Galois Theory Through Exercises
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Contents 1 Solving algebraic equations ............................................ 1 2 Field extensions ....................................................... 9 3 Polynomials and irreducibility ......................................... 13 4 Algebraic extensions ................................................... 17 5 Splitting fields. Finite fields ............................................ 23 6 Automorphism groups of fields. Galois groups ......................... 29 7 Normal extensions ..................................................... 35 8 Separable extensions ................................................... 39 9 Galois extensions ...................................................... 43 10 Cyclotomic extensions ................................................. 51 11 Galois modules ........................................................ 57 12 Solvable groups ........................................................ 63 13 Solvability of equations ................................................ 67 X Contents 14 Geometric constructions ............................................... 71 15 Computing Galois groups .............................................. 75 16 Supplementary problems .............................................. 83 17 Proofs of the theorems ................................................ 93 18 Hints and answers ..................................................... 135 19 Examples and selected solutions ....................................... 163 APPENDIX: Groups, rings and fields ..................................... 221 A.1 Equivalence relations................................................. 221 A.2 Groups............................................................. 222 A.3 Rings.............................................................. 231 A.4 Fields.............................................................. 237 A.5 Chinese Remainder Theorem.......................................... 241 A.6 Polynomial rings.................................................... 241 A.7 Modules over rings.................................................. 243 A.8 Group actions on sets................................................ 246 A.9 Permutations....................................................... 251 A.10 Some arithmetical functions.......................................... 254 A.11 Symmetric polynomials.............................................. 257 A.12 Roots of unity...................................................... 259 A.13 Transitive subgroups of permutation groups............................. 259 A.14 Zorn's Lemma...................................................... 262 A.15 Dual abelian groups................................................. 263 References ................................................................. 264 List of notations ........................................................... 265 Index ...................................................................... 268 Contents XI 1 Solving algebraic equations An algebraic equation of degree n with complex coefficients is an equation: n n−1 f(X) = a0X + a1X + ··· + an−1X + an = 0; where ai 2 C, n ≥ 0 and a0 6= 0 (if n = 0, f(X) is a constant polynomial). According to the Fundamental theorem of algebra (proved by C.F. Gauss1 in 1799), if f(X) is non- constant polynomial, then the equation f(X) = 0 has n complex solutions (roots) x1; : : : ; xn and f(X) = a0(X − x1) ··· (X − xn). Some of the xi may be equal and the number of occurrences of xi as a root of f(X) = 0 is called its multiplicity. One of the main mathematical problems before 18th century was to express solutions of al- gebraic equations through their coefficients using the four arithmetical operations (addition, subtraction, multiplication, division) and radicals. Such possibility was known for n = 1; 2 since antiquity, and for n = 3; 4 since 15th century. Finding similar expressions for equa- tions of degree 5 was an unachievable task. The idea that such formulae may not exist was a motivation for the works of many mathematicians during 18th century and finally lead to the results, which are a part of the theory developed by Evariste´ Galois in the beginning of 19th century. In this chapter, we look at some methods of solving the equations when general formulae exist. We can always assume that a0 = 1 (if not, divide both sides of the equation by a0 6= 0). We recall how to solve quadratic equations and afterwards discuss the cases of cubic and quartic equations. 1 Johann Carl Friedrich Gauss (1777 { 1855) was a German mathematician { one of the most prominent in the history of the subject. He left very important contributions in many parts of mathematics and many other natural sciences including number theory, algebra, statistics, anal- ysis, differential geometry, geodesy, geophysics, mechanics, electrostatics, astronomy and optics. 