Galois Group: Find the Galois group for x^3-2

[Steven G]

EDIT: I figured it out! I had the wrong roots for x^3-2. They are 2^1/3, w2^1/3 and w²2^1/3, where w= e^2pi*i/3


I think that I am doing something wrong and am hoping that someone can point it out to me.

Find the Galois group for x^3-2

Step 1: the roots are 2^1/3, w and w², where w= e^2pi*i/3

Step 2: The splitting feels is Q(2^1/3, w)

Step 3a: The min polynomial of 2^1/3 is x^3-2

Step 3b: The roots of that polynomial are 2^1/3, w and w²

Step 4a: The min polynomial for w is x^2+x+1

Step 4b: The roots of that polynomial are w and w²

Step 5: 2^1/3 can only be mapped to 2^1/3, w and w²

Step 6: w can only be mapped to w and w²

This gives 6 automorphisms

A=ID: 2^1/3 goes to 2^1/3 and w goes to w

B: 2^1/3 goes to w and w goes to w

C: 2^1/3 goes to w² and w goes to w

D: 2^1/3 goes to 2^1/3 and w goes to w²

E: 2^1/3 goes to w and w goes to w²

F: 2^1/3 goes to w² and w goes to w²

Here is what bothers me:

Note that B² sends 2^1/3 to w and w to w. But that is B!! This happens with the other mappings as well.

What am I doing wrong. I numbered my steps so that we can communicate better.

 

[blamocur]

How can B be an automorphism if it maps two different elements, [imath]2^{1/3}[/imath] and [imath]w[/imath], to the same element [imath]w[/imath] ?

 

[blamocur]

Another point: I believe that any automorphism from the group must map [imath]2^{1/3}[/imath] to itself. Can you see why?

 

[Steven G]

Here is the updated version.

Find the Galois group for x^3-2

Step 1: the roots are 2^1/3, w2^1/3 and w²2^1/3, where w= e^2pi*i/3

Step 2: The splitting feels is Q(2^1/3, w)

Step 3a: The min polynomial of 2^1/3 is x^3-2

Step 3b: The roots of the polynomial are 2^1/3, w2^1/3 and w²2^1/3

Step 4a: The min polynomial for w is x^2+x+1

Step 4b: The roots of that polynomial are w and w²

Step 5: 2^1/3 can only be mapped to 2^1/3, w and w²

Step 6: w can only be mapped to w and w²

This gives 6 automorphisms

A=ID: 2^1/3 goes to 2^1/3 and w goes to w

B: 2^1/3 goes to 2^1/3w and w goes to w

C: 2^1/3 goes to 2^1/3w² and w goes to w

D: 2^1/3 goes to 2^1/3 and w goes to w²

E: 2^1/3 goes to 2^1/3w and w goes to w²

F: 2^1/3 goes to 2^1/3w² and w goes to w²

 

[Steven G]

How can B be an automorphism if it maps two different elements, [imath]2^{1/3}[/imath] and [imath]w[/imath], to the same element [imath]w[/imath] ?

You got me good with that one! That's at least a week in the corner. Yuck!

 

[Steven G]

Another point: I believe that any automorphism from the group must map [imath]2^{1/3}[/imath] to itself. Can you see why?

Hmm, I don't see it yet. Can you convince me that it is true?

 

[blamocur]

Hmm, I don't see it yet. Can you convince me that it is true?

Not anymore -- I somehow assumed that the roots are [imath]2^{1/3}, w, w^2[/imath]

 

[blamocur]

Your post #4 looks good to me. You do realize that you got all 6 permutations of the 3 roots of [imath]x^3-2[/imath], don't you ?

 

[blamocur]

Your post #4 looks good to me.

Actually, almost: I don't agree with Step 5, which looks like a typo: e.g., [imath]2^{1/3}[/imath] can never be mapped to [imath]w[/imath].

 

[khansaheb]

You got me good with that one! That's at least a week in the corner. Yuck!

Not anymore -- I somehow assumed that the roots are 21/3,w,w22^{1/3}, w, w^221/3,w,w2

Both of you need to be sent to the "corner" - and "no talking".

 

[Steven G]

Actually, almost: I don't agree with Step 5, which looks like a typo: e.g., [imath]2^{1/3}[/imath] can never be mapped to [imath]w[/imath].

Yes, I missed that one when I updated my original post.