How can I prove that $\sqrt{2}$ is not inside Q($\sqrt{5 + \sqrt{5}}$)?

Question

I want to prove it, but although it is quite intuitive I don't know how to prove it mathematically. I have tried using the tower rule but I haven't got anything crear. Thanks in advance.

Answer 1

Note that $\sqrt{5+\sqrt{5}}$ is a degree 4 algebraic number, and so your extension has degree 4. But $\sqrt{2} + \sqrt{5+\sqrt{5}}$ can be computed to have minimal polynomial of degree 8, and so it cannot live in this extension.

Answer 2

To expand on Michael's answer: $$\sqrt 2 + \sqrt{5+\sqrt 5} = \sqrt{2+5+\sqrt 5 + \sqrt{10+2\sqrt 5}} =\sqrt{7+\sqrt 5 + \sqrt{10+2\sqrt 5}} $$ Which is pretty clearly degree 8(note that $\sqrt{10+\sqrt {20}}$ is undenestable because $10^2-20=80$ isn't a perfect square. see https://en.wikipedia.org/wiki/Nested_radical)

Answer 3

Another approach to the answer: first show that the Galois group of $\mathbb{Q}(\sqrt{5+\sqrt{5}})$ is isomorphic to $\mathbb{Z}/4\mathbb{Z}$. In particular $\mathbb{Q}(\sqrt{5})$ is the unique degree-2 subfield, by the Galois correspondence, since $\mathbb{Z}/4\mathbb{Z}$ only has one subgroup of order $2$.

So that reduces us to showing that $\mathbb{Q}(\sqrt{2}) \ne \mathbb{Q}(\sqrt{5})$. And this is true since $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{5})$ each have a unique (up to rational multiple) element of trace $0$, and $\sqrt{2}, \sqrt{5}$ can't be rational multiples of each other since $2/5$ isn't a square.

Still, it takes a computation to show that the Galois group is $\mathbb{Z}/4\mathbb{Z}$ (and that $\mathbb{Q}(\sqrt{5+\sqrt{5}})$ is a splitting field). One proof that the Galois group is cyclic is as follows: there must be an automorphism $\phi$ taking $\sqrt{5+\sqrt{5}}$ to its Galois conjugate $\sqrt{5-\sqrt{5}}$ (both are roots of $x^4 - 10x^2 + 20$, which is irreducible by Eisenstein with $p=5$). Squaring both sides of the equation $\phi(\sqrt{5+\sqrt{5}}) = \sqrt{5-\sqrt{5}}$ shows $\phi(\sqrt{5}) = -\sqrt{5}$. Next,

$$\sqrt{5+\sqrt{5}} \sqrt{5-\sqrt{5}} = \sqrt{20} = 2\sqrt{5},$$

so we see that $$\phi(\sqrt{5-\sqrt{5}}) = \phi(\frac{2\sqrt{5}}{\sqrt{5+\sqrt{5}}})=-2\sqrt{\frac{5}{5-\sqrt{5}}}=-\sqrt{5+\sqrt{5}}.$$

From this it follows that $\phi^2 = -\mathrm{id}$ (on roots of $f$), so $\phi$ has order $4$.

Answer 4

$\sqrt{5+\sqrt5}$ is a root of $(x^2-5)^2-5$ which is Eisenstein at $5$ whence also at the maximal ideal $5\Bbb{Z}[\sqrt2]$ of $\Bbb{Z}[\sqrt2]$.

This proves that $(x^2-5)^2-5\in \Bbb{Q}(\sqrt2)[x]$ is irreducible so that $$[\Bbb{Q}(\sqrt{5+\sqrt5},\sqrt2):\Bbb{Q}]= [\Bbb{Q}(\sqrt{5+\sqrt5},\sqrt2):\Bbb{Q}(\sqrt2)][\Bbb{Q}(\sqrt2):\Bbb{Q}]=8 $$