Irreducibility of polynomials over some quadratic fields Prove the following statements:
a) The polynomial $x^3-10$ is irreducible over $\mathbb{Q}(\sqrt{2})$
b) The polynomial $x^3-10$ is irreducible over $\mathbb{Q}(\sqrt{-3})$
c) The polynomial $x^3-x-1$ is irreducible over $\mathbb{Q}(\sqrt{-23})$
Proof:
a) Since $[\mathbb{Q}(\sqrt[3]{10}):\mathbb{Q}]=3$ and $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$ then $[\mathbb{Q}(\sqrt[3]{10},\sqrt{2}):\mathbb{Q}]=6$ because $\text{gcd}(2,3)=1$. Then it follows that $[\mathbb{Q}(\sqrt{2},\sqrt[3]{10}):\mathbb{Q}(\sqrt{2})]=3$ then it follows that $x^3-10$ is irreducible over $\mathbb{Q}(\sqrt{2})$ since it has the right degree.
b) The solution of this part is basically the same. Since $[\mathbb{Q}(\sqrt[3]{10}):\mathbb{Q}]=3$ and $[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}]=2$ then $[\mathbb{Q}(\sqrt[3]{10},\sqrt{-3}):\mathbb{Q}]=6$ because $\text{gcd}(2,3)=1$. Then it follows that $[\mathbb{Q}(\sqrt{-3},\sqrt[3]{10}):\mathbb{Q}(\sqrt{-3})]=3$ then it follows that $x^3-10$ is irreducible over $\mathbb{Q}(\sqrt{-3})$ since it has the right degree.
c) By rational root test it follows that $x^3-x-1$ is irreducible over $\mathbb{Q}$. Let $\alpha$ be its root then $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$ and $[\mathbb{Q}(\sqrt{-23}):\mathbb{Q}]=2$. So, $[\mathbb{Q}(\alpha,\sqrt{-23}):\mathbb{Q}]=6$ and $[\mathbb{Q}(\alpha,\sqrt{-23}):\mathbb{Q}(\sqrt{-23})]=3$ then the polynomial $x^3-x-1$ is irreducible over $\mathbb{Q}(\sqrt{-23})$ since it has a right degree.
Is my reasoning correct? I am sure that it's OK but I want to be sure.
Would be thankful for any comment!
 A: In your first two solutions, you use that $[\Bbb{Q}(\sqrt[3]{10}):\Bbb{Q}]=3$ and $[\Bbb{Q}(\sqrt{d}):\Bbb{Q}]=2$ for $d\in\{2,-3\}$ to claim
$$[\Bbb{Q}(\sqrt[3]{10},\sqrt{d}):\Bbb{Q}]=6.$$
But this presupposes that $\Bbb{Q}(\sqrt[3]{10})$ is a field, and hence that $x^3-10$ is irreducible over $\Bbb{Q}$.
Moreover, the fact that $[\mathbb{Q}(\sqrt{d},\sqrt[3]{10}):\mathbb{Q}(\sqrt{d})]=3$ does not imply that $x^3-10$ is irreducible over $\Bbb{Q}(\sqrt{d})$. The same conclusion fails in your third solution.
To illustrate; the polynomial $x^3-8$ is clearly reducible over $\Bbb{Q}$ but the quotient
$$R:=\Bbb{Q}[x]/(x^3-8),$$
is still a $3$-dimensional vector space over $\Bbb{Q}$. So $[R:\Bbb{Q}]=3$ and
$$[R(\sqrt{2}):\Bbb{Q}]=6
\qquad\text{ and }\qquad
[R(\sqrt{2}):\Bbb{Q}(\sqrt{2})]=3,$$
but $x^3-8$ is not irreducible over $\Bbb{Q}(\sqrt{2})$.

In stead, irreducibility can be proved much more directly in each case:


*

*The polynomial is Eisenstein at $p=5$. You might want to prove that $5$ is prime in $\Bbb{Q}(\sqrt{2})$.

*The polynomial is Eisenstein at $p=5$. You might want to prove that $5$ is prime in $\Bbb{Q}(\sqrt{-3})$.

*By the rational root theorem, every root divides $1$ and hence is a unit in the ring of integers of $\Bbb{Q}(\sqrt{-23})$. Show that $\pm1$ are the only units there, and conclude that the polynomial is irreducible over $\Bbb{Q}(\sqrt{-23})$.

