":::i too big ) aren't CI ( Galois groups [ E : F) 31gal ( - - PowerPoint PPT Presentation

i
SMART_READER_LITE
LIVE PREVIEW

":::i too big ) aren't CI ( Galois groups [ E : F) 31gal ( - - PowerPoint PPT Presentation

Galois Extensions " big " ? Galois group ' when I : are " Independence of " Chohuters , Part I ":::i too big ) aren't CI ( Galois groups [ E : F) 31gal ( EIHL i F) sis . Then ( E Suppose (


slide-1
SLIDE 1

Galois

Extensions

I :

when

are

Galois group'

" big

" ?
slide-2
SLIDE 2

" Independence of

Chohuters

, Part I

"

CI ( Galois

groups aren't

too big)

Suppose

( E

i F) sis . Then

[E : F) 31gal (EIHL

"÷÷÷÷÷÷:÷÷::÷÷÷i÷÷

> [ E

: E

Gal (E"")

z

( Gal ( ElF) I

.

k④

Note :

I Gal LEIF) l

is

"

Max

size

" iff

E

Ga' CE't )

= f
slide-3
SLIDE 3

Def

' n ( Galois

Extension

  • r Normal

Extension )

AT

extension

Elf

is

called

a

Galois

extension

if

[E : F ]

  • l Gal LEIF) l
.

Nate

EIF

is

Galois

iff

EGAKEIF)

f

.

Today

's

theme : characterize

Galois

extensions

Nate

we

already know that if

E

is

the

splitting

field

for

a superable

polynomial

th) c- FIX?

Then

[E : F) =L Gal LEIF)l

.
slide-4
SLIDE 4

Thin (Equivalent

characterizations

for

Galois

extensions) suppose

[ E :F) co

and let

G

  • Gal CE tf) . Then

the

following

are

equivalent

:

Cis F

  • EG

(2) for all

irreducible

p G) C-FIX) with

some

root in E,

we

have

plx)

is separable

and

plx) splits in E

E

is the

splitting field for

some gamble fide ECD

(4) Elf is

Galois

slide-5
SLIDE 5

PI

( we'll

use

the

"

square of happiness

"

City

⇒ of

y

④4) ⑤ 13)

(CH ⇒ L2))

we

get to

assume

F-

  • EG
.

WTS

: if

plx) EF

is

irreducible

and

has

some neat in

E

, then

pk)

is

separable

and

splits

in E .

let

plxteflx

)

be

irreducible

with

a-E satisfying plant

  • O.

We

can

assume

WLOG that

plx)

is

manic,

so

plx)

a irrp Ca ) .

slide-6
SLIDE 6

let

  • { da)
: re Gal (EIF)) = { a , .
  • sik)
  • rbit
  • f

a

under

no repeats

Gal LEIF)

consider

glx)

= II ( x
  • ri )

E E (x)

.

Claim : glx) @ FIX)

.

let

c- Gal LLEIF )

be

given

.

Then

*gun)? #(II. Ix

  • ai))
  • II. Ix
  • IIK-ait.gl#epobyrrke
slide-7
SLIDE 7

Since F

=

E Gall E'F)

, we

can

have

T

* (glx))

  • glx)

fer

all T

c-Gal LEIF)

iff

the

coefficients of glx)

all

come tram F

iiff glx) EF Cx)

.

Upshot

:

we

have

polynomial glx)

  • ITH
  • ai) EFG]
.

So

:

x

is

a

root of glx)

.

We

know

{ hlx) c- FED

: hk)
  • O)
= ( irrr.CN/--lplxD .

So : plxllglx)

.

Hence

roots of plxl

are from {a,

. . , Ga) .
slide-8
SLIDE 8

So

since

glx)

has

no

repeated

roots

and

plx)

is

a

" subproduct

"

  • f

glx) : IT ( x

  • a:) ,

we

have

plxl

has

no

repeated

roots

.

So :

plx)

is

separable

.

Also :

since

glx

)

splits in E

, so

does plx)

.

(Lii) ⇒ Ciii)) we get to

use :

any

irreducible plx) c- FIX)

with

a

root

in

E

is separable

and

splits

in

E

.

WTS :

E

is The

splitting field of

a separable polynomial

.

from FAT .

slide-9
SLIDE 9

We

know E

  • Fla,
.
  • pan)

for some algebraic

  • X. ,
  • -in EE
.

We

claim

that

E

is the

splitting

field

for II. irrplai)

.

why is this

separable ?

By (2)

,

since

irrflai)

is

irreducible

and

E

has the

root ai

,

we

know

irrp ki)

is

separable

.

By def'n

  • f

separable for

non

  • irreducible polynomials

,

we get

  • II. irrp (ai)

is

separable

.

Rest of proton homework