Èvariste Galois and the resolution of equations by radicals

Èvariste Galois and the resolution of equations by radicals A Math Club Event Department of Mathematical Sciences Florida Atlantic University Boca Raton, FL 33431 November 9, 2012

Outline 1 The Problem 2 3 4 5 6

Solving Equations The set of real numbers is a strange, messy, murky and unreal type of world. As the Greeks discovered, and Eudoxus made clear, we can know when two numbers are different. But equality is harder; two numbers are equal if they are not different; TRICHOTOMY RULES! What does it mean to know a number?

Solving Equations Let us start with the integers; we all know the integers of course. We can add, subtract, and multiply them, and still have integers. Now construct all numbers you can construct from the integers by division; you get the rational numbers. Next add roots; square roots, cubic roots, fourth roots, all roots. Also roots of numbers you get by adding or subtracting roots. If the real numbers were the ocean, all the numbers obtained by the previous steps are not even a drop of ocean water. The vast majority of numbers is still out there, forever unknown.

Solving Equations If we allow calculus methods, all algebraic equations can be solved. In fact, Newton s method, occasionally covered in Calculus 1 courses, is an efficient method for solving algebraic equations. And there are many others. The trick is to solve them using only the operations from the steps mentioned before: Addition, subtraction, multiplication, division, extraction of roots. Come up with a method, an algorithm that works. This is what is meant by solving equations by radicals. To be precise, the solution should be obtained in a finite number of steps by the operations of addition, subtraction, multiplication, division, and extraction of roots, starting with the integers (or the rationals) and the coefficients of the equation.

Solving Equations The Babylonians already figured out how to solve the quadratic equation. The solution of the equation ax 2 + bx + c = 0 is given by x = b ± b 2 4ac. 2a Apart from the standard operations of addition, subtraction, multiplication and division, all that is involved is taking square roots.

Solving Equations Around 1500, the cubic and quartic equations were solved by Scipio dal Ferro, Tartaglia, Cardano, Ferrari. The search for a formula to solve quintic equations was on. Who would discover it? And when?

A Slightly Different Radical Idea Instead of extracting roots, think of adjoining roots. To solve by radicals can also be interpreted as follows: We start with the rational numbers, and the coefficients of the equation, and (at least mentally) close this set with respect to the four basic operations: addition, subtraction, multiplication and division. I ll call this set (for now) our starting field K 0. If we can show that we ll always have the roots of the equations we are considering in K 0, our work is done. Except for first order equations, we can t. Next, we adjoin some roots of (a finite number of) elements of K 0 to K 0, close again by the four basic operations to get a second field of activity, the field K 1. Can we show that the roots are in K 1? Yes, for quadratic equations; we only need to adjoin b 2 ac.

Adjoining Radicals Around 1800, three mathematicians, Ruffini, Abel, and Galois, had similar ideas on how to analyze what happens when you keep adjoining radicals. It involved looking at what turned out to be a group of permutations of the roots. When a radical was adjoined to the field of numbers you had so far, the group could get smaller, or remain the same. If it got smaller, it did so in a very specific way: If G was the group before adjunction, and after adjunction it shrank to a subgroup G 1, then in modern day parlance (don t worry to much, I ll hardly use it again), G 1 had to be a normal subgroup of G and G/G 1 had to be commutative.

When Was The Equation Solved? As we shall see, if the group G of the equation had a single permutation; in other words just the identity permutation, the equation was solved. And conversely. So, to solve the equation one needed an algorithm by which one can construct a sequence G 0 G 1 G N of subgroups of G such that G 0 = G, G N = {e} (the identity), G i+1 is a normal subgroup of G i and G i /G i+1 is commutative (i = 0,..., N 1). A group for which such a sequence exists is called solvable Is every group of permutations that can appear as group of an equation solvable?

