Solutions for Problem Set 6

Transcription

1 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 previous problem set, it was proved that [K : Q] = 4 and that K = Q( 2, i). As discussed in class, it follows thsat every element in Gal(K/Q) has order 1 or 2. Therefore, we can say that Gal(K/Q) = Z/2Z Z/2Z. There are three elements of order 2 and therefore three subgroups of order 2. Those subgroups have index 2 in Gal(K/Q) By the Fundamental Theorem of Galois Theory, it follows that K contains exactly three subfields L such that [L : Q] = 2. The three subfields L 1 = Q( 2), L 2 = Q(i), L 3 = Q( 2i) = Q( 2) are distinct. We see this by using the simple criterion discussed in class, noting that 2 ( 1) = 2, 2 ( 2) = 4, ( 1) ( 2) = 2 are not perfect squares. Hence those three fields must be all of the subfields of K which have degree 2 over Q. In addition, Gal(K/Q) obviously has exactly one subgroup of order 1 and exactly one subgroup of order 4. The corresponding subfields of K are K itself and Q. B: Let K be the splitting field over Q for f(x) = x 4 2. We know that K = Q( 4 2, i) and that [K : Q] = 8. Determine all subfields F of K such that [F : Q] = 4. In addition, determine which of those subfields F is a Galois extension of Q. SOLUTION. As discussed in class, we know that K/Q is a Galois extension and that G = Gal(K/Q) = D 4, the dihedral group of order 8. Now the dihedral group has four rotations, one of which has order 2. The dihedral group has four reflections, all of which have order 2. Thus, the dihedral group has five elements of order 2 and hence five subgroups of order 2. The same will be true for Gal(K/Q) since an isomorphism of groups will preserve the number of elements of any given order. The index of a subgroup of order 2 is 4. The corresponding intermediate fields will be the intermediate fields which have degree 4 over Q. Every subfield of K automatically contains Q and will be an intermediate field. These remarks show that the number of intermediate fields F for the extension K/Q is five. Exactly one of the subgroups of G of order 2 will be a normal subgroup, namely the subgroup generated by the rotation of order 2. One can check easily that the each

2 reflection is conjugate to another distinct reflection. The correspond subgroups cannot be normal subgroups. And so the Fundamental Theorem of Galois Theory predicts that there are five intermediate fields for K/Q which have degree 4 over Q and exactly one of those intermediate fields will be a Galois extension of Q. The four roots of x 4 2 in C are: θ 1 = 4 2, θ 2 = i 4 2, θ 3 = 4 2, θ 4 = i We can regard these four roots as the vertices of a square in the complex plane C. The polynomial x 4 2 is irreducible over Q and is the minimal polynomial over Q for each of the four roots. As discussed in class, we can identify Gal(K/Q) with a subgroup of S 4 of order 8, namely the dihedral subgroup which permutes the vertices of the above square. It will be convenient to denote the elements of Gal(K/Q) by the corresponding permutation of the vertices. Note that if H is any subgroup of G of order 2, then the fixed field F = K H has degree 4 over Q. If we can find an element β F such that [Q(β) : Q] = 4, then it follows that F = Q(β). We will use that observation below. Let F 1 = Q(θ 1 ) and F 2 = Q(θ 2 ). We know that those fields have degree 4 over Q. Those fields are the fixed fields for the subgroups H 1 generated by the reflection (2 4) and H 2 generated by the reflection (1 3), respectively. We are using the observation in the previous paragraph. Now consider α = θ 1 + θ 2 K. That element is fixed by the subgroup H 3 of G generated by the reflection (1 2)(3 4). We have α = θ 1 + θ 2 = (1 + i) 4 2 and α 4 = 2(1 + i) 4 = 8 and (2/α) 4 = 2. Let F 3 = Q(α). We have F 3 K H 3. Let β = 2/α. Notice that F 3 = Q(β) and that β is a root of x The Eisenstein Criterion for p = 2 implies that x is irreducible over Q and hence is the minimal polynomial for β over Q. Thus [F 3 : Q] = 4 and therefore F 3 must be the fixed field for the subgroup H 3 of G. Let H 4 be the subgroup generated by the remaining reflection (1 4)(2 3). Consider the element γ = θ 1 + θ 4 of K. This element is fixed by H 4. Let F 4 = Q(γ). We have F 4 K H 4. Now γ = θ 1 + θ 4 = (1 i) 4 2 and γ 4 = 2(1 i) 4 = 8 and (2/γ) 4 = 2. Just as for F 3, it follows that F 4 = Q(γ) = Q(δ), where δ = 2/γ is a root of x As above, [F 4 : Q] = 4 and it follows that F 4 = K H 4.

