"Find the degree of $\mathbb{Q}(\sqrt[\leftroot{-2}\uproot{2}12]{2},\sqrt[\leftroot{-2}\uproot{2}8]{2})$ as an extension of $\mathbb{Q}$ and find basis for such an extension"

Let $L = \mathbb{Q}(\sqrt[\leftroot{-2}\uproot{2}12]{2},\sqrt[\leftroot{-2}\uproot{2}8]{2})$

The minimum polynomials of $K_1 =\mathbb{Q}(\sqrt[\leftroot{-2}\uproot{2}12]{2})$ and $K_2 = \mathbb{Q}(\sqrt[\leftroot{-2}\uproot{2}8]{2})$ to have degrees $12$ and $8$ respectively.

But as $gcd(8,12) = 4$ then $[L:\mathbb{Q}]=[L:K_1][K_1:\mathbb{Q}]<96$ (by the tower law)

That's as far as I've gotten. I tried to work out the basis (using a grid) and thus the degree and got 42, but that doesn't seem right at all.

My assumption is that I divide $96$ by $4$ giving a degree of $24$

What am I getting wrong?

  • $\begingroup$ Your field must also contain $\dfrac{1}{2}(\sqrt[12]{2}\sqrt[8]{2})^5=\sqrt[24]{2}$. $\endgroup$ – David Peterson Oct 17 '18 at 23:47
  • $\begingroup$ All the fields in here have infinitely many elements so I replaced the finite-fields tag with a more appropriate one. $\endgroup$ – Jyrki Lahtonen Oct 18 '18 at 5:58

call the field $F$. then:

$$ \mathbb{Q}(\sqrt[24]{2}) \subset F $$

because: $$ \sqrt[24]{2} = \frac{\sqrt[8]{2}}{\sqrt[12]{2}} $$

on the other hand: $$ F \subset \mathbb{Q}(\sqrt[24]{2}) $$

because $$ \sqrt[12]{2} = \bigg(\sqrt[24]{2} \bigg)^2 \\ $$ and $$ \sqrt[8]{2} = \bigg(\sqrt[24]{2} \bigg)^3 $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.