I : Algebra : prove goal First Theorem of Algebra ) them ( - - PowerPoint PPT Presentation

SLIDE 1

Applications

I

:

Fundamental theorem of

Algebra

SLIDE 2

First

goal

: prove

them (Fundamental

Theorem of Algebra )

All

nonconstant plate EW

have

some

root

in

a .

GI Any

plx) e Glx)

splits completely

.

If

By

induction

, using the

fact that

p G) = O

Mearns

plx )

• ( x -x) g

Cx) EECXT .

µz

This

means

Q

is

" algebraically

closed

"

.

This

mate

,

¢

The

" algebraic

closure

" of

IR

.

SLIDE 3

To

prove this

, we

need

a

few

preliminary

lemmas

.

Unmeet

Any

quadratic plxteclx)

splits completely

.

PI if

plx) = axztbxtc ,

we

know

roots of

p must

take

form

• betrayal

a-

.

Are these roots

in

¢?

This

comes

down

to : does

E

contain

some

8 with

82=62-4 ac

.

If

b' -Yao

= xtiy ← rectangular cords

= reit ← polar cord ,

so

r = rFty2

Consider

r

• rreik.com#eiaem.i*o=arctuPlxl

SLIDE 4

Core

G

has

no

extension of degree

2

.

Pf If

Ck:&)

• 2 ,

then

K

• Ela)

where

Hincks)=2 .

But

we

knew

That

no

quadratic

new

E is irreducible

.

DADA

tenma

( Real

• dd degree

polynomials

have! real

root)

If

flx) EIR Cx)

with

self)

• 2kt I

, Then f

has

a

root

• n

IR

.

Cos

IRN

has

no

• dd degree

, non

• linear irreducible .

SLIDE 5

PI (of

lemma 2)

idea : for such

an f

,

we

get fl huge negative) Hh.gr positive) so

,

~

so

by

intermediate

value theorem

we

have

some

c with

ftc)

.

• O

.

{⑦ If

you

want technical details

:

°

WLOG ,

can assume

flx) - Iota,Xt

• -tan.,xn
• '

txn

+

Show

t

• I * E lait

has

Ht) > O .

X

ft

• t) so

.

DA

SLIDE 6

Pf ( Fundamental

Theorem at

Algebra)

• structure

:

① argue

we

can

assume

WLOG

that

plx)HRW

② consider the splitting field

Elk of ith)pk) .

(Nde REQ - Rli) E E)

.

Use

Galois they

to argue

[ E

: IR) =L

This

will force Q

• E

, so

plxl splits

in Cl

.

SLIDE 7

① we

prove

pH

can

be assumed

to

be

in

Rex]

.

let fled =p (x) flx)

, where

it

plx)

• {

⇒

an x

"

we

have pix) ⇐ E.oaixn

, where

In

• conjugate of an

( if I

is complex

conjugation , then

f

Cx)

= I * (pm) )

why

is

flx) ERIN ?

we know

flxtelklx)

iff

T* (flx))

• ffx)

= f (x)

. But

I

* (HH) - T

*(plxtplx))

• (E Han) xn) (Etta ) xn)

ftp.p-MHPH-flxl

.

SLIDE 8

Now : if f

has

some

root

in

G,

do

we

knew

p

has

some

root

in G ? Know : if fla)

• O

,

Then

pk) plat

• O

.

If

plat

• O

,

Then

p

has

a

complex seat . (yay !)

If Fla)

• O , then I End

" =D

.

Apply

z

:

O

• TCO)
• Ethan) In

= E

an E

" = phi)

So

a- EQ

is

a

root

at

p

.

(yay!)

SLIDE 9

For

④ ,

assume

plxl.cl/2lx3

and

let

E

be

th

splitting

field for 1×2+1 ) pix)

.

This

forces

IREQEE

.

Since

Iti) pH

is

separable

( since

char ( E)

• O) , we

get

EHR

is

Galois

.

let

G

• Gul LEHR)

.

By the

degree formula

we

have

CE

: IRI

2mk=lGal LEARN

for

some m > I

and

K

is

• dd

.

pg lit , we

get that

Some fancy

group Theory

(Sylow theorem)

there

exists

HE Gulf EHR)

with

Htt - 2M

.

By

Fundamental Theorem of Galois,

we

have [FKH) : R)

• R .

SLIDE 10

So

FLA)

is

an

• dd

degree

extension

• f

IR

.

If

K

> I , then

FLH) t IR

.

let

LEECH) VR

.

we

get

[ IRK)

: 11231 [ FIH) : IRI

• k

,

and

so

d( irrp.la) )

is

a

paper

divisor

• f

k .

ie

,

dlirr.pk))

is

an

• dd

number

bigger than

1

.

→←

(to Lemmer 2)

So

k -

• I ,

and heme

IGWILEHRH - 2

"

.

We

already

knew

m3l ,

but

we

want

m=l

.

SLIDE 11

[ am,

If

m > I

, Then

I Gall Ele) I - 2M

"

, and

¢

some

more fry

group they

says

12

there

exists

HE GallEla)

with

R Htt

• 2M£

.

Galois correspondence says

we

have

[ Flit) : e)

• lGYYI=Y÷
• 2

.

But

we

know

G

has

no

degree

2 extensions !

→←

Hence

m

• I

, and

CE

: 1123=2

.

Since

IREEEE

,

the

degree

formula

gives

LE :

=L

, se

E - G .

ppg