NAP 2017. Module 4. Homework 2 Friday, July 14, 2017. These excercises are due July 21, 2017, at 10 pm. Nepal time. Please, send them to nap@rnta.eu, to laurageatti@gmail.com and schoo.rene@gmail.com. Contact us i you have any question! 1. Let p be a prime and let be an irreducible degree n polynomial in F p [X]. Show that the Galois group o is contained in the alternating group A n i and only i n is odd. Sol.: The Galois group o the splitting ield o is a cyclic group o order n, generated by the Frobenius automorphism. It can be identiied with the group (12... n) in S n. One has that (12... n) is even, and thereore contained in A n, i and only i n is odd. 2. Let H be a transitive subgroup o the symmetric group S n. Suppose that H contains a 2-cycle and an (n 1)-cycle. Show that H = S n. (See Milne Lemma 4.32) 3. Determine the Galois groups over Q o the polynomials (they are all irreducible) x 4 10x 2 + 1, x 4 8x 2 + 3, x 4 2x 2 + 25. Sol.: (a) Since (x) = x 4 10x 2 + 1 irreducible over Q, its Galois group is a transitive subgroup o S 4 and is contained in A 4 (one has disc() = 147456 = (384) 2, which is a square). Its resolvent cubic is g(x) = x 3 + 10x 2 4x 40 = (x 2)(x + 2)(x + 10) is completely reducible over Q. By the classiication, G = V 4. Remarks. The resolvent cubic g is completely reducible over Q: consequently Q g = Q. On the other hand, to each o the three subgroups o V 4 H 1 = 1, (12)(34), H 2 = 1, (13)(24), H 3 = 1, (14)(23) there is associated an intermediate quadratic extension o Q Denote by α 1 = Q Q H i Q, i = 1, 2, 3. 5 + 2 6, α 2 = 5 + 2 6, α 3 = 5 2 6, α 4 = 5 2 6 the our roots o and by the roots o g. α := α 1 α 2 + α 3 α 4, β := α 1 α 3 + α 2 α 4, γ := α 1 α 4 + α 2 α 3, Consider Q(α 1 α 2 ) = Q( 6). This is clearly an H 1 -invariant quadratic extension o Q, contained in Q = Q( 5 + 2 6, 5 2 6). More precisely, Q = Q( 5 + 2 6), since (5 + 2 6)(5 2 6) = 1, which is a square in Q and thereore in Q( 6). 1
Next consider the quantities (α 1 +α 3 ) and (α 2 +α 4 ), which are H 2 -invariant and satisy the degree two equation Y 2 +(α+γ) = Y 2 12 = 0. This means that Q( 3) is an H 2 -invariant quadratic extension o Q, contained in Q = Q( 5 + 2 6) (note that (5+2 6)(5 2 6) = 1, which is a square in Q, is also a square in Q( 3)). Finally consider the quantities (α 1 + α 4 ) and (α 2 + α 3 ), which are H 3 -invariant and satisy the degree two equation Y 2 + (α + β) = Y 2 8 = 0. This means that Q( 2) is an H 3 -invariant quadratic extension o Q, contained in Q = Q( 5 + 2 6) (note that (5 + 2 6)(5 2 6) = 1, which is a square in Q, is also a square in Q( 2)). Q = Q( 5 + 2 6) = Q( 6) Q H 2 = Q( 3) Q H 3 = Q( 2) Q g = Q Q H 1 (c) The discriminant o (x) = x 4 2x 2 + 25 is equal to 3686400 = (1920) 2, hence G is a transitive subgroup o S 4, contained in A 4. The resolvent cubic o is g(x) = x 3 + 2x 2 100x 200 = (x + 10)(x 10)(x + 2), which is completely reducible over Q. By the classiication, G = A 4. As in the previous case we have a diagram as ollows: Q = Q( 5 + i2 6, 5 i2 6) = Q( 5 + i2 6) Q H 1 = Q(i 6) Q H 2 = Q(i 3) Q g = Q Q H 3 = Q(i 2) (b) The discriminant o (x) = x 4 8x 2 + 3 is equal to 129792, which is not a square. Hence G is a transtive subgroup o S 4, not contained in A 4. Its resolvent cubic is g(x) = x 3 + 8x 2 12x 96 = (x + 8)(x 2 12). In this case we have to decide whether G = C4 or G = D4 ; equivalently whether the degree [Q : Q g ] is 2 or 4. Denote by α 1 = 4 + 13, α 2 = 4 + 13, α 3 = 4 13, α 4 = 4 13 the our roots o and by α := α 1 α 2 + α 3 α 4 = 8, β := α 1 α 3 + α 2 α 4 = 2 3, γ := α 1 α 4 + α 2 α 3 = 2 3, 2
