A Math Club Event Tomas Schonbekmath.fau.edu/schonbek/Galois/Galois121109.pdf · Solving Equations...

The ProblemThe End of the Quest

Galois First Memoir-PreliminariesThe Galois Group

ExamplesThe Main results

Evariste Galois and the resolution of equations byradicals

A Math Club Event

Tomas Schonbek

Department of Mathematical SciencesFlorida Atlantic UniversityBoca Raton, FL 33431

November 9, 2012

Tomas Schonbek EG

Outline

1 The Problem

2 The End of the Quest

3 Galois First Memoir-Preliminaries

4 The Galois Group

5 Examples

6 The Main results




Solving Equations

The set of real numbers is a strange, messy, murky and unrealtype of world.

As the Greeks discovered, and Eudoxus made clear, we canknow when two numbers are different.

But equality is harder; two numbers are equal if they are notdifferent; TRICHOTOMY RULES!

What does it mean to know a number?

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Solving Equations

Let us start with the integers; we all know the integers ofcourse. We can add, subtract, and multiply them, and stillhave integers.

Now construct all numbers you can construct from theintegers by division; you get the rational numbers.

Next add roots; square roots, cubic roots, fourth roots, allroots. Also roots of numbers you get by adding or subtractingroots.

If the real numbers were the ocean, all the numbers obtainedby the previous steps are not even a drop of ocean water. Thevast majority of numbers is still out there, forever unknown.

Tomas Schonbek EG




Solving Equations

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

Tomas Schonbek EG

Solving Equations

The Babylonians already figured out how to solve thequadratic equation.

The solution of the equation ax2 + bx + c = 0 is given by

x =−b ±

√b2 − 4ac

2a.

Apart from the standard operations of addition, subtraction,multiplication and division, all that is involved is taking squareroots.




Solving Equations

Around 1500, the cubic and quartic equations were solved byScipio dal Ferro, Tartaglia, Cardano, Ferrari.

The search for a formula to solve quintic equations was on.

Who would discover it? And when?

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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 theequation, and (at least mentally) close this set with respect tothe four basic operations: addition, subtraction, multiplicationand division. I’ll call this set (for now) our starting field K0. Ifwe can show that we’ll always have the roots of the equationswe are considering in K0, our work is done.Except for first order equations, we can’t. Next, we adjoinsome roots of (a finite number of) elements of K0 to K0, closeagain by the four basic operations to get a second field ofactivity, the field K1. Can we show that the roots are in K1?Yes, for quadratic equations; we only need to adjoin

√b2 − ac .

Tomas Schonbek EG




Adjoining Radicals

Around 1800, three mathematicians, Ruffini, Abel, and Galois,had similar ideas on how to analyze what happens when youkeep adjoining radicals.It involved looking at what turned out to be a group ofpermutations of the roots.When a radical was adjoined to the “field” of numbers youhad 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 thegroup before adjunction, and after adjunction it shrank to asubgroup G1, then in modern day parlance (don’t worry tomuch, I’ll hardly use it again), G1 had to be a normalsubgroup of G and G/G1 had to be commutative.

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When Was The Equation Solved?

As we shall see, if the group G of the equation had a singlepermutation; in other words just the identity permutation, theequation was solved. And conversely.

So, to solve the equation one needed an algorithm by whichone can construct a sequence G0 ⊃ G1 ⊃ · · · ⊃ GN ofsubgroups of G such that G0 = G , GN = {e} (the identity),Gi+1 is a normal subgroup of Gi and Gi/Gi+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 anequation solvable?

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The conclusion

To solve the general equation of degree n one had to assumethat its group was Sn, the group of ALL permutations of nelements.

S2,S3,S4 are solvable.

If n ≥ 5, then Sn is not solvable.

The general equation of order n ≥ 5 canNOT be solved byradicals.

Tomas Schonbek EG




g-rational Numbers

From now on we assume given an equation

amxm + am−1xm−1 + · · ·+ a1x + a0 = 0 (1)

to be known as the main equation.

The coefficients a0, . . . , am are complex numbers; am 6= 0.

