What is the Galois group of 27

Villaretq0

Villaretq0

Answered question

2022-06-24

What is the Galois group of 27 x 8 72 x 4 16 over the rationals?

Answer & Explanation

Eli Shaffer

Eli Shaffer

Beginner2022-06-25Added 16 answers

The roots are easily found to be
± 4 ( 3 + 2 3 ) 9 4 = ± a  (say) , ± i 4 ( 3 + 2 3 ) 9 4 , ± ( 1 ± i ) 2 3 3 9 4 = ± ( 1 ± i ) b  (say)
with a,b being positive real numbers.
The given polynomial is irreducible over Q and hence [ Q ( a ) : Q ] = 8. The splitting field L of the polynomial contains a , a i and hence i L. But i Q ( a ) and hence [ Q ( a , i ) : Q ] = 16
It can be proved with some effort that L = Q ( a , i ) is the splitting field of the polynomial and hence the Galois group is of order 16.
We can note that
a b = 12 4 3 = c 3  (say)
and next we show that c L. Since c is real we have in fact c Q ( a ). We can observe that
c 2 = 2 3 = 9 a 4 12 4 Q ( a )
and
9 a 4 = 4 c 2 + 12 = 4 c 2 + c 4 = c 2 ( c 2 + 4 ) = c 2 ( 4 + 2 3 ) = c 2 ( 1 + 3 ) 2
This implies
3 a 2 = c ( 1 + 3 )
ie
c = 6 a 2 2 + c 2 Q ( a )
It now follows that b = c / ( 3 a ) Q ( a ) L and L is the desired splitting field.
The Galois group is D 16 (dihedral group of order 16) as explained in another answer.

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