The conclusion To solve the general equation of degree n one had to assume that its group was S n, the group of ALL permutations of n elements. S 2, S 3, S 4 are solvable. If n 5, then S n is not solvable. The general equation of order n 5 cannot be solved by radicals.

g-rational Numbers From now on we assume given an equation a m x m + a m 1 x m 1 + + a 1 x + a 0 = 0 (1) to be known as the main equation. The coefficients a 0,..., a m are complex numbers; a m 0. We will assume that all its roots, which Galois denotes a, b, c,..., but I will denote r 1, r 2,..., r m, are distinct. Galois then calls a number rational if it is rational, or if it is a coefficient of the equation, or can be obtained from these first rationals by the basic operations +,,, /. But it is a fluid concept; once a number gets adjoined to the original rationals, it can be used to create more rationals. I will call a Galois rational a g-rational number.

g-rational Numbers, a Better Definition The Language of Fields Algebraists consider some very abstract fields; for us a field is just a subset K of the complex numbers that is closed under addition, subtraction, multiplication and division by non-zero elements. The smallest such set is Q, the field of rational numbers. If K is a field, α a complex number, then K(α) is the smallest field containing all the elements of K as well as α. If α is the zero of an irreducible polynomial of degree n, p(x ) = a n X n + + a 1 X + a 0 with coefficients in K, then one can show that K(α) = {c 0 + c 1 α + + c n 1 α n 1 : c 0,..., c n 1 K}. If α, β C, then K(α, β) = K(α)(β) is the smallest field containing all the elements of K, and α, and β. More generally, K(a 1,..., a n ) is the smallest field containing K as well as a 1,..., a n.

g-rational Numbers Let K= Q(a 0,..., a m ); Q extended by the coefficients of the main equation; just Q if these coefficients are rational. These are our first g-rational numbers. Every time a number α is adjoined to K, the set of g-rationals expands to include the elements of K(α). Eventually, K(α) becomes the new K.

How Galois Defined his Group Let r 1,..., r m be the distinct roots of the main equation. One can prove, it is not terribly hard, that one can find integers A 1,..., A m with the property that if we consider all the different m! linear combinations of the roots using these coefficients, they all take on different values. Precisely, if σ, τ S m, σ τ, then A 1 r σ(1) + A m r σ(m) A 1 r τ(1) + A m r τ(m). Let V = A 1 r 1 + + A m r m. This is a highly non-unique choice!

The Group of Galois, by Galois Let Q(X ) = σ S m X m A j r σ(j). This is a polynomial of degree m!. The coefficients of this polynomial are symmetric in the roots of the main equation; this implies that they are polynomials in the coefficients of the main equation; in other words they are in K. So Q is a polynomial of degree m! and Q(V ) = 0. Q(X ) is not necessarily irreducible over K; because Q(V ) = 0 there is an irreducible polynomial Q 0 with coefficients in K such that Q 0 (V ) = 0. This polynomial divides Q(X ). Let n be the degree of Q 0. j=1

The Group is Coming, but not here yet Denote the zeros of Q 0 (X ) by V, V,..., V n 1. Each V k is obtained from V by applying a permutation to the roots, since it also is a zero of Q(X ). Galois now proves that every root of the main equation can be expressed as a polynomial in V with coefficients in K. Denoting as one does the set of polynomials with coefficients in K by K[X ], there exist ϕ(x ), ϕ 1 (X ),..., ϕ m 1 (X ) K[X ] such that r 1 = ϕ(v ), r 2 = ϕ 1 (V ),..., r m = ϕ m 1 (V ). These polynomials can be assumed of degree at most n 1 (n the degree of Q 0 ). In this case they are unique.

Almost The Problem Galois now proves that for every root W of Q(X ), the values ϕ(w ), ϕ 1 (W ),..., ϕ m 1 (W ) are a rearrangement (permutation) of the roots r 1,..., r m and he sets up the following table:

! As Galois says: Let ϕv ϕ 1 V ϕ 2 V..... ϕ m 1 V be the roots of the proposed equation. Let us write down the following permutations of the roots: (V ) ϕv ϕ 1 V ϕ 2 V..... ϕ m 1 V (V ) ϕv ϕ 1 V ϕ 2 V..... ϕ m 1 V (V ) ϕv ϕ 1 V ϕ 2 V..... ϕ m 1 V........... (V (n 1) ) ϕv (n 1) ϕ 1 V (n 1) ϕ 2 V (n 1)..... ϕ m 1 V (n 1)

