MA3D5 Galois theory - Warwick

1 MA3D5 Galois theory Miles Reid Jan Mar 2004. printed Jan 2014. Contents 1 The theory of equations 3. Primitive question .. 3. Quadratic equations .. 3. The remainder theorem .. 4. Relation between coefficients and roots .. 5. Complex roots of 1 .. 7. Cubic equations .. 9. Quartic equations .. 10. The quintic is insoluble .. 11. Prerequisites and books .. 13. Exercises to Chapter 1 .. 14. 2 Rings and fields 18. Definitions and elementary properties .. 18. Factorisation in Z .. 21. Factorisation in k[x] .. 23. Factorisation in Z[x], Eisenstein's criterion .. 28. Exercises to Chapter 2 .. 32. 3 Basic properties of field extensions 35. Degree of extension .. 35. Applications to ruler-and-compass constructions .. 40. Normal extensions .. 46. Application to finite fields .. 51. 1. Separable extensions .. 53. Exercises to Chapter 3 .. 56. 4 Galois theory 60. Counting field homomorphisms .. 60. Fixed subfields, Galois extensions.

2 64. The Galois correspondences and the Main theorem .. 68. Soluble groups .. 73. Solving equations by radicals .. 76. Exercises to Chapter 4 .. 80. 5 Additional material 84. Substantial examples with complicated Gal(L/k) .. 84. The primitive element theorem .. 84. The regular element theorem .. 84. Artin Schreier extensions .. 84. Algebraic closure .. 85. Transcendence degree .. 85. Rings of invariants and quotients in algebraic geometry .. 86. Thorough treatment of inseparability .. 86. AOB .. 86. The irreducibility of the cyclotomic equation .. 86. Exercises to Chapter 5 .. 87. 2. 1 The theory of equations Summary Polynomials and their roots. Elementary symmetric functions. Roots of unity. Cubic and quartic equations. Preliminary sketch of Galois theory . Prerequisites and books. Primitive question Given a polynomial f (x) = a0 xn + a1 xn 1 + + an 1 x + an ( ). how do you find its roots? (We usually assume that a0 = 1.) That is, how do you find some solution with f ( ) = 0.

3 How do you find all solutions? We see presently that the second question isQ. equivalent to splitting f , or factoring it as a product of linear factors f = a0 ni=1 (x i ). Quadratic equations Everyone knows that f (x) = ax2 + bx + c has two solutions . b b2 4ac , = . ( ). 2a Set a = 1 for simplicity. You check that + = b, and = c, ( ). which gives the polynomial identity f (x) = x2 + bx + c (x )(x ). The relations ( ) imply that (f ) = ( )2 = ( + )2 4 = b2 4c. ( ). This gives the following derivation of the quadratic formula ( ): first, the argument of below (see Corollary ) proves directly the polynomial identity x2 + bx + c (x )(x ), hence equations ( ). Thus we have an equation for + and for = = , that yield ( ). The expression (f ) of ( ) is called the discriminant of f . Clearly, it is a polynomial in the coefficients of f , and is zero if and only if f has a repeated root. Over R, f has two distinct real roots if and only if > 0, and two conjugate complex roots if and only if < 0.

4 Compare Ex. 15. 3. The remainder theorem theorem (Remainder theorem ) Suppose that f (x) is a polynomial of degree n and a Then there exists an expression f (x) = (x )g(x) + c, where g(x) is a polynomial of degree n 1 and c is a constant. Moreover, c = f ( ). In particular, is a root of f if and only if x divides f (x). Proof The moreover clause follows trivially from the first part on sub- stituting x = . For the first part, we use induction on n. Suppose that f (x). is given by ( ). Subtracting a0 xn 1 (x ) from f (x) kills the leading term a0 xn of f (x), so that f1 (x) := f (x) a0 xn 1 (x ) has degree n 1. By induction, f1 (x) is of the form f1 (x) = (x )g1 (x) + const., and the result for f follows at once.. Corollary (i) Let 1 , ..Q. , k be distinct quantities. They are roots of f (x) if and only if f (x) = ki=1 (x i )g(x), where g(x) is a polynomial of degree n k. (ii) The number of roots of f (x) is n. (iii) If f (x) is monic (meaning that a0 = 1) of degree n and has n (distinct).

5 Roots then n Y. n n 1. f (x) = x + a1 x + + an 1 x + an (x i ). i=1. As discussed later in the course, we can always assume that f (x) of degree n has n roots (not necessarily distinct). For example, if the coefficients ai of f (x) are rational numbers, then the fundamental theorem of algebra . implies that f (x) has n complex roots i . The proof of the fundamental theorem is analytic, and is given in topology (winding number) or in complex analysis (contour integral). 1. Quantity is explained in Exercise below. For the moment, bear in mind the important special case ai Q and C. 4. Relation between coefficients and roots This section generalises the relations ( ). Suppose given n quantities 1 , .. , n . We eventually intend them as the n roots of a polynomial f (x), but in this section we only treat them in formal identities, so that we could also think of them as independent indeterminates. Definition The kth elementary symmetric function k of the i is de- fined by X k Y.

