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Galois Theory Exercises and Solutions, Exercises of Abstract Algebra

A series of exercises and solutions related to galois theory, a fundamental concept in abstract algebra. It covers topics such as galois extensions, closed subgroups, and the relationship between subgroups and subfields. Detailed solutions to each exercise, offering a valuable resource for students studying galois theory.

Typology: Exercises

2022/2023

Uploaded on 11/17/2024

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bg1
Assignment
in
infinite
Galis
extas
&
1
.
Let
HCG
=
4(4/K)
and
2
->
G
.
5600
:
2
***
-
(2)
.
ast
Solution
.
Let
-L
.
Then
2
20(xx)
=
x
-
for
all
+H
.
Hence
xs
=
-
/Tc(x)
.
We
show
>
(x)
+
L
.
Let
H
.
Then
It
EH-
58
en
(
(x))
=
xx)
E)
We
o
)
-
x
=H
Thus
-
2x)
->
.
Conversalylet
y
->
24.
.
Then
-
+
fteH
,
-
+
a
(0(y))
=
w+
(y)
=
c
(y)
.
SorsD=L
2
.
Let
2
=
=
G
(2/k)
and
K
<
EsL
bes
a
tower
of
fields
.
Show
&(20(E))
=
8
9
12/5)
=
solution
.
Essous
to
show
that
-14/020914)
.
L
Let
H
=
G(4/E)
.
Then
(UHC
-)
L
(E))
=
e)
=
1914/01E)
-
3
.
Let
H
,
H'
ba
closed
subgroups
of
3
and
[H
.
H'3
=
subgroup
generated
by
Hand
H'
show
that
(1)
Yes!
*
+
(2)
-
*
21 (
,
]
L
S
pf3
pf4
pf5

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Assignment in^ infinite

Galis extas &

(^1).^ Let^ HCG=^ 4(4/K)^ and^2 -> G (^).^5600 :^2 ***^ - (^) (2).

ast

Solution (^). Let -L (^). Then 2 20(xx)^ = x

  • for (^) all +H.^ Hence^ xs = (^) - /Tc(x) (^).^ We^ show > (^) (x) + L. Let H (^). Then It^ EH- 58 en ( (x))^ =^ xx)

E) We o^ )- x =H

Thus - (^) 2x) -> (^). Conversalylet y^ -> 24.. Then

fteH

,

    • a (0(y)) = w+^ (y) = c (^) (y). (^) SorsD=L

2.^ Let^2 =^ =^ G^ (2/k)^ and^ K^ <^ EsL^ bes^ a tower^ of

fields

. (^) Show (^) &(20(E))= 8 9 12/5)= solution. Essous to^ show^ that -14/020914) . L Let (^) H = (^) G(4/E). Then (UHC -)L^ (E))= e) = (^) 1914/01E)

3. Let^ H

,

H' ba^ closed

subgroups of^3 and [H (^).^ H'3^ = subgroup (^) generated by

Hand H'

show that (1) Yes!

  • (^) (2)
    • (^) 21 (, L (^) S ]

H2s 1 H solution.^ It^ is^ alcar^ that^ L^ -1 2^. Let x^ =^ +^ and^ y

  • (^) and (^) - -> Hist! Ther (^) r(xy) = 0(x)0(y) =

xy-

So (^) we have

the

equality . (2) By Galois Eresespradance -(4) 13) = G(4- +, 3)=^ +^13. Y and^ It' are Galo's^ extensions^ - Determine (^5) = 4)^ 4 2). Noter /Iis

a Galois extension . So 5 is a closed

subgroup (^) of -12/1)^. I 11 I I

  • (,^13 =

,^3

! (^) Ne (^) need to sbow rest's Et (^) :^ I xen (^) them ~

  • (^) [H , A']^.^ Hence^ <He ([

,'S

  1. (^) Get E , F (^) be interendiate subfacids of the Galis (^) extension (^) 2/K (^). Let (^) H , = 9(2/E) and Hg =^ 3(4/F)^

. Shar that

(1) 3(YEF)^ =^ G(y)E)^ +(v)=) (3372/15) =^ E, H-]

  1. (^) Show that^ an (^) subjers of

are-

(^3) 72F) whose^ I FSK is^ a Finsite extension in L. =^ Solution G .Let t^ bepai^ Theto his to (^) , thus a is digit wi o cosets is "Sida (^) Fixite une

of left^

costs

of

H . Stance [4 :^ H] Say.

Thro /

H = 4-

I finitely maxy left

cosets is

closed So^ - (4/() = E =^ H^ and

(^24) /I is a te (^) extension since A has firs finite

index in 4.

7) show^ that a closed^ subgeng H

of

" is^ moral

) K is^ a^ Galves^ extension^. solution. Let (^) H be closed and mesial. Then ~(i)=^ &Hst fro

all ~74. Thus ~(2)= L

are H &^ G.^ Hence^ 24Km Galas :

Conversely let / be^ Golois (^). Then och")=2"

fo all or (^) G (^).^ Is -(2) =^2 =^ asso↑ Stense (^) &(b) () =^ F=^ H^ = G(Y-+ )) = xxE! Thus H^ & (^) G (^).

  1. show^ that^ an open subject^ is normal in a iff will^ is^ a finite calous extension^. solution (^). Let (^) #4G be apar . By (6) (^) 4K is a

finite extension^

. Since HOG , /K^ is

alocs .

④ Let^ E be as intermediate subfield

I

bis .

show that the Kell

tapology on (^) G(2/E) is induced^ fro

the Errall

topology on (^3). Solution (^). Let H= (^) G1YE) (^) · Open sets in the

subspace topology^

elt one^ ton^ refuse

WC (^) - is (^) an oper

sat.^ We^ show^ that^ WiH

is (^) opens in the^ Kell topology m H^. Let 1 -> WhA- As^ U is (^) open 15 ⑦ 7 ↓ ata (^) e

  • (^) -12/5) where (^) FIK is (^) a

finite

& -^ )s^ exte

H 3/45) =

SEEG

/- (^) fixes and^ += 2 on = [ = e4(4/E)(z = (^) 2m (^) + 3 => [ +^ =^ 3(4E)([= (^) 0m = (^) F 3

Stence 0 & (4/EF) G + &H . Note

k

that is a fixita Galins^ extension. I

EFE (^) can in Krall

topology

Hexat Ho^ W^ an (^) E f G(2/EC.^ /E

Let va^ G^ E^ I

/4) be^ pare^ in^ the^ -

we say & (^) ssume K Krall^ Palis topology e (^) Zer v (^) F %. 2e^ I^ -^ (r/E) <^ V fer · Finite Gabris extr^ FE

⑦ Since (^) &CK/IF(F Lim j (^) L-- G(k/(Fpe) r

  • (^) lim a /ex

2

the

isomrphism is (^) established (^). Since K

7 IF (^) , (K/Ep) = 3(k)k(x)) =T: