MA 553: Homework 1
Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA
1 Composition series
1.1 First part of Jordan Holder Theorem Theorem 1.1 (Jordan Holder Theorem: Part One). Every finite group has a composition series.
Proof. Let G1 be a maximal normal subgroup of G; then G/G1 is simple. Let G2 be a maximal normal subgroup of Gl, and so on. Since G is finite, this process must end with Gn = 1. Thus G>G1 > ··· >Gn = 1 is a composition series.
1.2 Composition series Proposition 1.1. If G is a finite group and H ⊳ G. Then there is a composition series of G, one of whose term is H. Proof. On one hand, since H is also a finite group, then by Theorem 1.1, H has a composition series H>H1 > ··· >Hn = 1. On the other hand, consider the quotient group G/H. If G/H is not simple, then it has a nontrivial normal subgroup, which gives a group K1 between G and H extending the normal series H ⊳ K1 ⊳ G. Similarly we can consider G/K1. It indicates that we can find a normal series H = K0 ⊳ K1 ⊳ ··· ⊳ Km = G such that for each j ∈ [0, m], the quotient group Hj+1/Hj is simple. Now we find a composition series as
1= Hn < ···
∗E-mail address: [email protected]; Tel: 765 237 7149
1 Yingwei Wang Abstract Algebra
2 Normal subgroup
Proposition 2.1. Let A,B,C be subgroups of a group G with A ⊳ B and C ⊳ G. Then
CA ⊳ CB.
Proof. ∀ c1a ∈ CA, c2b ∈ CB, where c1,c2 ∈ C and a ∈ A, b ∈ B, then
−1 −1 −1 (c2b)(c1a)(c2b) = c2bc1ab c2 −1 −1 −1 = c2(bc1b )(bab )c2 −1 −1 −1 = c2c˜1ac˜ 2 , wherec ˜1 = bc1b ∈ C, a˜ = bab ∈ A, −1 −1 = c2c˜1(˜ac2 a˜ )˜a −1 −1 = c2c˜1c˜2a,˜ wherec ˜2 =˜ac2 a˜ ∈ C,
=˜ca˜ ∈ CA, wherec ˜ = c2c˜1c˜2 ∈ C.
It implies that CA ⊳ CB.
3 Application of Isomorphism Theorem
Proposition 3.1. Let G be a finite group with G =6 1. Suppose G has two composition series
1= N0 < N1 < ··· < Nr = G,
1= M0 < M1 < ··· < Ms = G.
(I) Show for i,j > 0 that Ni−1(Mj−1 ∩ Ni) ⊳ Ni−1(Mj ∩ Ni). (II) Use the Second Isomorphism Theorem to establish isomorphisms
Ni− (Mj ∩ Ni) Mj ∩ Ni (Mj ∩ Ni)Mj− 1 ∼= ∼= 1 . (3.1) Ni−1(Mj−1 ∩ Ni) (Mj−1 ∩ Ni)(Mj ∩ Ni−1) (Mj ∩ Ni−1)Mj−1
Proof. (I) For each i, j, we know that Mj Mj−1 ∩ Ni < Ni, Mj ∩ Ni < Ni. Besides, since Mj−1 ⊳ Mj, then Mj−1 ∩ Ni ⊳ Mj ∩ Ni. 2 Yingwei Wang Abstract Algebra Furthermore, we also know that Ni−1 ⊳ Ni. By the Proposition 2.1, we can conclude that Ni−1(Mj−1 ∩ Ni) ⊳ Ni−1(Mj ∩ Ni). (II) Define the map (Mj ∩ Ni)Mj−1 φ : Mj ∩ Ni → , (Mj ∩ Ni−1)Mj−1 by φ(y)= y(Mj ∩ Ni−1)Mj−1. It is easily seen to be a group homomorphism since Mj−1 ⊳Mj and Nj−1 ⊳Nj. Moreover, (Mj ∩Ni)Mj−1 since Mj ∩ Ni− 6 Mj ∩ Ni, it is easy to see that the image of φ is . 1 (Mj ∩Ni−1)Mj−1 Now we need to prove that ker φ =(Mj−1 ∩ Ni)(Mj ∩ Ni−1). On the one hand, if y ∈ Mj ∩ Ni−1, then φy is the identity coset (Mj ∩ Ni−1)Mj−1; similarly, if y ∈ Mj−1 ∩ Ni, then φy = y(Mj ∩ Ni−1)Mj−1 ⊆ Mj−1(Mj ∩ Ni−1)Mj−1 ⊆ (Mj ∩ Ni−1)Mj−1 since Mj−1 ⊳ Mj. Therefore, (Mj−1 ∩ Ni)(Mj ∩ Ni−1) ⊆ ker φ. On the other hand, if y ∈ ker φ then y ∈ (Mj ∩ Ni−1)Mj−1, so we can write y = ax with a ∈ Mj ∩ Ni−1 and x ∈ Mj−1. Now y ∈ Ni, x ∈ Ni, so x ∈ Ni and therefore y ∈ (Mj−1 ∩ Ni)(Mj ∩ Ni−1), which indicates that ker φ ⊆ (Mj−1 ∩ Ni)(Mj ∩ Ni−1). Then by the first isomorphism theorem, we can know that Mj ∩ Ni (Mj ∩ Ni)Mj− ∼= 1 . (Mj−1 ∩ Ni)(Mj ∩ Ni−1) (Mj ∩ Ni−1)Mj−1 In the same way we can prove another part of Eq.(3.1). 4 Second part of Jordan-Holder Theorem Theorem 4.1 (Jordan Holder Theorem: Part Two). Let G be a finite group with G =6 1. Then the composition factors in a composition series are unique, namely, if 1= N0 < N1 < ··· < Nr = G, 1= M0 < M1 < ··· < Ms = G. are two composition series for G, then r = s and there is some permutation π of {1, 2, ··· ,r} such that ∼ Mπ(i)/Mπi−1 = Ni/Ni−1. (4.1) 3 Yingwei Wang Abstract Algebra Proof. We refine the first series by inserting between each of the r pairs Ni−1 < Ni the following: Ni−1 = Ni−1(Ni ∩ M0) ⊳ Ni−1(Ni ∩ M1) ⊳ ··· ⊳ Ni−1(Ni ∩ Ms) ⊳ Ni. Note that the subquotient Ni/Ni−1 is replaced by the s subquotients N − (N ∩ M ) i 1 i j . (4.2) Ni−1(Ni ∩ Mj−1) Similarly, the second series can be refined to replace each of the s subquotient Mj/Mj−1 by the n subquotients (N ∩ M )M − i j j 1 . (4.3) (Ni−1 ∩ Mj)Mj−1 By Eq.(3.1), we know that (4.2) is isomorphic to (4.3). So if we delete any repetition in the refinements, we know that the sequences of subquotient are isomorphic after some permutation. 5 Unique factorization Theorem 5.1 (Fundamental Theorem of Arithmetic). Let n> 1 be a positive integer. Then there exist unique primes p1 >p2 < ··· n = |G| = |G/G1||G1/G2|···|Gm−1/Gm||Gm| = p1p2 ··· pm−1(1). By Jordan-Holder theorem (Thm 4.1), we can get the uniqueness of the prime decom- position of n. 4 MA 553: Homework 2 Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA 1 About S4 and D8 Question: Exhibit explicitly a Sylow 2-subgroup of the symmetric group S4, and show that it is (up to isomorphism) the dihedral group D8. Solution: Let H be a Sylow 2-subgroup of the symmetric group S4, then we can choose H as follows H = {(1), (1234), (13)(24), (1432), (24), (13), (12)(34), (14)(23)}. Let D8 be the dihedral group of order 8, then 2 3 2 3 D8 = {e, r, r ,r ,s,sr,sr ,sr }. Define the isomorphism π : D8 → H by π(e) = (1), π(r) = (1234), π(r2) = (13)(24), π(r3) = (1432), π(s) = (24), π(sr) = (14)(23), π(sr2) = (13), π(sr3) = (12)(43). ∼ Then it is obvious that D8 = H. 2 Non-abelian group of order 8 Let G be a non-abelian group of order 8. ∗E-mail address: [email protected]; Tel: 765 237 7149 1 Yingwei Wang Abstract Algebra 2.1 (a) Question: Show that the center of G has order 2. Denote its generator by c. Solution: Let C be the center of G, then C ⊳ G and C is abelian. Since |G| = 23, by Theorem 8 (on DF, p.125), C is nontrivial. Besides, G is non-abelian. Then we know that the only possibility for |C| is |C| =2, 22. If |C| = 22, then G/C = 2. Hence G/C is cyclic. Let G/C = m n m+n n m xy = a c1a c2 = a c1c2 = a c2a c1 = yx. It implies that G is abelian, which is a contradiction. So |C| = 2. Denote C = {1,c}. 2.2 (b) Question: If G contains an element z =6 c of order 2, then G is isomorphic to a subgroup of S4. Thus by Section 2.1, we know that G = D8. Solution: Let H = {1, z}, S = {all of the left cosets of H}. Then |H| = 2, |S| = 4. Consider the action of G on S: G × S → S, g × xH → gxH ∈ S. Then we find a group homomorphism σ : G → Permutation on S defined by [σ(g)](xH)= gxH. ∼ It indicates that G is a subgroup of S4. Since |G| = 8, we know that G = D8. 2.3 (c) Question: If c is the only order 2 element in G, then G is generated by two elements j and k of order 4 such that j2 = k2 and jk = kj−1. Hence G is isomorphic to the quaternion group, i.e. the multiplicative group of complex matrices generated by 0 1 0 i & , −1 0 −i 0 2 Yingwei Wang Abstract Algebra where i2 = −1. Solution: For ∀g ∈ G,g =6 c, the possible order for g is 2, 4, 8. But in this case c is the only order 2 element, and G is non-abelian, so we can conclude that the possible order for g is 4. Let j be one element of order 4, then j2 is order 2, so j2 = c. It is obvious that j is not the only element of order 4 since |G| = 8. Let k =6 j be another element of order 4, then k2 = c. Let H = (kj)−1 = kj3 = kj−1 ⇒ kjkj−1 =1 ⇒ kj = jk−1. Now we can conclude that G = {k, j|k4 = j4 =1, kj = jk−1}. Let 0 1 0 i k = & j = , −1 0 −i 0 ∼ then Q8 = G. 2.4 (d) Question: Is the (order 8) group of matrices of the form 1 a b 0 1 c , (a, b, c ∈ Z/2Z) 0 0 1 “diheedral ”(i.e. as in (b)) or “quaternionic ”(i.e. as in (c))? Solution: Let 1 1 0 1 0 0 A = 0 1 1 , B = 0 1 1 , 0 0 1 0 0 1 and 1 0 0 I = 0 1 0 , 0 0 1 then it is easy to check that A4 = B2 = I, AB = BA−1. ∼ It implies that here G = D8. 3 Yingwei Wang Abstract Algebra 3 Non-abelian group of order p3 Let G be a non-abelian group of order p3 (p an odd prime) and let C be its center. 3.1 (a) Question: Show that the center of G/C is isomorphic to Zp × Zp where Zp is a group of order p. Solution: Similarly as Section 2.1, we know that the only possibility for |C| is |C| = p,p2. If |C| = p2, then G/C = p. Hence G/C is cyclic. Let G/C = 3.2 (b) Question: Prove that the map f : G → G defined by f(x)= xp is a group homomorphism. Proof: First, we claim that ∀x, y ∈ G, there is a z ∈ C such that yx = xyz. Since yx = xy(xy)−1yx = xy[y−1x−1yx], we just need to show that z = y−1x−1yx ∈ C. Let π : G/C → Zp × Zp. Suppose π(xC)=(m1, n1), π(yC)=(m2, n2), then π(y−1x−1yxC) = π(y−1Cx−1CyCxC) = π(y−1C)+ π(x−1C)+ π(yC)+ π(xC) = (−m2, −n2)+(−m1, −n1)+(m2, n2)+(m1, n1) = 0, 4 Yingwei Wang Abstract Algebra which indicates that y−1x−1yx ∈ C. Second, claim that the map f : G → G defined by f(x)= xp is a group homomorphism. ∀x, y ∈ G, there exists a z ∈ C, such that yx = xyz. Then we have f(xy) =(xy)p =(xy)(xy) ··· (xy), = x(yx)(yx) ··· (yx)y, = x(xyz)(xyz) ··· (xyz)y, = x2(yx) ··· (yx)y2zp−1 = ··· = xpypzp(p−1)/2 = xpyp = f(x)f(y). 3.3 (c) Question: Prove that f(G) ⊂ C, and deduce that G has at least p2 − 1 elements of order p. Proof: First, we claim that ∀x ∈ G, xp ∈ C. Since yxp = xpyy−1x−pyxp = xpy[y−1x−pyxp], we just need to show that y−1x−pyxp = 1. We know that ∀x, y ∈ G, there exists a z ∈ C, such that yx = xyz. So y−1x−pyxp = y−1x−p(yx)xp−1, = y−1x−p(xyz)xp−1, = y−1x−p+1yxp−1z, = y−1x−p+2yxp−2z2, = ··· , = y−1x−p+pyxp−pzp, = y−1yzp, = zp =1. Second, claim that G has at least p2 − 1 elements of order p. Since f : G → G is a group homomorphism, we know that G/ ker f ∼= f(G). 5 Yingwei Wang Abstract Algebra Further, f(G) ⊂ C and |C| = p, so we know that |f(G)| ≤ p. It follows that | ker f| ≥ p2. Besides, ∀x ∈ ker f, x =6 1, we know that the order of x is p. So G has at least p2 −1 elements of order p. 3.4 (c) Question: Prove that G has subgroups H and K of orders p2 and p respectively, with H ∩ K = e. Proof: Choose K = C and let H′ = G/C. It is easy to know that H′ ∼= H, where H is a subgroup of G and H ∩ C = e. 4 Subgroup of index 2 4.1 (a) Lemma 4.1. Let G be a finite group and let π : G → SG be the left regular representation. Prove that if x is an element of G of order n and |G| = mn, then π(x) is a product of m |G| n-cycles. Deduce that π is an odd permutation if and only if |x| is even and |x| is odd. Proof. If x is an element of order n, then [π(x)](1) = x, [π(x)](x)= x2, ··· , [π(x)](xn−1)= xn =1. In other words, π(x) should include this n-cycle. (1, x, x2, ··· , xn−1). Similarly, for g ∈ G, g∈ / [π(x)](g)= xg, [π(x)](xg)= x2g, ··· , [π(x)](xn−1g)= xng = g, which implies that π(x) should include this n-cycle (g, xg, x2g, ··· , xn−1g). Since |G| = mn, we can conclude that π(x) is a product of m n-cycles. The Proposition 25 on Page 110, DF, says that “The permutation σ is odd if and only if the number of cycles of even length in its cycle decomposition is odd. ”By this proposition and the conclusion above, we can directly know that π is an odd permutation if and only if |G| |x| is even and |x| is odd. 6 Yingwei Wang Abstract Algebra 4.2 (b) Lemma 4.2. Let G and π be the same as in the preceding Lemma. Prove that if π(G) contains an odd permutation then G has a subgroup of index 2. Proof. First, claim that for any subgroup of Sn, either all of its elements are even permuta- tions or the number of even permutations is the same as the number of odd permutations. If σ is an even permutation, then all of {σ, σ2, ··· , } are even permutations. It implies that it is possible that a subgroup of Sn only contains even permutations. However, if σ is an odd permutation, then σ2 is even, and further, σ3 is odd. It means that if a subgroup of Sn contains some odd permutations, then it has to contain the same number of even permutations. Second, since π : G → SG, we know that π(G) is a subgroup of Sn where n = |G|. So if π(G) contains an odd permutation, then it has the same number of odd and even ′ −1 ′ permutation. Since A = An ∩ π(G) is a subgroup of π(G) of index 2, we can take π (A ) as the subgroup of G of index 2. 