# Help understanding how to show a field extension is a Galois extension .

I have the field extension $$\Bbb Q(\sqrt{2},i)$$ 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 $$\Bbb Q(\sqrt{2},i)=\Bbb Q(\sqrt{2}+i)$$ and so set $$x=\sqrt{2}+i$$ and then say $$x-\sqrt{2}=i$$ square it to get rid of the i then do a similar technique to get rid of the $$\sqrt{2}$$. 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 $$\sqrt2$$ must have simple roots as if it didn't it would imply $$\sqrt{2}$$ would be a root which it cant be as its not an adjoined element in the extension then we could factor out $$x-\sqrt2$$ 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

• Your use of $i$ as an index is unfortunate. – Servaes Mar 10 at 13:38
• 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. – Noel Lundström Mar 10 at 17:34

## 1 Answer

Since the characteristic is zero, the extension is separable. It is the splitting field of $$X^8-2$$ as you have remarked.