I : Theorem Great Galois ' when polynomial a theorem that - - PowerPoint PPT Presentation

SLIDE 1

Applications

I

:

statement

• f

Galois

'

Great

Theorem

SLIDE 2

today

write

down

a theorem that

expresses when

a

polynomial

has

roots that are

expressible

in something

analogous to

the gvntmhi

formula

Motivation for

this

: quadratic formula, cubic formula,quartic

formula

iwmdn

:

roots of

ax'tbxtc

• O

Cin

Q) are

x

• b ± ME
• Z a

Note :

roots

are

expressible

in form of

held elements, felt ops,

and

not extraction

SLIDE 3

Cubic

formula

• a

root at

ax't bx'texted

is

✓qtT t Vq-Tt tp

where

p

• %a

,

r

• %a ,

q= p't bcjaazd Note:

formula

is

in terms

• f

field

elements, field operations,

and

root extinctions

Quartic formula

same takeaway

:

roots

are

in terms of

• field elements, field operating,

root extractions

SLIDE 4

To

generalize these

nations

• f

" solvable

by

a formula

"

Def

( Radical Extension)

A field extension

Elf

is

called

a

radical

extension

if

there

exist

intermediate

fields -

F

= Eo E E , E Ez E

• E Eee , E Ee

= E

so that

for

all

Kiel

we

have

some

ei E Ei - i

and

some

ni EIN

with

Ei

• Ei - i ( nine; )

( where

hifi

is just some

root of

x

"

• ei EE; -did)

SLIDE 5

EI

Ut

E

be

the splitting field of

x'-2€

Ix)

. we

know

E

= Q ( Vz

, wt )

where

w= -11¥

=

( Vz

,

w )

Vz)

=

(RE , Fs)

a

:÷÷÷÷÷:i÷÷÷÷::*:*

..

SLIDE 6

Define ( solvable by

radicals)

A polynomial

flxltflx)

is

solvnbkbgmdicab if

its splitting field

Kf

is

contained

in

some

radical extension

.

÷i÷÷÷÷÷iem¥÷÷÷÷

:

its

roots

are

expressible in

terms at

field

elements , field openheirs

• and

root extinctions

.

SLIDE 7

Exe

X3 -2E

ID

is

solvable

by

radicals

since

Kf

• Qfsryfs)

where

we've

seen

④(vz, Fs)1a is

a

radical extn

.

EI

suppose

char (F) Hn

, and

let

ctf

be given

.

Then

f-(x)

• X

"

• c

is

soluble

by

radicals

since

Kf

• = FIVE

, Wn )

and

Flora , un)

is

a

radical

extension it

( same

basic

idea

even works if

church In ,

but

need

some

car

with

roots of unity)

SLIDE 8

EI

Evey

quadratic , cubic, quartic polynomial

are

solvable

by

radicals ( if

charles -42,33)

because

we

have formulas

That

express the

cook

in

terms at

field

element, field

• perates ,

and

root

extinction)

Whit

about

the

" other

side of the

coin

" ?

Ie, if

f

' is

not solvable by

medicals, does Thet

mean

f

has

not that

aren't expressible

in

terms

• f

field element, field

• ps, and

root extractions?

SLIDE 9

( radical

extensions

are

cloud

under

composition)

If

E

and

k

are

radical

extensions of

F

, Tha

Ev K

is

a radical

extension

• f

F.

Een

'te .

emirate.im,

• ÷:÷,
• " -

*

÷

Ez =

,E

, (Vez)

Ei

• FIVE

, )

¥

• Ii

:*,

• I

:*,

• iii.Imai .

SLIDE 10

Cos If f

is

not soluble radicals, then

There

is

no analog

• f

quadratic

formula

for polynomials of degree self)

.

PI If

There

were

such

a

formula , Thu

all

roots

• f f

would

be

expressible

in

terms of

F ,

Held

• pontian

, and

root

• extinctions. Ie

,

if

a is

a

root

• f

f

,

then

FK)

is

a subfield of some

medical extension.

But by

• ur

last

result,

this

means

ke =¥÷Ik

)

is

also

in

some

radical extension . Ie , f

is solvable

by

indicaIs ,

BB

SLIDE 11

So :

looking fr

higher degree analogs of quadratic formula

amounts

to

asking

:

when

is f

solvable by radicals ?

Higginson

How

can figure

• ut

if f

is

soluble

by

medicals

w/o

looking fer

a formula for its

roots ?

To

answer this

, we

need to take

an

excursion into

group Theory

.

SLIDE 12

Deth

(solvable group)

A group

G

is

called soluble

it

there

is

a sequence

• f

subgroups

geg)

⇐ Hos HEHE .

