Lctur 6.4: Galois groups Matthw Macauly Dpartmnt of Mathmatical Scincs Clmson Univrsity http://www.math.clmson.du/~macaul/ Math 4120, Modrn Algbra M. Macauly (Clmson) Lctur 6.4: Galois groups Math 4120, Modrn Algbra 1 / 7
Th Galois group of a polynomial Dfinition Lt f Z[x] b a polynomial, with roots r 1,..., r n. Th splitting fild of f is th fild Q(r 1,..., r n). Th splitting fild F of f (x) has svral quivalnt charactrizations: th smallst fild that contains all of th roots of f (x); th smallst fild in which f (x) splits into linar factors: f (x) = (x r 1)(x r 2) (x r n) F [x]. Rcall that th Galois group of an xtnsion F Q is th group of automorphisms of F, dnotd Gal(F ). Dfinition Th Galois group of a polynomial f (x) is th Galois group of its splitting fild, dnotd Gal(f (x)). M. Macauly (Clmson) Lctur 6.4: Galois groups Math 4120, Modrn Algbra 2 / 7
A fw xampls of Galois groups Th polynomial x 2 2 splits in Q( 2), so Gal(x 2 2) = Gal(Q( 2)) = C 2. Th polynomial x 2 + 1 splits in Q(i), so Gal(x 2 + 1) = Gal(Q(i)) = C 2. Th polynomial x 2 + x + 1 splits in Q(ζ), whr ζ = 2πi/3, so Gal(x 2 + x + 1) = Gal(Q(ζ)) = C 2. Th polynomial x 3 1 = (x 1)(x 2 + x + 1) also splits in Q(ζ), so Gal(x 3 1) = Gal(Q(ζ)) = C 2. Th polynomial x 4 x 2 2 = (x 2 2)(x 2 + 1) splits in Q( 2, i), so Gal(x 4 x 2 2) = Gal(Q( 2, i)) = V 4. Th polynomial x 4 5x 2 + 6 = (x 2 2)(x 2 3) splits in Q( 2, 3), so Gal(x 4 5x 2 + 6) = Gal(Q( 2, 3)) = V 4. Th polynomial x 3 2 splits in Q(ζ, 3 2), so Gal(x 3 2) = Gal(Q(ζ, 3 2)) = D 3??? M. Macauly (Clmson) Lctur 6.4: Galois groups Math 4120, Modrn Algbra 3 / 7
Th towr law of fild xtnsions Rcall that if w had a chain of subgroups K H G, thn th indx satisfis a towr law: [G : K] = [G : H][H : K]. Not surprisingly, th dgr of fild xtnsions obys a similar towr law: Thorm (Towr law) For any chain of fild xtnsions, F E K, [K : F ] = [K : E][E : F ]. W hav alrady obsrvd this in our subfild lattics: [Q( 2, 3) : Q] = [Q( 2, 3) : Q( 2) ][ Q( 2) : Q ] = 2 2 = 4. }{{}}{{} min. poly: x 2 3 min. poly: x 2 2 Hr is anothr xampl: [Q(ζ, 3 2) : Q] = [Q(ζ, 3 2) : Q( 3 2) ][ Q( 3 2) : Q ] = 2 3 = 6. }{{}}{{} min. poly: x 2 +x+1 min. poly: x 3 2 M. Macauly (Clmson) Lctur 6.4: Galois groups Math 4120, Modrn Algbra 4 / 7
Primitiv lmnts Primitiv lmnt thorm If F is an xtnsion of Q with [F : Q] <, thn F has a primitiv lmnt: som α Q for which F = Q(α). How do w find a primitiv lmnt α of F = Q(ζ, 3 2) = Q(i 3, 3 2)? Lt s try α = i 3 3 2 F. Clarly, [Q(α) : Q] 6. Obsrv that α 2 = 3 3 4, α 3 = 6i 3, α 4 = 18 3 2, α 5 = 18i 3 4 3, α 6 = 108. Thus, α is a root of x 6 + 108. Th following ar quivalnt (why?): (i) α is a primitiv lmnt of F ; (ii) [Q(α) : Q] = 6; (iii) th minimal polynomial m(x) of α has dgr 6; (iv) x 6 + 108 is irrducibl (and hnc must b m(x)). In fact, [Q(α): Q] = 6 holds bcaus both 2 and 3 divid [Q(α): Q]: [Q(α): Q] = [Q(α): Q(i 3)] [Q(i 3): Q], [Q(α): Q] = [Q(α): Q( 3 2)] [Q( 3 2): Q]. }{{}}{{} =2 =3 M. Macauly (Clmson) Lctur 6.4: Galois groups Math 4120, Modrn Algbra 5 / 7
An xampl: Th Galois group of x 4 5x 2 + 6 Th polynomial f (x) = (x 2 2)(x 2 3) = x 4 5x 2 + 6 has splitting fild Q( 2, 3). W alrady know that its Galois group should b V 4. Lt s comput it xplicitly; this will hlp us undrstand it bttr. W nd to dtrmin all automorphisms φ of Q( 2, 3). W know: φ is dtrmind by whr it snds th basis lmnts {1, 2, 3, 6}. φ must fix 1. If w know whr φ snds two of { 2, 3, 6}, thn w know whr it snds th third, bcaus φ( 6) = φ( 2 3) = φ( 2) φ( 3). In addition to th idntity automorphism, w hav { φ2( 2) = 2 ( 3) = 3 { φ3( 2) = 2 ( 3) = 3 { φ4( 2) = 2 ( 3) = 3 Qustion What gos wrong if w try to mak φ( 2) = 3? M. Macauly (Clmson) Lctur 6.4: Galois groups Math 4120, Modrn Algbra 6 / 7
An xampl: Th Galois group of x 4 5x 2 + 6 Thr ar 4 automorphisms of F = Q( 2, 3), th splitting fild of x 4 5x 2 + 6: : a + b 2 + c 3 + d 6 a + b 2 + c 3 + d 6 : a + b 2 + c 3 + d 6 a b 2 + c 3 d 6 : a + b 2 + c 3 + d 6 a + b 2 c 3 d 6 : a + b 2 + c 3 + d 6 a b 2 c 3 + d 6 Thy form th Galois group of x 4 5x 2 + 6. Th multiplication tabl and Cayly diagram ar shown blow. y 3 2 2 3 x Exrcis Show that α = 2 + 3 is a primitiv lmnt of F, i.., Q(α) = Q( 2, 3). M. Macauly (Clmson) Lctur 6.4: Galois groups Math 4120, Modrn Algbra 7 / 7