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

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1 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 Basic Theory and Field Extensions Algebraic Extensions Classical Straightedge and Compass Constructions Splitting Fields and Algebraic Closures Separable and Inseparable Extension Cyclotomic Polynomials and Extensions Field Theory 13.1 Basic Theory and Field Extensions Exercise 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 (1 + θ) 1 = θ θ 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 )

2 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 ). Combining these two equations we obtain So, 8 9 (θ + θ + 1)(θ 1)( θ θ ) = 1. (θ + θ + 1) 1 = 8 θ (θ 1)( 9 θ ) = θ 3 + θ , where we used θ 3 = θ + again. Therefore, Exercise θ = (1 + θ)( θ 1 + θ + θ 3 + θ ) = θ 3 + θ Since = 1 and = 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 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 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 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 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.

3 13.1 Basic Theory and Field Extensions Exercise This is straightforward. If a n α n + a n 1 α n 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 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 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

4 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 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 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 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 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 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 x x + 1 x 0 x x + 1 x x x + 1 x x The multiplication table for F 3 /(g(x)) is 0 1 x x + 1 x + x x + 1 x x x + 1 x + x x + 1 x x x + x + 1 x x + x + 1 x 0 x x x x + 1 x + x + x 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 x x + 1 x + x + x 1 x + 1 x x + 0 x + x + 1 x + 1 x 1 x x + 4

5 13. Algebraic Extensions In both cases, x is a generator of the cyclic group of nonzero elements. Exercise 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 First, note that (+ 3) = = Let θ = + 3. Then θ 4θ = = 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 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) = 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 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 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 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 = and θ 4 = = 1 (θ3 9θ), 3 = 1 (11θ θ3 ) and θ 4 = Therefore Q(θ) and 3 Q(θ), so Q(, 3) Q( + 3). The equality follows. Hence, [Q( + 3) : Q] = [Q(, 3) : Q] = 4. 5

6 13. Algebraic Extensions We also have θ 4 10θ = ( ) 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 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 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

7 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 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 Note that 3 + = 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) = +1. Therefore, Q( 3 + ) = Q( ) and the degree of the extension Q( 3 + ) over Q is. Exercise (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 = 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 + i. Furthermore, we find 3 4i = 1 i. Therefore, 3 + 4i + 3 4i = 4, i.e., 3 + 4i + 3 4i Q. (b) Let θ = Then θ = ( ) = (1 + 3) + ( 1 + 3) + (1 3) = 6. Since x 6 is irreducible over Q (Eisenstein p = ), then θ has degree over Q. 7

8 13. Algebraic Extensions Exercise 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 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 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 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 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 a 1 ). Since a i F R and r R, we have r 1 R. 8

9 13. Algebraic Extensions Exercise 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 (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 (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 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 = 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 = and 9 A β =

10 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 = Exercise 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 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

11 13.3 Classical Straightedge and Compass Constructions 13.3 Classical Straightedge and Compass Constructions Exercise 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 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 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

12 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 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 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 Splitting Fields and Algebraic Closures Exercise 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 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 13.5 Separable and Inseparable Extension Exercise 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 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) = 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 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 (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] Separable and Inseparable Extension Exercise 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 (a + b )x + (a 1 + b 1 ) = D x ( f (x)) + D x (g(x)).. 13

14 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 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

15 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 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 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 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 = 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 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 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

16 13.5 Separable and Inseparable Extension Exercise 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 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 ( 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 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 x n ) p = x p 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 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

17 13.6 Cyclotomic Polynomials and Extensions (1 + x p ) n = over F p, hence ( pn pi ) ( n i ) (mod p). Exercise 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 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 x n ) p = x p 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 a γ1,...,γ n x γ xγ n n ) p = a p γ 1,...,γ n (x γ xγ n n ) p = (a γ1,...,γ n x γ 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] Cyclotomic Polynomials and Extensions Exercise 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 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 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 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

18 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 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 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 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 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 ). 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

19 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, , mod 7, hence we have an irreducible factor of degree 6. Exercise 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 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 ), so n m. Exercise 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 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 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

20 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 ζ i ζ J i = i ζ i ζ i for some 0 i m 1. We already know the Jordan canonical form is given by Exercise J J 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 ) 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

21 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 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 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 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 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 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 , or at website. positron080@mail.com 1