We will assume that all its roots, which Galois denotesa, b, c , . . ., but I will denote r1, r2, . . . , rm, are distinct.

Galois then calls a number rational if it is rational, or if it is acoefficient 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 theoriginal “rationals,” it can be used to create more rationals.

I will call a Galois rational a g-rational number.

Tomas Schonbek EG

g-rational Numbers, a Better DefinitionThe Language of Fields

Algebraists consider some very abstract fields; for us a field isjust a subset K of the complex numbers that is closed underaddition, subtraction, multiplication and division by non-zeroelements.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 smallestfield containing all the elements of K as well as α.If α is the zero of an irreducible polynomial of degree n,p(X ) = anX n + · · ·+ a1X + a0 with coefficients in K , thenone can show that

K (α) = {c0 + c1α + · · ·+ cn−1αn−1 : c0, . . . , cn−1 ∈ K}.

If α, β ∈ C, then K (α, β) = K (α)(β) is the smallest fieldcontaining all the elements of K , and α, and β.More generally, K (a1, . . . , an) is the smallest field containingK as well as a1, . . . , an.




g-rational Numbers

Let K= Q(a0, . . . , am); Q extended by the coefficients of themain equation; just Q if these coefficients are rational. Theseare our first g-rational numbers.

Every time a number α is adjoined to K , the set of g-rationalsexpands to include the elements of K (α).

Eventually, K (α) becomes the new K .

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How Galois Defined his Group

Let r1, . . . , rm be the distinct roots of the main equation.

One can prove, it is not terribly hard, that one can findintegers A1, . . . ,Am with the property that if we consider allthe different m! linear combinations of the roots using thesecoefficients, they all take on different values. Precisely, ifσ, τ ∈ Sm, σ 6= τ , then

A1rσ(1) + · · ·Amrσ(m) 6= A1rτ(1) + · · ·Amrτ(m).

Let V = A1r1 + · · ·+ Amrm. This is a highly non-uniquechoice!

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The Group of Galois, by Galois

Let

Q(X ) =∏σ∈Sm

X −m∑j=1

Aj rσ(j)

.

This is a polynomial of degree m!.

The coefficients of this polynomial are symmetric in the rootsof the main equation; this implies that they are polynomials inthe coefficients of the main equation; in other words they arein K . So Q is a polynomial of degree m! and Q(V ) = 0.

Q(X ) is not necessarily irreducible over K ; because Q(V ) = 0there is an irreducible polynomial Q0 with coefficients in Ksuch that Q0(V ) = 0. This polynomial divides Q(X ). Let nbe the degree of Q0.

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The Group is Coming, but not here yet

Denote the zeros of Q0(X ) by V ,V ′, . . . ,V n−1. Each V k isobtained 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 beexpressed as a polynomial in V with coefficients in K .Denoting as one does the set of polynomials with coefficientsin K by K [X ], there exist ϕ(X ), ϕ1(X ), . . . , ϕm−1(X ) ∈ K [X ]such that

r1 = ϕ(V ), r2 = ϕ1(V ), . . . , rm = ϕm−1(V ).

These polynomials can be assumed of degree at most n− 1 (nthe degree of Q0). In this case they are unique.

Tomas Schonbek EG




Almost

Galois now proves that for every root W of Q(X ), the valuesϕ(W ), ϕ1(W ), . . . , ϕm−1(W ) are a rearrangement(permutation) of the roots r1, . . . , rm and he sets up thefollowing table:

Tomas Schonbek EG

The Galois Group!As Galois says: ‘Let

ϕV ϕ1V ϕ2V . . . . . ϕm−1V

be the roots of the proposed equation.Let us write down the following permutations of the roots:

(V ) ϕV ϕ1V ϕ2V . . . . . ϕm−1V

(V ′) ϕV ′ ϕ1V ′ ϕ2V ′ . . . . . ϕm−1V ′

(V ′′) ϕV ′′ ϕ1V ′′ ϕ2V ′′ . . . . . ϕm−1V ′′

. . . . . . . . . . .

