. Let p be an odd prime dividing Φ m (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 F p = p 1, this implies m divides p 1, that is, p 1 (mod m). Exercise 13.6.17. By Exercise 14, there are infinitely many primes dividing Φ m (1), Φ m (), Φ m (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). Please send comments, suggestions and corrections by e-mail, or at website. https://positron080.wordpress.com positron080@mail.com 1