• ⇐ Hn -f- HE G

so that

His Hit ,

fer all

• eisn

and

Hitya ,

is

cyclic

.

⇐ All finite

abelian

groups

are

solvable

. By

Fundamental

theorem , GI kn

,

④ Renato

. -

• ④Ink

for

some

4, ne,

. > ne HN

and

so

{e) skatoleHo

• -⑦let Eka.to?LnztofeHo--tOlelE---EKn.to.--tOKnk

SLIDE 13

EL

S,

is

solvable

since

{e) E ( ( 123))

E S3

siteI

size 3

size 6

Sime

Iska) > 1=-115,1

,

we

have

412317053

and

53/4123,7=12

Also

{e)a ( CRH)

and

4123%3=83

SLIDE 14

EI

Sy

is

solvable

toe :

{e) s ( ( 12713477 E le , 1127134)

, 1137124), 1147123)) E Ay s Su ,

Fun

exercise :

this

chain

• f subgroups Sats bees

desired

properties

Exe

Dn ( symmetries of

a

• gon)

is

solvable

because

{ e) E { rotations) E Dn

fits

the

bill .

SLIDE 15

Thin

( As

is

not

solvable)

As

is

not

solvable .

If ( sketch)

⑨ As

is

generated

by

3

• cycles, which

means

As

= { o , .

• ok

:

KEN

and

Tiff is

some

3

• cycle}

⑤ all

3

• cycles

are

conjugate in

As

,

ie

if

r,

• lab c) c-As

and rz

• lxyz) c-As

thin

there is

Sonu

F

c-As so

that

F

, = FTZF

"

.

SLIDE 16

④ if

Ho Ag

and

Atle)

,then

H

has

some

3

• cycle

why

is this

enough ?

If

H As

.

and H

• le)

, then

Ayy

not cyclic

Hence

fee As to

be

solvable,

we

need

some

Hotty

with

H t le)

subgroups in As

H

contains

some

3- '5k

w/

"consecutive

II.

'

Ii :

tins

. !

1¥

. torments . " "

{e) DA 5

pm

SLIDE 17

Deep

• ish theorem tem

group theory

• If

G

is

solvable, then

evey

subgroup of G

and

every quotient at

G is

solvable too

.

CI If

n > 5. Then

Sn

and

A.

are not

solvable .

Pf

• One

can

find

a

subgroup isomorphic to

As

in

Sn

• r

An

( namely

" S

's

" ↳ Sn

since

{Gesu : Olli) - i

and

tnke the As

inside this)

.

for all i >5)

SLIDE 18

Now FOR THE

BIG

THEOREM !

• Thon ( Galois

'

Great Theorem )

Tt

chart F)

• O

, let flat flex) , and

let Kf

be

its splitting field

.

Then

f

is

solvable by

radicals

iff

Gal ( KH F)

is

solvable

.

PI

Next

time

.

SLIDE 19

A few takeaways

• Abel

was first

to prove not all quirks

are

solvable

°

Abel

could prove

that

certain

polynomials for

[ Ya

,

"

. "!! """

"

hnd.am?i::neiemic

• we

knew

Sa

, 53, Sy

are

solvable,

so

ay poly

at degree

E 4

is

solvable

by

radicals

SLIDE 20

EX

In

1962 ,

Feit

• Thompson

proved

that

all

• dd
• rder

groups

are

solvable . (proof

was

n 250 pgs)

Hence

it

164 ( KHE) )

is

• dd , then

f- is

solvable

by

indicants

.

SLIDE 21

EI

There exist

" unsolvable quintic

"

let

flx) E

Ix)

be

any

quintic polynomial

that's irreducible

and

with

precisely

3 real

root

.

Claim

: f

is not

solvable

by

radicals .

Ff :

514 Gal ( KHAN

, so

group they says That God ( KHQ)

has

an

element r of

• rder

5 .

Since Gul LKFIQ) ↳ 55

, we

get

that

that r

is

a

5

• cycle

(thought of in It).

SLIDE 22

But

complex

conjugation

exchanges the

2

complex

roots but keeps

the

3

rat

roots fixed,

so

its

a

transposition

.

So

Gul ( Kha)

(thought of inside

Ss )

has

a

5

cycle

and

a

2

• cycle

.

Group Thay faut

: ay

subgroup of

Si

with

a 5.gale

and

a

2- cycle

is

all

• f

Ss

. So ' . Gul (KH@HSI

so

is

not

solvable

. So f

not solvable by

• radicals .

SLIDE 23

For

example

:

x

5

• 4xt2