(V (n−1)) ϕV (n−1) ϕ1V (n−1) ϕ2V (n−1) . . . . . ϕm−1V (n−1)




Properties, Independent Characterization

Galois calls it a “group,” but it is likely that he used the wordas a synonym for set. But, as he remarked, it is a set ofpermutations closed under composition, thus a subgroup ofSm.The permutations in this group satisfy:

If F (X1, . . . ,Xm) is a polynomial in m-variables withcoefficients in K and if F (r1, . . . , rm) = c ∈ K , thenF (rτ(1), . . . , rτ(m)) = c for all permutations in this group.If F (X1, . . . ,Xm) is a polynomial in m-variables withcoefficients in K that satisfiesF (rτ(1), . . . , rτ(m)) = F (r1, . . . , rm) for all permutations τ ofthe group, then F (r1, . . . , rm) ∈ K

As a corollary one obtains: A permutation τ ∈ Sm is in thegroup if and only if it satisfies: Whenever F (X1, . . . ,Xm) is apolynomial in m-variables with coefficients in K such thatF (r1, . . . , rm) = 0, then F (rτ(1), . . . , rτ(m)) = 0. This lastproperty shows that the definition of the group is independentof the choice of V and that it is, in fact, a subgroup of Sm.

Tomas Schonbek EG




A modern perspectiveIgnore if confusing

Given abstract up in the cloud fields F ,E with F a subfield ofE , one considers the set of all automorphisms of E that leaveF fixed. One says E is a normal extension of F if it is alsotrue that if an element of E is left fixed by all theseautomorphisms, then it has to be in F . If E is a normalextension of F , then the order of the group of automorphismsof E leaving F fixed equals the dimension [E : F ] of E as avector space over F .If F ,E are fields, if p(X ) is a polynomial with coefficients in Fand E is a minimal field extending F and containing all rootsof p(X ), then E is called a splitting field for p(X ) over F .One proves all splitting fields are isomorphic, so one talks ofthe splitting field of p(X ).One proves: E is a normal extension of F if and only if it isthe splitting field of some polynomial p(X ) having norepeated roots.If p(X ) is a polynomial with coefficients in F and E is asplitting field for p(X ) over F , then the group ofautomorphisms of E leaving F fixed is the Galois group ofp(X ). This group can be identified with a subgroup of thegroup of permutations of the roots and, as such, is the sameas the group defined by Galois.

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In Fact,

The zeros of Q(X ), hence also of Q0(X ) are all distinct.

K (V ) is the smallest field containing V and K .

Because all the roots of the main equation are polynomials inV , K (V ) contains them all.

Because all other zeros of Q(X ) and of Q0(X ) are linearcombinations of the roots of the main equation, all thesezeros are also in K (V )

It follows that K (V ) is a splitting field of both the polynomialdefined by the main equation and for Q0(X ), hence a normalextension of K .

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Example 1.

Details are left out.

Lets look at x3 − 2 = 0, with K = Q. Withω = e2πi/3 = −1

2 +√32 i , its roots are

3√

2,3√

2 ω,3√

2 ω2.

It is easy to see that a working choice of A1,A2,A3 isA1 = 1,A2 = 2,A3 = 0.

With this choice, V =√

3 3√

2 i .

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x3 − 2 = 0 continued

The irreducible polynomial over Q satisfied by V isx6 + 108 = 0.

The other roots of this polynomial are

V ′ = −V = − 3√

2√

3i , V ′′ =

(−3

2+

√3

2i

)3√

2, V ′′′ = −V ′′ =

(3

2−√

3

2i

)3√

2,

V (4) = V ′′ =

(−3

2−√

3

2i

)3√

2, V (5) = −V (4) =

(3

2+

√3

2i

)3√

2

A bit of arithmetic/algebra shows that we have to take

ϕ(x) =1

18x4, ϕ1(x) =

1

2x − 1

36x4, ϕ2(x) = −1

2x − 1

36x4.