6 K = ij . 1 i1 <i2 < <ik n j=1. In other words, take the sum of all products of k distinct choices of the i , starting with 1 2 k . Thus X. 1 = i = 1 + 2 + + n ;. 1 i n X. 2 = i j = 1 2 + ;. 1 i<j n Yn n = i . i=1. These quantities are defined in order to provide the polynomial identity n Y n X. (x + i ) n i xi . i=1 i=0. Or, more relevant to our context n Y. n n 1 n 1 n f (x) = x 1 x + + ( 1) n 1 x + ( 1) n (x i ). i=1. We set 0 = 1 by convention (a single choice of the empty product, if you like that kind of thing). Corollary Suppose that f (x) is a monic polynomial of degree n, having n roots 1 , .. , n . Then the coefficient ak of xn k in f (x) is equal to ( 1)k times k , the kth elementary symmetric function of the i . 5. theorem (Symmetric Polynomials) Let P ( 1 , .. , n ) be a polyno- mial expression that is symmetric in the i . Then P ( 1 , .. , n ) can be written as a polynomial in 1 , .. , n . The elementary symmetric polynomials i are an important ingredient in many different areas of math, and give rise to many useful calculations.

7 P 3. Example What is i ? Write 13 = ( 1 + 2 + + n )3. = 13 + 3 12 ( 2 + + n ) + 3 1 ( 2 + + n )2 + ( 2 + + n )3. X X X. = i3 + 3 i2 j + 6 i j k . i6=j i<j<k i2 j ? P. So what is X X. 1 2 = ( 1 + + n )( 1 2 + ) = i2 j + 3 i j k ;. i<j<k note the coefficient 3:Peach term, say 1 2 3 occurs as 1 ( 2 3 ), 2 ( 1 3 ). and 3 ( 1 2 ). Thus i2 j = 1 2 3 3 , and finally X. i3 = 13 3 1 2 + 3 3 . These computations get moderately cumbersome to do by hand. They provide lots of fun exercises in computer algebra (see Ex. 5 and Ex. 14). Proof of theorem A polynomial is a sum of monomials b = ibi ;. Q. introduce the lex order (dictionary order) on these monomials, in which 1 < 1 < 12 < 13 < 12 2 < 1 2 3 < etc. More formally, write each monomial b as a word 1 1 2 2 n n 1, | {z } | {z } | {z }. b1 b2 bn adding 1 as an end-of-word marker, with 1 < 1 < 2 . A word beats another if and only if it beats it the first time they differ. The leading term 6.

8 Of P ( 1 , .. , n ) is its first term in lex order. Obviously, the leading term b of a symmetric polynomial P has b1 b2 bn . Now consider the polynomial 1c1 2c2 ncn . Its leading term is the prod- uct of the leading terms in each factor, that is 1c1 +c2 + +cn 2c2 + +cn ncn . Thus we can hit the leading term of P by choosing ci = bi bi 1 . Then P. minus a scalar multiple of 1c1 2c2 ncn is a symmetric polynomial that is a sum of monomials that is later in the lex order. An induction completes the proof. Complex roots of 1. The equation xn = 1 and its roots are important for several reasons. As everyone knows, its complex roots are the nth roots of unity 2 ai 2 a 2 a exp = cos + i sin for a = 0, .. , n 1. n n n These form a subgroup of the multiplicative group of complex numbers n . C that is cyclic of order n, generated by exp 2 i n . Example (Cube roots of 1) Write . 2 i 2 2 1 3. = exp = cos + i sin = . 3 3 3 2. Then 3 = 1.

9 In fact x3 1 = (x 1)(x2 + x + 1), with satisfying 2 + + 1 = 0. There are 3 complex cube roots of 1, namely 1, and 2 = , and the equation 2 + + 1 = 0 says that these add to 0. You can think of this geometrically (see Figure ): the 3 cube roots of 1 are the vertexes of a regular triangle centred at 0. Clearly xn 1 factors as (x 1)(xn 1 + + x + 1); if n = p is prime, it is known (and proved in below) that the polynomial p = xp 1 + +x+1. is irreducible in Q[x]. It is called the pth cyclotomic polynomial. Definition If n is composite, the nth roots of 1 include the mth roots for different factors m | n, satisfying xm = 1. We say that is a primitive nth root of unity if n = 1 but m 6= 1 for any m < n, or equivalently, if it generates the cyclic group n . 7. '$.. H 1. HH. H.. &%.. Figure : Three cube roots of 1. The primitive nth roots of 1 in C are exp 2 ai n with a coprime to n. Remark The number of primitive roots of 1 is given by the Euler phi function of elementary number theory : n o (n) = a [0, n] a is coprime to n Yp 1.

10 =n . p p|n That is n = pai i has (n) = pai i 1 (pi 1). Q Q. The primitive nth roots of 1 are roots of a polynomial n , called the nth cyclotomic polynomial (see Ex. ). It is determined by factorising xn 1. as a product of irreducible factors, then deleting any factors dividing xm 1. for some m < n. (See also Ex. ) We are mainly concerned with the case n = p a prime, although other cases will occur as examples at several points. We finally prove that n is irreducible of degree (n) in One reason for the importance of nth roots of 1 is as follows. Suppose that we already own a full set of nth roots of 1; equivalently, that our field contains a primitive root of unity , or that xn 1 splits into linear factors: n 1. Y. n x 1= (x a ). a=0.. Then if we manage to find one nth root = n a of any quantity a, we automatically get all n of them without any further ado; in other words, xn a also splits into linear factors n 1. Y. xn a = (x a ).