II are : created ? Gal LEIF ) Lastine Gull Elf ) - f Ift F } Idf ) - - PowerPoint PPT Presentation

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II are : created ? Gal LEIF ) Lastine Gull Elf ) - f Ift F } Idf ) - - PowerPoint PPT Presentation

Dec 30, 2023 227 likes •359 views

The Galois Group elements How in II are : created ? Gal LEIF ) Lastine Gull Elf ) - f Ift F } Idf ) ={ rebut - iff - idf de re Gal ( EIF ) ie , - and th ) C- FIX ) has temma If a- Gal ( EIF ) LEE , Then f- ( rt ) ) too . - O f- (a) =D

slide-1
SLIDE 1

The Galois

Group

II

:

How

are

elements

in

Gal LEIF )

created ?

slide-2
SLIDE 2

Lastine

GullElf)

={ rebut

Idf)

  • f IftF}

ie,

re Gal (EIF )

iff de

  • idf

temma If

a- Gal ( EIF)

and th) C- FIX)

has

f- (a) =D

for

LEE, Then f-(rt))

  • O
too . ie , automorphisms

permute

roots of F - polynomials
slide-3
SLIDE 3

COI If

ffx) e F [x)

has a

distinct rook and

E

is

its

splitting

field, then

Gal CEle) s Sn .

Building

  • n groundwork from

Gabby

& Rachel

Corn (order of

Galois group for splitting

: field)

If

E

is a splitting field ofTtx

FED, then

I Gal CEIFH
  • CE
'
  • FI
. thisapp.lu?edthYfI:EtnQ
slide-4
SLIDE 4
slide-5
SLIDE 5

iYYu

Hmhg

slide-6
SLIDE 6

lemme If

E

  • Fla ,
", an) for algebraic elements

4 ,

. . . , an . , then an element re Gal CEH) is determined

by

its action
  • n

a ,

. . ., an .

Ie ,

if

we know the values

Ha ,) ,

. . . , ok . ), Then we know how

t

acts
  • n

any

EEE

.

PI

An

F

  • basis

for

E

is given

by

{ he

' . . - men : oeeisdlirf-ca.ai.im))
slide-7
SLIDE 7

(This

is an extension of the ideas That

Gabby

gave

us in The

proof

  • f The

degree

term . la

.)

This

means that

every

element EEE

can be

expressed

uniquely

as an

F - combination

at

this basis : e = a?÷m,

te

.. . ... en die '
  • - an

"

.

Hence

slide-8
SLIDE 8 So weget r(e) =o⇐fq,

te

.. . ... en die '
  • - an

")

=

Eap.muts

t ( te

,
  • e. he
'
  • men)
=

Exponents

rlfe

. ..-en) da ,)e '
  • - - Hagen
= Z

te

. . . - en

r la , )e

' .
  • - r kn )

"

. DM
slide-9
SLIDE 9

Ey

what

is

Gul ( QK , )/Q ) where

4=352 ?

Notation ,

irala,)=x3 -2

(you

saw this
  • n Hwy
  • ther
roots :

q=

w T2 and 43
  • WZVZ
  • It F3
where w
  • 2-EQ
and w2=
  • t = J
Note : w is a " 3rd root of unity " since w3=1 .
slide-10
SLIDE 10

On homework

6

you

show

That

92,234

la )

.

Nate

re Gall Okla)1a)

has

run ,)E{ a .az , as }.

On

the

  • ther

hand

, we know

re Aut IQK, ))

and

hence
  • ld,) EQ (a)
.

Hence

  • k , )
  • a .
. Nate a Q
  • basis for
Okla ) is

{ 1,4

, ai}
slide-11
SLIDE 11

and

by

  • ur

last

result we knew

That

a

given

at Gal (⑥ (a)1a)

is determined by its

action

  • n
a . .

Huie

any

rt Gal Cala ) 1a)

must be idek .,

de)

= r ( q , . It qz
  • a, t q,-42)
= Hq.) r IH tha) r Kil tr ( q , ) ok . )' = q ,
  • I tqz.at E3 'd ,
' a e .
slide-12
SLIDE 12

So

:

Gal ( Okla )1a )

  • Lida.ca ,}
.

Nele : This

doesn't contradict

  • ur

result that

I Gal (El F) I

= ( E : F) because

this

result

applies

  • nly
to

splitting

fields

.

We

actually find

that

④ (a) you

is

not

a

splitting field

.