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x3 − 2 = 0 continued

Galois’ table looks as follows:

(V ) ϕ(V ) = 3√

2 ϕ1(V ) = 3√

2ω ϕ2(V ) = 3√

2ω2

(V ′) ϕ(V ′) = 3√

2 ϕ1(V ′) = 3√

2ω2 ϕ2(V ) = 3√

2ω

(V ′′) ϕ(V ′′) = 3√

2ω2 ϕ1(V ′′) = 3√

2ω ϕ2(V ′′) = 3√

2

(V ′′′) ϕ(V ′′′) = 3√

2ω2 ϕ1(V ′′′) = 3√

2 ϕ2(V ′′′) = 3√

2ω

(V (4)) ϕ(V (4)) = 3√

2ω ϕ1(V (4)) = 3√

2ω2 ϕ2(V (4)) = 3√

2

(V (5)) ϕ(V (5)) = 3√

2ω ϕ1(V (5)) = 3√

2 ϕ2(V (5)) = 3√

2ω2

As we can see, all permutations occur. The Galois group is(isomorphic to ) S3.

Tomas Schonbek EG




Example 2

We consider again x3−2 = 0, but after adjoining the root 3√

2.

V = 3√

2√

3 i as before, but now the irreducible polynomial itsatisfies is

X 2 + 33√

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; itconsists of the identity and one transposition.

Tomas Schonbek EG




Example 3

We consider the equation x4 + x3 + x2 + x + 1 = 0, withroots r , r2, r3, r4, where r = e2πi/5 = cos(2π/5) + i sin(2π/5),over Q.

Because r , r2, r3, r4 are independent over Q, any four distinctA1,A2,A3,A4 will work.

For example, A1 = 1, A2 = 2, A3 = −2, A4 = −1, giving

V = r + 2r2 − 2r3 − r4 = 2

(sin

2π

5+ 2 sin

π

5

)i .

Tomas Schonbek EG




Example 3 contd., x4 + x3 + x2 + x + 1 = 0

With some effort (and help from Maple) one gets that thepolynomials ϕ,ϕ1, ϕ2, ϕ3 are

ϕ(X ) = − 1

25x3 − 1

10x2 − 1

2x − 3

2, ϕ(V ) = r , (2)

ϕ1(X ) =1

50x3 +

1

10x2 +

1

2x + 1, ϕ1(V ) = r2, (3)

ϕ2(X ) = − 1

50x3 +

1

10x2 − 1

2x + 1, ϕ2(V ) = r3,(4)

ϕ3(X ) =1

25x3 − 1

10x2 +

1

2x − 3

2, ϕ(V ) = r4. (5)

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Example 3 concluded, x4 + x3 + x2 + x + 1 = 0

Here is how the table of permutations looks like.

ϕ(V ) = r ϕ1(V ) = r2 ϕ2(V ) = r3 ϕ3(V ) = r4

ϕ(V ′) = r4 ϕ1(V ′) = r3 ϕ2(V ) = r2 ϕ3(V ′) = rϕ(V ′′) = r3 ϕ1(V ′′) = r ϕ2(V ′′) = r4 ϕ3(V ′′) = r2

ϕ(V ′′′) = r2 ϕ1(V ′′′) = r4 ϕ2(V ′′′) = r ϕ3(V ′′′) = r3

If τ denotes the permutationr → r3, r2 → r , r3 → r4, r4 → r2, then G is the cyclic group{e, τ, τ2, τ3}.

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The heart 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 theGalois 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 whatnowadays is called the Fundamental Theorem of GaloisTheory.Proposition IV is a bit of a digression.Proposition V gives the necessary and sufficient condition foran equation to be solvable by radicals.Propositions VI-VIII expand on this matter.

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Propositions II and III

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

Tomas Schonbek EG




Solving Equations

An equation is considered solved if all its roots have becomeg-rational, are in the field K in which we are working.

In this case V ∈ K ,the irreducible equation satisfied by V isx − 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 objectsalready in the field, to size 1?

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Equations that hate radicals, and don’t allow them to dotheir work

Adjoining radicals means adjoining things like k√

a, wherea ∈ 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) rootsof 1; since we can also just adjoin them. So adjoining radicalsbecomes equivalent to adjoining ALL roots of an irreducibleequation of the form xp − a, where p is prime.This means that IF the original group G can be reduced, itcan also be reduced to a normal subgroup G0 such that G/G0

is prime. Groups of prime order are cyclic, in particularcommutative.