3 The four fields F 1, F 2, F 3, and F 4 are distinct because the subgroups H 1, H 2, H 3, and H 4 are distinct. None of those subgroups of G is a normal subgroup and hence none of those fields are Galois extensions of Q. It remains to find the one intermediate field of degree 4 over Q which is Galois over Q. Notice that K contains 2 and i. Hence K contains F 5 = Q(i, 2). Actually, this field is Q(ζ 8 ). We easily verify that [F 5 : Q] = 4. It is also clear that F 5 is Galois over Q because it is the splitting field over Q for (x 2 2)(x 2 + 1) (or alternatively for x 8 1). Thus, F 5 is the unique intermediate field for K/Q which has degree 4 over Q and is Galois over Q. C: Suppose that r Q. Let β = cos(rπ). Prove that β is algebraic over Q. Let K = Q(β). Prove that Q(β) is a Galois extension of Q and that Gal(K/Q) is an abelian group. SOLUTION: Let n be the denominator of that rational number r/2. Then r = 2k/n, where k is an integer. Let ω = cos(2π/n) + sin(2π/n)i. Thus, ω is a root of the polynomial x n 1. Therefore, Q(ω) is a finite extension of Q. Let M = Q(ω). In fact, M contains all of the roots of x n 1 and hence M is the splitting field over Q for x n 1. Thus, M is a finite, Galois extension of Q. Note that M contains ω k + ω k = ( cos(2kπ/n) + sin(2kπ/n)i ) + ( cos(2kπ/n) sin(2kπ/n)i ) and therefore β M. = 2cos(2kπ/n) = 2cos(rπ) = 2β Since M is a finite extension of Q and β M, it follows that β is algebraic over Q. We are using proposition 1 on the handout about Field Extensions and their Degrees. Thus, Q K M. Recall that Gal(M/Q) is abelian (as proven in class for Gal(Q(ζ n )/Q) for any positive integer n). It follows that every subgroup of Gal(M/Q) is a normal subgroup. In particular, Gal(M/K) is a normal subgroup of Gal(M/Q). This implies that K is a Galois extension of Q. Furthermore, we have Gal(K/Q) = Gal(M/Q) / Gal(M/K). Since Gal(M/Q) is an abelian group, the quotient group Gal(M/Q) / Gal(M/K) must also be abelian. Hence Gal(K/Q) is indeed an abelian group. D: Let K = Q(ω), where ω = cos( 2π 2π ) + sin( )i. Prove that K contains a unique subfield L such that [L : Q] = 8. Prove that L is a Galois extension of Q. Find an element β L such that L = Q(β).

4 SOLUTION. Note that 17 is a prime. We proved in class that there is a group isomorphism χ 17 : Gal(K/Q) ( Z/17Z ). However, one can check easily that ( Z/17Z ) is a cyclic group of order 16. (To see this, one can verify that Z is an element of ( Z/17Z ) of order 16.) A cyclic group of order 16 will have a unique subgroup H of order 2. Thus, Gal(K/Q) has a unique subgroup of order 2 and therefore a unique subgroup of index 8. Thus, there will be a unique intermediate field L for K/Q such that [L : Q] = 8. As proven in class, any subfield of K contains Q and hence is an intermediate field for the extension K/Q. Note that Gal(K/Q) is a cyclic group of order 16 and hence is abelian. Therefore, every subgroup of Gal(K/Q) is a normal subgroup. In particular, Gal(K/L) is a normal subgroup of Gal(K/Q). As proven in class, it follows that L/Q is indeed a Galois extension. Finally, consider β = ω + ω 1 = 2cos(2π/17). Then β K. Let M = Q(β). Thus, M is a subfield of K. Also, β R and hence M R. Since K R, we have M K. Hence [K : M] 2. Now ω is a root of the following polynomial f(x) = (x ω)(x ω 1 ) = x 2 (ω + ω 1 )x + ωω 1 = x 2 βx + 1 Note that f(x) M[x] and that K = M(ω). It follows that [K : M] 2. Since we also have [K : M] 2, it follows that [K : M] = 2. Therefore, M is a subfield of K and [M : Q] = [K : Q] / [K : M] = 16/2 = 8. Since L is the unique subfield of K which has degree 8 over Q, we must have M = L. Thus, we have L = Q(β) = Q ( cos(2π/17) ). E: Prove the existence of a Galois extension K of Q such that Gal(K/Q) = Z/14Z. SOLUTION. Consider the splitting field M for x p 1 over Q, where p is a prime. As shown in class, we have group isomorphism χ : Gal(M/Q) (Z/pZ). Let p = 29. The group (Z/29Z) is a cyclic group of order 28. It is generated by Z. To verify this, it suffices to show that (mod 29) and (mod 29), which we

