Hints for finding a Galois closure

[Mystic998]

Homework Statement

Find the Galois closure of the field $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$, where $$\alpha = \sqrt{1 + \sqrt{2}}$$.

Homework Equations

Um...the Galois closure of E over F, where E is a finite separable extension is a Galois extension of F containing E which is minimal.

The Attempt at a Solution

I've found that the minimal polynomial of $\alpha$ is $x^{4} - 2x^{2} - 1$ (at least I think. I'm having a heck of a time showing it's irreducible over the rationals but I'm willing to just leave that as an exercise for the reader, so to speak), and it's separable, so I know a Galois closure exists. But I can't figure out a good way to generate it.

There are no examples of actually doing this explicitly in the book, though the proof of existence implies that you should take the intersection of all Galois extensions of F containing E. Or would finding explicitly the splitting field of the minimal polynomial be enough in this case? If so, could you guys help me out with that? This is kind of tricky.

 

[morphism]

Yes, finding the splitting field will be enough. You're just going to throw in the conjugates of $\alpha$, and that will make your extension normal. Can you guess what they are? One of them is obviously $-\alpha$. What about the other 2? Considering $\alpha^2$ will be helpful.

 

[Mystic998]

I believe the splitting field is just the original extension adjoin i. So that's handy. But it seems like that really shouldn't be the Galois closure. Why would any Galois extension of the rationals that contains $\mathbb{Q}(\alpha)$ have to contain i? It seems like it would only have to be a subset of the reals.

 

[morphism]

There's no reason it should be real at all. Think about it: any normal extension of $\mathbb{Q}$ that contains $\alpha$ will contain the conjugates of $\alpha$, so the minimal such extension is the one you get by adjoining these elements to $\mathbb{Q}(\alpha)$. And there's nothing saying $\alpha$ can't have non-real conjugates, just because it itself is real.

 

[Mystic998]

I think I understand... The splitting field is the minimal field that contains all the roots of the minimal polynomial, and anything that's Galois over the rationals contains all the conjugates of $\alpha$ (i.e., the roots of the minimal polynomial). So it contains the splitting field. Thanks!

Edit: Er...Galois over the rationals and contains alpha. Heh.