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Equations that hate radicals, and don’t allow them to dotheir work, contd.

To solve the general equation of order m, because thecoefficients could all be transcendental, unrelated to eachother, we must assume its Galois group is Sm.One thing that happens in Sm if m ≥ 5 is that you can havetwo 3-cycles having exactly one element in common. A3-cycle denoted by (abc), where a, b, c ∈ {1, . . . ,m} is thepermutation σ such that σ(a) = b, σ(b) = c , σ(c) = a, andσ(k) = k if k 6= a, b, c.Suppose G is a subgroup of Sm that contains all 3-cycles. Letσ = (abc) be one of the 3-cycles; a 6= b 6= c 6= a. If and onlyif m ≥ 5, we can find d , e so a, b, c , d , e are 5 distinctelements and form the 3-cycles τ = (dac) and λ = (ceb)Suppose H is a normal subgroup of G and suppose G/H isabelian.

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Equations that hate radicals, and don’t allow them to dotheir work, contd.

Because G/H is abelian (commutative),

idG/H = (τH)−1(λH)−1(τH)(λH) = (τ−1λ−1τλ)H,

implying τ−1λ−1τλ ∈ H.

Going from right to left (usual composition order)

τ−1λ−1τλ = (dca)(cbe)(dac)(ceb) = (abc) = σ.

This proves: If a subgroup G of Sm, m ≥ 5, contains all3-cycles, every normal subgroup H such that G/H iscommutative, must also contain all three cycles.

This severely reduces the amount by which G can be reduced.It shows that S5 is not solvable.

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Solvable equations

But why should an equation whose group G has the propertythat one can find a sequence of normal subgroupsG0 = G ⊃ G1 ⊃ G2 ⊃ · · · ⊃ {e} such that Gi1/Gi is of primeorder, be solvable by radicals?

Here is Galois’ argument.

We must show that if the group is G , and G has a normalsubgroup 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 = e2πi/p.

Assume the equation has m distinct roots r1, . . . , rm.

Tomas Schonbek EG




Galois’ argument continued

Let f ∈ K [X1, . . . ,Xm] be a polynomial in m-variables suchthat if for σ ∈ Sm we setσ · f (X1, . . . ,Xm) = f (Xσ(1), . . . ,Xσ(m)), then

σ · f (r1, . . . rm) = f (r1, . . . rm) ∀ σ ∈ H,

σ · f (r1, . . . rm) 6= f (r1, . . . rm) ∀ σ ∈ G\H,

Let σ ∈ G be such that σH is a generator of G/H.

Let θ = f (r1, . . . , rm), θ1 = σ · f (r1, . . . , rm),θ2 = σ2 · f (r1, . . . , rm),. . . , θp−1 = σ · f (r1, . . . , rm).

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The argument cont’d

Let ξ =(θ + αθ1 + · · ·+ αp−1θp−1

)p.

If there is time, I will argue at the board that ξ ∈ K , ξ1/p /∈ Kand if we adjoin ξ1/p to K , then H becomes the new group ofthe equation.

Tomas Schonbek EG




Second degree equations

Suppose the equation is ax2 + bx + c = 0, with roots r , s.The group S2 = {e, σ} with σ(r) = s, σ(s) = r . We have toreduce it to {e}.In this case p = 2, α = −1 (Actually, this is always the firstreduction). We need a polynomial in two variables leftinvariant by {e} (any polynomial will do) but not by σ. Thesimplest choice is f (X1,X2) = X1 − X2.The quantity Galois says should be adjoined is

√ξ where

ξ = (f (r , s)− f (s, r))2 = (2(r − s))2 = 4(r2 + s2 − 2rs)

= 4((r + s)2 − 4rs

)= 4

((b

a

)2

− 4c

a

)=

4(b2 − 4ac)

a2.

Tomas Schonbek EG




The End

THE ENDThanks for listening

Tomas Schonbek EG

A Math Club Event Tomas Schonbekmath.fau.edu/schonbek/Galois/Galois121109.pdf · Solving Equations...

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