# Degree extension $[\mathbb{Q}(\sqrt{2},\sqrt{2}):\mathbb{Q}(\sqrt{2})]$

Find degree of extension $$[\mathbb{Q}(\sqrt{2},\sqrt{2}):\mathbb{Q}(\sqrt{2})]$$.

My approach was the following:

Consider the polynomial $$x^2-2\in \mathbb{Q}[x]$$ and $$\sqrt{2}$$ is its root;

this shows that $$[\mathbb{Q}(\sqrt{2},\sqrt{2}):\mathbb{Q}(\sqrt{2})]\leq 2$$.

I think that this equals $$2$$, but I cannot prove it rigorously.

I would be very grateful if anyone can show how to solve this problem.

BTW, please do not use any Galois theory, because I am not familiar with it yet.

• If the degree were less than $2$ then it would have to be $1$, so it would necessarily be the case that $\sqrt{2}\in\mathbb Q(\sqrt{2})$.
– Dave
Feb 19, 2019 at 21:03
• @Dave, i have already considered it. If $\sqrt{2}\in \mathbb{Q}(\sqrt{2})$ so what is the contradiction?
– RFZ
Feb 19, 2019 at 21:05
• There's a few answers below, in particular Dietrich Burde's answer is along this line.
– Dave
Feb 19, 2019 at 21:07

$$\Bbb Q(\sqrt2,\sqrt2)=\Bbb Q(\sqrt2)$$, so has degree $$6$$ over $$\Bbb Q$$ (Eisenstein). But $$\Bbb Q(\sqrt2)$$ has degree $$3$$ over $$\Bbb Q$$.

• Let me ask you one question: how did you get $\sqrt{2}$?
– RFZ
Feb 19, 2019 at 21:09
• $\sqrt2=\sqrt2/\sqrt2$. Feb 20, 2019 at 2:16
• Sorry but could you help with this topic? math.stackexchange.com/questions/3120335/…
– RFZ
Feb 20, 2019 at 18:31

The only alternative would be that $$[\mathbb{Q}(\sqrt{2},\sqrt{2}):\mathbb{Q}(\sqrt{2})]=1$$, i.e., that $$\sqrt{2}\in \Bbb Q(\sqrt{2})$$ and hence $$\Bbb Q(\sqrt{2}\subseteq \Bbb Q(\sqrt{2})$$. However, this is impossible, because of $$3=[\Bbb Q(\sqrt{2}):\Bbb Q]=[\Bbb Q(\sqrt{2}):\Bbb Q(\sqrt{2})]\cdot [\Bbb Q(\sqrt{2}):\Bbb Q].$$

• How to get contradiction? I guess i should it raise to the 2nd power and 3rd?
– RFZ
Feb 19, 2019 at 21:12
• Yes, exactly. But of course the abstract argument is better, i.e., that $2=[\Bbb Q (\sqrt{2}):\Bbb Q]$ does not divide $3=[\Bbb Q (\sqrt{2}):\Bbb Q]$. Feb 20, 2019 at 9:35

Use the tower law. You know that $$[\mathbb{Q}(\sqrt{2},\sqrt{2}):\mathbb{Q}]=6$$.

This is because $$[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=3$$

and $$[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$$, and 2 and 3 are coprime, which in turn implies $$[\mathbb{Q}(\sqrt{2},\sqrt{2}):\mathbb{Q}]=6$$.

Now use the tower law to deduce the degree is 2.