How to prove that $\sqrt{2-\sqrt{2}} \in \mathbb{Q}(\sqrt{2+\sqrt{2}})$ I am trying to prove a statement about the decomposition field of a polynomial that has both $\sqrt{2-\sqrt{2}}$ and $\sqrt{2+\sqrt{2}}$ as roots. I cannot find a way to prove that $\sqrt{2-\sqrt{2}} \in \mathbb{Q}(\sqrt{2+\sqrt{2}})$. I have tried writing it in the basis $1,\sqrt{2+\sqrt{2}}(\sqrt{2+\sqrt{2}})^2,(\sqrt{2+\sqrt{2}})^3$ but nothing works and without this I cannot prove that the decomposition field of $t^4-4t^2+2$ is $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ over $\mathbb{Q}$
 A: Hint: what is $\sqrt{2 - \sqrt 2} \cdot \sqrt{2 + \sqrt 2}$ ?

 It's $\sqrt 2$ ! Therefore $\sqrt{2 - \sqrt 2} = \frac{\sqrt 2}{\sqrt{2 + \sqrt 2}} = \frac{\left(\sqrt{2 + \sqrt 2}\right)^2 - 2}{\sqrt{2 + \sqrt 2}}$

A: Using the hint given by Bart Michels, we have$$
\sqrt{2-\sqrt{2}}\sqrt{2+\sqrt{2}}=\sqrt{2}\\
\sqrt{2+\sqrt{2}}(\sqrt{2-\sqrt{2}})-\sqrt{2}=0
$$
And so the minimial polynomial of $\sqrt{2-\sqrt{2}}$ is linear in $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ and so it must be an element of the field.
A: Alternatively (without using conjugates),
Observe that,

*

*If $\sqrt{2-\sqrt{2}}=x$, then

$$(2-x^2)^2=2$$

*

*If $\sqrt{2+\sqrt{2}}=x$, then

$$(x^2-2)^2=2$$
Finally, we can observe that both polynomial equations are equal to each other:
$$(2-x^2)^2=2 \iff (x^2-2)^2=2$$

This implies, if $\sqrt{2-\sqrt{2}}$ is the root of your minimal polynomial over $\mathbb Q$, then $\sqrt{2+\sqrt{2}}$ also must be root of the polynomial.
So, your minimal polynomial equals to: ($x_{1,2}=\sqrt{2±\sqrt{2}}$)
$$(x^2-2)^2=2$$
which gives
$$x^4-4x^2+2=0.$$
