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Find degree of extension $[\mathbb{Q}(\sqrt[3]{2},\sqrt{2}):\mathbb{Q}(\sqrt[3]{2})]$.

My approach was the following:

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

this shows that $[\mathbb{Q}(\sqrt[3]{2},\sqrt{2}):\mathbb{Q}(\sqrt[3]{2})]\leq 2$.

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.

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  • $\begingroup$ If the degree were less than $2$ then it would have to be $1$, so it would necessarily be the case that $\sqrt{2}\in\mathbb Q(\sqrt[3]{2})$. $\endgroup$
    – Dave
    Feb 19, 2019 at 21:03
  • $\begingroup$ @Dave, i have already considered it. If $\sqrt{2}\in \mathbb{Q}(\sqrt[3]{2})$ so what is the contradiction? $\endgroup$
    – RFZ
    Feb 19, 2019 at 21:05
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    $\begingroup$ There's a few answers below, in particular Dietrich Burde's answer is along this line. $\endgroup$
    – Dave
    Feb 19, 2019 at 21:07

3 Answers 3

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$\Bbb Q(\sqrt[3]2,\sqrt2)=\Bbb Q(\sqrt[6]2)$, so has degree $6$ over $\Bbb Q$ (Eisenstein). But $\Bbb Q(\sqrt[3]2)$ has degree $3$ over $\Bbb Q$.

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  • $\begingroup$ Let me ask you one question: how did you get $\sqrt[6]{2}$? $\endgroup$
    – RFZ
    Feb 19, 2019 at 21:09
  • $\begingroup$ $\sqrt[6]2=\sqrt2/\sqrt[3]2$. $\endgroup$
    – Angina Seng
    Feb 20, 2019 at 2:16
  • $\begingroup$ Sorry but could you help with this topic? math.stackexchange.com/questions/3120335/… $\endgroup$
    – RFZ
    Feb 20, 2019 at 18:31
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The only alternative would be that $[\mathbb{Q}(\sqrt[3]{2},\sqrt{2}):\mathbb{Q}(\sqrt[3]{2})]=1$, i.e., that $\sqrt{2}\in \Bbb Q(\sqrt[3]{2})$ and hence $\Bbb Q(\sqrt{2}\subseteq \Bbb Q(\sqrt[3]{2})$. However, this is impossible, because of $$ 3=[\Bbb Q(\sqrt[3]{2}):\Bbb Q]=[\Bbb Q(\sqrt[3]{2}):\Bbb Q(\sqrt{2})]\cdot [\Bbb Q(\sqrt{2}):\Bbb Q]. $$

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  • $\begingroup$ How to get contradiction? I guess i should it raise to the 2nd power and 3rd? $\endgroup$
    – RFZ
    Feb 19, 2019 at 21:12
  • $\begingroup$ Yes, exactly. But of course the abstract argument is better, i.e., that $2=[\Bbb Q (\sqrt{2}):\Bbb Q]$ does not divide $3=[\Bbb Q (\sqrt[3]{2}):\Bbb Q]$. $\endgroup$
    – Dietrich Burde
    Feb 20, 2019 at 9:35
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Use the tower law. You know that $[\mathbb{Q}(\sqrt[3]{2},\sqrt{2}):\mathbb{Q}]=6$.

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

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

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

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