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I'm in the middle of a proof where I'd like to show that $\sqrt{2 - \sqrt{2}} \notin \mathbb{Q}(\sqrt{2 + \sqrt{2}})$

The only way I can think of involves finding an explicit set representation for $\mathbb{Q}(\sqrt{2 + \sqrt{2}})$.

At first I tried showing $\mathbb{Q}(\sqrt{2 + \sqrt{2}}) = \{a + b\sqrt{2 + \sqrt{2}}: a,b\in\mathbb{Q}\}$

and then realised this is probably false, as it doesn't look like it contains $\sqrt{2}$.

I figured I could try $\mathbb{Q}(\sqrt{2 + \sqrt{2}}) = \{a + b\sqrt{2 + \sqrt{2}} + c\sqrt{2}: a,b,c\in\mathbb{Q}\}$

but this method seems really long-winded. I'm pretty sure there'll be plenty of shorter methods, but I don't know any method to show this.

Any pointers appreciated!

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  • $\begingroup$ Wouldn't it be enough to find minimal polynomials for each number and then show that they have no common factors? $\endgroup$ – MJD Nov 15 '12 at 16:21
  • $\begingroup$ @MJD Don't they have the same minimal polynomial, $(x^2-2)^2-2$? $\endgroup$ – Thomas Andrews Nov 15 '12 at 16:24
  • $\begingroup$ @MJD: What would it imply if that were the case? $\endgroup$ – maliky0_o Nov 15 '12 at 16:26
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    $\begingroup$ I'm not even sure that this is true - isn't $\frac{2}{2+\sqrt{2}}=2-\sqrt{2}$, so $\frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}} = \sqrt{2-\sqrt{2}}$? Since $\sqrt{2}\in\mathbb Q[\sqrt{2+\sqrt{2}}]$, your statement seems false. $\endgroup$ – Thomas Andrews Nov 15 '12 at 16:26
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    $\begingroup$ From my personal point of view, I would be happy to see this one not deleted, as I have learned something from it - but of course you are entitled to do as you prefer. $\endgroup$ – Old John Nov 15 '12 at 16:54
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Actually, this statement is not true.

Using $(2+\sqrt{2})(2-\sqrt{2})=2$, taking square roots and dividing, we see that $$\sqrt{2-\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}}$$

So, since $\sqrt{2}\in\mathbb Q(\sqrt{2+\sqrt{2}})$, $\sqrt{2-\sqrt{2}}\in\mathbb Q(\sqrt{2+\sqrt{2}})$.

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  • $\begingroup$ Ah, thanks. That makes it easier to see how you derived $\sqrt{2 - \sqrt{2}} \in \mathbb{Q}(\sqrt{2 + \sqrt{2}})$. But this trick wouldn't work for, say, telling if $\sqrt{3 - \sqrt{7}} \in \mathbb{Q}(\sqrt{3 + \sqrt{7}})$, since $(3 + \sqrt{7})(3 - \sqrt{7}) = 2$ doesn't give you anything. $\endgroup$ – maliky0_o Nov 16 '12 at 12:13
  • $\begingroup$ True, but (1) that wasn't the problem given, and (2) this indicates that you can't just make a blanket statement blind about expression of this sort. You have to use something specific about the numbers. $\endgroup$ – Thomas Andrews Nov 16 '12 at 13:47
  • $\begingroup$ For example, it becomes clear that, if $a\neq b$, then $\sqrt{\sqrt{a}-\sqrt{b}}\in \mathbb Q\left(\sqrt{\sqrt{a}+\sqrt{b}}\right)$ if and only if $\sqrt{a-b}\in \mathbb Q\left(\sqrt{\sqrt{a}+\sqrt{b}}\right)$. $\endgroup$ – Thomas Andrews Nov 16 '12 at 14:13
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$\begin{eqnarray}{\rm\bf Hint}\ \ w-1 && =\ \ (w+1)\,(w-1)^2\ \ \ {\rm by}\ \ \ w^2-1 = 1\ \ \ {\rm for}\ \ \ w := \sqrt{2} \\ \Rightarrow\ \ \ \ 2-w && =\ \ (2+w)\, (w-1)^2\ \ \ {\rm by}\ \ \ w * {\rm prior,\ \ using}\,\ \ w^2 = 2 \\ \Rightarrow\ \sqrt{2-w} && =\, \sqrt{2+w}\,\ (w-1) \ \ \ \ \ {\rm by\ taking\ \sqrt{prior}} \\ \Rightarrow\ \sqrt{2-w} && \in\Bbb Q(\sqrt{2+w}) = {\rm R}\ \ \ \ \ \ {\rm by}\ \ \sqrt{2+w}^{\,2}\! =\, 2+w\in {\rm R}\:\Rightarrow\: w \in\rm R \end{eqnarray} $

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These expressions show up when computing the sine and cosine of $\pi/8 = 22.5^\circ$. Since the corresponding extensions are cyclotomic, hence normal (and even abelian), any extension of the automorphism sending $\sqrt{2}$ to $-\sqrt{2}$ must fix the extension.

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Let $\alpha = \sqrt{2 + \sqrt{2}}$ and $\beta = \sqrt{2 - \sqrt{2}}$. I want to find out if $\mathbb Q(\alpha)$ is equal to $\mathbb Q(\beta)$ or not. They are equal but I was wondering about proving it with Galois theory instead of direct computation.

Since $\alpha,-\alpha,\beta,-\beta$ are the roots of the polynomial $P(x) = x^4 - 4x^2 + 2$ we could try building the splitting field of $P$ (which has degree 4 since the polynomial is irreducible with degree 4) by a chain of extensions like this:

enter image description here

the bottom ones all have degree 2 and the full tower has degree 4, so I think that implies the greyed out ones at the very top have degree 1 so it's a trivial extension, proving $\mathbb Q(\alpha) = \mathbb Q(\beta)$.

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  • $\begingroup$ Consider the extension $S/\mathbb{Q}(\sqrt{2})$ instead. It is a degree $2$ extension by Galois theory/tower of fields. Now $\mathbb{Q}(\alpha)\neq \mathbb{Q}(\sqrt{2})$, similarly for $\mathbb{Q}(\beta)$ (since they are degree $4$ extensions of $\mathbb{Q}$ whereas $\mathbb{Q}(\sqrt{2})$ is of degree $2$. Hence both equal $S$. $\endgroup$ – fretty Nov 16 '12 at 8:02
  • $\begingroup$ @sperners-lemma: Is there a result that tells you that the degree of a splitting field of a polynomial p is the degree of p if p is irreducible? $\endgroup$ – maliky0_o Nov 16 '12 at 12:07
  • $\begingroup$ @maliky0_o, i guess there isn't there I don't know how to fix this... $\endgroup$ – sperners lemma Nov 16 '12 at 12:40
  • $\begingroup$ all we know is degree of splitting field of degree n polynomial divides n! $\endgroup$ – sperners lemma Nov 16 '12 at 13:06

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