# The degree of a field extension

"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?

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

call the field $$F$$. then:
$$\mathbb{Q}(\sqrt{2}) \subset F$$
because: $$\sqrt{2} = \frac{\sqrt{2}}{\sqrt{2}}$$
on the other hand: $$F \subset \mathbb{Q}(\sqrt{2})$$
because $$\sqrt{2} = \bigg(\sqrt{2} \bigg)^2 \\$$ and $$\sqrt{2} = \bigg(\sqrt{2} \bigg)^3$$