# A quicker way to decide if this extension is Galois.

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

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 $$1+\sqrt{2}$$.