Oss1: Your extension is equivalent to $\mathbb{Q}(i\sqrt{2})/\mathbb{Q}$
Oss2: What you can say about $\mathbb{Q}(i\sqrt{2})/\mathbb{Q}$? Is normal? You can find a polynomial such that $i\sqrt{2}$ is one of its roots? Is $\mathbb{Q}(i\sqrt{2})$ its splitting field?
Finally I recall to you that a finite Galois extension is normal.
And I think this ask you answer...
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Bumbble Comm
On
It is the splitting field of $x^2-2x+9$.
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Bumbble Comm
On
The field is the same as $Q(\sqrt{-2})$ which is the splitting field of $x^2+2$.
Oss1: Your extension is equivalent to $\mathbb{Q}(i\sqrt{2})/\mathbb{Q}$
Oss2: What you can say about $\mathbb{Q}(i\sqrt{2})/\mathbb{Q}$? Is normal? You can find a polynomial such that $i\sqrt{2}$ is one of its roots? Is $\mathbb{Q}(i\sqrt{2})$ its splitting field?
Finally I recall to you that a finite Galois extension is normal.
And I think this ask you answer...