Properties, Independent Characterization Galois calls it a group, but it is likely that he used the word as a synonym for set. But, as he remarked, it is a set of permutations closed under composition, thus a subgroup of S m. The permutations in this group satisfy: If F (X 1,..., X m ) is a polynomial in m-variables with coefficients in K and if F (r 1,..., r m ) = c K, then F (r τ(1),..., r τ(m) ) = c for all permutations in this group. If F (X 1,..., X m ) is a polynomial in m-variables with coefficients in K that satisfies F (r τ(1),..., r τ(m) ) = F (r 1,..., r m ) for all permutations τ of the group, then F (r 1,..., r m ) K As a corollary one obtains: A permutation τ S m is in the group if and only if it satisfies: Whenever F (X 1,..., X m ) is a polynomial in m-variables with coefficients in K such that F (r 1,..., r m ) = 0, then F (r τ(1),..., r τ(m) ) = 0. This last property shows that the definition of the group is independent of the choice of V and that it is, in fact, a subgroup of S m.

A modern perspective Ignore if confusing Given abstract up in the cloud fields F, E with F a subfield of E, one considers the set of all automorphisms of E that leave F fixed. One says E is a normal extension of F if it is also true that if an element of E is left fixed by all these automorphisms, then it has to be in F. If E is a normal extension of F, then the order of the group of automorphisms of E leaving F fixed equals the dimension [E : F ] of E as a vector space over F. If F, E are fields, if p(x ) is a polynomial with coefficients in F and E is a minimal field extending F and containing all roots of p(x ), then E is called a splitting field for p(x ) over F. One proves all splitting fields are isomorphic, so one talks of the splitting field of p(x ). One proves: E is a normal extension of F if and only if it is the splitting field of some polynomial p(x ) having no repeated roots. If p(x ) is a polynomial with coefficients in F and E is a splitting field for p(x ) over F, then the group of automorphisms of E leaving F fixed is the Galois group of p(x ). This group can be identified with a subgroup of the group of permutations of the roots and, as such, is the same as the group defined by Galois.

In Fact, The Problem The zeros of Q(X ), hence also of Q 0 (X ) are all distinct. K(V ) is the smallest field containing V and K. Because all the roots of the main equation are polynomials in V, K(V ) contains them all. Because all other zeros of Q(X ) and of Q 0 (X ) are linear combinations of the roots of the main equation, all these zeros are also in K(V ) It follows that K(V ) is a splitting field of both the polynomial defined by the main equation and for Q 0 (X ), hence a normal extension of K.

Example 1. The Problem Details are left out. Lets look at x 3 2 = 0, with K = Q. With ω = e 2πi/3 = 1 2 + 3 2 i, its roots are 3 2, 3 2 ω, 3 2 ω 2. It is easy to see that a working choice of A 1, A 2, A 3 is A 1 = 1, A 2 = 2, A 3 = 0. With this choice, V = 3 3 2 i.

x 3 2 = 0 continued The irreducible polynomial over Q satisfied by V is x 6 + 108 = 0. The other roots of this polynomial are V = V = 3 2 3i, V = V (4) = V = ( 3 ) 3 2 2 i ( 3 2 + 3 2 i ) 3 2, V (5) = V (4) = A bit of arithmetic/algebra shows that we have to take 3 2, V = V = ϕ(x) = 1 18 x 4, ϕ 1 (x) = 1 2 x 1 36 x 4, ϕ 2 (x) = 1 2 x 1 36 x 4. ( ) 3 3 2 + 2 i (

x 3 2 = 0 continued Galois table looks as follows: (V ) ϕ(v ) = 3 2 ϕ 1 (V ) = 3 2ω ϕ 2 (V ) = 3 2ω 2 (V ) ϕ(v ) = 3 2 ϕ 1 (V ) = 3 2ω 2 ϕ 2 (V ) = 3 2ω (V ) ϕ(v ) = 3 2ω 2 ϕ 1 (V ) = 3 2ω ϕ 2 (V ) = 3 2 (V ) ϕ(v ) = 3 2ω 2 ϕ 1 (V ) = 3 2 ϕ 2 (V ) = 3 2ω (V (4) ) ϕ(v (4) ) = 3 2ω ϕ 1 (V (4) ) = 3 2ω 2 ϕ 2 (V (4) ) = 3 2 (V (5) ) ϕ(v (5) ) = 3 2ω ϕ 1 (V (5) ) = 3 2 ϕ 2 (V (5) ) = 3 2ω 2 As we can see, all permutations occur. The Galois group is (isomorphic to ) S 3.

