E=Q(sqrt{1-\sqrt{2}}) is Galois over Q.

apopiw83

apopiw83

Answered question

2022-11-11

Q ( 1 2 ) is Galois over Q
I am trying to show that E = Q ( 1 2 ) is galois over Q. The extension has the minimal polynomial
( x 1 2 ) ( x + 1 2 ) ( x 1 + 2 ) ( x + 1 + 2 )
but I can't manage to show using elementary arithmetic that 1 + 2 is in E (i.e. multiplying, adding etc.) so that its the splitting field.
The trick would be to use that 1 + 2 = i 1 2 but we only know that i 2 1 is in the field, we dont know if we have i.
Can someone give a hint on how to proceed, maybe there is a theorem I could use that I'm not thinking of.

Answer & Explanation

ontzeidena8a

ontzeidena8a

Beginner2022-11-12Added 17 answers

Step 1
The given field E = Q ( 1 2 ) is not a galois extension. If it were it would contain 1 + 2 as stated in the question and then Q ( 1 2 ) would contain Q ( 1 + 2 ) which is an extension of degree 4 as is Q ( 1 2 ) (the minimal polynomial is X 4 2 X 2 1 for both numbers), thus this inclusion would be an equality which is not the case because Q ( 1 + 2 ) R and 1 2 C but 1 2 R .

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