The University of Hong Kong DEPARTMENT OF MATHEMATICS MATH3302/MATH4302 Algebra II Suggested Solution to Assignment 2 In the following, E, F, K are fields. 1. Let G be an abelian group and a, b ∈ G with ord (a) = m, ord (b) = n. (a) Prove that: i. ii. iii. iv. Let gcd(m, n) = 1. Then ord (ab) = mn. Let r|m with postive integers r. Then ord (ar ) = m r. There exists c ∈ G such that ord (c) = lcm (m, n). Suppose for all g ∈ G, ord (g) ≤ m. Then for all g ∈ G, ord (g)|m. (b) Let G be a finite subgroup of the multiplication group F ∗ := F − {0} of a field F . Prove that G is cyclic. (c) Concluding a finite extension of a finite field must contain a primitive elements. Solution: (a) i. Let t = ord (ab). On the one hand, (ab)mn = amn bmn = e ⇒ t|m. On the other hand, (ab)tn = atn btn = atn = e ⇒ m|tn. Since gcd(m, n) = 1, m|t. Similarly, we also have n|t. Thus mn|t. Therefore mn = t. m ii. Write k = ord (ar ). On the one hand (ar ) r = e ⇒ k| m r ⇒ r )k = e ⇒ m|rk ⇒ m ≤ rk. k≤ m . One the other hand (a r Thus k = m r. iii. By unique factorization of integers, m = pk11 · · · pknn , n = k kj 1 ···p jn−m p p si 1 ···psim jn−m ps11 · · · psnn . Set a0 = a j1 , b0 = b i1 im where {i1 , . . . , im } = {i, ki ≥ si }, {j1 , . . . , jn−m } = {j, kj < sj }. Then ord (a0 ) = ki sj k sj n−m pi1 1 · · · pimim , ord (b0 ) = pj11 · · · pjn−m , so that gcd(ord (a0 ), ord (b0 )) = 0 0 1. Let c = a b . We have ord (c) = ord (a0 )ord (b0 ) = lcm (m, n). iv. For any g ∈ G, write ord (g) = kg . Since ∃c ∈ G such that m ≤ lcm (m, kg ) = ord (c) ≤ m. Thus kg |lcm (m, kg ) = m. (b) Let u ∈ G − {e} be an element of maximal order. We claim that G =< u >. Write ord (g) = kg and ord (u) = m. For any g ∈ G, kg |m. Thus g is a root of f (x) := xm − e. Consider f (x) ∈ A[x] for some algebraically closed field A, we see by the 1 factor theorem that (x−g) is a factor of f (x) for any g ∈ G. Thus h(x) = Πg∈G (x − g) is a factor of xm − e so that |G| = deg(h) ≤ deg(f ) = m. Hence |G| = m. (c) If [Fq : Fp ] = n, then |Fq | = pn . Then F∗q =< u > for some u ∈ F∗q . Thus Fq = Fp (u). 2. (a) Prove that FLT(F ) = ax + b a, b, c, d ∈ F, ad − bc 6= 0 cx + d is a group with function composition. (b) Prove that ϕ : GL(2, F) → FLT(F ) a b ax+b 7→ cx+d c d is a surjective group homomorphism. (c) Find Ker (ϕ). (d) Define PGL(2, F ) = GL(2, F )/Ker (ϕ). Prove that PGL(2, F ) is isomorphic to FLT(F ). (e) Find a subgroup of order 6 of FLT(F ). Solution: (a) The closure of operation follows from the condition that ad−bc 6= 0. The identity element is the identity function x and for each ax+b −dx+b cx+d , the inverse is cx−a . The associativity follows from the associativity of the composition of functions. (b) Sujectivity follows from the condition ad−bc 6= 0. To show that ϕ is homomorphism, let A = (aij ), B = (bij ), C = (cij ) ∈ GL(2, F ), P2 where cij = k=1 aik bkj . Then P2 P a b x+ 2 a1k bk2 P2k=1 1k k1 Pk=1 , 2 k=1 a2k bk1 x+ k=1 a2k bk2 a11 x+a12 P P2 b11 a x+a +b12 a1k bk1 x+ 2k=1 a1k bk2 21 22 P2 = Pk=1 . a11 x+a12 2 b21 a x+a +b22 k=1 a2k bk1 x+ k=1 a2k bk2 P ϕ(AB) = ϕ((cij )) = ϕ( 2k=1 aik bkj ) = ϕ(A)ϕ(B) = a11 x+a12 a21 x+a22 ◦ b11 x+b12 b21 x+b22 = 21 (c) Ker ϕ = {(aij ) ∈ GL(2, F ) | a11 x+a12 a21 x+a22 22 = x} = {λI | λ ∈ F \ {0}}. (d) By the fundamental theorem of homomorphism, GL(2, F )/Ker ϕ ∼ = ϕ(GL(2, F )). Since ϕ is surjective, ϕ(GL(2, F )) = FLT(F ). 2 1 x (e) {x, x1 , 1 − x, 1−x , 1 − x1 , x−1 }. 3. Let x be non-algebraic over F . 