Mathematics Asked by user462999 on January 18, 2021
$omega$ is the primitive $4^{th}$ root of unity, then find $Galleft(mathbb{Qleft(sqrt{2},omegaright)/mathbb{Q}}right).$
First i calculate order of $Galleft(mathbb{Qleft(sqrt{2},omegaright)/mathbb{Q}}right)$ which is $4$ by use of minimal polynomial.
$left[mathbb{Q}left(sqrt{2},omegaright):mathbb{Q}right]=left[mathbb{Q}left(sqrt{2},omegaright):mathbb{Q}left(omegaright)right]left[mathbb{Q}left(omegaright):mathbb{Q}right]$
$left[mathbb{Q}left(sqrt{2},omegaright):mathbb{Q}left(omegaright)right]=2$
$left[mathbb{Q}left(omegaright):mathbb{Q}right]=2$ because $x^2+1$ is minimal polynomial.
Therefore $left[mathbb{Q}left(sqrt{2},omegaright):mathbb{Q}right]=2times 2=4$.
(I am not sure that i calculate right)
I don’t know how elements of $Galleft(mathbb{Q}left(sqrt{2},omegaright)/mathbb{Q}right)$ look like.
I think it is of the type
$varepsilon=omega rightarrow omega,~sqrt{2}rightarrow sqrt{2}$
$alpha=omegarightarrow omega^2,~sqrt{2}rightarrow sqrt{2}$
$beta=omegarightarrow omega^3,~sqrt{2}rightarrow sqrt{2}$
$gamma=omega rightarrow omega,~sqrt{2}rightarrow omegasqrt{2}$.
Four elements complete here and i think i did wrong.
Any hint how to find element.
Thank you.
$mathbb Q(sqrt 2, omega)$ is the compositum of two Galois fields extensions, $mathbb Q(sqrt 2)$ and $Q(omega)$ and hence also Galois over $mathbb Q$.
Note that $mathbb Q(sqrt 2) cap mathbb Q(omega) =mathbb Q$.
This induces a group homomorphism.
$$operatorname{Gal}(mathbb Q(sqrt 2, omega)/ mathbb Q)rightarrow operatorname{Gal}(mathbb Q(sqrt 2)/mathbb Q) times operatorname{Gal}(mathbb Q(omega )/mathbb Q)cong mathbb Z_2 times mathbb Z_2 $$ $$sigmamapsto (sigma|_{mathbb Q(sqrt 2)}, sigma|_{mathbb Q(omega)} ) $$
This is injective and hence by cardinality arguments, an isomorphism. So we get $operatorname{Gal}(mathbb Q(sqrt 2, omega)/ mathbb Q) cong mathbb Z_2 times mathbb Z_2 $
Answered by Baidehi on January 18, 2021
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