4.3 (c) Lemma 4.3. Prove that if |G| =2k where k is odd then G has a subgroup of index 2. Proof. Recall the Cauchy’s Theorem: Theorem 4.1 (Cauchy). Let G be a finite group and p be a prime. If p divides |G|, then G has an element of order p. Here, 2 is a prime and 2 divides |G| =2k, so G has an element of order 2, called x. By Lemma 4.1, we know that π(x) is an odd permutation. And by Lemma 4.2, G has subgroup of index 2. 7 MA 553: Homework 3 Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA Note: in this paper, let p> 0 be a prime number. 1 Sylow p-subgroups Question: For any finite group H, PH denotes the set of all Sylow p-subgroups of H. Let N be a normal subgroup of the finite group G. Prove that PH = {P ∩ N|P ∈ PG}, (1.1) PG/N = {P N/N|P ∈ PG}. (1.2) Proof. First, let us prove Eq.(1.1). Choose P ∈ PG, and P1 = P ∩ N. It is obvious that P1 is a p-subgroup of N. Now, consider the index of P1 in N. [N : P1] = [N :(P ∩ N)] = [P N : P ], (1.3) [G : P ] = [G : P N][P N : P ]. (1.4) Since P ∈ PG, we know that [G : P ] has no factor p. By (1.4), [P N : P ] has not factor p; and by (1.3), [N : P1] has not factor p. It follows that P1 is the Sylow p-subgroups of N. Second, let us prove Eq.(1.2). m n n−m Suppose |N| = p t, |G| = p ts, (p,st) = 1, then |G/N| = p s. Choose P ∈ PG, then |P | = pn, |P ∩ N| = pr, r ≤ m. On one hand, by isomorphism theorem, P N/N ∼= P/(P ∩ N), ⇒ |P N/N| = |P/(P ∩ N)| = pn−r, where n − r ≥ m − n. (1.5) ∗E-mail address: [email protected]; Tel: 765 237 7149 1 Yingwei Wang Abstract Algebra On the other hand, it is easy to know that |G/N| = pn−mt, (1.6) |P N/N| | |G/N|. (1.7) By Eq.(1.5)-(1.7), we know that |P N/N| = pn−m. It implies that P N/N ∈ PG/N . 2 Normal p subgroup of G Question: Let G be a finite group and H a normal p subgroup of G. Prove the following assertions 2.1 (a) H is contained in each Sylow p-subgroup of G. Proof. Let P be any Sylow p-subgroup of G. By Sylow theorem, ∃g ∈ G such that H ⊂ gPg−1, ⇒ g−1Hg ⊂ P. Since H ⊳ G, we know that g−1Hg = H ⊂ P. 2.2 (b) If K is any normal p subgroup of G, then HK is a normal p subgroup of G. Proof. Suppose |G| = pnt, (p, t) = 1 and |H| = pr, |K| = ps, |HK| = px, where r,s,x < n. By Sylow theorem, HK 2 Yingwei Wang Abstract Algebra 2.3 (c) The subgroup Op(G) generated by all normal p subgroup of G is equal to the intersection of all the Sylow p-subgroup of G. Proof. Let Op(G)=< ∪Hi >, Hi be the normal p subgroup of G, S = ∩Pj, Pj be the Sylow p subgroup of G. On one hand, by Section 2.1, we know that Hi ∈ ∩Pj, for each i. So Op(G) ⊂ S. On the other hand, ∀g ∈ S, S ⊂ Op(G). Now we can conclude that S = Op(G). 2.4 (d) Op(G) is the unique largest normal p-subgroup of G. Proof. Section 2.3 tells us that every normal p-subgroup of G is contained in Op(G), which means Op(G) is the largest normal p-subgroup of G. Besides, if H˜ is another “largest ”normal p-subgroup of G, then |H˜ |≥|Op(G)|. But H˜ ⊂ Op(G), so H˜ = Op(G). It means that Op(G) is the unique largest normal p-subgroup of G. 2.5 (e) Op(G¯)= {1}, where G¯ = G/Op(G). ′ ′ ′ Proof. Suppose Op(G¯)= J =6 {1}, then there exists a J such that J ⊳G and J = J /Op(G). Besides, |J| = pk1 , ′ k2 |J | = |Op(G)||J| = p . ′ ′ It follows that J is also a normal p-subgroup of G and |J | > |Op(G)|, which contradicts with Section 2.4. 3 Yingwei Wang Abstract Algebra 3 Groups of order 20 3.1 (a) Show that the center of the group of transformations x 7→ ax + b, (a, b ∈ Z5, a =6 0) (3.1) is trivial. Proof. Denote G as all of the transformations in the form (3.1). Let C(G) be the center of G. Then ∀g ∈ G, g(x)= ax + b; ∀c ∈ C(G), c(x)= acx + bc. We have gc = cg, ⇒ gc(x)= cg(x), ∀x, ⇒ g(acx + bc)= c(ax + b), ∀x, ⇒ aacx + abc + b = aacx + acb + bc, ∀x, ⇒ abc + b = acb + bc, ⇒ (a − 1)bc = b(ac − 1), ∀a, b ∈ Z5, a =06 , ⇒ ac =1, bc =0, ⇒ c = identity. Besides, let r ∈ G such that r(x) = x + 1 and s ∈ G such that s(x)=2x. Then it is 5 4 −1 2 easy to check that r = 1, s = 1, and srs = r . So in this case, G ∼= F20. 3.2 (b) Let ζ = e2πi/5, a fifth root of unity. Is the group generated by the complex matrices 0 i ζ 0 and − i 0 0 ζ 1 isomorphic to the group in (a)? Proof. No. Let 0 i ζ 0 A = , and B = − . i 0 0 ζ 1 It is easy to check that A4 = I, B5 = I, A−1BA−1 = B−1. It follows that this group ∼= Z5 ⋊ Z4. 4 Yingwei Wang Abstract Algebra 3.3 (c) Show that the Sylow 5-subgroup of a group of order 20 is normal. Proof. Let G be a group and |G| =20=4 ∗ 5. Suppose the number of Sylow 5-subgroup of G is k, then by Lagrange theorem and Sylow theorem, we know that k | 4∗5 and k =5s+1. Since (5s +1, 5) = 1, it follows that 5s +1 | 4 ⇒ s =0 ⇒ k = 1. Now we know that there are only one Sylow 5-subgroup in G. By Corollary 20 on Page 142,DF, we can conclude that this Sylow 5-subgroup is normal. 3.4 (d) Show that there are exactly five distinct groups of order 20. Proof. By the theorem 3 on page 158 DF, there are 2 abelian group of order 20: Z2 ×Z2 ×Z5 and Z20. But for the non-abelian groups, we have to use the “semidirect products”to find them. Let |G| =20=22 ∗ 5, H be the Sylow 2-subgroup of G, and K be the Sylow 5-subgroup of G. Besides, we know that K ⊳ G. Case I: H ∼= Z4 and K ∼= Z5. G should be in this form Z5 ⋊ Z4. Consider ∗ ϕ : Z4 → Aut (K)= Z5 = {1, 2, 3, 4}. If ϕ1 is trivial, then Z5 ⋊ Z4 = Z5 × Z4 = Z20. 4 5 Suppose ϕ2(1) = 2. Let x = (0, 1),y = (1, 0), then x = y = 1. Besides, xy = (0, 1)(1, 0) = (0+ ϕ2(1)(1), 1+0) = (2, 1), y2x = (1, 0)(1, 0)(0, 1) = (1+ ϕ2(0)(1), 0+ 0)(0, 1), = (2, 0)(0, 1), = (2+ ϕ2(0)(0), 0+1), = (2, 1), 2 −1 2 which means that xy = y x ⇒ xyx = y . So in this case, G ∼= F20. Suppose ϕ3(1) = 3. Then it is easy to check that in this case, G ∼= F20. 5 Yingwei Wang Abstract Algebra 4 5 Suppose ϕ4(1) = 4. Let x = (0, 1),y = (1, 0), then x = y = 1. Besides, xy = (0, 1)(1, 0) =(0+ ϕ4(1)(1), 1+0) = (4, 1), yxy = (1, 0)(4, 1) =(1+ ϕ4(0)(4), 0+1), =(1+4, 1), = (0, 1), −1 −1 which means yxy = x ⇒ x yx = y . So in this case G ∼= Z5 ⋊ Z4. Case II: H ∼= Z2 × Z2 and K ∼= Z5. G should be in this form Z5 ⋊ (Z2 × Z2). Consider ∗ ϕ : Z2 × Z2 → Aut (K)= Z5 = {1, 2, 3, 4}. If ϕ1 is trivial, then Z5 ⋊ (Z2 × Z2)= Z5 × (Z2 × Z2)= Z10 × Z2. Let a = (1, 0), b = (0, 1), then H = Z2 × Z2 =< a,b >. Suppose ϕ2(a)=2,ϕ2(b) = 3. Choose r = (1, 1, 0),s = (0, 1, 0), then it is easy to know that r10 = 1 and s2 = 1. Besides, sr = (0, 0, 1)(1, 1, 0) =(0+ ϕ2(1, 0)(1), 1+1, 0+0) = (2, 0, 0), rsr = (1, 1, 0)(2, 0, 0) =(1+ ϕ2(1, 0)(2), 1+0, 0+0), =(1+4, 1, 0), = (0, 1, 0), −1 which means rsr = s ⇒ rs = sr . So in this case G ∼= D12. It is easy to know that for any other ϕ, no new kinds of group appear. In summary, the distinct groups of order 20 are as follows. Let Z5 = 1. Z20; 2. Z10 × Z2; ⋊ −1 3. Z5 ϕ1 Z4 where ϕ1(x)(y)= y ; ∼ ⋊ 2 4. F20 = Z5 ϕ2 Z4 where ϕ2(x)(y)= y ; 6 Yingwei Wang Abstract Algebra ∼ −1 5. D20 = Z5 ⋊ψ (Z2 × Z2) where ψ(a)(y)= y and ψ(b)(y)= y. 4 Groups of order 30 Show that the four different groups of order 30 have 1, 3, 5, 15 elements of order 2, respec- tively. Proof. By the structure theorem, there is only one abelian group of order 30. Now G has 1 or 10 Sylow 3-subgroups and 1 or 6 Sylow 5-subgroups. If it has 10 Sylow 3-subgroups and 6 Sylow 5-subgroups, it has 45 elements, so it must have a unique Sylow 3- or 5-subgroup. If there is a unique Sylow 3-subgroup, let c be an element of order 3 so |C(c)| = 15 or 30. If |C(c)| = 15, then C(c)is generated by an element of order 15. Otherwise we have an element of order 5 commuting with an element of order 3 to get our element of order 15. If there is a unique Sylow 5-subgroup, then |C(c)| = 10, 15, or 30. This similarly gives us an element of order 15. Now we get three possible groups: < a, b | a15 = b2 =1, ba = a4b >, < a, b | a15 = b2 =1, ba = a11b >, < a, b | a15 = b2 =1, ba = a−1b > . 5 Groups of order 12 There are 5 distinct groups of order 12, namely Z12,Z2 × Z2 × Z3, A4,D12 and Z3 ⋊ Z4. Similarly as 2(d), Let Z3 = 1. Z12; 2. Z6 × Z2; ∼ 3. A4 = (Z2 × Z2) ⋊φ Z3 where φ(y)(a)= b and φ(y)(b)= a; −1 4. Z3 ⋊ϕ Z4 where ϕ(x)(y)= y ; ∼ −1 5. D12 = Z3 ⋊ψ (Z2 × Z2) where ψ(a)(y)= y and ψ(b)(y)= y. For each of the following groups, determine which of the groups in that example it is isomorphic to. 7 Yingwei Wang Abstract Algebra 5.1 (a) The multiplicative group of matrices of the form a b (a, b, c ∈ Z3, ac =6 0). 0 c Proof. Let 2 2 1 0 r = , and s = . 0 2 0 2 Then it is easy to check that r6 = s2 =1, rs = sr−1. It implies that this group ∼= D12. 5.2 (b) The multiplicative group generated by the complex matrices 0 i ω 0 , and (i2 =1,ω3 =1,ω =6 1). i 0 0 ω2 Proof. Let 0 i ω 0 A = , and B = . i 0 0 ω2 Then it is easy to check that A4 = I, B3 = I, A−1BA = B−1. It follows that this group ∼= Z3 ⋊ Z4. 5.3 (c) The transformation of the form x 7→ ax + b, (a =6 0), of the field F4 into itself. Proof. This group ∼= A4. See the classification of groups of order 12. 5.4 (d) The dihedral group D12 ∼ Proof. D12 = Z3 ⋊ψ (Z2 × Z2). See the classification of groups of order 12. 8 Yingwei Wang Abstract Algebra 5.5 (e) A non-abelian semidirect product of a group of order 4 by a group of order 3. ∼ Proof. This group = Z3 ⋊ϕ Z4. See the classification of groups of order 12. 6 Groups of order p3 There are two non-abelian groups of order p3. The one is p p2 −1 1+p G1 =< x,y | x = y =1, xyx = y > . The other one is p p p −1 −1 G2 =< x, a, b | x = a = b =1, xa = ax, xb = bx, x = aba b > . Let p be an odd prime. Consider the group, of order p3, consisting of all matrices 1 a b 0 1 c 0 0 1 with a, b, c ∈ Z/p. To which of the two non-abelian groups intentioned above is this group isomorphic? Proof. Let 1 1 0 1 0 0 1 0 1 A = 0 1 0 , B = 0 1 1 , X = 0 1 0 . 0 0 1 0 0 1 0 0 1 Then it is easy to know that Xp = Ap = Bp = I, XA = AX, XB = BX,X = ABA−1B−1. It follows that this group ∼= G2. 