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I have the field extension $\Bbb Q(\sqrt[8]{2},i)$ over $\Bbb Q$. I want to show that this is a Galois extension. I know that I can do this by showing that It is an extension which is both normal and separable.

1) for separability I'm pretty sure it can be shown by saying $\Bbb Q(\sqrt[8]{2},i)=\Bbb Q(\sqrt[8]{2}+i)$ and so set $x=\sqrt[8]{2}+i$ and then say $x-\sqrt[8]{2}=i$ square it to get rid of the i then do a similar technique to get rid of the $\sqrt[8]{2}$. And then finally you'd be left with a large degree polynomial. The next step would then be to show that each irreducible factor has simple roots. I think then you could show by contradiction that the root $\sqrt[8]2$ must have simple roots as if it didn't it would imply $\sqrt[16]{2}$ would be a root which it cant be as its not an adjoined element in the extension then we could factor out $x-\sqrt[8]2$ and try and continue in this manner .

My first question: this seems like a really long and awkward way to show separability is there a better way to do it ?

2)for normality I think we need to consider the extensions separately . first $x^2+1$ with roots $\pm i$ is the minimal polynomial and splits over $\Bbb Q(i)$. $x^8 -2$ is the minimal polynomial for the roots $w^i\alpha, i=0,1,2,3,4,5,6,7.$ w is a prim $8^th $ root of 1 $w=\tfrac{1+i}{\sqrt2}$. $\alpha ^4=\sqrt2$ so $ w=\tfrac{1+i}{\alpha^4}\in \Bbb Q(\alpha, i )$ also $w^2=i$ so $\Bbb Q(\alpha, i )=\Bbb Q(\alpha,w)$

my second question is that correct ?

so the extension is normal

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  • $\begingroup$ Your use of $i$ as an index is unfortunate. $\endgroup$ – Servaes Mar 10 at 13:38
  • $\begingroup$ Since the field is of charachteristic zero it is seperable and an extension is normal if it is a splitting field of some polynomial. For example try showing the polynomial (x^8 -2)(x^2 +1) splitting field is the one in your question. $\endgroup$ – Noel Lundström Mar 10 at 17:34
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Since the characteristic is zero, the extension is separable. It is the splitting field of $X^8-2$ as you have remarked.

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