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I am trying to show that $E = \mathbb{Q}\left(\sqrt{1-\sqrt{2}}\right)$ is galois over $\mathbb{Q}$. The extension has the minimal polynomial


but I can't manage to show using elementary arithmetic that $\sqrt{1+\sqrt{2}}$ is in $E$ (i.e. multiplying, adding etc.) so that its the splitting field.

The trick would be to use that $\sqrt{1+\sqrt{2}} = \frac{i}{\sqrt{1-\sqrt{2}}}$ but we only know that $i\sqrt{\sqrt{2}-1}$ is in the field, we dont know if we have $i$.

Can someone give a hint on how to proceed, maybe there is a theorem I could use that I'm not thinking of.

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Maybe I am mistaken, but if $E$ was galois this would imply that $E = \mathbb{Q}\left(\sqrt{1+\sqrt{2}}\right)$, right? But $\mathbb{Q}\left(\sqrt{1+\sqrt{2}}\right) \subset \mathbb{R}$ which is not the case for $E$. – Matthias Klupsch Nov 28 '11 at 5:09
Perhaps the problem is flawed? It was made up by my teacher so this is a possibility.. – AnonymousCoward Nov 28 '11 at 5:27
I think there are issues with the problem, though I might be wrong. The minimal polynomial of $\sqrt{1 - \sqrt{2}}$ is $f(x) := x^{4} - 2x^{2} - 1$ which has 2 real roots and 2 imaginary roots. If $E/\mathbb{Q}$ was a Galois extension, it would be the splitting field of $f$, but $E \subset \mathbb{R}$ and hence $E$ can't be the splitting field of $f$, a contradiction. – Shayla Nov 28 '11 at 6:00
My uneducated guess is that your teacher meant to write down something like $\sqrt{2 - \sqrt{2}}$. – Dylan Moreland Nov 28 '11 at 6:11
@GottfriedLeibniz: Adjoining $\sqrt{1+\sqrt{2}}$ gives you an extension of degree 4 as does adjoining with $\sqrt{1 - \sqrt{2}}$. If $\mathbb{Q}(\sqrt{1-\sqrt{2}})$ were to contain all four roots you would have $\mathbb{Q}(\sqrt{1+\sqrt{2}}) \subset \mathbb{Q}(\sqrt{1-\sqrt{2}})$ and by equality of degrees you have equality of the extensions. – Matthias Klupsch Nov 28 '11 at 7:04
up vote 12 down vote accepted

The given field $E = \mathbb{Q}\left(\sqrt{1-\sqrt{2}}\right)$ is not a galois extension. If it were it would contain $\sqrt{1 + \sqrt{2}}$ as stated in the question and then $\mathbb{Q}\left(\sqrt{1-\sqrt{2}}\right)$ would contain $\mathbb{Q}\left(\sqrt{1+\sqrt{2}}\right)$ which is an extension of degree 4 as is $\mathbb{Q}\left(\sqrt{1-\sqrt{2}}\right)$ (the minimal polynomial is $X^4-2X^2-1$ for both numbers), thus this inclusion would be an equality which is not the case because $\mathbb{Q}\left(\sqrt{1+\sqrt{2}}\right) \subset \mathbb{R}$ and $\sqrt{1-\sqrt{2}} \in \mathbb{C}$ but $\sqrt{1-\sqrt{2}} \notin \mathbb{R} $.

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