1 (a) Show that F (x) = F ( x1 ) = F (1 − x) = F (1 − x1 ) = F ( 1−x ) = x F ( x−1 ). (b) Let σ1 : g(x) 7→ g(x), σ2 : g(x) 7→ g( x1 ), σ3 : g(x) 7→ g(1 − x), σ4 : 1 1 x g(x) 7→ g( 1−x ), σ5 : g(x) 7→ g( 1−x ), σ6 : g(x) 7→ g( x−1 ) for g(x) ∈ F (x). Show that for i = 1, . . . , 6, σi are automorphisms of F (x) 2 −x+1)3 (ie., σi (I(x)) = I(x)). over F fixing I(x) := (x x2 (x−1)2 ) (c) Show that [F (x) : F (I(x))] ≤ 6. (d) Show that there are only 6 distinct automorphisms of F (x) over F (I(x)). (e) Hence show that F (x)/F (I(x)) is separable. Solution: (a) It is easy to see that F (x) = F ( x1 ) = F (1 − x) and {x, x1 , 1 − x} generates the rest. (b) The check for σi , i = 1, 2, 3 is straight forward. The rest follows from that fact that they are generated by σi , i = 1, 2, 3. (c) Consider f (t) = (t2 − t + 1)3 − I(x)(t2 (t − 1)2 ) ∈ F (I(x))[t]. As x is a root of f , [F (x) : F (I(x))] ≤ deg(f (t)) = 6. (d) Since 6 = |{σi , i = 1, . . . , 6}| ≤ |Aut (F (x)/F (I(x))| ≤ [F (x) : F (I(x)] ≤ 6. This forces |Aut (F (x)/F (I(x))| = 6. (e) Note that σi ’s are actually embeddings of F (x) over F (I(x)). Thus |Emb (F (x)/F (I(x))| = [F (x) : F (I(x))]. 4. Prove that the following two statements are equivalent: (a) F is a separable field. (b) Char (F ) = 0; or Char (F ) = p > 0 and for all a ∈ F , there exists b ∈ F such that bp = a. Solution: The case that Char F = 0 is clear. We only need to consider Char F = p > 0. 3 Suppose F is separable. Let a ∈ F and consider f (x) = xp − a ∈ F [x]. Then in some extension field E over F , we can find some α ∈ E such that a = αp . Then in f (x) = xp −αp = (x−α)p ∈ E[x]. Since minimal polynomials are irreducible, we have MinPoly F (α)|f (x). Since F is separable, any irreducible polynomial in F [x] has no multiple roots. Thus MinPoly F (α) = x − α and α ∈ F . Conversely, suppose every element in F has a p-th root but F is not separable, then ∃u ∈ F such that MinPoly F (u) has a multiple root. It follows that MinPoly F (u) is a polynomial in xp . Writting the coefficients of MinPoly F (u) as p-th roots, one notice that n MinPoly F (u) = (x − u)p . This is a contradiction to the irreducibility of MinPoly F (u). 5. (a) Let K/E and E/F be separable. Prove that K/F is separable. (b) Let K/F be separable, and E/F any extension. KE/E is separable. Prove that (c) Let K/F and E/F are separable. Prove that KE/F is separable. Solution: Define |K|F to be the number of distinct embeddings of K over F . Note that for a finite extension K/F , it is separable if and only if |K|F = [K : F ]. (a) |K|F = |K|E |E|F = [K : E][E : F ] = [K : F ]. (b) Since K/F is finite and separable, we may assume K = F (α) for some α ∈ K. Then KE = E(α) and |KE|E = |α|E . Note that K/F is separable implies that α is separable over F and hence separable over E. Thus |KE|E = |α|E = [E(α) : E] = [KE : E]. (c) [KE : F ] = [KE : E][E : F ] = |KE|E [E : F ] = |KE|E |E|F = |KE|F . 6. (a) Let K/F be algebraic. Show that any σ ∈ Emb (K/F ) induces an automorphism of K. (b) Give an example σ ∈ Emb (K/F ) such that σ : K → K over F does not induce an automorphism of K (ie., σ(K) 6= K). Solution: (a) If u ∈ K and all the distinct roots of MinPoly F (u) are S := {α1 , . . . , αr }, then the embedding σ is injective on the finite set S → S. Thus σ is also surjective on S. This means that σ is surjective on K. 4 (b) Let σ : Q(π) → Q(π), f (π) 7→ f (π 2 ). Then σ is homomorphism √ √ but [Q( π) : Q(π)] = 2. Thus π ∈ / Q(π), which implies that π has no inverse image, ie., π ∈ / σ(Q(π)). Hence σ is not surjective. 5
© Copyright 2024