II Relative : tasting Galois computation of An a explicit - - PowerPoint PPT Presentation

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II Relative : tasting Galois computation of An a explicit - - PowerPoint PPT Presentation

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Galois Group The Galois theory II Relative : tasting Galois computation of An a explicit group Newstead E ' FEKEE , how If Gul 'Elk ) : : " " " . . . . other ? each te subgroups ) Lemma ( Tep sub

slide-1
SLIDE 1

The

Galois Group

II

:

Relative

Galois theory

slide-2
SLIDE 2

tasting

An

explicit

computation of

a

Galois group

slide-3
SLIDE 3

Newstead

E

If

FEKEE , how

Gul'Elk)

÷:÷÷¥÷:

'

.

"""

...

te

each

  • ther ?
slide-4
SLIDE 4

Lemma ( Tep

sub extensions

are

subgroups)

IF FE KEE ,

then

Gal (Elk ) EGAICEIF)

.

PI

we

need

to

check

GallElk) C- GullElf)

.

Now

regal (Elk ) implies

re Aut CE)

with

Mk

'

  • idk
.

Hence

  • ff

= folk ) If = ideate

  • idf
.

Therefore

it Gal ( Elf )

.

IDK

slide-5
SLIDE 5

Nhlvral

question

:

are

" bottom

subextensions

" also

subgroups ? No

.

=

E

If

TE Gall KIF) , do

we

I have TE Gal ( EIF ) ?

k

Ip}

TEGAKKIF) Recall

T

'
  • K -5k , and

hence

it

isn't

defined

  • n

all

  • f

E .

slide-6
SLIDE 6

Can

we

make

rt Gul (EIF )

an

element of

Gal ( KIF ) .

Notimnediatdy

: r

is

a fucken en E , not k .

However ,

we

do

know

4k

: K → E

.

In

  • rder

for

rlk

to

be

an

element of

Gall KIF )

, we

need

ion ( rha)

  • K
.

When

does

this

happen ?

slide-7
SLIDE 7

Neaexampte

het

E

be

the

splitting

field

for

x ' -2 ,

and let

K

  • ( Vz )
.

( Here

:

F

  • Q1 )
.

Suppose

we

take

TE Gall Ela)

with

  • ld , )

=L,

and

rlxz )

= as

.

Is

imlrlk )

=

k ?

from

homework

we

know

da)

  • * * ④ (a)
  • K
.
slide-8
SLIDE 8

Thy ( when restrictions

are

" nice " )

If

FE KEE ,

where

k

is

the splitting field of

gcx) c- FIX)

and

E

is

the splitting field of

f-Ix)

e F EXT , then

for all re Gal ( EIF)

we

have

im ( Nk )

  • K
.

PI let

pi ,

. ., pmek

be

the

roots

  • f glx)
.

we've

seen then that { 13,

"

' - - Palm

: a-eiadlirfep.

is

an

F - basis for

k .

slide-9
SLIDE 9

key fact

: r

permutes

{ is. .

...

. pm }

.

First let's

show

im ( Hk ) E K

.

Let

KEK

be

given

,

so

k

  • Z te

,

. . -em Pie '
  • - ' Prem
.

Observe

ruck)

= r(Efe.

..-em Pie

' - - Bim)

= Efe.

. . -em r ( pile

'

  • - r ( Pm)"

E K

.

If I

Hence in (Hk) Ek .

Similar

argument

resolves

" 2

"

.

④

slide-10
SLIDE 10

Core

( Restrictions to splitting field

are

"nice ")

If

FE KEE

where

k

is

The splitting

field

for

glx) EF Ix)

and

E

is th sphlty field

far

ffx) EF Ix) ,

then f

: GalletF) → Gal ( KIF)

given

by

YG)

= rly

is

a

homomorphism with

Kerl 4)

= Gal ( Elk)

.
slide-11
SLIDE 11

PI

we know

4

is

well

  • defined
. Operation

preserving

is

r

, re Ik

=

  • , Ik k Ik
.

Now

Ker (4)

= { TE Gall EIF )

: Hk

  • idk}

= { re Aut (E)

: elk

= idk}

=

Gal ( Elk )

.

FE

slide-12
SLIDE 12

Core (

Galois Quotients)

If

FEKEE

where k

is th splitting field for

separable gcx)

c- FIX)

and E

is the splitting field for

separable

f-Ix) c- FIX

, then

Gul ( Elk)a Gullet)

and

Gal ( Elf)Kal CE Ik)

=

Gal ( KIF ) .

PI

we only need to

check

that y from the

last

result

is

surjective

.
slide-13
SLIDE 13

We

know

I Gul (EIF) I

  • IE : F)

and

( Gul (KIF) I

= [ K

'
  • FT
.

separable

Observe that

E

is

the

splitting field fernflxleklx)

,

and

I Gul ( Elk) I

= [ E

: KJ .

So

we get

limits

  • I'

"" "'

cent

  • Y;:Y¥i
  • = [E :KIK :F)

T enay

=

Ik :F)

= ( Gal (HEH

  • I codomain

I