Example 2 The Problem We consider again x 3 2 = 0, but after adjoining the root 3 2. V = 3 2 3 i as before, but now the irreducible polynomial it satisfies is X 2 + 3 3 4 = 0. V = V is the second root of the equation satisfied by V. The new group reduces to the first two lines of the old one; it consists of the identity and one transposition.

Example 3 The Problem We consider the equation x 4 + x 3 + x 2 + x + 1 = 0, with roots r, r 2, r 3, r 4, where r = e 2πi/5 = cos(2π/5) + i sin(2π/5), over Q. Because r, r 2, r 3, r 4 are independent over Q, any four distinct A 1, A 2, A 3, A 4 will work. For example, A 1 = 1, A 2 = 2, A3 = 2, A4 = 1, giving ( V = r + 2r 2 2r 3 r 4 = 2 sin 2π 5 + 2 sin π ) i. 5

Example 3 contd., x 4 + x 3 + x 2 + x + 1 = 0 With some effort (and help from Maple) one gets that the polynomials ϕ, ϕ 1, ϕ 2, ϕ 3 are ϕ(x ) = 1 25 x 3 1 10 x 2 1 2 x 3, ϕ(v ) = r, (2) 2 ϕ 1 (X ) = 1 50 x 3 + 1 10 x 2 + 1 2 x + 1, ϕ 1(V ) = r 2, (3) ϕ 2 (X ) = 1 50 x 3 + 1 10 x 2 1 2 x + 1, ϕ 2(V ) = r 3,(4) ϕ 3 (X ) = 1 25 x 3 1 10 x 2 + 1 2 x 3 2, ϕ(v ) = r 4. (5)

Example 3 concluded, x 4 + x 3 + x 2 + x + 1 = 0 Here is how the table of permutations looks like. ϕ(v ) = r ϕ 1 (V ) = r 2 ϕ 2 (V ) = r 3 ϕ 3 (V ) = r 4 ϕ(v ) = r 4 ϕ 1 (V ) = r 3 ϕ 2 (V ) = r 2 ϕ 3 (V ) = r ϕ(v ) = r 3 ϕ 1 (V ) = r ϕ 2 (V ) = r 4 ϕ 3 (V ) = r 2 ϕ(v ) = r 2 ϕ 1 (V ) = r 4 ϕ 2 (V ) = r ϕ 3 (V ) = r 3 If τ denotes the permutation r r 3, r 2 r, r 3 r 4, r 4 r 2, then G is the cyclic group {e, τ, τ 2, τ 3 }.

The heart The Problem of the matter Galois divides his Memoir into Preliminaries (Four Lemmas) and eight Propositions. Propositions can be Theorems (Propositions I, II, III, IV, VIII), Problems (Propositions V, VII), or Lemmas (Proposition VI). The construction and basic properties of what we now call the Galois group of an equation is the content of Proposition I (and what precedes it) in the Memoir. Propositions II and III are the main results; very roughly what nowadays is called the Fundamental Theorem of Galois Theory. Proposition IV is a bit of a digression. Proposition V gives the necessary and sufficient condition for an equation to be solvable by radicals. Propositions VI-VIII expand on this matter.

Propositions II and III In the notes I am writing on this Memoir, I have it the way Galois stated it. But here I have to be brief, and use modern terms. Suppose the group of the equation, as constructed above, has n permutations. Proposition II states that if we extend the field by adjoining the root of an irreducible polynomial of degree p, two things can happen: Nothing; the group of the equation with respect to the extended field is the same as before, or the new group is a subgroup of size n/p of the previous group. Implicit in this is that p has to divide n for the second option to be possible. Proposition III in modern language says precisely that if we add ALL (Galois emphasis) roots of an irreducible equation to the field, then new group has to be a normal subgroup of the previous group.

Solving Equations An equation is considered solved if all its roots have become g-rational, are in the field K in which we are working. In this case V K,the irreducible equation satisfied by V is x V = 0 of degree 1, so that the Galois group is of order (size) 1; reduced to the identity. Can the Galois group be reduced, by adjoining roots of objects already in the field, to size 1?

