Find the degree of the extension $\mathbb{Q}(\sqrt{11},\sqrt[3]{11}):\mathbb{Q}(\sqrt{11})$

Question

I am trying to find the degree of the following field extension:

$\mathbb{Q}(\sqrt{11},\sqrt[3]{11}):\mathbb{Q}(\sqrt{11})$.

I know it must be at most three since $\sqrt[3]{11}$ is a root of $x^3-11$. I have also proven that it cant be 1 because if it was, that would mean there would exist two rational numbers a and b such that $11=(a+ b\sqrt{11})^3$ which is not true for any rational a,b. I assume that I then have to prove it can not be two either, which leaves me with the answer, 3. How can I prove it is not 2?

Answer 1

Write $\alpha = \sqrt[3]{11}$ and $\beta = \sqrt{11}$. Then you have the following inclusions of fields, namely $\mathbb Q \subseteq \mathbb Q(\alpha) \subseteq \mathbb Q(\alpha, \beta)$ and $\mathbb Q \subseteq \mathbb Q(\beta) \subseteq \mathbb Q(\alpha, \beta)$.

Using the tower formula, you know that $[\mathbb Q(\alpha, \beta) : \mathbb Q] = [\mathbb Q(\alpha, \beta) : \mathbb Q(\alpha)] \cdot [\mathbb Q(\alpha) : \mathbb Q] $ and similarly for $\beta$. As you know that $\mathbb Q(\alpha) : \mathbb Q= 3$, we find that $3$ divides $[\mathbb Q(\alpha, \beta) : \mathbb Q]$ and hence also $[\mathbb Q(\alpha, \beta) : \mathbb Q(\beta)]$. Along with the fact that $[\mathbb Q(\alpha, \beta) : \mathbb Q(\beta)]\leq 3$ this suffices.

Answer 2

Notice that $x^3-11$ is irreducible over $\mathbb{Q}(\sqrt{11})$. It's obvious that $[\mathbb{Q}(\sqrt[3]{11}, \sqrt{11}): \mathbb{Q}(\sqrt{11})]=3$.