Generalized Rationalization in Finite Radical Field Extensions In the square root case of a radical extension of, say, $\mathbb{Q}$, we have that $\mathbb{Q}(\sqrt{2}) = \{a + b \sqrt{2} | a, b \in \mathbb{Q} \}$.
The only semi-hard axiom to prove is that inverses exist.  We reason that this is a field because the inverse of $a + b \sqrt{2}$ can be found by rationalizing the denominator.  Specifically:
$$ \frac{1}{a + b \sqrt{2}} = \frac{a - b \sqrt{2}}{a^2 - 2b^2} = \frac{a}{a^2 - 2b^2} - \frac{b}{a^2 - 2b^2} \sqrt{2},$$ which is clearly of the form $x + y \sqrt{2}$, for $x, y \in \mathbb{Q}$, by the closure of the rationals on arithmetic operations.
Let's say we want to consider $\mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$.  My intuition says that this should be something similar -- at least we know that clearly $\{a + b \sqrt{2} + c \sqrt[3]{2} | a, b, c \in \mathbb{Q} \} \subseteq \mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$ (though I'm not sure exactly). 
I'm just learning field extensions, so I'm not sure if this is right, but I believe that you can say $[\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}): \mathbb{Q}]= 2 \cdot 3$ because the degree of the square root extension is 2, and then you can show that $x^3 - 2$, a polynomial of degree 3, is irreducible over $\mathbb{Q}(\sqrt{2})$.  
Can we generalize some form of "rationalization" to help us prove that inverses exist "directly" in the way we do from the square root case?  At the very least, does the existence of a finite degree field extension prove that algebraic rationalizations of this form exist?
 A: Rationalization is an instance of Euclid's algorithm, as used to obtain Bezout's identity.
Let $E$ be a field, and let $\alpha \in E$ be algebraic over a field $F \subseteq E$. Then the minimal polynomial $f$ of $\alpha$ over $F$ is irreducible in $F[x]$.
Let $0 \ne \beta \in F(\alpha)$. Then $\beta = g(\alpha)$ for some $g \in F[x]$. Clearly $g$ is not divisible by $f$, as $\beta = g(\alpha) \ne 0$. Since $f$ is irreducible, $\gcd(g, f) = 1$. Euclid yields $u, v \in F[x]$ such that
$$
1 = g u + f v.
$$
Evaluate for $x = \alpha$ to get
$$
1 = g(\alpha) u(\alpha) + f(\alpha) v(\alpha) = g(\alpha) u(\alpha).
$$
So
$$
\frac{1}{\beta} = \frac{1}{g(\alpha)} = {u(\alpha)}
$$
is the required rationalization.
A: Here is a method that works in your specific case and also generalizes.
We know that $\mathbb Q$ is a field, so $\mathbb Q[x]$ is a PID. We know that $x^2-2$ is irreducible over $\mathbb Q$ and has $\sqrt 2$ as a root. It is easy to see it divides every polynomial that has $\sqrt 2$ as a root. 
Consider the homomorphism $\phi: \mathbb Q[x] \rightarrow \mathbb Q [\sqrt 2]$ sending $x$ to $\sqrt 2$. By a standard theorem on homomorphisms, we have 
$$\frac{ Q [x]}{\ker \phi} \cong \mathbb Q[\sqrt 2].$$
We know that the kernel is an ideal. Because $\mathbb Q[x]$ is a PID, it is generated by a single element, and by the considerations above we see this element is $x^2-2$. So 
$$\frac{ Q [x]}{(x^2-2)} \cong \mathbb Q[\sqrt 2].$$
Now note that since $\mathbb Q[x]$ is a PID and $x^2-2$ is irreducible, $(x^2-2)$ generates a maximal ideal. So $Q[\sqrt 2]$ is a field and $\mathbb Q[\sqrt 2]=\mathbb Q(\sqrt 2)$. This shows we don't need to add anything more than polynomials in $\sqrt 2$ to get a field.
This argument can be repeated for $\sqrt[3] 2$ over $Q(\sqrt 2)$ almost word for word. You don't actually need to find the polynomial that generates the kernel. All you need to show is that the one that does is irreducible, and I leave that to you. Let me know if you want more detail.