9 MA 553: Homework 4 Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA 1 Normal Sylow p-subgroup Lemma 1.1. Let H be a normal subgroup of a finite group G, and let N ⊳ H be a Sylow subgroup of H. Prove that N ⊳ G. Proof. Since H ⊳ G, we know that gHg−1 = H, ∀g ∈ G. Besides, N ⊂ H, we can get gNg−1 ⊂ gHg−1 = H, ∀g ∈ G. By Corollary 20 on Page 142 DF, that N ⊳ H is a Sylow p-subgroup of H means that N is the unique Sylow subgroup of H. Besides, by Sylow theorem, all of the Sylow p-subgroups of H are conjugate in H. Then gNg−1 = N, ∀g ∈ G. It follows that N ⊳ G. 2 Nilpotent groups Lemma 2.1. Let G be a finite group. Let N be a normal subgroup such that G/N is nilpotent. Suppose that for every Sylow subgroup P of G, P N is nilpotent. Prove that G is nilpotent. ∗E-mail address: [email protected]; Tel: 765 237 7149 1 Yingwei Wang Abstract Algebra Proof. First, claim that P N ⊳ G. Since P is a Sylow subgroup of G and N ⊳ G, we know that P N/N is a Sylow subgroup of G/N. Besides, G/N is nilpotent, so we know that P N/N ⊳G/N. It follows that P N ⊳G. Second, claim that P is a Sylow subgroup of P N and P ⊳ P N. Since PN 3 Abelian groups Lemma 3.1. Let n be an integer > 1. Prove the following classification: every group of α1 α2 αr order n is abelian if and only if n = p1 p2 ··· pr , where p1,p2, ··· ,pr are distinct primes, αj αi =1 or 2 for all i ∈{1, ··· ,r} and pi does not divide pj − 1 for all i and j. α1 α2 αr Proof. First, suppose |G| = n = p1 p2 ··· pr where p1,p2, ··· ,pr are distinct primes, αj αi = 1 or 2 for all i ∈{1, ··· ,r} and pi does not divide pj − 1 for all i and j. We want to show that |G| is abelian. Let us do induction by r. α1 If r = 1, then |G| = p1 , where α1 =1or 2. If α1 = 1, then G is cyclic; if α1 = 2, then ∼ 2 ∼ G = Zp1 or G = Zp1 × Zp1 . Anyway, here G is abelian. Suppose that every group of width at most k which satisfies the above conditions for |G| is abelian. Let G be a group of width r + 1 such that |G| = n satisfies the above conditions. Now every proper subgroup of G has width r at most, so that by the induction hypothesis, every proper subgroup of G is abelian. We know that every finite group whose every proper subgroup is abelian is solvable. So G is solvable. Since G is solvable, we can find H 2 Yingwei Wang Abstract Algebra α2 generality) that p1 divides p2 . If α2 = 1, then we may construct a nonabelian group Q of 1 2 α3 αr 2 order p p . Then Q×Zp3 ···pr is a nonabelian group of order n. If a = 2, we may construct a nonabelian group of order pq2. By taking the direct product with a cyclic group, there αj exists a nonabelian group of order n. It implies that pi ∤ pj − 1 for all i and j. Now we can conclude that every group of order n is abelian, then n should satisfy the conditions enumerated in the problem statement. 4 Classification of finite order Question: Describe explicitly some groups of order 1, 163, 225 such that every group of that order is isomorphic to precisely one of those you have described. Solution: Since 1, 163, 225= 52 × 7 × 172 × 23, the elementary factors are {5, 5, 7, 17, 17, 23}, {5, 5, 7, 172, 23}, {52, 7, 17, 17, 23}, {52, 7, 172, 23}. The invariant factors are {5 × 17, 5 × 7 × 17 × 23}, {5, 5 × 7 × 172 × 23}, {17, 52 × 7 × 17 × 17 × 23}, {52 × 7 × 172 × 23}. The invariant factor decompositions of the abelian groups of order 1, 163, 225 are as follows Z85 × Z13685, Z5 × Z232645, Z17 × Z68425, Z1163225. Besides, by Lemma 3.1, we know that there is no nonabelian group of order 1, 163, 225. 3 Yingwei Wang Abstract Algebra 5 Simple groups 5.1 (a) Lemma 5.1. Show that a simple group which has a subgroup of index n> 2 is isomorphic to a subgroup of the alternating group An. Proof. Let G be a simple group and H 5.2 (b) Question: What is the smallest index [An : G] occurring for a subgroup G & An? Solution: For n =1, 2, An is trivial. For n = 3, A3 only has trivial subgroup. For n = 4, A4 has a proper normal subgroup K ⊳ A4, where K = {(1), (12)(34), (13)(24), (14)(23)}. It follows that [A4 : K] = 3. For n ≥ 5, we know that An is simple. Suppose An has a subgroup of index k, then by Lemma 5.1, An is isomorphic to a subgroup of Ak. It is obvious that |An|≤|Ak|⇒ n ≤ k. So For n at least 5, the index of a subgroup of An is at least n. 5.3 (c) Question: Show that there is not simple group of order 112. Solution: Let G be a simple group of order 112. We know that 112 = 24 × 7. Suppose there are n2 Sylow 2 subgroup of G. Sylow’s theorems imply that n2 =7. (n2 =6 1 since G 4 Yingwei Wang Abstract Algebra is simple.) Let H be one of the Sylow 2 subgroups of G. By Lemma 5.1, G is isomorphic to a subgroup of A7. 4 3 However, since |G| = 2 × 7 and |A7| = 2 × 3 × 5 × 7, we know that |G| ∤ |A7|, which means G can not be the subgroup of A7. Therefore there is no simple group of order 112. 5.4 (d) Question: Show that there is not simple group of order 120. Solution: Let G be a simple group of order 120. We know that 120 = 23 ×3×5. Suppose there are n5 Sylow 5 subgroup of G. Sylow’s theorems imply that n5 = 6. Let H be one of the Sylow 5 subgroups of G, and its normalizer is NG(H). Then [G : N(H)] = 6. By Lemma 5.1, G is isomorphic to a subgroup of A6. On one hand, since [G : NG(H)] = 6, then |NG(H)| = 20. On the other hand, NG(H) is a subgroup of A6. Claim that there is no subgroup of A6 of order 20. By the table of groups of small order on page 168 DF, we know that there are 5 distinct groups of order 20, namely Z20,Z10 × Z2, D20, Z5 ⋊ Z4 and F20. It is easy to check that none of them is the subgroup of A6. This contradiction proves that there is no subgroup of A6 of order 20 and therefore there is no simple group of order 120. 5.5 (e) Question: Is every group of order 120 solvable? Solution: No. There are solvable as well as non-solvable groups of this order. All the non-solvable groups have alternating group A5 and cyclic group Z2 as the composition factors in their composition series. For example, S5 is a non-solvable of order |S5| = 120. We know that {e} < A5 5 Yingwei Wang Abstract Algebra 6 Groups of order 2800 Question: Find the distinct abelian groups of order 2800 having different number of elements of order 28. Solution: Since 2800 = 24 × 52 × 7, there are 10 distinct abelian groups. We will discuss them respectively. 1. Z16 × Z25 × Z7, the number of elements of order 28 is 2 × 6 = 12; 2. Z16 × Z5 × Z5 × Z7, the number of elements of order 28 is 2 × 6 = 12; 3. Z2 × Z8 × Z25 × Z7, the number of elements of order 28 is 2 × 2 × 6 = 24; 4. Z2 × Z8 × Z5 × Z5 × Z7, the number of elements of order 28 is 2 × 2 × 6 = 24; 5. Z4 × Z4 × Z25 × Z7, the number of elements of order 28 is 12 × 6 = 72; 6. Z4 × Z4 × Z5 × Z5 × Z7, the number of elements of order 28 is 12 × 6 = 72; 7. Z2 × Z2 × Z4 × Z25 × Z7, the number of elements of order 28 is 2 × 2 × 2 × 6 = 48; 8. Z2 × Z2 × Z4 × Z5 × Z5 × Z7, the number of elements of order 28 is 2 × 2 × 2 × 6 = 48; 9. Z2 × Z2 × Z2 × Z2 × Z25 × Z7, the number of elements of order 28 is 0; 10. Z2 × Z2 × Z2 × Z2 × Z5 × Z5 × Z7, the number of elements of order 28 is 0. In conclusion, there are exactly 2 distinct abelian groups of order 2800 having exactly 12 elements of order 28; there are exactly 2 distinct abelian groups of order 2800 having exactly 24 elements of order 24; there are exactly 2 distinct abelian groups of order 2800 having exactly 48 elements of order 28; there are exactly 2 distinct abelian groups of order 2800 having exactly 72 elements of order 28; there are exactly 2 distinct abelian groups of order 2800 having exactly 0 elements of order 28. 6 MA 553: Homework 5 Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA Note: in this paper, let M be a fixed commutative monoid with cancelation; [a, b] means the least common multiple (LCM) of a and b; (a, b) means the greatest common divisor (GCD) of a and b. 1 LCM and GCD Lemma 1.1. Prove that if [a, b] exists, then for all c, 1. [ac, bc] = [a, b]c. 2. (a, b) exists, and (ac, bc)=(a, b)c. Proof. Let d = [a, b], then a | d, ab | d ⇒ ac | dc, bc | dc ⇒ dc is a common multiple of ac and bc. Let x be any common multiple of ac and bc, then ac | x, bc | x, ⇒ x = ace = bcf, x x ⇒ | a, | b, c c x ⇒ d = [a, b] | , c ⇒ dc | x. It follows that dc = [a, b]c is the least common multiple of ac and bc. By the proposition in class, we know that if [a, b] exists, then ab (a, b)= . [a, b] ∗E-mail address: [email protected]; Tel: 765 237 7149 1 Yingwei Wang Abstract Algebra Let g =(a, b). Suppose y be any common divisor of ac and bc, then y | ac, y | bc. (1.1) Let us focus on y. On one hand, if y ∤ c, then by (1.1), we know that y | a, y | b, ⇒ y | g, ⇒ y | gc. On the other hand, if y | c, then it is obvious that y | gc. Anyway, we can get y | gc. It follows that gc =(a, b)c is the greatest common divisor of ac and bc. 2 LCM and GCD Lemma 2.1. Prove that [a, b] exists ⇔ (ac, bc) exists for all c. Proof. ⇒): By Lemma 1.1 (Part two), we know that if [a, b] exists then (a, b) exists and (ac, bc)=(a, b)c exists for all c. ⇐): Since for ∀c, (ac, bc) | ac and (ac, bc) | bc, we know that (ac, bc) | abc. Now let us denote abc d = . (2.1) (ac, bc) Since (ac, bc) | bc, we can rewrite d as bc d = a , (ac, bc) which means a | d. Similarly, we can get b | d. It follows that d is a common multiple of a and b. Choose c = 1, then we know that (a, b) exists. Besides, similarly as Lemma 1.1 (Part one), it is easy to know that (ac, bc)=(a, b)c for all c. Now we can rewrite d as ab d = . (2.2) (a, b) Let a = a1(a, b), b = b1(a, b), then, (a1, b1) = 1. 2 Yingwei Wang Abstract Algebra Besides, suppose x is any common multiple of a and b, then x = ae = bf, (2.3) ⇒ x = a1(a, b)e = b1(a, b)f, (2.4) ⇒ a1e = b1f. (2.5) Now we claim that a1 | f. Since a1 a1 b1f | a1, | , (a1, f) (a1, f) (a1, f) a1 b1f ⇒ | a1, =1 (a1, f) (a1, f) ⇒ a1 =(a1, f), ⇒ a1 | f. Now we know that f = a1f1, ⇒ x = bf = b1(a, b)a1f1. (2.6) Besides, ab d = = a1(a, b)b1. (2.7) (a, b) By Eqs.(2.6) and (2.7), we know that d | x. It implies that d is the least common multiple of a and b. 3 Prime Lemma 3.1. If a is prime and a does not divide b, then [an, b]= anb for all n. Proof. First, claim that if n = 1, then [a, b]= ab. It it obvious that ab is a common multiple of a and b. Suppose x is any common multiple of a and b, then x = ae = bf, ⇒ a | x = bf, ⇒ a | b, or a | f, (since a is prime) ⇒ a | f, (since a ∤ b) ⇒ abf | xf, (since bf = x) ⇒ ab | x. 3 Yingwei Wang Abstract Algebra It follows that ab is the least common multiple of a and b. Second, if [an, b]= anb, claim that [an+1, b]= a[an, b]= an+1b. Since [an+1, b] | [an+1, ab]= a[an, b], we know that a[an, b] is a common multiple of an+1 and b. Besides, suppose x is any common multiple of an+1 and b, then [an, b] | [an+1, b] | y, ⇒ anb | y. Since an+1 | y and a ∤ b, we know further that an+1b | y. It implies that an+1b is least common multiple of a and b. By induction, we can get the conclusion. 4 Unit Lemma 4.1. Suppose a is a unit or a product of prime. Prove that 1. (a, b) and [a, b] exist for all b. 2. If (b, c) exists then (ab, ac)=a(b,c). Proof. 