5 leave to the reader. It then follows that the order of Z divides 28, but does not divide 4 or 14. The order of Z must then be 28. Let G = Gal(M/Q). As explained above, G is a cyclic group of order 28. If H is the unique subgroup of G of order 2, then G/H is a cyclic group of order 14 and hence is isomorphic to Z/14Z. Thus, if K = M H, then we have Gal(K/Q) = G/H and hence Gal(K/Q) is isomorphic to Z/14Z. F: Prove that 2 Q( 3, 5). Let K = Q( 2, 3, 5). Prove that K is a Galois extension of Q, that [K : Q] = 8, and that Gal(K/Q) = Z/2Z Z/2Z Z/2Z. SOLUTION. Let F = Q( 3, 5). We know from previous homework problems that [F : Q] = 4 and that Gal(F/Q) = Z/2Z Z/2Z. Thus, Gal(F/Q) has exactly three elements of order 2 and hence exactly three subgroups of order 2. Those subgroups are the subgroups of Gal(F/Q) which have index 2. Therefore, by the Fundamental Theorem of Galois Theory, there must be exactly three intermediate fields for the extension F/Q which have degree 2 over Q. It is easy to find them because F contains 3, 5, and 15. The three fields in question are Q( 3), Q( 5), Q( 15). Those fields are distinct because 3 5, 3 15, and 5 15 are not perfect squares. We are using a criterion discussed in class. We can now prove that 2 F. Assume to the contrary that 2 F. Then F contains Q( 2). Thus, Q( 2) must be one of the above three fields. Using the same criterion discussed in class, we see that this is not so. The reason is that none of the following numbers are perfect squares: 2 3, 2 5, Now K = F ( 2). The polynomial x 2 2 is irreducible over F because it has degree 2 and none of its roots are in F, as shown above. Thus, the minimal polynomial for 2 over F must be x 2 2. Hence [F ( 2) : F ] = 2. Therefore, [K : Q] = [K : F ][F : Q] = 2 4 = 8. Furthermore, K/Q is a Galois extension since K is the splitting field over Q for the polynomial (x 2 2)(x 2 3)(x 2 5). Also, as explained in class, Gal(K/Q) has exponent 2. That group has order 8. It follows from group theory that Gal(K/Q) = Z/2Z Z/2Z Z/2Z

6 as stated. G. Suppose that K is a finite Galois extension of Q and that Gal(K/Q) = S 3. Prove that there exists a polynomial g(x) Q[x] such that g(x) has degree 3 and K is the splitting field for g(x) over Q. SOLUTION: Let G = Gal(K/Q). The group S 3 has three subgroups of order 2. None of those subgroups is a normal subgroup of S 3. Since G = S 3, the same statements are true about G. Let H be one of the subgroups of G of order 2. Then H is not a normal subgroup of G. Let L = K H. Then [K : L] = H = 2 and [L : Q] = [G : H] = 3. Thus L is a finite extension of Q of degree 3. Furthermore, since H is not a normal subgroup of G, L cannot be a Galois extension of Q. By the Primitive Element Theorem, L = Q(θ) for some θ L. (Actually, we can take θ to be any element of L which is not in Q because [L : Q] is a prime.) Let g(x) be the minimal polynomial for θ over Q. Then g(x) Q[x] and deg ( g(x) ) = [L : Q] = 3. Furthermore, as proved in class, all three roots θ 1, θ 2 and θ 3 of g(x) in C are in the field K. We can assume that θ 1 = θ. Let M = Q(θ 1, θ 2, θ 3 ). Then M is the splitting field over Q for g(x). Note that L M K and that M/Q is a finite Galois extension. Hence L M. But 2 = [K : L] = [K : M][M : L] is a prime and [M : L] > 1. It follows that [M : L] = 2 and that K = M. Therefore K is the splitting field over Q for the polynomial g(x). H: Suppose that E and F are finite extensions of Q and that E F = Q. By the Primitive Element Theorem, we know that E = Q(α) and F = Q(β) for certain elements α E and β F. Let K = Q(α, β). TRUE OR FALSE: [K : Q] = [E : Q][F : Q]. If true, give a proof. If false, give a counterexample. SOLUTION: The statement is false. We will give a counterexample. Let α and β be two distinct roots in C of the polynomial x 3 2. Let E = Q(α) and F = Q(β). We have discussed these fields early in this course. We know that α and β both have x 3 2 as their minimal polynomial over Q. Thus, [E : Q] = [F : Q] = 3. Furthermore, we verified in class that E F. Let L = E F. Then L is a subfield of E and hence [L : Q] must divide [E : Q] = 3. However, we cannot have [L : Q] = 3. To see this, note that if [L : Q] = 3,

7 then we must have L = E. It would then follow that E F. Since [E : Q] = [F : Q] = 3, it would follow that E = F. As mentioned above, this is not so. Therefore, we have shown that [L : Q] divides 3 and [L : Q] 3. Therefore, [L : Q] = 1 and hence L = Q. We have shown that E F = Q. Let K be the splitting field for x 3 2 over Q. In fact, K = Q(α, β). This is clear because if γ is the third root of x 3 2 in C, then (α + β + γ) is the the coefficient of x 2 in the polynomial x 3 2 and hence is in Q. (Of course, that coefficient is actually 0 in this example.). Hence γ Q(α, β). Observe that [K : Q] = 6 and [E : Q][F : Q] = 3 3 = 9 6 and so we have a counterexample to the statement in the problem.