the roots o g. We have Q = Q( 4 + 13, 4 13) and Q g = Q( 3). The ields Q((α 1 +α 3 ), (α 2 +α 4 )) and Q((α 1 +α 4 ), (α 2 +α 3 )) are intermediate extensions: they are between Q g and Q. One computes that Q((α 1 +α 3 ), (α 2 +α 4 )) = Q( 8 + 2 3) and Q((α 1 + α 4 ), (α 2 + α 3 )) = Q( 8 2 3). The numbers 8 ± 2 3 are not squares in Q( 3), because their norms are equal to 52, which is not a square in Q. This shows that both ields are degree 2 extensions o Q( 3). To see that they are distinct, we observe that the product (8+2 3)(8 2 3) = 52 is not a square in Q( 3) either. Indeed, the equation (x + y 3) 2 = 52 has no solutions x, y Q. This proves that G cannot be isomorphic to C 4, but it is necessarily isomorphic to D 4. 4. Let = x 5 x + 3 Z[X]. This is an irreducible polynomial. (a) Show that has three linear actors modulo 3. (b) Show that is irreducible modulo 5. (c) Show that the Galois group o over Q is S 5. Sol.: (a) In F 3 [x], the polynomial becomes where x 2 + 1 is an irreducible actor. (x) = x 5 x = x(x 1)(x + 1)(x 2 + 1), (b) In F 5 [x] the polynomial is irreducible: it has no linear actors, nor degree 2 actors... (c) By (a) and (b), the Galois group G o over Q contains a 2-cycle and a 5-cycle. Now Lemma 4.32 in [Milne] ensures that G = S5. 5. Let g = x 5 + 8x + 3 Z[X]. This is an irreducible polynomial. (a) Show that has three linear actors modulo 3. (b) Show that is the product o a linear polynomial and an irreducible polynomial o degree 4 modulo 2. (c) Show that the Galois group o over Q is S 5. Sol.: (a) In F 3 [x] the polynomial g becomes g(x) = x 5 x = x(x 1)(x + 1)(x 2 + 1), where x 2 + 1 is an irreducible actor. (b) In F 2 [x] the polynomial g becomes g(x) = x 5 1 = (x 1)(x 4 + x 3 + x 2 + x + 1). The degree 4 actor is irreducible because 2 is a primitive root in F 5 (c. Excercises 1, n.6). (c) By (a) and (b), the Galois group G o over Q contains a 2-cycle and a 4-cycle. Now Lemma 4.32 in [Milne] ensures that G = S5. 6. Let K be a ield. Show that the Galois group o X n 1 over K is commutative (Hint: without loss o generality one may assume that the characteristic o K does not divide n. Show that any automorphism σ o the splitting ield K o X n 1 is determined by σ(ζ), where ζ is a primitive n-th root o unity in K ). 3
Sol.: Assume that char(k) = p, with p prime and that n = mp r, with gcd(p, m) = 1. Then X n 1 = X mpr 1 = (X m 1) pr. This means that the splitting ield o X n 1 is the same as the splitting ield o (X) = X m 1. Hence, without loss o generality we may assume that the characteristic o K does not divide n. The roots o orm a inite and hence cyclic subgroup o K. So they are ξ, ξ 2,..., ξ m = 1, where ξ is a generator. Hence K = K(ξ). Any automorphism σ Aut(K /K) is determined by σ(ξ) (being an automoprhism implies σ(ξ k ) = σ(ξ) k ) and must be o the orm σ(ξ) = ξ r, or some r Z. Since also ξ r must be a primitive root o 1, then gcd(r, m) = 1. It ollows that the map that sends σ to r is a well deined group homomorphism ψ : G Z n Since ψ is injective, it ollows that G is abelian. 