1. If a is a unit, then it is obvious that (a, b)= a and [a, b]= b for all b. If a is a prime, then it is easy to know that b if a | b, [a, b]= ab if a ∤ b. If a = p1p2 ··· pn, where each pi is a prime, i =1, ··· , n, then claim that [a, b]= LCM {[p1, b], ··· , [pn, b]} , where b if pi | b, [pi, b]= pib if pi ∤ b. By induction, we know that [a, b] exists. By the proposition in class, we know that if [a, b] exists, then ab (a, b)= . [a, b] 4 Yingwei Wang Abstract Algebra 2. It is obvious that (b, c) | b, (b, c) | c, ⇒ a(b, c) | ab, a(b, c) | ac, ⇒ a(b, c) is a common divisor of ab and ac. If g is any common divisor of ab and ac, then ge = ab, gf = ac. (4.1) If a is a unit, then a−1ge = b, a−1gf = c, ⇒ a−1g | (b, c), ⇒ g | a(b, c), which implies that a(b, c) is the GCD of ab and ac. If a is a prime, then by Eq.(4.1), we know that either g | a or g | b&g | c. It follows that g | a(b, c), which means a(b, c) is the GCD of ab and ac. If a is a product of prime, then by induction, we can get the conclusion. 5 GCD Lemma 5.1. Assume that (x, y) exists for all x, y ∈ M. 1. Prove that if (a, b)=1, then (ai, bj)=1 for all (i, j). 2. Without assuming (a, b)=1, deduce from Part 1 that (a, b)n =(an, bn) for all n. 3. Prove that if (a, b)=1 and ab = cn then a ∼ (a, c)n and b ∼ (b, c)n. Proof. 1. In class, we know this lemma: Lemma 5.2. If (ac, bc) exists, and (a, b)=(a, c)=1, then (a, bc)=1. 5 Yingwei Wang Abstract Algebra Here, by Lemma 5.2, we know that (a, b)=1, ⇒ (a, b2)=1, ⇒ (a, bj)=1, ⇒ (a2, bj)=1, ⇒ (ai, bj)=1, for ∀i, j. 2. We know that a b , =1, (a, b) (a, b) an bn ⇒ , =1, (by Part 1) (a, b)n (a, b)n ⇒ (an, bn)=(a, b)n, for all n. 3. On one hand, by Part 2, we know that (a, c)n =(an,cn) ⇒ a | (a, c)n. (since a | an, a | cn) (5.1) n n n n−1 On the other hand, (a, c) =(a ,c )=(an, b)=(an, ab)= a(a , b). By Part 1, (a, b)=1, ⇒ (an−1, b)=1, ⇒ a(an−1, b) | a, ⇒ (a, c)n | a. (5.2) Now, by Eqs.(5.1) and (5.2), we can conclude that a ∼ (a, c)n. Similarly, we can conclude that b ∼ (b, c)n. 6 MA 553: Homework 6 Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA 1 Definition of ring In each case, decide whether the given structure forms a ring. If it is not a ring, determine which of the ring axioms hold and which fail: 1.1 (a) U is an arbitrary set, and R is the set of subsets of U. Addition and multiplication of elements of R are defined by the rules A + B = A B, ∪ A B = A B. · ∩ Solution: This is a ring. 1. (R, +) is an abelian group: A + B = A B = B A = B + A. ∪ ∪ 2. (R, ) is associative: · (A B) C = A B C = A (B C)= A (B C). · · ∩ ∩ ∩ ∩ · · 3. the distributive laws hold: A (B + C)= A (B C)=(A B) (A C)=(A B)+(A C), · ∩ ∪ ∩ ∪ ∩ · · (A + B) C =(A B) C =(A C) (B C)=(A C)+(B C). · ∪ ∩ ∩ ∪ ∩ · · ∗E-mail address: [email protected]; Tel: 765 237 7149 1 Yingwei Wang Abstract Algebra 1.2 (b) U is an arbitrary set, and R is the set of subsets of U. Addition and multiplication of elements of R are defined by the rules A + B =(A B) (A B), ∪ − ∩ A B = A B. · ∩ Solution: This is a ring. 1. (R, +) is an abelian group: A + B =(A B) (A B)=(B A) (B A)= B + A. ∪ − ∩ ∪ − ∩ 2. (R, ) is associative: · (A B) C = A B C = A (B C)= A (B C). · · ∩ ∩ ∩ ∩ · · 3. the distributive laws hold: A (B + C)= A [(B C) (B C))] = (A B) (A C) A B C, · ∩ ∪ − ∩ ∩ ∪ ∩ − ∩ ∩ A B + A C =(A B)+(A C)=(A B) (A C) A B C, · · ∩ ∩ ∩ ∪ ∩ − ∩ ∩ A (B + C)= A B + A C. ⇒ · · · Similarly, we can get (A + B) C =(A C)+(B C). · · · 1.3 (c) R is the set of continuous functions R R. Addition and multiplication are defined by the rules → [f + g](x)= f(x)+ g(x), [f g](x)= f(g(x)). ◦ Solution: This is not a ring. 1. (R, +) is an abelian group: [f + g](x)= f(x)+ g(x)= g(x)+ f(x) = [g + f](x). 2 Yingwei Wang Abstract Algebra 2. (R, ) is associative: · [(f g) h](x)= f(g(h(x))) = [f (g h)](x). ◦ ◦ ◦ ◦ 3. but the distributive laws do not hold: f [g + h](x)= f(g(x)+ h(x)), ◦ [f g + f h](x)= f(g(x)) + f(h(x)), ◦ ◦ f [g + h] = f g + f h. ⇒ ◦ 6 ◦ ◦ 2 Cubic root ring Let ζ = e2πi/3. 2.1 (a) Show that Z[ζ] := a + bζ a, b Z { | ∈ } is a subring of the field C of complex numbers. Solution: It is easy to know that 1 √3 ζ = e2πi/3 = + i, −2 2 1 √3 ζ2 = e4πi/3 = i = 1 ζ. −2 − 2 − − Let z1 = a1 + b1ζ, z2 = a2 + b2ζ Z[ζ], then ∈ z1 + z2 =(a1 + a2)+(b1 + b2)ζ Z[ζ], ∈ and z1z2 = (a1 + b1ζ)(a2 + b2ζ), 2 = a1a2 + b1b2ζ +(a1b2 + a2b1)ζ, 1 √3 = a1a2 + b1b2 i +(a1b2 + a2b1)ζ, −2 − 2 ! 1 √3 = a1a2 b1b2 b1b2 + i +(a1b2 + a2b1)ζ, − − −2 2 ! = (a1a2 b1b2)+(a1b2 + a2b1 b1b2)ζ Z[ζ]. − − ∈ It follows that Z[ζ] is a subring of C. 3 Yingwei Wang Abstract Algebra 2.2 (b) Let the norm of s Z[ζ] be defined to be ∈ N(s) := ss,¯ wheres ¯ is the complex conjugate of s. Show that for s, t Z[ζ], ∈ N(st)= N(s)N(t). Solution: It is easy to know that ζζ¯ = ζ 2 =1, | | ζ + ζ¯ = 1. − Let s = a1 + b1ζ, t = a2 + b2ζ Z[ζ], then ∈ N(s)= ss,¯ = (a1 + b1ζ)(a1 + b1ζ¯), 2 2 2 = a1 + b1 ζ + a1b1(ζ + ζ¯), 2 2| | = a1 + b1 a1b1. − Similarly, we can get 2 2 N(t)= a2 + b2 a2b2. − By Part (a), we know that st =(a1a2 b1b2)+(a1b2 + a2b1 b1b2)ζ. − − It follows that 2 2 N(st)= (a1a2 b1b2) +(a1b2 + a2b1 b1b2) (a1a2 b1b2)(a1b2 + a2b1 b1b2), −2 2 2 − 2 − 2 − 2 2 − 2 = (a1a2) +(a1b2) +(a2b1) +(b1b2) a1b1a2 a1b1b2 a1a2b2 a2b2b1 + a1a2b1b2, 2 2 2 2 − − − − = (a1 + b1 a1b1)(a2 + b2 a2b2), − − = N(s)N(t). 2.3 (c) Show that s Z[ζ] s¯ Z[ζ]. ∈ ⇒ ∈ 4 Yingwei Wang Abstract Algebra Solution: Let s = a + bζ Z[ζ], then ∈ s¯ = a + bζ,¯ 1 √3 = a + b i , −2 − 2 ! 1 √3 = (a b) b + i , − − −2 2 ! = (a b) bζ Z[ζ]. − − ∈ 2.4 (d) Show that s is a unit in Z[ζ] if and only if N(s) = 1. Solution: On one hand, if ss−1 = 1, then by part (b), N(1) = N(ss−1)= N(s)N(s−1), 1= N(s)N(s−1), ⇒ 1 N(s−1)= . ⇒ N(s) But both N(s) and N(s−1) are in N, which implies N(s)= N(s−1) = 1. On the other hand, if N(s) = 1, let s = a + bζ Z[ζ], then ∈ N(s)= a2 + b2 ab =1. (2.1) − So 1 s−1 = , a + bζ 1 = , (a b/2) + √3b/2i − (a b/2) √3b/2i = − − , (a b/2)2 +3b2/4 − a b b = − ζ. a2 + b2 ab − a2 + b2 ab − − By Eq.(2.1), we can get s−1 =(a b) bζ Z[ζ]. − − ∈ 5 Yingwei Wang Abstract Algebra 2.5 (e) Show that the group of units in Z[ζ] consists of all 6-th roots of unity in C. Solution: First, claim that all 6-th roots of unity are in Z[ζ]. pi/3 1 √3 ξ1 = e = + i =1+ ζ Z[ζ], 2 2 ∈ 2pi/3 ξ2 = e = ζ Z[ζ], ∈ ξ3 = 1 Z[ζ], −4pi/∈3 2 ξ4 = e = ζ = 1 ζ Z[ζ], − − ∈ 5pi/3 1 √3 ξ5 = e = i = ζ Z[ζ], 2 − 2 − ∈ ξ6 =1 Z[ζ]. ∈ Second, by part (d), just need to show that if ξ is one of the 6-th roots of unity in C, then N(ξ) = 1. We know that N(ξ )= ξ ξ¯ = ξ 2 =1, j =1, 2, 3, 4, 5, 6. j j j | j| 3 Nilpotent An element x of a ring R is called nilpotent if some power of x is zero. Prove that if x is nilpotent, then 1 + x is a unit in R. Solution: If x is nilpotent, then p N, such that ∃ ∈ xp =( x)p =0. − For 1+ x, suppose (1 + x)y = 1, then 1 y = , 1+ x ∞ = ( x)k, − k=0 Xp−1 = ( x)k R. =0 − ∈ Xk It implies that 1 + x is a unit in R. 6 Yingwei Wang Abstract Algebra 4 Unit ideal Prove or disprove: If an ideal I contains a unit, then it is the unit ideal. Solution: Yes. Since ring R contains a unit u, then we know that u−1 is in R. Since u I and u−1 R, then uu−1 =1 I. ∈ ∈ ∈ But then for any r R, 1 I, and thus 1r = r I. Thus I is all of R. ∈ ∈ ∈ Note: That is why the ideal of R which is all of R is called the “unit ideal ”– it is the only ideal that could possibly contain a unit. 5 Polynomial ring Prove that if two rings R, R′ are isomorphic, then so are the polynomial rings R[x] and R′[x]. Solution: Suppose we have an isomorphic φ : R R′, → such that r φ(r), r R. → ∀ ∈ Now we can define a map φ˜ : R[x] R′[x], → n k such that for any p = k=0 akx , P n k φ˜(p)= φ(ak)x . =0 Xk An easy verification proves that φ˜ is a ring isomorphism. 6 Automorphism Let R be a ring, and let f(y) R[y] be a polynomial in one variable with coefficients in R. Prove that the map R[x, y]∈ R[x, y] defined by → φ : x x + f(y), y y, (6.1) → → 7 Yingwei Wang Abstract Algebra is an automorphism of R[x, y]. Solution: It is obvious that the map φ : R[x, y] R[x, y] defined by (6.1) is homo- → morphism. We just need to define the inverse: φ−1 : x x f(y), y y, (6.2) → − → which implies that φ−1 is also a homomorphism. It follows that φ is an automorphism. 7 Automorphism Determine all automorphisms of the polynomial ring Z[x]. Solution: Suppose φ : Z[x] Z[x] → be an automorphism of Z[x]. First, it is easy to know that φ(1) = 1, which means φ(c)= c for all constant c. Hence, φ is completely determined by φ(x), which can be regarded as an element in Z[x]. Second, suppose deg(φ(x)) = d, then for non-constant polynomials f(x) Z[x], ∈ deg(φ(f(x))) > d, since f(x) is a linear combination of the power of x. However, since φ is an automorphism, it must be surjective, so there exists non-constant f(x) Z[x] such that φ((f(x)) = x. Hence, it must be the case that d = 1, which is to say that∈ φ(x)= αx. Third, suppose α = pq where p, q Z, and ∈ φ(g(x)) = qx, then φ(x)= αx = pqx, p = φ(p), qx = φ(g(x)), φ(x)= φ(p)φ(g(x)) = φ(pg(x)). ⇒ However, the only values of p and g for which this could be satisfied are p = ,g(x)= x. ± ± Finally, there are only two automorphisms of Z[x], φ1 and φ2, where φ1(x)= x, φ2 = x. − 8 Yingwei Wang Abstract Algebra 8 Quotient ring Find a simpler description for each of the following rings. 8.1 (a) Z[x]/(x2 3, 2x + 4). − Solution: My answer is 2 Z[x]/(x 3, 2x + 4) = a + bi a =0, 1 & b Z . − ∼ { | ∈ } n For any p(x) = anx + + a1x + a0 Z[x], if deg p(x) = n 2, we can choose n−2 ··· ∈ ≥ q1(x)= anx such that 2 p(x)= q1(x)(x 3) + r1(x), (8.1) − where deg r1(x) n 1. ≤ − Keep doing this process, we can get 2 p(x)=(q1 + q )(x 3) + r (x), (8.2) ··· k − k where deg r (x) 1. Suppose k ≤ rk(x)= a1x + a0. If a1 =0, 1, then it is fine; if a1 2, then we can get ≥ a1 =2s + t, where t =0, 1. So we have r (x)= s(2x +4)+ tx +(a0 4s), (8.3) k − where t =0, 1. By Eqs.(8.2)-(8.3), we can get 2 p(x)=(q1 + q )(x 3) + s(2x +4)+ tx +(a0 4s). (8.4) ··· k − − It is implies that 2 Z[x]/(x 3, 2x + 4) = a + bi a =0, 1 & b Z . − ∼ { | ∈ } 9 Yingwei Wang Abstract Algebra 8.2 (b) Z[i]/(2 + i). Solution: My answer is Z[i]/(2 + i) ∼= Z5. For any a + bi Z[i], we know that ∈ a + bi = b(2 + i)+(a 2b). (8.5) − Besides, we know that (2 i)(2 + i)=5. − If (a 2b) < 5, then it is fine; if (a 2b) 5, then we can do further: − − ≥ a 2b =5q + r =(2+ i)(2 i)q + r, (8.6) − − where r =0, 1, 2, 3, 4. By Eqs.(8.5)-(8.6), we can get a + bi =(b +2q qi)(2 + i)+ r, (8.7) − where r =0, 1, 2, 3, 4. Now it is clear that Z[i]/(2 + i) ∼= Z5. 10 MA 553: Homework 7 Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA 1 Fractions of ring Let p be a prime ideal in an integral domain R, and let M consist of all elements in R lying outside p (so that M is a multiplicative submonoid of R). In this case it is customary to denote the ring of fractions RM by Rp . 1.1 (a) Show that if q is any prime ideal in Rp, then q ∩ R is a prime ideal in R, contained in p; and that one obtains in this way a one-one correspondence between all prime ideals in Rp and those prime ideals in R which are contained in p. Solution: We know that the homomorphism θ : R → Rp is given by r θ(r)= . 