Solutions for Field Theory Problem Set 5

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Amberly Burns
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1 Solutions for Field Theory Problem Set 5 A. Let β = i. Let K = Q(β). Find all subfields of K. Justify your answer carefully. 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). Furthermore, it was pointed out in class that Gal(K/Q) = U(Z/8Z) and that this group is isomorphic to the Klein 4-group. Thus, every element in Gal(K/Q) has order 2, except for the identity element. Therefore, Gal(K/Q) has exactly three elements of order 2 and hence exactly three subgroups of order 2. Therefore, by the Fundamental Theorem of Galois Theory, it follows that K contains exactly three subfields L such that [K : L] = 2. Those are the subfields L of K such that [L : Q] = [K : Q] [K : L] = 4 2 = 2. The three subfields L 1 = Q( 2), L 2 = Q(i), L 3 = Q( 2i) = Q( 2) are distinct (see below) and hence must be all of the subfields of K of 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. One could prove directly that the three fields L 1, L 2, and L 3 are distinct. But we can also use the following lemma which was proved in the solution for problem B in problem set 3. We will give the proof again. Lemma: Suppose that d 1 and d 2 are nonzero integers. Then Q( d 1 ) = Q( d 2 ) if and only if d 1 d 2 is the square of an integer. Proof. One direction is straightforward. Suppose that d 1 d 2 = n 2, where n Z. We are assuming that d 1 and d 2 are nonzero. Any subfield of C containing d 1 will also contain n( d 1 ) 1 and hence will contain d 2. Similarly, any subfield of C containing d 2 will also contain d 1. All subfields of C contain Q. Thus, assuming that d 1 d 2 is the square of an integer, it follows that Q( d 1 ) = Q( d 2 ). Now assume that Q( d 1 ) = Q( d 2 ). This field coincides with Q if and only if x 2 d 1 and x 2 d 2 are reducible over Q. This means that those polynomials have a root in Q. By

2 the rational root test, any rational root of those polynomials will have denominator dividing 1. Hence those roots will be integers and both d 1 and d 2 will be squares of integers. Thus, their product will be a square of an integer. We must now consider the case where Q( d 1 ) = Q( d 2 ) and this field is an extension of Q of degree 2. Let K denote that field. Then K is the splitting field for the polynomial x 2 d 1 over Q and [K : Q] = 2. Hence Aut(K/Q) is a group of order 2. Let σ denote its nontrivial element. For d = d 1 or d = d 2, σ( d) must be a root of x 2 d different from d itself. Hence we must have σ( d) = d. Thus, we have σ( d 1 ) = d 1, σ( d 2 ) = d 2. Let η = d 1 d2. Then η K. Furthermore, σ(η) = σ( d 1 d2 ) = σ( d 1 )σ( d 2 ) = ( d 1 )( d 2 ) = d 1 d2 = η. We will show that η Q. To see this, assume to the contrary that η Q. Then Q(η) is a subfield of K and [Q(η) : Q] > 1. It would follow that [Q(η) : Q] = 2 and that Q(η) = K. Since σ is determined by σ(η) and σ(η) = η, it follows that σ is the identity map. This contradicts the fact that σ is the nontrivial element in Aut(K/Q). It follows that η Q. Note that η 2 = d 1 d 2. Thus η is a root of the polynomial x 2 d 1 d 2. By the rational root test, the denominator of η divides 1. Thus, η is actually an integer. Hence, assuming that Q( d 1 ) = Q( d 2 ), it follows that d 1 d 2 is the square of an integer. We have proved the lemma. The lemma tells us that L 1, L 2, and L 3 are distinct because 2 ( 1), 2 ( 2), and ( 1) ( 2) are not squares in Z. B. Let f(x) = x 4 + x 3 + x 2 + x + 1. Let ω be a root of the polynomial f(x) in C. Let K = Q(ω). Find all subfields of K. Justify your answer carefully. SOLUTION. We proved in class that [K : Q] = 4 and that Gal(K/Q) is a cyclic group of order 4. Thus, Gal(K/Q) has exactly three subgroups, namely the trivial subgroup, a subgroup of order 2, and Gal(K/Q) itself. Therefore, by the Fundamental Theorem of Galois Theory, there are exactly three distinct intermediate fields for the extension K/Q. Every subfield of K will be one of those three fields. We also showed in class that K contains L = Q( 5). Some other subfields of K are K itself and Q. These subfields of K, namely the fields Q, L, and K, are clearly distinct and hence must be the three distinct subfields of K.

