Computing $[\mathbb{Q}(a):\mathbb{Q}]$ (minimal polynomial)

Question

Let $a=\sqrt{3+\sqrt[3]{3}}$.

How to compute the minimal polynomial of $a$ over $\mathbb{Q}$ and $[\mathbb{Q}(a):\mathbb{Q}]$?

My way was:

$a=\sqrt{3+\sqrt[3]{3}} \Rightarrow a^2=3+\sqrt[3]{3} \Rightarrow (a^2-3)^3=3 \Rightarrow a^6-9a^4+27a^2-30=0$

So $a$ is a root of the polynomial $f(x)=x^6-9x^4+27x^2-30$.

With Eisenstein and $p=3$, $f(x)$ is irreducible, so it's the minimal polynomial of $a$ over $\mathbb{Q}$.

Now I'm not sure how to compute $[\mathbb{Q}(a):\mathbb{Q}]$.

My idea was to use the Tower law:

$[\mathbb{Q(\sqrt{3+\sqrt[3]{3}})}:\mathbb{Q}]=[\mathbb{Q(\sqrt{3+\sqrt[3]{3}})}:\mathbb{Q}(\sqrt{3})][\mathbb{Q(\sqrt{3})}:\mathbb{Q}]$, but it doesn't work.

How to find $[\mathbb{Q}(a):\mathbb{Q}]$?

Answer 1

When you extend a field $F$ with the root of an irreducible polynomial $p(x) \in F[x]$, you are not just adding an algebraic element to $F$; you can see the extension as a quotient of the polynomial ring of $F$, $F[x]$, for the ideal generated by the irreducible polynomial: $$F(a) \cong F[x]/(p(x))$$ Doing so, the degree of the extension is equal to the degree of $p(x)$. In your case, you have a polynomial of degree 6 over $\mathbb{Q}[x]$, which you have already shown to be irreducible. Thus, $[\mathbb{Q}(a):\mathbb{Q}]=6$.

Also, the tower extension you were using is wrong: the correct one is $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt[3]{3}) \subseteq \mathbb{Q}(a)$.

Answer 2

You're going a few steps too far. $[\Bbb Q(a):\Bbb Q]$ is equal to the degree of the minimal polynomial of $a$ over $\Bbb Q$. You have shown that that degree is $6$.