1 It implies that for any r ∈ R, we can regard r as an element in Rp. Suppose q is any prime ideal in Rp, then ∀x, y ∈ q ∩ R, either x ∈ q or y ∈ q. But x, y ∈ R, so either x ∈ q ∩ R or y ∈ q ∩ R. It implies that q ∩ R is a prime ideal in R. Besides, since the kernel of θ is p, so q ∩ R is contained in p. Now we have a one-one correspondence between all prime ideals in Rp and those prime ideals in R which are contained in p: φ : q ↔ q ∩ R. ∗E-mail address: [email protected]; Tel: 765 237 7149 1 Yingwei Wang Abstract Algebra 1.2 (b) Definition 1.1. A commutative ring R is called a local ring if it has a unique maximal ideal. Show that Rp is a local ring. Solution: We know that every maximal ideal is prime. But by part (a), we also know that any prime ideal q in Rp, q ∩ R is a prime ideal in R and contained in p. −1 Then φ (p ∩ R) is the unique maximal ideal in Rp, which means Rp is a local ring. 2 Category Let R be a commutative ring and let M and N be commutative monoids. With coordi- natewise multiplication, M × N is then also a commutative monoid. 2.1 (a) Consider the category T of triples (S,µ,ν) such that S is a commutative R-algebra and µ : M → S and ν : N → S are monoid homomorphisms, maps between such triples being defined in the obvious way. Find monoid homomorphisms µ1 : M → (R[M])[N] and ν1 : N → (R[M])[N], µ2 : M → R[M ×N] and ν2 : N → R[M ×N], such that both ((R[M])[N],µ1, ν1) and (R[M × N],µ2, ν2) are initial objects in T ; and deduce that there is an R-algebra isomorphism α :(R[M])[N] → R[M × N], such that α ( rmnm)n = rmn(m, n). n∈N m∈M ! ∈ × X X (m,nX) M N Solution: First, we have ϕ1 : M → R[M], in which ϕ1(m) = polynomial with value 1 at m and 0 eleswhere Second, we also have φ1 : R[M] → (R[M])(N), 2 Yingwei Wang Abstract Algebra in which φ1(r)= fr, where fr(e)= r, fr(n)=0, n =6 e. Now just define µ1 : M → (R[M])[N] in a natural way µ1 = φ1 ◦ ϕ1, which means µ1(m)= m. (2.1) Note that the m in the right hand side of the Eq.(2.1) should be regarded as an element in (R[M])[N]. Similar definition can be done to ν1 : N → (R[M])[N] in which ν1(n)= n. (2.2) Note that the n in the right hand side of the Eq.(2.2) should be regarded as an element in (R[M])[N]. Besides, we have these homomorphisms in natural way ϕ2 : M → M × N, and φ2 : M × N → R[M × N]. So we can define µ2 : M → R[M × N] as µ2 = φ2 ◦ ϕ2, which means µ2(m)=(m, 1N ). (2.3) Similarly, we can define ν2 : N → R[M × N], which means ν2(n)=(1M , n). (2.4) Suppose the universal property of monoid algebras holds here, for any S in this cate- gory T, we can define f :(R[M])[N] → S as f ( rmnm)n = rmnµ1(m)ν2(n). (2.5) n∈N m∈M ! m∈M,n∈N X X X Besides, we can define g : R[M × N] → S as g r (m, n) = r (µ (m), ν (n)). (2.6) mn mn 2 2 ∈ × m∈M,n∈N (m,nX) M N X 3 Yingwei Wang Abstract Algebra It follows that both ((R[M])[N],µ1, ν1) and (R[M × N],µ2, ν2) are initial objects in T. That is to say there exists a unique R-algebra homomorphism α1 :(R[M])[N] → R[M × N], and a unique R-algebra homomorphism α2 : R[M × N] → (R[M])[N]. −1 It follows that α1 = α2 , which means there is an R-algebra isomorphism α :(R[M])[N] → R[M × N]. 2.2 (b) Explain carefully how the isomorphism α specializes to give isomorphisms of polynomial rings, such as (R[W, X])[Y,Z] → R[W,X,Y,Z]. Solution: By part(a), we just need to choose M = WX and N = YZ. and define α in this way α rwxwx yz = rwxyz(wx, yz), y∈Y,z∈Z w∈W,x∈X ! ! w∈W,x∈X,y∈Y,z∈Z X X X which means α :(R[W, X])[Y,Z] → R[WX × YZ] is an isomorphism. Besides, we know that there is a natural isomorphism π : WX × YZ → WXYZ, in which π(wx, yz)= wxyz. Then we have an isomorphism π˜ : R[WX × YZ] → R[W,X,Y,Z] 4 Yingwei Wang Abstract Algebra defined by π˜ rwxyz(wx, yz) = rwxyzπ(wx, yz). w∈W,x∈X,y∈Y,z∈Z ! w∈W,x∈X,y∈Y,z∈Z X X Now we know that β :(R[W, X])[Y,Z] → R[WX × YZ] → R[W,X,Y,Z], where β =π ˜ ◦ α is an isomorphism. 2.3 (c) Let φ : M → N be a monoid homomorphism, and let θ : R[M] → R[N] be the cor- responding R-algebra homomorphism (given by the universal property of R[M], so that θ( rmm) = rmφ(m)). Show that the kernel of θ is generated by the set of elements of the form 1.m − 1.m with φ(m)= φ(m). P P Solution: Since 0 is obvious in the kernel of θ, we can assume rm =6 0 such that θ rmm = rmφ(m)=0. X X Let f = rmm ∈ R[M]. For each n ∈ N, if there are two element m, m′ such that φ(m)= φ(m′), then P rmθ(m)= (rm + rm′ )θ(m). It follows that in order to getXθ(f)=0 ∈ RX[N], we should let rm + rm′ =0, which means ′ f = rm(m − m ). Besides, if there are three elements m1, m2, m3 ∈ M such that φ(m1) = φ(m2) = φ(m3)= n, then similarly, in order to get θ(f)=0 ∈ R[N], we have rm1 + rm2 + rm3 =0 ⇒ rm3 = −rm1 − rm2. So f = rm1m1 + rm2m2 + rm3m3 = rm1(m1 − m3)+ rm2(m2 − m3). By induction we know that the kernel of θ is generated by the set of elements of the form 1.m − 1.m with φ(m)= φ(m). 5 Yingwei Wang Abstract Algebra 3 Commutative monoid with cancellation Show that a monoid M is a commutative monoid with cancellation if and only if there exists an injective monoid homomorphism from M into an abelian group. Solution: On one hand, suppose there exist an injective monoid homomorphism φ : M → R, where R is an abelian group. Then φ(M) is a submonoid of R. Now φ : M → φ(M) ⊂ R, is an isomorphism. Since R is abelian, we know that M is a commutative monoid with cancellation. On the other hand, suppose M is a commutative monoid with cancellation, we just need to do the “localization ”with respect to M. Find a ring R such that M is a multiplication submonoid of R. Define a equivalence relation on pairs (r, m) ∈ R × M, (r1, m1) ≡ (r2, m2) if r1m2 = r2m1. Denote r/m := equivalence class of (r, m). Then define r r r m + r m 1 + 2 = 1 2 2 1 , m1 m2 m1m2 r r r r 1 · 2 = 1 2 . m1 m2 m1m2 Now define λ : R → RM by r λ(r)= , 1 such that λ(m) is a unit for every element m ∈ M. Now λ is a ring injective homomorphism. 6 MA 553: Homework 8 Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA 1 Quadratic norm-Eculidean domain 1.1 (a) Let α and β be rational numbers with α 1/2, and let m> 0 be an integer such that | | ≤ α2 mβ2 = 1 δ, (1.1) − − − where 0 δ < 1. Set ≤ 1, if α 0 ǫ = ≥ (1.2) 1, if α< 0. − Show that if m is not of the form 5n2(n Z) then ∈ (α + ǫ)2 mβ2 < 1. (1.3) | − | Solution: It is easy to know that (α + ǫ)2 mβ2 = α2 +2αǫ + ǫ2 mβ2 = 2αǫ δ . (1.4) | − | | − | | − | By the definition of ǫ, we know that if 0 α< 1/2, then ≤ 2αǫ =2α (0, 1), ∈ (2αǫ δ) ( 1, 1), ⇒ − ∈ − 2αǫ δ < 1. ⇒ | − | ∗E-mail address: [email protected]; Tel: 765 237 7149 1 Yingwei Wang Abstract Algebra Similarly, if 1/2 <α< 0, then − 2αǫ = 2α (0, 1), − ∈ (2αǫ δ) ( 1, 1), ⇒ − ∈ − 2αǫ δ < 1. ⇒ | − | Besides, if α = 1 , thanks to the fact that m is not of the form 5n2(n Z), then ± 2 ∈ α2 mβ2 = 1 δ, where δ (0, 1) (note that here δ = 0). Now we can also get the inequality− (1.3).− − ∈ 6 1.2 (b) Deduce that Z[ω] is norm-Euclidean when ω = √6 or ω = √7. Solution: Let ω2 q = 0, then q = 6 or 7. For any α, β Q, e, f Z, then − ∈ ∈ N((α + βω) (e + fω))=(α e)2 q(β f)2. (1.5) − − − − First, choose e such that α e 1/2; second, choose f such that 1 | − 1| ≤ (α e )2 q(β f)2 = 1 δ, − 1 − − − − where 0 δ < 1; third, choose ǫ such that ≤ 1, if α e 0 ǫ = − 1 ≥ 1, if α e1 < 0. − − Let e = e ǫ N. Then by part (a), we can get 1 − ∈ (α e)2 q(β f)2 < 1, | − − − | where q = 6 or 7. It follows that Z[ω] is norm-Euclidean when ω = √6 or ω = √7. 1.3 (c) Deduce that Z[ω] is norm-Euclidean when ω2 ω + q = 0 with q = 4, 5, 7. − − − − 2 Yingwei Wang Abstract Algebra 2 Primes in polynomial ring Let R be a UFD, with fraction field K. Suppose you already have computer algorithms for factoring into primes in R and in the polynomial ring K[X]. Describe briefly how you would instruct a computer to factor into primes in R[X]. Solution: Choose f(x) R[X], and factor f(x) into primes in K[X], we can get ∈ f(x)= c(x a /b )(x a /b ) (x an/bn). − 1 1 − 2 2 ··· − In order to have f(x) R[X], we should have ∈ n c = d bj. Yj=1 Now it is clear that f(x)= d(b x a )(b x a ) (bnx an). 1 − 1 2 − 2 ··· − Factoring d into primes in R, we can get d = d d dn. 1 2 ··· Now, we can factor f(x) into primes in R[X], f(x)= d d dn(b x a )(b x a ) (bnx an). 1 2 ··· 1 − 1 2 − 2 ··· − Note that aj and bj should not have any common factor for any j. 3 Prime and irreducible polynomial Let k be a field, x, y, and z indeterminates. 3.1 (a) Let f(x) and g(x) be relatively prime polynomials in k[x]. Show that in the polynomial ring k(y)[x], f(x) yg(x) is irreducible. − Proof. We know that k(y)[x] ∼= k(x)[y]. So we can see f(x) yg(x) as an element in k(x)[y]. If it is not irreducible, then − f(x) yg(x)=(a(x)y + b(x))c(x), − f(x)= a(x)c(x), g(x)= b(x)c(x), ⇒ which contradicts with the fact that f(x) and g(x) are relatively prime. 3 Yingwei Wang Abstract Algebra 3.2 (b) Prove that in k(y, z)[x], the polynomial f(x, y, z)= x4 yzx3 +(y2z2 y)x2 +(y2z y)x + y2z, − − − is irreducible. Proof. We can rewrite f(x, y, z) as f(x)= x2y2z2 +(xy2 + y2 yx3)z yx2 yx + x4, − − − = (xy)2z2 + y[(x + 1)y x3]z x[(x + 1)y x3], 2 − − − = a2z + a1z + a0 Choose P be the ideal generated by (x+1)y x3. By part (a), P is a prime ideal. Besides, 2 − a1, a0 P while a2 / P and a0 / P . By Eisenstein theorem, we know that f(x, y, z) is irreducible.∈ ∈ ∈ 4 Polynomial ring Let R be an integral domain with fraction field K, let R[X] be a polynomial ring, and let a and b be nonzero elements in R. 4.1 (a) If R is a UFD and P R[X] is a prime ideal with P R = (0), then P is a principal ideal. ⊂ ∩ Proof. There is a natural map π : R[X] K[X] defined by → π(f(x)) = f(x). If P is a prime ideal in R[X], then P is a principal ideal in K[X], which means f(x) P = , d where f(x) R[X]. ∈ 4 Yingwei Wang Abstract Algebra It implies that hinP , ∀ f(x) h(x) = g(x), d g(x) = f(x) . d g(x) Since f, h R[X], we know that d R[X]. It follows that P is generated by f(x). Hence, P is a principle∈ ideal. ∈ 4.2 (b) aR bR = abR iff the ring R[X]/(aX b) is an integral domain. ∩ − Proof. On one hand, suppose aR bR = abR. Then we know that aR bR = [a, b]R so ab = [a, b], which means (a, b) = 1.∩ It follows that aX b is irreducible. ∩ − Let P =(aX b). Then for any fg P , f(X)g(X)=(aX b)h(X). So we have either aX b f(X) or−aX b g(X). It implies∈ that P is a prime ideal.− Hence, R[X]/(aX b) − | − | − is an integral domain. On the other hand, if R[X]/(aX b) is an integral domain, then (aX b) is a prime ideal. So aX b is irreducible. It follows that− (a, b) = 1, which implies that aR− bR = abR. − ∩ 4.3 (c) If c = aq = bp is a nonzero common multiple of a and b then c is an l.c.m. of a and b iff pX q is a prime element in R[X]. − Proof. On one hand, if c = [a, b], then (p, q) = 1. By part(b), we know that R[X]/(pX q) − is an integral domain. It follows that pX q is a prime element in R[X]. On the other hand, if (pX q) is a prime− ideal. So pX q is irreducible. It follows that − − (p, q) = 1, which implies that c = aq = bp = [a, b]. 