3 C. Let K = Q(ω), where ω = cos( 2π 17 )+sin(2π )i. Prove that K contains a unique subfield 17 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). SOLUTION. We proved in class that there is a group isomorphism Gal(K/Q) = U(Z / 17Z). However, one can check easily that U(Z / 17Z) is a cyclic group of order 16. (To see this, verify that 3+17Z is an element of U(Z / 17Z) of order 16.) A cyclic group of order 16 will have a unique subgroup H of order 2. Thus, there will be a unique intermediate field L for K/Q such that [K : L] = 2. Thus, [L : Q] = [K : Q] [K : L] = 16 2 = 8. Thus, it is clear that L is the unique intermediate field with [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 abelian and hence every subgroup of Gal(K/Q) is a normal subgroup of Gal(K/Q). In particular, Gal(K/L) is a normal subgroup of Gal(K/Q). As discussed in class, it follows that L/Q is 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) ). D. Suppose that K is a finite Galois extension of Q and that Gal(K/Q) = S 4. Prove that there exists a polynomial g(x) Q[x] such that g(x) has degree 4 and K is the splitting field for g(x) over Q. SOLUTION. We are given that K is a finite, Galois extension of Q. Let G = Gal(K/Q). By assumption, we have G = S 4. Now S 4 contains a subgroup of order 6, namely { g S 4 g(4) = 4 },

4 a subgroup of S 4 which is isomorphic to S 3. It follows that G has a subgroup H such that H = S 3. We let L denote K H. Note that [L : Q] = [G : H] = G H = 24 6 = 4. Thus, by the Primitive Element Theorem, L = Q(β) for some β C. Since [L : Q] = 4, it follows that β has degree 4 over Q. Let m(x) be the minimal polynomial for β over Q. We have m(x) Q[x] and deg ( m(x) ) = 4. Let M denote the splitting field for m(x) over Q. Since K is a Galois extension of Q and β K, it follows that all of the complex roots of m(x) are also in K. We proved this in class. Furthermore, it follows that M K. Also, L M. Thus, we have L M K. We have Gal(K/L) = H = S 3. Now [K : L] = Gal(K/L) = 6. Therefore, we have [K : L] = 6. By the degree formula, we have [K : M][M : L] = [K : L] = 6. Thus, [K : M] divides 6. Since M is a Galois extension of Q, it follows that Gal(K/M) is a normal subgroup of G = Gal(K/Q). But Gal(K/M) = [K : M] divides 6. Recall that S 4 has normal subgroups only of orders 24, 12, 4, and 1. Since G = S 4, the same statement is true for G. Since Gal(K/M) is a normal subgroup of G and its order divides 6, the only possibility is that Gal(K/M) = 1. Hence Gal(K/M) = {id G }. This means that M = K. That is, the splitting field over Q for m(x) is indeed the field K. E. Suppose that K is a finite Galois extension of Q and that Gal(K/Q) = S 5. Show that K cannot contain 5 2. SOLUTION. Assume to the contrary that 5 2 K. We know that the minimal polynomial for 5 2 over Q is m(x) = x 5 2. As explained in class, since K is a Galois extension of Q which contains one root of m(x), it follows that K contains all of the roots of m(x). Let F be the splitting field for x 5 2 over Q. The above remark shows that F K. We also know that [F : Q] = 20. This was proved in a previous problem set. Furthermore, since F/Q is a Galois extension, it follows that there is a surjective group homomorphism r : Gal(K/Q) Gal(F/Q) and that Ker(r) = Gal(K/F) is a normal subgroup of Gal(K/Q) and has order equal to [K : F] = [K : Q] / [F : Q] = 120/20 = 6.

5 We obtain a contradiction because Gal(K/Q) = S 5 and S 5 has no normal subgroups of order 6. F. 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. Let n bethe denominator of that rational number r/2. Then r = 2k/n, where k is aninteger. 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. Thus, Q K M. Recall that Gal(M/Q) is abelian. (See problem C in problem set 3.) 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.

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