Equations that hate radicals, and don t allow them to do their work Adjoining radicals means adjoining things like k a, where a K. But all roots can be obtained by taking prime roots. For example, 45 a = 3 3 5 a. We can also assume that our field contains all (complex) roots of 1; since we can also just adjoin them. So adjoining radicals becomes equivalent to adjoining ALL roots of an irreducible equation of the form x p a, where p is prime. This means that IF the original group G can be reduced, it can also be reduced to a normal subgroup G 0 such that G/G 0 is prime. Groups of prime order are cyclic, in particular commutative.

Equations that hate radicals, and don t allow them to do their work, contd. To solve the general equation of order m, because the coefficients could all be transcendental, unrelated to each other, we must assume its Galois group is S m. One thing that happens in S m if m 5 is that you can have two 3-cycles having exactly one element in common. A 3-cycle denoted by (abc), where a, b, c {1,..., m} is the permutation σ such that σ(a) = b, σ(b) = c, σ(c) = a, and σ(k) = k if k a, b, c. Suppose G is a subgroup of S m that contains all 3-cycles. Let σ = (abc) be one of the 3-cycles; a b c a. If and only if m 5, we can find d, e so a, b, c, d, e are 5 distinct elements and form the 3-cycles τ = (dac) and λ = (ceb) Suppose H is a normal subgroup of G and suppose G/H is abelian.

Equations that hate radicals, and don t allow them to do their work, contd. Because G/H is abelian (commutative), id G/H = (τh) 1 (λh) 1 (τh)(λh) = (τ 1 λ 1 τλ)h, implying τ 1 λ 1 τλ H. Going from right to left (usual composition order) τ 1 λ 1 τλ = (dca)(cbe)(dac)(ceb) = (abc) = σ. This proves: If a subgroup G of S m, m 5, contains all 3-cycles, every normal subgroup H such that G/H is commutative, must also contain all three cycles. This severely reduces the amount by which G can be reduced. It shows that S 5 is not solvable.

Solvable equations But why should an equation whose group G has the property that one can find a sequence of normal subgroups G 0 = G G 1 G 2 {e} such that G i1 /G i is of prime order, be solvable by radicals? Here is Galois argument. We must show that if the group is G, and G has a normal subgroup subgroup H such that G/H is of order p, p a prime, then by adjoining a radical (a p-th root it will turn out to be), the new group of the equation is H. So assume all this, and let α be a primitive p-th root of 1; α = e 2πi/p. Assume the equation has m distinct roots r 1,..., r m.

Galois argument continued Let f K[X 1,..., X m ] be a polynomial in m-variables such that if for σ S m we set σ f (X 1,..., X m ) = f (X σ(1),..., X σ(m) ), then σ f (r 1,... r m ) = f (r 1,... r m ) σ H, σ f (r 1,... r m ) f (r 1,... r m ) σ G\H, Let σ G be such that σh is a generator of G/H. Let θ = f (r 1,..., r m ), θ 1 = σ f (r 1,..., r m ), θ 2 = σ 2 f (r 1,..., r m ),..., θ p 1 = σ f (r 1,..., r m ).

The argument cont d Let ξ = ( θ + αθ 1 + + α p 1 θ p 1 ) p. If there is time, I will argue at the board that ξ K, ξ 1/p / K and if we adjoin ξ 1/p to K, then H becomes the new group of the equation.

Second degree equations Suppose the equation is ax 2 + bx + c = 0, with roots r, s. The group S 2 = {e, σ} with σ(r) = s, σ(s) = r. We have to reduce it to {e}. In this case p = 2, α = 1 (Actually, this is always the first reduction). We need a polynomial in two variables left invariant by {e} (any polynomial will do) but not by σ. The simplest choice is f (X 1, X 2 ) = X 1 X 2. The quantity Galois says should be adjoined is ξ where ξ = (f (r, s) f (s, r)) 2 = (2(r s)) 2 = 4(r 2 + s 2 2rs) = 4 ( (r + s) 2 4rs ) ( (b ) ) 2 = 4 4c = 4(b2 4ac) a a a 2.

The End The Problem THE END Thanks for listening