4.4 (d) b An l.c.m. [a, b] exists iff the kernel of the R-homomorphism φ : R[X] R[ a ] K taking X b → ⊂ to a is a principal ideal. 5 Yingwei Wang Abstract Algebra Proof. If f(X) ker(φ), then ∈ φ(f(X)) = f(b/a)=0, which means b/a is a root of f(x) in K. So f(X)=(aX b)g(X). (4.1) − Now we can prove the conclusion. a b On one hand, if [a, b] exists, then (a, b) exists, and (a,b) , (a,b) = 1. By Eq.(4.1), we can get f(X) =(aX b)g(X), − a b = X (a, b)g(x), (a, b) − (a, b) =(pX q)˜g(x), − a b where p = (a,b) , q = (a,b) . By part (c), we know that the kernel of φ is a principle ideal. On the other hand, if the kernel of φ is a principle ideal, which is generated by (pX q) − (Since aX b ker(φ), so the generator should be in this form.) By part (b) and (c), we have (p, q)=1.− ∈ So aX b = c(px q), − − a = cp, b = cq. ⇒ Now cpq = [a, b]. 5 Quotient ring 5.1 (a) Prove that if x = 0 and y are elements in a UFD such that x2 divides y2, then x divides y. 6 Proof. In a UFD, we have x = pα1 pα2 pαn , 1 2 ··· n y = pβ1 pβ2 pβn , 1 2 ··· n where pj are primes and αj, βj 0. { } ≥ 6 Yingwei Wang Abstract Algebra Now x2 y2 means | 2αj 2βj, ≤ αj βj, ⇒ ≤ x y. ⇒ | 5.2 (b) Let k be a field. In the quotient ring R = k[X,Y,Z]/(Y 2 X2Z) let x = X¯ and y = Y¯ be − the natural images of X and Y . Show that x2 divides y2 in R, but x does not divide y. Proof. By the definition of the quotient ring, we know that x = X¯ = f(X,Y,Z)(Y 2 X2Z)+ X, y = Y¯ = g(X,Y,Z)(Y 2 X2Z)+ Y, − − x2 =(Y 2 X2Z)(f 2Y 2 fX2Z +2f)+ X2, − − ⇒ y2 =(Y 2 X2Z)(g2Y 2 gX2Z +2g)+ Y 2 =(Y 2 X2Z)(g2Y 2 gX2Z +2g +1)+ X2Z, − − − − x2 y2. ⇒ | But it is obvious that x can not divides y since Y can not be written in the form (Y 2 X2Z)q + rX. − 5.3 (b) Is R an integral domain? (Why?) Proof. No. Since if R is an integral domain, then the cancelation law holds. But by part (b), we know that it does not holds. 6 Fermat equation Find all solutions in positive integers of the equation y3 = x2 +4. (6.1) 7 Yingwei Wang Abstract Algebra 6.1 (a) a + bi is divisible by 1 + i a b is even. ⇔ − Proof. On one hand, suppose a + bi is divisible by 1 + i, then a + bi =(1+ i)(c + di)=(c d)+(c + d)i, − a b =2d. ⇒ − which implies that a b is even. On the other hand,− if a b is even, then it is easy to know that a + b is also even. We can suppose that − a b =2k, − a + b =2l, which implies that a = l k − b = l + k. Now we can find l + ki such that (1 + i)(l + k)=(l k)+(l + k)i = a + bi, − which means a + bi is divisible by 1 + i. 6.2 (b) If y3 = x2 +4, (x, y Z), then ∈ 1 if x is odd, (x +2i, x 2i)= , (6.2) − (1 + i)3 if x is even, Proof. If x =2k + 1, then (x +2i, x 2i)=(2k +1+2i, 2k 1 2i)=1. − − − If x =2k, then (x +2i, x 2i)=(2k +2i, 2k 2i)= 2+ i = (2i)(1 + i)=(1+ i)3. − − − 8 Yingwei Wang Abstract Algebra 6.3 (c) If y3 = x2 + 4, then x +2i = in(a + bi)3, for some n, a, b. Proof. By part (b), we know that (x +2i, x 2i)= z3, − where z =1 or z =1+ i. It follows that x +2i x 2i , − =1, z3 z3 Now we can get y3 =(x +2i)(x 2i), − y 3 x +2i x 2i 2 = 3 −3 , ⇒ a z z x +2i =(a + bi)3, ⇒ z3 x +2i =(z(a + bi))3 = in(a + bi)3. ⇒ Let n = 0, then x +2i =(a + bi)3 =(a3 3ab2)+(3a2b b3)i, − − 3a2b b3 =2, ⇒ − b(3a2 b2)=2, ⇒ − b =1, a = 1; or b = 2, a = 1, ⇒ ± − ± x =2,y =2; or x = 11,y =5. ⇒ So the equation (6.1) has two solutions over positive integers: x =2, y =2, or x = 11, y =5, 9 MA 553: Homework 9 Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA 1 Quadratic norm-Eculidean domain Let ω C satisfy ω2 pω + q = 0 where p and q are integers such that p2 4q is not the square∈ of an integer. The− norm of a + b Z[ω] is − ∈ N(a + bω): =(a + bω)(a + bω¯), =(a + bω)(a + b(p ω)). − It was shown that if (a, b) = 1, then the natural map is an isomorphism Z/(N(a + bω))Z Z[ω]/(N(a + bω))Z[ω]. → Prove that for any a, b Z, N(a + bω) is the cardinality of Z[ω]/(N(a + bω))Z[ω] ∈ | | Proof. First, I want to show that φ : Z Z Z[ω] ⊕ → is a group isomorphism. Just define φ as φ(a, b)= a + bω, where (a, b) Z Z, a + bω Z[ω]. Then it is∈ easy⊕ to know that∈ φ is a group isomorphism. Second, I want to show that ϕ : Z[ω] (a + bω)Z[ω] → ∗E-mail address: [email protected]; Tel: 765 237 7149 1 Yingwei Wang Abstract Algebra is also a group isomorphism. Here, for any x + yω Z[ω], we define ∈ ϕ(x + yω) =(a + bω)(x + yω), = ax + byω2 +(ay + bx)ω, = ax + by(pω q)+(ay + bx)ω, − =(ax bqy)+(bpy + ay + bx)ω. − Note that a and b does not equal to 0 at the same time. It is obvious that ϕ is a homomorphism. Besides, since (a + bω)Z[ω] Z[ω], we just need to show that ϕ is injective, which means ker ϕ = 0. For any x + yω ker⊂ ϕ, ∈ ϕ(x + yω)=(a + bω)(x + yω)=0, x + yω =0, ⇒ since a + bω = 0. 6 Now we have this group isomorphism: Z Z = Z[ω] = (a + bω)Z[ω]. (1.1) ⊕ ∼ ∼ Third, we know that Z[ω] Z[ω] (a/(a, b)+ bω/(a, b))Z[ω] = (1.2) (a + bω)Z[ω] (a/(a, b)+ bω/(a, b))Z[ω] · (a + bω)Z[ω] a b Since (a,b) , (a,b) = 1, we know that Z[ω] N(a + bω) cardinality of = | |. (1.3) (a/(a, b)+ bω/(a, b))Z[ω] (a, b)2 Besides, by the isomorphism (1.1), we know that a b (a/(a, b)+ bω/(a, b))Z[ω] = Z Z, ∼ (a, b) ⊕ (a, b) (a + bω)Z[ω] = aZ bZ. ∼ ⊕ It follows that a b (a/(a, b)+ bω/(a, b))Z[ω] (a,b) Z (a,b) Z cardinality of = cardinality of ⊕ =(a, b)2. (a + bω)Z[ω] aZ bZ ⊕ (1.4) 2 Yingwei Wang Abstract Algebra By Eqs.(1.2)-(1.4), we can get Z[ω] cardinality of , (a + bω)Z[ω] Z[ω] (a/(a, b)+ bω/(a, b))Z[ω] = cardinality of cardinality of , (a/(a, b)+ bω/(a, b))Z[ω] × (a + bω)Z[ω] N(a + bω) = | | (a, b)2 = N(a + bω) . (a, b)2 × | | 2 Square in Z/π Assume that Z[ω] (as in section 1) is a UFD. Let π be a Z-prime. Suppose there are integers a and b, not both divisible by π, such that π divides a2 + pab + qb2. Show that there are integers c and d such that π = (c2 + pcd + qd2). Deduce from this that if e =1 or e = 2, then π is of the form x2 + ey2 ± e is a square in Z/π. ⇔ − Proof. We know that π a2 + pab + qb2, | π (a + bω)(a + bω¯), but π ∤ a + bω, π ∤ a + bω,¯ ⇒ | π is not a prime, ⇒ π is not irreducible, ⇒ π is reducible, ⇒ π = (c + dω)(c + dω¯), ⇒ ± π = (c2 + pcd + qd2). ⇒ ± Furthermore, if π = 2, then for e = 1, we have 2= x2 + y2, x =1,y =1, ⇔ 1 is a square in Z/2. ⇔ − Similarly, for e = 2, we have 2= x2 +2y2, x =0,y =1, ⇔ 2 is a square in Z/2. ⇔ − 3 Yingwei Wang Abstract Algebra For the case that π 3, we claim that π = x2 + ey2 a2 + b2, but not both a and b divisible by π, where e =≥ 1. For e = 1, we know that x2 + y| 2 4x2 +4y2. So we can choose | a =2x, b =2y, then it is fine. Similarly, for π 3 and e = 2, we can choose a =2x, b =2y, then π = x2 + ey2 a2 + b2, ≥ | but not both a and b divisible by π. By previous conclusion, we know that x2 + ey2 e is a square in Z/π. ⇔ − 3 Primes Let ω = 1 be a complex number satisfying ω3 = 1. We showed in class that Z[ω] is a Euclidean6 − domain. − 3.1 (a) Let p> 3 be an odd prime in Z. Show that: p 1( mod 6) 1 has three cubic roots in Z/π 3 is a square in Z/π. (3.1) ≡ ⇔ − ⇔ − Proof. p 1( mod 6), ≡ (Z/π)∗ divided by 6, ⇔ | | (Z/π)∗ has an element of order 6, ⇔ 1 has three cubic roots in Z/π. (3.2) ⇔ − Besides, p 1, ( mod 6), ≡ p =6n +1, ⇔ (Z/π)∗ has an element of order 6n, ⇔ 3 is a square in Z/π, since ( 3)3n 1, ( mod p) (3.3) ⇔ − − ≡ By Eq.(3.2)-(3.3), we can get (3.1). 4 Yingwei Wang Abstract Algebra 3.2 (b) Prove that every prime p > 0 in Z of the form p = 6n + 1 can be represented in the form p = a2 + ab + b2(a>b> 0) in one and only one way. Proof. Consider the UDF Z[ω] where ω satisfies ω2 + ω +1=0. By part (a), we know that 3 is a square in Z/p. Hence for some a, b Z, − ∈ p = a2 + ab + b2. Furthermore, if we set a>b> 0, then the a and b should be unique. 3.3 (c) Prove that every prime p > 0 in Z of the form p = 6n + 1 can be represented in the form p = a2 +3b2(a,b > 0) in one and only one way. Proof. Consider the UDF Z[ω] where ω satisfies ω2 +3=0. Since 12 = 3 22 and 3 is a square in Z/p, we know that for some a, b Z, − − × − ∈ p = a2 +3b2. Furthermore, if we set a>b> 0, then the a and b should be unique. 3.4 (d) Prove that every odd prime p in Z factors into primes in Z[√ 3]. What about p = 2? − Proof. Let p be a prime and p =2n + 1. Consider n by different cases. 1. If n =3k, then p =6k + 1. By part (c), we know that p factors into primes in Z[√ 3]. − 2. If n =3k + 1, then p =6k +3= 3(2k + 1). Now p is not a prime, which is impossible. 3. If n =3k + 2, then p =6k + 5. Now p can also factor into primes in Z[√ 3]. − 5 MA 553: Homework 10 Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA 1 Field extension Question: Let K L be a field extension of finite degree, and let f K[X] be an irreducible polynomial whose⊂ degree is relatively prime to [L : K]. Show that f∈is irreducible in L[X]. Proof. Suppose f is reducible in L[X]. But we know that f is irreducible in K[X]. So α L ∃ ∈ such that f(x) is the minimal polynomial of α in K. Consider F = K(α), then K F L. (1.1) ⊂ ⊂ Let [L : K]= m, deg(f)= n, then [F : K]= n, by (1.1), m = [L : F ]n, which contradicts with (m, n) = 1. Thus f should be irreducible in L[X]. 2 Factor into irreducibles 2.1 (a) 16 Question: Factor X X into irreducible polynomials over F4 and over F8. − 16 Solution: In the textbook, we know that over F2, the X X can be factored into irreducible polynomials as − X16 X = X(X 1)(X2 + X +1)(X4 + X +1)(X4 + X3 +1)(X4 + X3 + X2 + X +1). (2.1) − − In order to get the factorization over F4, we just need to do factoring based on (2.1). ∗E-mail address: [email protected]; Tel: 765 237 7149 1 Yingwei Wang Abstract Algebra 2 Suppose F4 = 0, 1,α,α . It is easy to know that over F4, { } X2 + X +1=(X α)(X α2), (2.2) − − X4 + X +1=(X2 X + α)(X2 + X + α2), (2.3) − X4 + X3 +1=(X2 + αX + α)(X2 α2X + α2), (2.4) − X4 + X3 + X2 + X +1=(X2 αX + 1)(X2 + α2X + 1). (2.5) − Now, we know that over F4, X16 X − = X(X 1)(X α)(X α2)(X2 X + α)(X2 + X + α2) − − − − (X2 + αX + α)(X2 α2X + α2)(X2 αX + 1)(X2 + α2X + 1). − − For F8, we know that 3 2 F8 = F2[X]/(X + X + 1), = 0, 1, x, x +1, x2, x2 +1, x2 + x, x2 + x +1 . { } 2.2 (b) Question: Use Maple to factor X80 1 mod 3. How many irreducible polynomials of degree − 4 are there in Z3[X]? What about degree 3? Solution: The answer is 71 X80 1= X(X + 2) 1+ Xk . − =5 ! Xk There are 18 irreducible polynomials of degree 4 are there in Z3[X]. And there are 8 irreducible polynomials of degree 4 are there in Z3[X]. 2.3 (c) Question: Display the set of all irreducible polynomials of degree 2 over a field of cardinality 9. Solution: 2 Yingwei Wang Abstract Algebra 3 Irreducible polynomials over finite field 3.1 (a) Question: Show that the polynomials f(X) = X3 + X2 + 1 and g(X) = X3 + X + 1 are irreducible over the field F4 with four elements. Proof. We know that F2 = 0, 1 , and it is easy to check that both f(X) and g(X) are { } irreducible in F2[X], whose degrees are 3. Besides, [F4 : F2] = 2. According to Problem 1, we know that f(X) and g(X) are irreducible in F4[X]. 