:i extensions ) characterizations for Galois Thin ( Equivalent - - - PowerPoint PPT Presentation

SLIDE 1

Galois Extension,

I :

Intermediate

Galois

Extensions

SLIDE 2

lame

Elf

is Galois 'tf l Gal LEIF ) t.EE if] Thin (Equivalent characterizations for Galois extensions) suppose [ E :F) so and let G

• Gal (EIF) . Then

the following are equivalent :

:÷÷÷÷÷÷÷÷÷÷÷÷÷¥÷÷÷÷

:÷÷i÷÷

(4) Elf is Galois

SLIDE 3

long time ago

:

Cory (

Galois Quotients) me and Elf are Galois

"

÷÷÷÷÷:¥÷÷÷

.it#i::iEii:::D

and Gal ( E #Kaye , k, =

Gall KIF ) .

WTNwHti ,

this

' ' iff " ? when are intermediate extensions Galois ?

SLIDE 4

Def

' n

( Conjugate

intermediate fields )

Tet Elf

be

Galois

,

and

supper

FEKEE

and

FE KE E

. If there exists some

isomorphism

T : k→I so

That

Tle

• side

, Then we say

I

is

conjugate to k .

Thin (

"conjugate to " is an equivalence relation)

The

relation

Kai

given

by

" I is conjugate to

K

"

is

an

quintana

relation

.

PI

Go

Through The

motions .

SLIDE 5

Thin (Alternate description

for conjugate fields)

let EIF

be Galois , and

let

conj ( K)

• { I

: I is

conjugate to

K)

. Then

conj ( K)

• = { t ( K)

:

retral ( EIF)}

.

PI For

" Z " ,

• a. k fr

ay

regal LEIF) ,

we hat

4k

: K -7 idk) is an isomorphisms with (Mk)Ip

• idp .

For ' ' e " , let

II

be

conjugate

to

k

with T : ktk .

By Thur 51 ,

since

Elk

and

Elli

are Galois we know

t

extends to some re Gal LEIA . So rlk) .

• TWI

. 1B

SLIDE 6 So : If

conj ( K)

• { K) , Then

we have

HK)

• k

fr

all

re Gal ( EIF)

.

Ex let

E be

splitting field

• f

x' -Le

XT , and let

k , - ④ ( Vz)

and

ka

• ④ ( wt)

and

Ks

• ⑥(WEE)

Each is

degree

3

• ur

, so as vector spaces They are

isomorphic?

Are Thy

isomorphic

as

fields ? yes

. we know

that

we can build

f

: arr) → ④(wer)

by

sending

Vc to

any

root "' f

f

• f idfirra 1352)) . idk
• 2) 'x'-2

Q

Ff

Q

SLIDE 7

Likewise

we can build

4,3

:

( Va) →

⑥ ( w't)

that

extends

idea :

④ → Q .

Hence

we

have

Ceuj ( ki)

= { K , .kz , Ks}

D Hew

is

Conjlk)

related

to

"

Galoisness

"

?

E.g.

,

• ur

" Galois quotients "

says if

KIF

is

Galois ,Then

Gall Elk ) # Gal LEIF)

SLIDE 8

Thy ( Equivalent

ChanduEakins of

intermediate Galois ness)

let

EIF

be Galois , and let

FE KEE

. Then

the

following

are egvinlent .

H conj (K)

= { K) (" for

all

• f Gal (EIF) ,

we have

rly C- Gull Klf)

(iii) Gal (Elk) A Gal ( EIF)

(iv ) Khs is Galois .

tf

use

" square of

happiness

" ⇒

'

(a) ⇐ Ciii)

SLIDE 9 4) ⇒ Cii) Given :

(oujlk)

• (K}

want :

foetal LEIF)

, we have

4k

c- Gal ( KIF) . Note :

tha

is

always

an

isomorphism them

k to Hk)

that

fixes

F

pointwise

. We just need to show

HK)

• K

for

all

rt Gal LEIF) . But

{ K)

• Coujlk)

= lock) : rt Gal LEIF)) .

1¥

SLIDE 10

Ci) ⇒ Ciii)

Given :

firebrat LEIF)

we

have

the C-Gall KIF)

want

: Gull Elk) s Gal (Elf)

By

• ur

hypothesis

. we have f : Gul LEIF) -' GUICHE)

given

by

Xlr)=Hk

is a

homomorphism

.

Further

Kerl 4) = Lothal LEIF)

: olk ' idk } = Gaulle Ik) .

By

305 ,

Gal ( Elk ) o Gal LEIF) . Dad

SLIDE 11 Ciii) ⇒ Liu) Given :

Gal ( Elk) s Gal LEIF)

want :

KIF

is

Galois .

we'll

prove this by showing i for

all

irreducible pileFck)

with

• ne

root m

k, then

plx)

is

separable

and splits in KUD . Since

Elf

is Galois , we already know Simon

plx

)

has a root in

E

, then

plus)

is

Leprnble

and

all roots

are in E . Assume t th carthy that plxl has a irreducibly fucker in

KID

• f

degree at

least

2

.

let

a be

the

root we knew about in

KCH,

SLIDE 12 and

let

13,8

be mob

at the

non

• linear fuhr
• f

plx

)

in

KCXT .

By separability

,

• a. 13,8

are

all distinct .

Since

Elf

is

Galois , we know

there

is Sean TE Gul LEIF) so

that ok)

• p

.

By

hwk 9,

Elk

is

Galois

, and so

there is

some

TE Gall Elk)

with

Tlp)

• 8

. Since we assume Gul ( Elk) o Gal ( EIF) , we know

for

ay

lothal LEIF)

and

pe Gal CEIK) , we

have

4pct

• ' e Gal ( Elk)

. In

particular , 9pct

' fixes

K .

SLIDE 13

Consider

r

• '

Te (o " )

• '

.

It

should fix all

KEK

. Since

AEK ,

we should have

(r

• ' T @

"5) Ca)

• a .

Nate

(r

• ' t (r ")
• 1) (d)

= r

• ' ( t ( ok )))

=

• ' ( t LPD
• r
• ' (8)

* a

where

the

last inequality

fellows swim

r (a) =p,

so

r

• ' ( p)
• d

. →a-

SLIDE 14 Civ) ⇒ Ci) Given :

KIF

is Galois want :

conj ( K)

= { K) . Note

that

fer

all

TE Gul ( KIF) is an isom

• phis'm

from

k to

k

that

extends

idf

: F - SF.

By

Theorem

51

, we know for

each

TEGAILKIF)

we have

[ Eik]

may

extensions r :

E -7 E

• f

T .

Evey

such extension is an element of

Gal LEIF?

How my

demark at

Gal LEIF) take this form ? I Gul (Klett CE :k) = (K : FILE :K)

• LE : F) - IGNKEIFH
• choices for

T # of extras

SLIDE 15 Since eade

retral (Elf)

is an extension at some TE Gall KIF) , we

have

HK)

• K .

So

Coujlk)

• { K)

.

1¥