First of all, I know that the splitting field for this particular polynomial will be the field of the rationals adjoined the cube-root of 2 and a primitive third root of unity: Q(root{3}{2}, gamma), where gamma^3=1 and gamma^2=1-gamma. I also know that since the degree of this extension is 6, the order of my Galois group will also be 6.

Andreasihf

Andreasihf

Answered question

2022-09-12

First of all, I know that the splitting field for this particular polynomial will be the field of the rationals adjoined the cube-root of 2 and a primitive third root of unity: Q ( 2 3 , γ ), where γ 3 = 1 and γ 2 = 1 γ. I also know that since the degree of this extension is 6, the order of my Galois group will also be 6.
I know that my Galois group will consist of the following automorphisms:
σ i = identity
σ 1 : 2 3 2 3 γ and γ is fixed.
σ 2 : 1 3 2 3 γ 2 and γ is fixed.
σ 3 : γ γ 2 and 2 3 is fixed.
σ 4 : 2 3 2 3 γ and γ γ 2 .
σ 5 : 1 3 2 3 γ 2 and γ γ 2 .
If I didn't miss something or make some error.
My specific question is, when constructing my automorphisms, why exactly is it that I can't send 2 3 to something like 4 3 or γ or some other random thing? Why are these automorphisms so strictly defined?

Answer & Explanation

Isaiah Haynes

Isaiah Haynes

Beginner2022-09-13Added 16 answers

Explanation:
By the Fundamental Theorem of Galois Theorem, if the Galois group were abelian, then all subfields would be normal extensions. But this cannot be, as Q ( 2 3 ) / Q is not a normal extension.
sooxicyiy

sooxicyiy

Beginner2022-09-14Added 3 answers

Step 1
let j = 1 2 ( 1 + i 3 ). Clearly 2 3 , 2 3 j and 2 3 j 2 are roots of f ( x ) = x 3 2. We claim E = Q ( 2 3 , j ) is splitting field f over Q because
E = Q ( 2 3 , 2 3 j , j 2 2 3 ) Q ( 2 3 , j )
and
j = 1 2 2 3 ( 2 3 ) 2 j Q ( 2 3 , j ) E = Q ( 2 3 , 2 3 j , j 2 2 3 )
Step 2
The polynomial f is not irreducible over Q as a result minimal polynomial of 2 3 is f and [ Q ( 2 3 ) : Q ] = 3 on the other hand
[ Q ( 2 3 , j ) : Q ] = [ Q ( 2 3 , j ) : Q ( 2 3 ) ] [ Q ( 2 3 ) : Q ] = 2 × 3 = 6
in the other words
[ E : Q ] = 6
so the Galois group is order 6. let
A: The metamorphism which fixes j and maps j j 2
B: The metamorphism which fixes 2 3 and maps 2 3 2 3 j
The Galois group is thus:
A , B | A 2 , B 3 , A B = B 1 A D 6

Do you have a similar question?

Recalculate according to your conditions!

New Questions in College Statistics

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?