3.2 (b) Question: Describe explicitly an isomorphism between the field F4[X]/(f(X)) and F4[Y ]/(g(Y )), both of which have cardinality 64. Solution: The isomorphism is given by Y = X +1. We just need to check that α(x) = x +1 is a root of f(X) = X3 + X2 + 1 in the field F4[X]/(g(X)). f(x +1) =(x + 1)3 +(x + 1)2 +1, = x3 +4x2 +5x +3, = x3 + x +1, =0. 4 Minimal polynomial Question: Determine the minimal polynomial for α = √3+ √5 over each of the following: 4.1 (a): Q. Solution: α = √3+ √5, α2 =8+2√15, ⇒ α2 8=2√15, ⇒ − α2 16α2 +64=60, ⇒ − α2 16α2 +4=0, ⇒ − which means the minimal polynomial of α over Q is X4 16X2 +4. − 3 Yingwei Wang Abstract Algebra 4.2 (b): Q(√5). 2 Solution: Suppose α + aα + b = 0, where a = a1 + a2√5, b = b1 + b2√5. It is easy to find that a1 =0, a2 = 2, b1 =2, b2 = 0. Then the minimal polynomial of α over Q√5 is − X2 2√5X +2. − 4.3 (c): Q(√10). Solution: This case is similary to part (a). The minimal polynomial of α over Q√10 is X4 16X2 +4. − 4.4 (d): Q(√15). Solution: Since α2 =8+2√15, the minimal polynomial of α over Q√15 is X2 8 2√15. − − 5 Algebraic extension Let F be a field, and let α be an element which generates a field extension of F of degree 5. Prove that α2 generates the same extension. Proof. First, claim that α2 / F and further [F (α2): F ] > 1. 2 ∈ 2 2 If α F , then f(X) = X α would be the minimal polynomial of α over F , which implies [F∈(α): F ] = 2. It contradicts− with the fact that F (α) is a field extension of F of degree 5. Seccond, since αinF (α) and F (α) is a field, then α2 F (α). So F (α2) F (α). Now we have ∈ ⊂ F F (α2) F (α) ⊂ ⊂ [F (α): F ] = [F (α): F (α2)][F (α2): F ], ⇒ 5 = [F (α): F (α2)][F (α2): F ]. ⇒ Since [F (α2): F ] > 1, we can get [F (α): F (α2)]=1, [F (α2): F ]=5. It implies that α2 generates the same extension as α. 4 Yingwei Wang Abstract Algebra 6 Algebraic extension Decide whether or not i = √ 1 is in the field − 6.1 (a): Q(√ 2). − No. 6.2 (b): Q(√4 2). − No. 6.3 (c): Q(√4 4). − Yes. Since (√4 4)2 = √ 4=2√ 1, − − − 1 4 4 √ 1= (√ 4)2 Q(√ 4). ⇒ − 2 − ∈ − 6.4 (b): Q(α) where α3 + α = 1. − No. Since we know that [Q(α): Q] = 3 and [Q(√ 1) : Q]=2. If √ 1 Q(α), then we have − − ∈ Q Q(√ 1) Q(α), ⊂ − ⊂ [Q(α): Q] = [Q(α): Q(√ 1)][Q(√ 1) : Q], ⇒ − − 3 = [Q(α): Q(√ 1)] 2, ⇒ − which is impossible. 7 Algebraic extension Question: Let α, β be complex numbers of degree 3 over Q, and let K = Q(α, β). Determine the possibilities for [K : Q]. Solution: It is easy to know that [K : Q] 9. Besides, ≤ Q Q(α) Q(α, β), ⊂ ⊂ [K : Q] = [Q(α, β): Q(α)][Q(α): Q], ⇒ [K : Q] = [Q(α, β): Q(α)] 3. ⇒ It follows that the possibilities for [K : Q] is 3, 6, 9. 5 Yingwei Wang Abstract Algebra 8 Straightedge and compass construction Determine whether or not the regular 9-gon is constructible by ruler and compass. Solution: It is imposible to consctruct the regular 9-gon. To construct a regular 9-gon is the same as constructing the complex number ζ = e2π/9. Then the minimal polynomial for ζ over Q is f(X)= X6 + X3 +1. It follows that [Q(ζ): Q]= ϕ(9) = 6. But we can not find any integer k such that 2k =6. So it is imposible to consctruct the regular 9-gon. 9 Finite field 9.1 (a) Let F be a finite field of characteristic p, and let ϕ : F F be the map defined by ϕ(x)= xp. → Show that ϕ is an automorphism of F . Proof. First, it is easy to know that x, y F , ∀ ∈ ϕ(x + y)=(x + y)p = xp + yp = ϕ(x)+ ϕ(y), ϕ(xy)= xpyp = ϕ(x)ϕ(y). It implies that ϕ is a homomorphism. Second, since ϕ(x)= xp =0 x = 0, we know that ker ϕ = 0. ⇒ Now we can conclude that ϕ is an automorphism of F . 9.2 (b) Show that every automorphism of F is a power of ϕ. Proof. First, considering only the multiplicative structure, claim that any automorphism must be of the form xxa, for some fixed a< F . Second, write a| =| peb with (p, b) = 1. Claim that every element of F is a root of the polynomial e (Xp + 1)b Xa 1 − − (of degree ¡ F ). Further, we can conclude that b = 1. | | 6 MA 553: Homework 11 Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA 1 Splitting field Determine the splitting field and its degree over Q. 1.1 x4 2. − We know that 4 4 4 4 f(x)= x4 2=(x √2)(x + √2)(x i√2)(x + i√2). − − − 4 4 Let K = Q(√2, i)= Q(√2)(i). On one hand, K contains all the roots of f(x)= x4 2. Hence it contains the splitting field of f(x). − 4 4 4 i √2 On the other hand, since i√2 and √2 are contained in the splitting field, then i = 4 √2 should also be contained in the splitting field, which means the splitting field contains K 4 since K is the smallest field containing Q, i and √2. 4 Now we can conclude that K = Q(√2, i) is the splitting field of f(x)= x4 2. 4 2 − Besides, the minimal polynomial of i over Q(√2) is p1(x)= x + 1; while the minimal 4 4 polynomial of √2 over Q is p2(x)= x 2. Now we have − 4 4 4 [K : Q] = [Q(√2, i): Q(√2)][Q(√2) : Q]=2 4=8. × ∗E-mail address: [email protected]; Tel: 765 237 7149 1 Yingwei Wang Abstract Algebra 1.2 x4 +2. Suppose α is a root of g(x)= x4 + 1, then π π √2 α = cos + i sin = (1 + i). 4 4 2 It is easy to check that α2 = i and (α3)2 = i. We know that − 4 4 4 4 f(x)= x4 +2=(x2 i√2)(x2 + i√2)=(x √2α)(x + √2α)(x √2α3)(x + √2α3). − − − 4 4 Let K = Q(√2α, i)= Q(√2α)(i) and S = the splitting field of f(x)= x4 + 2. On one hand, K contains all of the roots of f(x), which means K S. On the other hand, ⊇ 4 4 √2α, √2α3 S, ∈ √4 2α3 = α2 = i S, ⇒ √4 2α ∈ 4 K = Q(√2α, i) S, ⇒ ⊆ 4 since K is the smallest field containing Q, √2α and i. 4 Now we can conclude that K = Q(√2α, i) is the splitting field of f(x)= x4 + 2. 4 4 Besides, the minimal polynomial of √2α over Q(i) is p1(x)= x + 2; while the minimal 2 polynomial of i over Q is p2(x)= x + 1. Now we have 4 [K : Q] = [Q(√2α, i): Q(i)][Q(i): Q]=4 2=8. × 1.3 x4 + x2 +1. We know that f(x)= x4 + x2 +1=(x2 + x + 1)(x2 x + 1). − 1 √3 2 1 √3 2 Suppose ω = 2 + 2 i, then ω = 2 2 i. It is easy to know that ω and ω are roots 2 − − − of g1(x)= x + x + 1. 1 √3 2 1 √3 2 Besides, w = 2 2 i and ω = 2 + 2 i. It is easy to know that ω and ω are roots of g (x−)= x2 −x + 1. − − − 2 − Now we know that f(x)=(x ω)(x ω2)(x + w)(x + w2). − − 2 Yingwei Wang Abstract Algebra It is easy to check that the splitting field of f(x) is K = Q(ω). It follows that [K : Q]=2. 1.4 x6 4. − We know that f(x)= x6 4=(x3 2)(x3 + 2). − − Let ω = 1 + √3 i, then ω2 = 1 √3 i. It is easy to know that ω3 = 1. Then − 2 2 − 2 − 2 3 3 3 3 3 3 f(x)=(x √2)(x ω√2)(x ω2√2)(x + √2)(x + ω√2)(x + ω2√2). (1.1) − − − 3 Let K = Q(√2,ω). On one hand, since K contains all roots of f(x), it must contain the splitting field of f(x). 3 3 3 ω √2 On the other hand, the splitting field contains √2 and ω√2; hence ω = 3 is contained √2 in the splitting field. It follows that the splitting field contains K since K is the smallest 3 field containing Q, √2 and ω. 3 Now we know that K = Q(√2,ω) is the splitting field of f(x)= x6 4. 3 2− Furthermore, the minimal polynomial of ω over Q(√2) is p1(x)= x + x + 1; while the 3 3 minimal polynomial of √2 over Q is p2(x)= x 2. Now we have − 3 3 3 [K : Q] = [Q(√2)(ω): Q(√2)][Q(√2) : Q]=2 3=6. × 2 Splitting field Lemma 2.1. Let K be a finite extension of F . Prove that K is a splitting field over F if and only if every irreducible polynomial in F [x] that has a root in K splits completely in K[x]. Question: Let K1 and K2 be finite extensions of F contained in the field K, and assume both are splitting field over F . 3 Yingwei Wang Abstract Algebra 2.1 (a) Prove that their composite K1K2 is a splitting field over F . Proof. Since K1 and K2 are finite extensions of F , we can suppose K = F (α ,α , ,αn), 1 1 2 ··· K = F (β , β , , βm). 2 1 2 ··· Then it is easy to know that K K = F (α ,α , ,αn, β , β , , βm). 1 2 1 2 ··· 1 2 ··· By Lemma 2.1, we just need to show that every irreducible polynomial in F [x] that has a root in K1K2 splits completely in K1K2[x]. Suppose there exists a polynomial f(x) F [x] which has a root αiβj K K , but it ∈ ∈ 1 2 can not splits completely in K1K2[x]. Then f(x)= (x αiβj)f˜(x) Y − Then we can find an irreducible polynomial f˜(x) F [x] such that f˜ has a root in K ∈ 1 but can not split in K1[x], which contradicts with the fact that K1 is a splitting field over F . Therefore, K1K2 is a splitting field over F . 2.2 (b) Prove that K K is a splitting field over F . 1 ∩ 2 Proof. Let f(x) be irreducible in F [x] which has a root γ K K . By Lemma 2.1, ∈ 1 ∩ 2 f(x) splits completely over both K1 and K2. Therefore, f(x) splits splits completely over K K , which means K K is a splitting field over F . 1 ∩ 2 1 ∩ 2 3 Wilson Theorem Prove that an integer p> 1 is prime if and only if (p 1)! 1( mod p). − ≡ − 4 Yingwei Wang Abstract Algebra n n Proof. Let F be a finite field with char(F) = p and F = p . So F∗ = p 1 where | | | | − F∗ = F 0 . Let \{ } pn 1 f(x)= x − 1. (3.1) − n On one hand, by Lagrange’s theorem, if α F∗, then the order of α divides p 1. So pn 1 pn 1 ∈ − α − =1 or α − 1 = 0. It implies that α F∗, f(α) = 0. On the other hand,− since the degree of ∀f(x∈) is pn 1, then f(x) has at most pn 1 roots. − − No we can conclude that f(x) splits completely over the field F, which means f(x)= (x α). (3.2) ∗ − αYF ∈ Let x = 0 in (3.2), we can get n 1, p =2 α =( 1)p = (3.3) ∗ − 1, p is odd . αYF ∈ − Now we can prove the conclusion that an integer p> 1 is prime if and only if (p 1)! 1( mod p). − ≡ − If p = 2, the it is obviously true. If p is odd and n = 1, then F∗ ∼= Zp∗ = 1, 2, 3, ,p 1; mod p . By (3.3), we know that { ··· − } (p 1)! 1( mod p). (3.4) − ≡ − Conversely, if (3.4) is true, then kp 1=(p 1)!. − − If p> 3, then (p 1)! is even, so p must be odd. If p is not a prime,− say p = rs. Then rs (p 1)!, | − since r 5 Yingwei Wang Abstract Algebra 4 Even polynomials Let f(X) Q[X] be an irreducible polynomial of degree n> 2 such that ∈ f(X)= f( X). (4.1) − Prove that the Galois group of f is not the symmetric group Sn. Proof. By the (4.1), we know that the degree of f(X) is an even number, say n = 2k. Besides, since the f(X) is irreducible over Q, then the roots of f(X) should appear in pairs, say S = w , w ,w , w , ,wk, wk , { 1 − 1 2 − 2 ··· − } where S = n =2k. | | Let G be the Galois group of f, then G Sn. If σ G such that σ (w )= w (or we ⊂ ∈ 1 1 2 say if σ Sn contains a two cycle (12)) then σ should satisfy σ( w ) = w since σ is ∈ − 1 − 2 an automorphism of the splitting field of f. It implies that the permutation (12) Sn but ∈ (12) / G, which means G $ Sn. ∈ 5 Galois group Let K be the splitting field over Q of f(x)= x4 2x2 1. − − 5.1 (a) Determine the Galois group of K/Q. Proof. The resolvent cubic for f(x) is g(x) = x3 +4x2 +8x, = x(x2 +4x + 8), = x(x +2+2i)(x +2 2i). − The discriminant of g(x) is D = 322. − Just need to consider whether f(x) is reducible over Q(32i)= Q(i). 6 Yingwei Wang Abstract Algebra We know that f(x) = x4 2x2 1, − − =(x2 +(√2 1))(x2 (1 + √2)), − − = x +(√2 1)i x (√2 1)i x 1+ √2 (x + 1+ √2 , − − − − q q where i = √ 1. − It is clear that f(x) is irreducible over Q(i). It follows that the Galois group of K/Q is D = 1, (12)(34), (13)(24), (14)(23), (12), (34), (1324), (1423) . 