Solutions for Field Theory Problem Set 1

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Claribel McCoy
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1 Solutions for Field Theory Problem Set 1 FROM THE TEXT: Page 355, 2a. ThefieldisK = Q( 3, 6). NotethatK containsqand 3and = 2. Thus, K contains the field Q( 2, 3). In fact, those two fields are the same. To see this, it suffices to prove that Q( 2, 3) contains K. This is true because Q( 2, 3) is a field containing Q and 3 and 2 3 = 6. Thus, we have K = Q( 3, 2) = Q( 3)( 2) = F( 2), where F = Q( 3). Note that the minimal polynomial for 3 over Q is x 2 3. Hence [F : Q] = 2 and a basis for F over Q is {1, 3 }. Furthermore, the minimal polynomial for 2 over F is x 2 2. This is so because 2 is a root of the polynomial x 2 2, but that polynomial has no roots in F (as proven in problem D below). Thus, we have [F( 2) : F] = 2 and a basis for K = F( 2) over F is { 1, 2 }. Since [K : F] = 2 and [F : Q] = 2, it follows that [K : Q] = [K : F][F : Q] = 2 2 = 4. A basis for K over Q is { 1 1, 1 2, 3 1, 3 2 } = { 1, 2, 3, 6 }. 355, 2e. This question concerns the field K = Q( 2, 3 2) = F( 3 2), where F = Q( 2). We know that [F : Q] = 2. Furthermore, 3 2 F. To verify this, assume to the contrary that 3 2 F. Then we would have Q( 3 2) F. However, as discussed in class, [Q( 3 2 : Q] = 3. Thus, Q( 3 2) is a 3-dimensional vector space over Q and cannot be a subspace of F because F is a 2-dimensional vector space over Q. Therefore, 3 2 F. The other roots of x 3 2 cannot be in F because those roots are not in R whereas F is a subfield of R. It follows that x 3 2 has no roots in F, and since its degree is 3, that polynomial is irreducible over F. Therefore, x 3 2 is the minimal polynomial for 3 2 over F. It follows that [F( 3 2) : F] = 2. That is, [K : F] = 3. A basis for F over Q is {1, 2}. A basis for K over F is {1, 3 2, }. It follows that a basis for K over Q is We have [K : Q] = 6. { 1 1, 1 3 2, , 2 1, 2 3 2, }.

2 Page 355, 2f. Note that 8 = 2 2. It is clear that 8 Q( 2) and hence Q( 8) Q( 2). Furthermore, 2 = Q( 8) and hence we have Q( 2) Q( 8). It follows that Q( 8) = Q( 2). In particular, [Q( 8) : Q( 2)] = 1. A basis for Q( 8) over Q( 2) is {1}. Page 355, 2h. Let F = Q( 5). Let K = Q( 2+ 5). We first point out that F K. To verify this, note that ( 2+ 5)( 2 5) = = 2 5 = 3 It follows that 2 5 = ( 3) ( 2+ 5) 1 K and therefore 5 = 1 2( ( 2 5) ( 2 + 5) ) is also in K. Therefore, Q( 5) K. Thus, we have F K as stated. Furthermore,notethatK = F( 2+ 5). OneseeseasilythatK = F( 2). Furthermore, 2 F = Q( 5). This statement can be verified in a similar way to the proof that 2 Q( 3) given in problem D below. We omit that verification. It follows that the minimal polynomial for 2 over F is x 2 2. Consequently, we have [K : F] = 2 and a basis for K over F is { 1, 2 }. Page 355, 3a. We have x 4 10x = (x 2 3)(x 2 7). The four roots are ± 3, ± 7. The splitting field for that polynomial over Q is Q(± 3, ± 7) = Q( 3, 7). Page 355, 3b. If η is a root of x 4 +1 in C, then η 4 = 1 and hence η 8 = 1. Therefore, η is an 8th root of unity, but not a 4th root of unity. Note that η 3 has the same properties, and so η 3 is also a root of the polynomial x Also, η 5 and η 7 are roots of x 4 +1 in C. The four roots of x 4 +1 in C are η,η 3,η 5,η 7. Therefore, the splitting field for x 4 +1 over Q is Q(η, η 3, η 5, η 7 ). Note that the field Q(η) actually contains all powers of η. Hence η 3, η 5, and η 7 are in Q(η). Therefore, the splitting field for x 4 +1 over Q can be specified more simply as Q(η). One can take η = i.