1.1 Quadratic equations. A quadratic equation f(X) = 0 has 2 roots x1; x2 and f(X) = 2 a0X + a1X + a2 = a0(X − x1)(X − x2). Comparing the coefficients on the left and right in this equality, we get the Vieta formulae2: a1 x1 + x2 = − a0 a2 x1x2 = a0 Denoting a1 = p and a2 = q, we get an equivalent equation: a0 a0 X2 + pX + q = 0 p and solve it by transforming to an equation with roots xi + 2 whose sum is 0, that is, replacing the equation above by the equivalent expression in which X2 + pX is completed to a square: p2 p2 X + + q − = 0: 2 4 Solving, we get the two solutions: r r p p2 p p2 x = − − − q and x = − + − q: 1 2 4 2 2 4 2 2 2 The number ∆ = p − 4q = (x1 + x2) − 4x1x2 = (x1 − x2) is called the discriminant of 2 the polynomial X + pX + q (see p. 258). It is non-zero if and only if the roots x1; x2 are different. 1.2 Cubic equations. A cubic equation f(X) = 0 has 3 roots x1; x2; x3 and f(X) = 3 2 a0X + a1X + a2X + a3 = a0(X − x1)(X − x2)(X − x3). Comparing the coefficients on the left and right in this equality, we get the Vieta formulae: a1 x1 + x2 + x3 = − ; a0 a2 x1x2 + x2x3 + x1x3 = ; (1.1) a0 a3 x1x2x3 = − ; a0 2 Franciscus Vieta, in Franch Fran¸coisVi`ete,(1540{1603) was a Franch mathematician who intro- duced modern notation in connection with algebraic equations. 2 a1 2 The equation having the roots xi + , where i = 1; 2; 3, has the coefficient of x equal to 0, 3a0 since the sum of these 3 numbers is zero. Technically, we get such an equation substituting X = Y − a1 in the given cubic equation: 3a0 3 2 a0X + a1X + a2X + a3 = 3 2 a1 a1 a1 3 a0 Y − + a1 Y − + a2 Y − + a3 = a0(Y + pY + q) 3a0 3a0 3a0 where p and q are easily computable coefficients (it is not necessary to remember these formulae, since it is easier to remember the substitution and to transform each time, when it is necessary). Thus we can start with an equation: X3 + pX + q = 0: (1.2) We compare the last equality with the well-known identity: (a + b)3 − 3ab(a + b) − (a3 + b3) = 0 (1.3) and choose a; b in such a way that: p = −3ab; q = −(a3 + b3): (1.4) If a; b are so chosen, then evidently x = a + b is a solution of the equation (1.2). Since a3 + b3 = −q; p3 a3b3 = − ; 27 a3; b3 are solutions of the quadratic equation: p3 t2 + qt − = 0: 27 Solving this equation gives a3; b3. Then we choose a; b, which satisfy (1.4) and get x = a + b. An (impressing) formula, which hardly needs to be memorized (it is easier to start \from the beginning", that is, from the identity (1.3), than to remember it) is thus the following: 3 s s r 2 3 r 2 3 3 q q p 3 q q p x = a + b = − + + + − − + (1.5) 1 2 4 27 2 4 27 We leave a discussion of different choices of signs of the roots to exercises, but notice that in order to solve a cubic equation, it is sufficient to find one root x1 and solve a quadratic equation after dividing the cubic by x − x1. The above expression (in different notations) was first published by Gerolamo Cardano3 in his book \Ars Magna" in 1545 and is known today as Cardano's formula. However, the history surrounding the formula is very involved and several Italian mathematicians knew the method before Cardano (notably, Scipione del Ferro, Antonio Fiore and Niccol´oTartaglia about 20 years earlier that the appearance of Cardano's book ). 3 2 2 The discriminant of the cubic (1.2) is ∆ = −(4p + 27q ) = [(x1 − x2)(x2 − x3)(x3 − x1)] . See p. 258 and Ex. 1.3 below. 4 3 2 1.3 Quartic equations. Such an equation f(X) = a0X + a1X + a2X + a3X + a4 = a0(x − x1)(X − x2)(X − x3)(X − x4) has 4 roots x1; x2; x3; x4. Similarly as for quadratic and cubic equations, we can write down the Vieta formulae expressing relations between the roots and the coefficients of the equation (we leave it as an exercise). Solving a quartic equation (we assume a0 = 1 and denote the remaining coefficients avoiding indices): X4 + mX3 + pX2 + qX + r = 0; (1.6) m it is not important to assume that m = 0 (but it could be achieved substituting X − 4 instead of X similarly as in the quadratic and cubic cases). We simply complement X4+mX3 in such a way that we get a square of a quadratic polynomial and when it is done, we find a parameter λ so that the quadratic polynomial in square brackets m λ2 m2 mλ λ2 X2 + X + − + λ − p X2 + − q X + − r = 0 (1.7) 2 2 4 2 4 is a square of a first degree polynomial. This is achieved when the discriminant of the quadratic polynomial in the square bracket equals 0 (see Ex. 1.8), that is, mλ 2 m2 λ2 − q − 4 + λ − p − r = 0 (1.8) 2 4 4 This is a cubic equation with respect to λ. Solving it, we get a root λ, which gives a possibility to factorize the quartic polynomial (1.7) into a product of two quadratic polynomials and 3 Girolamo Cardano (1501 { 1576) was an Italian mathematician