8 { } 5.2 (b) Show that the only three subfields of K having degree 2 over Q are Q(√ 1), Q(√2) and − Q(√ 2). − Proof. Let r =(√2 1)i = √ 2 √ 1, 1 − − − − r = (√2 1)i = √ 2+ √ 1, 2 − − − − − r = 1+ √2, 3 q r = 1+ √2. 4 −q Let σ = (1324), τ = (12). It is easy to know that D8 has only three subgroup of order 4, S = 1, σ, σ2, σ3 , { } T = 1, σ2,τ,σ2τ , { } U = 1, σ2,στ,σ2τ . { } The corresponding subfields are KS = Q(√ 1), − KT = Q(√2), KU = Q(√ 2). − 7 Yingwei Wang Abstract Algebra 6 Resolvent cubic Let k be a field of characteristic = 2, and let f k[X] be an irreducible polynomial of 6 ∈ degree 4. If r1,r2,r3 and r4 are the roots of f (in some splitting field), then the polynomial g whose roots are p1 = r1r2 + r3r4, p2 = r1r3 + r2r4, p3 = r1r4 + r2r3. is called the resolvent cubic of f. 6.1 (a) Show that the the discriminant of f is the same as that of g. Proof. It is easy to get that p p =(r r )(r r ), (6.1) 1 − 2 1 − 4 2 − 3 p p =(r r )(r r ), (6.2) 1 − 3 1 − 3 2 − 4 p p =(r r )(r r ). (6.3) 2 − 3 1 − 2 3 − 4 By the definition of the discriminant, we can know that discriminant of f is the same as that of g. 6.2 (b) Let G S be the galois group of f , and let V ⊳ S be the unique normal subgroup of ⊂ 4 4 order 4. Prove that the fixed field T of V G is a splitting field of g. ∩ Proof. It is easy to know that V = (1), (12)(34), (13)(24), (14)(23) . { } Since G V G ⊳ G. ∩ Let L = k[r1,r2,r3,r4], then G V T = L ∩ = x L gx = x, g G V . { ∈ | ∀ ∈ ∩ } 8 Yingwei Wang Abstract Algebra Let S = k[p1,p2,p3] be the splitting field of g over k and denote the galois group of S/k as GS. On one hand, by the definition of V , we know that any permutation in V fixes p1,p2,p3, so GS G V . On the other hand, if σ G but σ / G V , then σ moves at least one of ⊇ ∩ ∈ ∈ ∩ the p1,p2,p3, which means σ / GS. Therefore, GS = G V . By fundamental theorem of∈ Galois theory, we can conclude∩ that T = S. 6.3 (c) Let t = [T : k] (see (b)). Prove that G = S4, A4 or V according as t =6, 3 or 1. What are the possibilities for G when t = 2? Proof. First of all, by the fundamental theorem of Galois theory, we know that t = [T : k] = [G : V G]. ∩ Now let us consider t =6, 3, 1 or 2 respectively. 1. t = 6 or 3. In these two cases, since G = t V G , 3 is a divisor of G . Besides, 4 is also a | | | ∩ | | | divisor of G. Hence, G is a multiply of 12. Notice that S = 24. So G must be S | | | 4| 4 or A4. Furthermore, [S : S V ] = [S : V ]=24/4=6, 4 4 ∩ 4 [A : A V ] = [A : V ]=12/4=3. 4 4 ∩ 4 It follows that if t = 6 then G = S4 while if t = 3 then G = A4. 2. t = 1. In this case, G = G V , so G 3. If t = 2, then G = D8 or Z4. In this case, G =2 G V . Since V = 4, we have G V =1, 2, 4. | | | ∩ | | | | ∩ | (a) If G V = 1, then G = 2, which contradicts with the face that G is a multiple| ∩ of| 4. | | | | (b) If G V = 2, then G = 4, which implies that G = Z . | ∩ | | | 4 9 Yingwei Wang Abstract Algebra (c) If G V = 4, then G = 8. But a subgroup of S4 of order 8 is a Sylow 2-subgroup,| ∩ | and all of| such| subgroup are conjugate and therefore isomorphic. One of theses subgroups is D8. 6.4 (d) Can the roots of f(X)= X4 + X 5 Q[X] be constructed with ruler and compass? − ∈ Solution: The resolvent cubic of f(X) is g(X)= X3 + 20X +1, which is irreducible over Q. Besides, the discriminant is D = 32027, which is a prime. So the galois group of f(x) is S4. Let L be the splitting field of f over Q, then [F : Q] = 24 is not a power of 2. It follows that roots of f(X) can not be constructed with ruler and compass. 6.5 (e) Lemma 6.1. Suppose that f(x) = x4 + ax2 + b Q[x] is irreducible and G is its galois group. Then ∈ 1. If √b Q, then G = V , the Klein 4-group; ∈ ∼ 2 2. If √a 4b√b Q, then G = C, the cyclic group Z4; − ∈ ∼ 3. otherwise, G ∼= D8. Proof. By Problem 4 of this homework, we know that G = S . Furthermore, G = A since 6 4 6 4 we have shown that (12) / G. The possible G’s are V,C,D8. Suppose the roots of f(x) are α, β, α, β which satisfy∈ the following relations: − − αβ = √b, α2 β2 = √a2 4b, − − α3β βα3 = √a2 4b√b. − − Now we have the discussion here. 10 Yingwei Wang Abstract Algebra √ 1. If √b Q, then αβ Q. Let σ G be such that σ(α) = β, then σ(β) = b = α. ∈ ∈ ∈ σ(α) Similarly, if σ(α) = β, then σ( β) = α; Finally, if σ(α) = α, then σ(β) = β. − − − − Thus every element of G has order 2. It implies that G ∼= V , the Klein 4-group. 2. If √a2 4b√b Q, then α3β βα3 Q. Let σ G be such that σ(α) = β. If σ(β)=−α, then ∈ − ∈ ∈ σ(α3β βα3)= β3α αβ3, − − which is impossible. Therefore σ(β) = α. It implies that σ is of order 4, which − means G ∼= C, the cyclic group Z4. 3. We know that the splitting field must contain Q(√b), Q(√a2 b) and Q(√a2 4b√b). − − The irreducibility of the polynomial implies that √a2 4b / Q . Therefore if √b, − ∈ √a2 4b√b / Q, the splitting field contains at least three subfields of degree 2. − ∈ Hence the either G ∼= K4 or G ∼= D8. However, if G ∼= K4, then αβ is fixed by any element of G. Since √b / Q, the only possibility is G = D8. ∈ ∼ Using the Lemma 6.1, we can determine the galois group for the minimal polynomial over Q of these numbers. 6.5.1 3+2√2 p x = 3+2√2, q x2 =3+2√2, ⇒ x2 3=2√2, ⇒ − (x2 3)2 =8, ⇒ − x4 6x2 +1=0. ⇒ − So the minimal polynomial is x4 6x2 + 1. − Besides, since 1 Q, the galois group is V , the Klein 4-group. ∈ 11 Yingwei Wang Abstract Algebra 6.5.2 7+2√10 p x = 7+2√10, q x2 =7+2√10, ⇒ x2 7=2√10, ⇒ − (x2 7)2 = 40, ⇒ − x4 14x2 +9=0. ⇒ − So the minimal polynomial is x4 14x2 + 9. − Besides, since √9 Q, the galois group is V , the Klein 4-group. ∈ 6.5.3 5+2√5 p x = 5+2√5, q x2 =5+2√5, ⇒ x2 5=2√5, ⇒ − (x2 5)2 = 20, ⇒ − x4 10x2 +5=0. ⇒ − So the minimal polynomial is x4 10x2 + 5. 2 − Besides, since √10 4 5√5=20 Q, the galois group is C, the cyclic group Z4. − × ∈ 6.5.4 5+2√21 p x = 5+2√21, q x2 =5+2√21, ⇒ x2 5=2√21, ⇒ − (x2 5)2 = 84, ⇒ − x4 10x2 49=0. ⇒ − − So the minimal polynomial is x4 10x2 49. − 2 − Besides, since √ 49 / Q and √10 +4 49√ 49 / Q, the galois group is D8. − ∈ × − ∈ 12 MA 553: Homework 12 Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA 1 Problem 40(c), Page 622 Question: Express the symmetric function as a polynomial in the elementary symmetric functions. 2 2 2 2 2 2 f(x1, x2, x3)= x1x2 + x1x3 + x2x3. Solution: Since n = 3, we know that s1 = x1 + x2 + x3, s2 = x1x2 + x1x3 + x2x3, s3 = x1x2x3. By the procesure in Exercise 38, we know that the first step is to compute 0 2 f(x1, x2, x3) s1s2 2 2 2 2 − 2 2 2 = x1x2 + x1x3 + x2x3 (x1x2 + x1x3 + x2x3) , 2 2 − 2 = 2x1x2x3 2x1x2x3 2x1x2x3. − − − The second step is 0 2 0 f(x1, x2, x3) s1s2 +2s1s2s3, 2 − 2 2 = 2x1x2x3 2x1x2x3 2x1x2x3 + 2(x1 + x2 + x3)x1x2x3, − − − = 0. Now we know that 2 f(x1, x2, x3)= s2 2s1s3. − ∗E-mail address: [email protected]; Tel: 765 237 7149 1 Yingwei Wang Abstract Algebra 2 Problem 48, Page 623 Determine the splitting field K for the polynomial f(x)= x6 2x3 2, − − over Q. 2.1 (a) Prove that f(x) is irreducible over Q with roots the three cube roots of 1 √3. ± Proof. By Eisenstein’s criterion, we know that f(x) is irreducible over Q. Besides, it is easy to get f(x)=(x3 1 √3)(x3 1+ √3). (2.1) − − − 3 3 1 √3 Let α = √3+1, β = √3 1 and ω = 2 + 2 i, then p p − − f(x)=(x α)(x ωα)(x ω2α)(x + β)(x + ωβ)(x + ω2β). (2.2) − − − 3 (b) Prove that K contains the field Q(√ 3) of 3rd roots of unity and containsQ(√3), and hence − contains the biquadratic field F = Q(i, √3). Furthermore, conclude that K is an extension of the field L = Q(√3 2, i, √3). Proof. First of all, 3 α √3+1 = =2+ √3 K, β √3 1 ∈ − Q(√3) K. ⇒ ∈ Second, α,ωα K, ∈ ωα 1 √ 3 = ω = + − K, ⇒ α −2 2 ∈ Q(√ 3) = Q(√3i) K. ⇒ − ∈ Now we know that K F = Q(i, √3). ⊃ 2 Yingwei Wang Abstract Algebra Furthermore, 3 3 αβ = (√3+1)(√3 1) = √2 K, − ∈ q3 L = Q(√2, i, √3) K, ⇒ ∈ 3.1 (c) Prove that [L : Q] = 12 and that K is obtained from L by adjoining the cube root of an element in L, so that [K : Q] = 12 or 36. Proof. It is easy to know that 3 [L : Q] = [Q(√2, i, √3) : Q(i, √3)][Q(i, √3) : Q(i)][Q(i): Q], =3 2 2=12. × × Besides, since K is obtained from L by adjoining the cube root of 1 + √3 L, then [K : Q] = 12 or 36. ∈ 3.2 (d) Prove that if [K : Q] = 12 then K = Q(√3 2, i, √3) and that Gal(K/Q) is isomorphic to the direct product of the cyclic group of order 2 and S3. Furthermore, if [K : Q] = 12 then there is a unique real cubic subfield in K, namely √3 2. Proof. We know that K L Q and [L : Q] = 12. So if [K : Q] = 12 then K = L = Q(√3 2, i, √3). ⊃ ⊃ It is easy to know that L is the splitting field of g(x) over Q, where g(x) defined by g(x)=(x3 2)(x2 3). − − 3 2 The galois group of x 2 is S3 while the galois group of x 3 is Z2. It follows that − − Gal (K/Q)= Z2 S3. × 3 3 Furthermore, [(Z2 S3):(Z2 Z2)] = 3 and the fixed field of Z2 Z2 is Q(√2). So √2 is the unique real cubic× subfield in K× if [K : Q] = 12. × 3 Yingwei Wang Abstract Algebra 3.3 (e) 3 3 3 Show that 2+ √3, 2 √3 R are both elements of K. Show that γ = 2+ √3+ 3 − ∈ 2 √3 isp a real rootp of the irreducible cubic equation x3 3x 4, whose discriminantp is p224−4. Conclude that the galois closure of Q(γ) contains Q(i−) so in− particular Q(γ) = √3 2. − 6 Proof. First of all, α 3 β 3 = 2+ √3 K, = 2 √3 K, β q ∈ α q − ∈ 3 3 γ = 2+ √3+ 2 √3 K. ⇒ q q − ∈ Besides, the minimal polynomial of γ over Q is h(x)= x3 3x 4. − − It is easy to check that h(x) has one real root and two complex roots. So the Galois closure of Q(γ) contains Q(i). In particular, Q(γ) = √3 2. 6 3.4 (f) Determine all the elements of G = Gal(K/Q) explicitly and in particular show that G is isomorphic to S3 S3. × Proof. Form part (e), we know that G = [K : Q] = 36. | | 4 Yingwei Wang Abstract Algebra 4 Finite fields Let L K be finite fields, c := K , and let f(X) K[X] be irreducible, of degree e dividing [L : K].⊃ Show that there is an a| |L such that in L∈[X], ∈ 2 e−1 f(X)=(X a)(X ac)(X ac ) (X ac ). − − − ··· − How many such a are there? Proof. First, suppose c = p is a prime. Then K = Fp, L = Fpm and L/K is a Galois extension. th p Besides, the Galois group Gal(L/K) is cyclic and a generator is the p power map:ϕp : t t . → Since e [L : K], the splitting field of f(x) over K is contained in L. Let a L is a root of f(x), then| other roots can be obtained from a by applying Gal(K(a)/K) to this∈ root. Since 2 the Galois group is generated by the pth power map, the roots of f(x) are a, ap, ap , . Once pe pe ···e we reach a , we have cycled back to the start a = a since K(a) ∼= K/f has order p . The polynomial f is separable since its roots lie in a Galois extension K(α)/K. Besides, since its degree is d, its different roots must be p p2 pe 1 a, a , a , , a − . ··· n Second, suppose c = p and [L : K]= d Then K = Fc and L = Fcd . Similarly, L/K is also a Galois extension and the Galois group GalL/K is cyclic with generator the cth power map q ϕc : t t . Similarly, we can show that if f(x) is irreducible with degree e d, then there is a L →such that a is a root of f and the full set of roots is | ∈ c c2 ce 1 a, a , a , , a − . ··· Since K(a) L and [K(a): K] = e, [L : K] = d, the number of such a should be d e + 1. ⊂ − 5