3 Page 355, 3c. Let F = Z/3Z. We write the elements of F as ã = a + 3Z, where a {0, 1, 2}. Let g(x) = 1x 3 + 2x+ 2, an element in F[x] of degree 3. One checks easily that g(x) has no roots in F. Since g(x) has degree 3, it follows that g(x) is irreducible over F. Let K = F[x]/M, where M = ( g(x) ). Since M is a maximal ideal in F[x], it follows that K is a field. As discussed in class, we can regard F as a subfield of K. We do this by identifying an element ã in F with the element ã+m in K. Furthermore, as discussed in class, we have K = F[β], where β = x+m. Also, β is a root of g(x). It follows that g(x) has at least one root in K, namely β. The field K is generated by β over F. As explained in class, the map ϕ : K K defined by ϕ(k) = k 3 for all k K is an automorphism of the field K and ϕ F is the identity map for F. It follows that ϕ(β) = β 3 is also a root of g(x) in K. We have ϕ(β) = β 3 = (x+m) 3 = x 3 +M = x 3 g(x)+m = 2x 2 = 1x+ 1+M = β+ 1 Thus, β + 1 is another root of g(x) in K. We can apply ϕ to this root, obtaining ϕ(β + 1) = ϕ(β)+ϕ( 1) = (β + 1) + 1 = β + 2 Thus, β, β + 1, and β + 2 are roots of g(x) and are all in the field K. We have K = F(β) = F(β, β + 1, β + 2) and hence K is indeed a splitting field for g(x) over F. Page 355, 3d. Let ω = i, which is a cube root of unity. The roots of x3 3 are 3 3, ω 3 3, and ω The splitting field for x 3 3 over Q is Q( 3 3, ω 3 3, ω ). This field can be specified more simply as Q( 3 3, ω). Page 355, problem 5. Let F = Z/2Z. The elements of F are 0 = 0+2Z and 1 = 1+2Z. Let g(x) = 1x 3 + 1x+ 1. Let M = ( g(x) ). Since g(x) has degree 3 and has no roots in F, it follows that g(x) is irreducible over F and that M is a maximal ideal in F[x]. Let K = F[x]/M. Then K is a field and K has 8 elements because [K : F] = 3. Let β = x+m. The elements of K are of the form e+fβ+gβ 2, where e,f,g F. Furthermore, we have β 3 +β + 1 = 0 and hence β 3 = β 1 = 1+β.

4 The group U(K) consists of the nonzero elements of K and has 7 elements. This group must be cyclic because 7 is a prime. The identity element is 1. Any other element will be a generator. Thus, β is a generator. We have β 0 = 1, β 1 = β, β 2 = β 2, β 3 = 1+β, β 4 = ββ 3 = β +β 2, β 5 = ββ 4 = β 2 +β 3 = 1+β +β 2, β 6 = ββ 5 = β +β 2 +β 3 = β +β β = 1+β 2. Thus, the distinct elements of U(K) are { β j 0 j 6 }. One uses the law of exponents to multiply those powers of β. Thus, the above equations give the multiplication table for U(K). ADDITIONAL PROBLEMS: A: Let p = (which happens to be a prime number). Let a = 571+pZ, a nonzero element in the field F = Z/pZ. Find the additive and multiplicative inverses of a in the field F. You should express your answers in the form r+pz, where 0 r < p. SOLUTION. We take p = The additive inverse of 571+pZ in the field Z/pZ can be found as follows. 571+pZ = p 571+pZ = pZ The multiplicative inverse can be found by the Euclidean algorithm. Therefore, we have = , 571 = , 364 = = , 157 = , 50 = = = 50 7 ( ) = = 22 ( ) = = ( ) = = 51 ( ) = = ( ) = Thus, (mod p) and hence (571+pZ) 1 = 1571+pZ.

5 B: Let F = Q[θ], where θ = 3 2 is the unique cube root of 2 in R. Find the multiplicative inverse of f = 2+3θ θ 2 in F. Express your answer in the form f 1 = a+bθ+cθ 2, where a,b,c Q. SOLUTION. We will use the Euclidean algorithm to find f 1. As explained in class, the minimal polynomial for θ over Q is p(x) = x 3 2. Let f(x) = 2 + 3x x 2 Q[x]. Then f(θ) = f. Since f 0, the polynomial f(x) is not divisible by p(x) in Q[x]. The two polynomials are relatively prime in Q[x]. The Euclidean algorithm will lead to polynomials m(x), n(x) Q[x] such that m(x)f(x)+n(x)p(x) = 1. It then follows that m(θ)f(θ) = 1. Hence m(θ) = f 1. We have p(x) = ( 3 x)f(x) x, f(x) = and therefore = f(x) = ( 1 ) ( 4+11x) Therefore, we can take m(x) to be ( 1 m(x) = That is, we can take It follows that = f(x) ) (3+x ) ) f(x) ) ( 4+11x) ) (3+x ) ) = m(x) = x x2. ) ( p(x) ( 3 x)f(x)) ) f 1 = m(θ) = θ θ2. p(x). ( x+ 1 ) 2 C: Let F be the same field as in question B. Suppose that ϕ is an automorphism of F. Carefully prove that ϕ(α) = α for all α F.