7. (Optional) The Möbius unction µ : N { 1, 0, +1} is deined by { ( 1) µ(n) = r ; i n is a product o r distinct primes, 0; otherwise. (a) Compute µ(10), µ(20) and µ(30). (b) Show that µ is multiplicative, i.e. show that µ(nm) = µ(n)µ(m) i gcd(n, m) = 1. (c) Let (n) = d n µ(d). Here the summation runs over the positive divisors d o n N. Show that is also a multiplicative unction. (d) (Möbius inversion) Suppose that the sequences a n, b n satisy a n = d n b d or all n 1. Show that b n = d n µ( n d )a d Sol.: (a) 10 = 2 5 and µ(10) = ( 1) 2 = 1; 20 = 2 2 5 and µ(20) = 0; 30 = 2 3 5 and µ(30) = ( 1) 3 = 1. (b) I either n or m contains a square, then so does nm and µ(nm) = 0 = µ(n)µ(m). I both n and m are square ree, then nm is square ree i and only i gcd(n, m) = 1. In this case the prime actors o nm are the disjoint union o the prime actors o n and those o m, implying that µ(nm) = µ(n)µ(m). (c) I gcd(n, m) = 1, then the divisors d o nm are o the orm d 1 d 2, where d 1 is a divisor o n and d 2 is a divisor o m, and gcd(d 1, d 2 ) = 1. Then, rom µ(d 1 d 2 ) = µ(d 1 )µ(d 2 ), we obtain (nm) = µ(d) = µ(d 1 d 2 ) = µ(d 1 ) µ(d 2 ). d nm d 1 n d 2 m d 1 d 2 nm d 1 n, d 2 m (d) We evaluate d n µ( n d )a d. It is equal to µ( n d ) d n e d b e. Changing the order o summation we get µ( n d )b e. e n e d n 4
In the second sum, d runs over the multiples o e that divide n. Writing d = d/e, this becomes a sum over the divisors d o n/e. We get µ( n/e e n d d ) b e. n e Since d m µ( n d ) = d m µ(d) is equal to 1 or m = 1 and zero otherwise, we wee that the second sum only gives a non-zero contribution or e = n. In other words, the sum is equal to b n as required. (By part (c) it suices to check that d m µ(d) is zero, or m > 1 a power o a prime) 8. (Optional) Let p be a prime and let N n denote the number o irreducible polynomials in F p [X] o degree n. (a) Compute N 4 and N 6 or any inite ield F p. (b) Show that d n dn d = p n or every n 1. (c) Show that N n = 1 n d n µ( n d )pd. (Use previous exercise) Sol.: We can count irreducible polynomials o degree n in F p [X] by counting the elements o F p n whose minimum polynomial has degree n and dividing the result by n (every such polynomial has n zeros in F p n). Recall that the elements o F p n whose minimum polynomial has degree less than n are those lying in proper subields o F p n, namely in ields F p d, with d n. (a) N 4 = 1 4 (#F p 4 #F p 2) = 1 4 (p4 p 2 ); N 6 = 1 6 (#F p 6 #F p 2 #F p 3 + #F p) = 1 6 (p6 p 2 p 3 + p); (b) The elements o F p n can be subdivided according to the degree d o their minimum polynomial (necessarily a divisor o n). The number o elements in F p n whose minimum polynomial has degree d is d times the number o irreducible polynomials o degree d in F p [X]. Hence p n = dn(d). d n (c) Let a n = p n and b n = nn(n). By part (b) we have a n = d n b d and hence by the previous exercice b n = d n µ( n d )a d. 5