Does the irreducibility of $x^6-2$ over $\mathbb{Q}$ follow from the fact that the roots of $x^2-2^{1/3}$ are not in $\mathbb{Q}(2^{1/3})$? In an attempt to prove that $x^{6}-2$ is irreducible over $\mathbb{Q}$ without using Eisenstein's criterion, I came up with the following incomplete argument.
Proving that $x^6-2$ is irreducible over $\mathbb{Q}$ is equivalent to proving that the degree of the extension field $\mathbb{Q} (2^{1/6})$ over $\mathbb{Q}$ is $6$.
Using the fact that extension degrees are multiplicative, we know that $$[\mathbb{Q}(2^{1/6}):\mathbb{Q}] = [\mathbb{Q}(2^{1/6}):\mathbb{Q}(2^{1/3})][\mathbb{Q}(2^{1/3}):\mathbb{Q}].$$
And we know that $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=3$ since $x^{3}-2$ is irreducible over $\mathbb{Q}$.
So we have to show that $[\mathbb{Q}(2^{1/6}):\mathbb{Q}(2^{1/3})]=2$, which is equivalent to showing that $x^{2}- 2^{1/3}$ is irreducible over $\mathbb{Q}(2^{1/3})$. But this in turn is equivalent to showing that that roots of $x^{2}-2^{1/3}$ are not in $\mathbb{Q}(2^{1/3})$.
Is my argument correct up to this point? And if it is, how do you argue that $\pm 2^{1/6}$ are not elements in $\mathbb{Q}(2^{1/3}) = \{a+b2^{1/3}+c2^{2/3}|a,b,c \in \mathbb{Q} \}?$
 A: Yes, your argument is correct: it is enough to prove that $t=\sqrt[3]{2}$ has no square root in $\mathbb{Q}(t)$.
To show this directly, we can calculate $(a+bt+ct^2)^2 = (a^2 + 4bc) + (2c^2 + 2ab)t + (b^2 + 2ac)t^2$.  This gives us three equations to be solved in rationals:
$$a^2 + 4bc = 0$$
$$2c^2 + 2ab = 1$$
$$b^2 + 2ac = 0$$
If $a=0$ or $b=0$, then the second equation is $2c^2 = 1$, which has no solution.
Otherwise, the first and third equations give us $c=-a^2/(4b) = -b^2 / (2a)$, so $a^3/b^3 = 2$, which has no solution.
A: Another argument uses the inclusions $\Bbb Q\subset\Bbb Q(\sqrt2)\subset\Bbb Q(\sqrt[6]2)$ and the “well-known” fact that $\Bbb Z[\sqrt2]$ is a principal ideal domain and therefore has unique factorization. Irreducibility of $X^3-\sqrt2$ over the quadratic field follows from the fact that the polynomial has no roots in $\Bbb Q(\sqrt2)$, and this follows from the usual argument that uses the uniqueness of factorization. For if $a,b\in\Bbb Z[\sqrt2]$ and $(a/b)^3=\sqrt2$, then when you clear of fractions, the $\Bbb Z[\sqrt2]$-equation $a^3=b^3\sqrt2$ has a number of appearances of the prime element $\sqrt2$ that’s divisible by three on the left, but prime to three on the right.
