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.

  • $\begingroup$ Maybe related $\endgroup$
    – PinkyWay
    Commented Apr 19, 2020 at 14:21
  • $\begingroup$ See $\endgroup$
    – PinkyWay
    Commented Apr 19, 2020 at 14:22
  • $\begingroup$ 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. $\endgroup$
    – user208649
    Commented Apr 19, 2020 at 15:41
  • $\begingroup$ Actually, basic abstract algebra is enough, i.e., to know what the field degree $[L:K]$ is. $\endgroup$ Commented Apr 19, 2020 at 16:44

2 Answers 2


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

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

  • $\begingroup$ That's a nice approach! I wonder whether the approach I took works at all? (although this one is surely quicker) $\endgroup$
    – Sigurd
    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 $\Bbb Q(\sqrt{2})$.


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


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