The Radically insolvable Quintic

Galois Theory and The Radically insolvable Quintic Phil Pollard

To Be Spoken: The purpose of this project is to present a brief introduction to Galois Theory and to illustrate it’s uses in focusing infinite situations into a finite workspace. These slides will focus mainly on the use of Galois Theory to prove the insolvability of the quintic polynomial by a general equation using radicals. Most of the descriptions of theorems and properties are paraphrased interpretations rather than rigorous proofs.

Things you should know by now (And if you don’t, you won’t know what the hell I’m talking about.) • Permutation Groups • Extension Fields • Splitting Fields

Etymology (Unless otherwise stated) • F := An arbitrary Field (assume char = 0) • E := An extension Field over F • ℚ := Field of Rational Numbers • F[x] := Polynomial Ring over F • F(a1,a2,…) := Extension of F by a1, a2,… • e := Identity (of current discussed group) • p.g. := “Proof Gist” expedited proof.

Review of Terms Degree of Extension Denoted [E:F] = n Where n is the dimension of E as a vector space over F. Automorphism A ring isomorphism from a Field onto itself Simple Group A group with no proper, non-trivial normal subgroup Field of Rational Functions f(x1, x2,…,xn) g(x1, x2,…,xn) Denoted F(x1, x2,…,xn) = { | f,g ∈ F[x1, x2,…,xn] }

New Definitions Galois Group Denoted Gal(E/F)[read Galois Group of E fixing F] The set of all automorphisms of E that “fix” F (take each element of F to itself) The Galois Group is a group under function composition. Fixed Field Denoted EH for H ⊆ Gal(E/F) EH = {x ∈ E | Φ(x)=x ∀ Φ∈H} Note: EGal(E/F) = F Intermediate Field F ≤ K ≤ E and K is a field then K is an intermediate field and Ģ(K) := Gal(E/K) Note: EĢ(K) = K

Theorem 1 If f(x) is irreducible in F[x] and E is a splitting field of f(x), Φ∈Gal(E/F) permutes the roots of f(x) p.g. If ai is a root of f(x), then f(ai) = Φ (f(ai)) = 0 because 0 is fixed by Φ = f(Φ(ai)) Operation Preserving But since ai ∉ F, it is not necessarily fixed Obvious Corollary 1 If f(x) ∈ F[x] has degree n, then Gal(E/F) is isomorphic to a subgroup of Sn (E being a splitting field of F)

Example 1: Consider E = ℚ(ω,3√2) where ω = -1+i√3 2 Then the elements of Gal(E/ℚ) are described as follows: Note: Gal(E/ ℚ ) is not Abelian. αβ ≠ βα

Lattices Subgroups of Gal(E/ ℚ) {e, α, β, β2, αβ, αβ2} 2 3 3 3 {e, β, β2} {e, α} {e, αβ} {e, αβ2} 3 2 2 2 {e} Subfields of E E = ℚ (ω,3√2) 3 2 2 2 ℚ (ω) ℚ (ω2 · 3√2) ℚ (3√2) ℚ (ω · 3√2) 2 3 3 3 ℚ

Solvable by Radicals Pure Extension An extension F(a)/F where am ∈ F (referred to as “type m”) Radical Extension An extension E/F such that F = K0 ⊆ K1 ⊆ ··· ⊆ Kn-1 ⊆ Kn = E creates a tower of fields in which each extension Ki+1/Ki is pure Solvable by Radicals f(x) ∈ F[x] is Solvable by Radicals if it has a splitting field E such that E/F forms a radical extension.

Solvable Group A group G that has a series of normal subgroups {e} = H0◁ H1◁ ··· ◁ Hn-1◁ Hn = G where for each 0 ≤ i ≤ n, Hi+1/Hi is Abelian Note 1: The following are easy to show: · Abelian Groups are solvable · Non-Abelian Simple Groups are not solvable · Subgroups of solvable groups are also solvable (slightly less easy) Note 2: We will eventually need the following Theorem: Theorem 2 For E, a splitting field for xn – a over F, Gal(E/F) is solvable. Proof by Smart Dead Guy ■

Suspiciously Convenient Example S5 is not a solvable group p.g. Through an exhaustive orders argument , we find that A5 is a simple group. This means that A5 itself is not a solvable group. And since subgroups of solvable groups are solvable, S5 and Sn for n ≥ 5 cannot be solvable. ■ (The viewer is spared the details, but a full proof of the simplicity of A5 can be found online or in almost any Algebra text.)

Solvable by Radicals impliesSolvable group For E = F(a1,…,an), a Radical Extension for f(x) ∈ F[x], Gal(E/F) is solvable This can be shown by induction. For the Base Case, E = F(a1), E splits some function f(x) = xn – a, and Gal(E/F) is therefore solvable by Theorem 2. For the complete induction proof, the viewer is referred to Gallien, page 556. Proof by Smart Dead Guy ■ Obvious Contraposition 1 If Gal(E/F) is not solvable, then f(x) is not solvable by radicals.

The Galois Group for Certain Polynomials of nth power (para. Herstein) Consider F(x1,…,xn) and let S be the field of symmetric rational functions over F. Consider the polynomial: f(y) = yn – a1yn-1 + a2yn-2 – ··· + (-1)n an in S[y] with a1,…,an ∈ S The roots of f(y) are not contained in S because a1,…,an must have independent, permutable variables But we find that F(x1,…,xn) is a splitting field of f(x) over S and f(y) = yn – a1 yn-1 + a2 yn-2 – ··· + (-1)n an = (y – t1)(y – t2)···(y – tn) ∈ F(x1,…,xn) [y] (ti in terms of x’s) And the roots t1, … ,tn are distinct. Thus, |Gal(F(x1,…,xn) / S)| = n! And by Obvious Corollary 1, Gal( F(x1,…,xn) / S) is isomorphic to a subgroup of Sn Hence, Gal(F(x1,…,xn) / S) is isomorphic to Sn

The Insolvability of the Quintic by Radicals We just found out that certain n-power Functions produce a Galois Group isomorphic to Sn And we know by our Suspiciously Convenient Example that Sn is not solvable for n ≥ 5 And we know by our Obvious Contraposition that if a Galois Group Gal(E/F) is not a solvable group, the equation split by E is not solvable by radicals. Thus, the Quintic Polynomial is not generally solvable by radicals. Fin.

References Contemporary Abstract Algebra, 7th Edition, Joseph Gallian. 2010, Brooks/Cole Advanced Modern Algebra, Joseph J. Rotman. 2003, Prentice Hall Galois Theory, 2nd Edition, Ian Stewart. 1989, Chapman & Hall/CRC Topics in Algebra, I.N. Herstein. 1965, Blaisdell Publishing Chapter 6 Notes – Galois Theory, University of Illinois, Urbana-Champaign http://www.math.uiuc.edu/~r-ash/Algebra/Chapter6.pdf

The Radically insolvable Quintic