6 SOLUTION. We will prove that ϕ(θ) = θ. This is sufficient because if α F, then we have α = a+bθ+cθ 2, where a,b,c Q. As explained in class, we have ϕ(r) = r for all r Q. Therefore, assuming that ϕ(θ) = θ, we obtain ϕ(α) = ϕ(a+bθ +cθ 2 ) = ϕ(a)+ϕ(b)ϕ(θ)+ϕ(c)ϕ(θ) 2 = a+bθ +cθ 2 = α as we wanted to prove. To prove that ϕ(θ) = θ, note that θ is a root of the polynomial p(x) = x 3 2 Q[x]. As proven in class, ϕ(θ) must also be a root of p(x). Since θ R, F is a subfield of R. The polynomial p(x) has only one root in R, namely θ. Hence θ is the only root of p(x) in F. Since ϕ(θ) F and ϕ(θ) is a root of p(x), we must indeed have ϕ(θ) = θ. This proves that ϕ(α) = α for all α F. That is, ϕ is the trivial automorphism of F. D: Let F = Q[ 2]. Let K = Q[ 3]. Both F and K are subfields of R. Carefully prove that F and K are not isomorphic. SOLUTION. Let θ = 2. Then θ F and θ is a root of the polynomial x 2 2. That is, θ 2 = 2. Suppose that a field isomorphism ϕ : F K does exist. We will derive a contradiction. As explained in class, ϕ(r) = r for all r Q. Let γ = ϕ(θ). Then γ K. Also θ 2 = 2 = ϕ(θ 2 ) = ϕ(2) = ϕ(θ) 2 = 2 = γ 2 = 2 and therefore K contains an element γ satisfying γ 2 = 2 We can write γ as γ = a+b 3, where a,b Q. Now and if γ 2 = 2, then we have γ 2 = (a 2 +3b 2 )+2ab 3 2ab 3 = 2 (a 2 +3b 2 ). That equation implies that 3 = ( 2 (a 2 +3b 2 ) )/ 2ab if ab 0. It would then follows that 3 Q since a,b Q. However we know that 3 is an irrational number. We can therefore conclude that ab = 0 and hence either a = 0 or b = 0. If a = 0, then γ = b 3 and γ 2 = 3b 2 = 2. This implies that b is a rational root of the polynomial 3x 2 2. Hence the denominator of b divides 3 and the numerator of b divides

7 2. One checks all the possibilities easily and finds that no such rational number b can exist. On the other hand, if b = 0, then γ = a Q and γ 2 = a 2 = 2. But this is also impossible because the polynomial x 2 2 has no root in Q. Therefore, we must have a 0 and b 0. We now have a contradiction. Hence no such field isomorphism ϕ : F K can exist. E: Is Q[x] / (x 4 +4) a field? Justify your answer carefully. SOLUTION. Let R = Q[x]/I, where I = ( x ). In fact, R is not a field. To prove that R is not a field, it suffices to show that I is not a maximal ideal of Q[x]. To show that I is not maximal, it is sufficient to show that x 4 +4 is a reducible polynomial over Q. Here is the verification of that fact: x 4 +4 = x 4 +4x x 2 = (x 2 +2) 2 (2x) 2 = (x x)(x x) = (x 2 +2x+2)(x 2 2x+2). Thus, it follows that x 4 +4 is reducible over Q and hence I is not a maximal ideal of Q[x]. F: Show that there exists a field with 64 elements. (HINT: The field in problem 5 on page 355 may be helpful. That field has 8 elements. ) SOLUTION. Let K be the field in problem 5 on page 355. Then K is a field with 8 elements. Suppose that we can find a polynomial g(x) in K[x] such that g(x) has degree 2 and such that g(x) is irreducible over K. Then K[x] /( g(x) ) will be a field containing K and will have degree 2 over K. Thus, it will be a field with 8 2 = 64 elements. It is enough to prove the existence of such a polynomial g(x). Consider all the polynomials of the form x 2 +ax+b, where a,b K. Obviously, there are 64 such polynomials. Now if x 2 +ax+b is reducible over K, then we have x 2 +ax+b = (x c)(x d) for some c,d K. There are 64 possible choices for c and d. However, if c d, then (x c)(x d) = (x d)(x c). The number of distinct polynomials of the form (x c)(x d) where c,d K and c = d is 8. The number of distinct polynomials (x c)(x d) where c,d K and c d is 8 7/2 = 28. Hence there are only 36 distinct polynomials of the form (x c)(x d), where c,d K. Therefore, the number of irreducible polynomials of the form x 2 + ax + b, where a,b K is = 28. Therefore, such irreducible polynomials do indeed exist. Hence a field with 64 elements does exist.

Solutions for Problem Set 6

Solutions for Problem Set 6 A: Find all subfields of Q(ζ 8 ). SOLUTION. All subfields of K must automatically contain Q. Thus, this problem concerns the intermediate fields for the extension K/Q. In a