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?
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