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Given $\alpha=\sqrt{1+\sqrt{2}}$, a min. poly for this element over $\Bbb Q$ is $x^4-2x^2-1$, as it's monic irreducible over $\Bbb Q$ and has $\alpha $ as a root.

The roots of this min. polynomial are $^+_-\sqrt{1^+_-\sqrt{2}}$.

Now consider the extensions

i) $\Bbb Q(\alpha)/\Bbb Q$

ii) $\Bbb Q(\alpha, i)/\Bbb Q$.

I know that i) is not a Galois extension because it's not normal , it's not normal because it only contains two of the roots of the min. poly of $\alpha$ namely $^+_-\sqrt{1+\sqrt{2}}$.

I want to decide if the second on is Galois and to do this I tried to tackle the problem by denesting the radical $^+_-\sqrt{1-\sqrt{2}}$ to see it was a complex number, which it is. But this is a very long and tedious method and I'm worried that in an exam I wouldn't have the time to compute it.

Is there a much faster way of deciding if ii) is a Galois extension ?

Edit: I realised that I made an error in calculation and that de-nesting is in fact impossible here

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Notice that $(1-\sqrt{2})(1+\sqrt{2})=-1$ and so a square root of $1-\sqrt{2}$ is $\pm i$ divided by a square root of $1+\sqrt{2}$.

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