# Is $\sqrt{2 + \sqrt{2}} \in \mathbb{Q}(\sqrt{2})$?

I am stuck at the following seemingly simple problem: is Is $$\sqrt{2 + \sqrt{2}} \in \mathbb{Q}(\sqrt{2})$$?

(Context: I want to show that $$\mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt{2 + \sqrt{2}})$$ is an extension of degree 2).

So far, I assumed it is not, and argued with contradiction. Suppose $$\sqrt{2 + \sqrt{2}} = a + b\sqrt{2}$$ for some $$a,b \in \mathbb{Q}$$. Then squaring both sides we have that $$2 + \sqrt{2} = a^2 + 2ab\sqrt{2} + 2b^2,$$ and hence $$2 - a^2 - 2b^2 = (2ab-1)\sqrt{2}.$$ Now if $$2ab-1 \neq 0$$, we get $$\sqrt{2} \in \mathbb{Q}$$, a contradiction. So suppose we are in the case $$2ab = 1$$, I don't know how to proceed to derive a contradiction in this case. Is there a general strategy to do this? I notice that then $$a^2 = 2-2b^2,$$ so $$a = \sqrt{2(1-b^2)} = \sqrt{2} \sqrt{1-b^2} \in \mathbb{Q}$$. Not sure how to proceed from here.

• Maybe related Commented Apr 19, 2020 at 14:21
• See Commented Apr 19, 2020 at 14:22
• Do you know any algebraic number theory? Your solution can definitely be made to work, but it saves time to use a few facts from ANT.
– user208649
Commented Apr 19, 2020 at 15:41
• Actually, basic abstract algebra is enough, i.e., to know what the field degree $[L:K]$ is. Commented Apr 19, 2020 at 16:44

## 2 Answers

Hint: $$(X^2-2)^2-2=0$$

$$P(X)=X^4-4X^2+2=0$$ Eisenstein is irreducible implies that $$\mathbb{Q}(\sqrt{2+\sqrt2}:\mathbb{Q}]=4$$

• That's a nice approach! I wonder whether the approach I took works at all? (although this one is surely quicker) Commented Apr 19, 2020 at 14:37

We know that $$[\Bbb Q(\sqrt{2+\sqrt{2}}):\Bbb Q]=4$$, but $$[\Bbb Q(\sqrt{2}):\Bbb Q]=2$$. Hence $$\sqrt{2+\sqrt{2}}$$ cannot be in $$\Bbb Q(\sqrt{2})$$.

Reference:

Galois Group of $\sqrt{2+\sqrt{2}}$ over $\mathbb{Q}$