Prove that $x^2 + \alpha x + {\alpha}^2$ is irreducible over $\Bbb{Q}(\alpha)$ where $\alpha ^3 = 2$. I was asked to show this. I am in the beginning of a Galois Theory course. My idea is to apply Eisenstein's Criterion, but I am not exactly sure how to use the generalized version of it. Is there a simpler way to solve this problem? Can someone give me a hint, please?
 A: Using essentially the comment by Mark Saving and also essentially the answer by Angelo, but with a lot less work:
Assume that $x^2 + \alpha x + \alpha^2$ is reducible over ${\mathbb Q}(\alpha)$. Then it has a root in ${\mathbb Q}(\alpha)$ and therefore its discriminant is a square in ${\mathbb Q}(\alpha)$. That discriminant is $-3 \alpha^2$, so that means $-3$ is a square in ${\mathbb Q}(\alpha)$, so ${\mathbb Q}(\sqrt{-3}) \subseteq {\mathbb Q}(\alpha)$. This is impossible by a degree argument: ${\mathbb Q}(\sqrt{-3})$ has degree $2$ over ${\mathbb Q}$, but ${\mathbb Q}(\alpha)$ has degree $3$ over ${\mathbb Q}$.
A: If $\;x^2+\alpha x+\alpha^2\,$ were reducible over $\,\mathbb Q(\alpha)\,,\,$ then it would have a root  $\,x_0\in\mathbb Q\left(\alpha\right)\,$ where $\,\alpha^3=2\,.$
Since $\;x_0^2+\alpha x_0+\alpha^2=0\quad,\quad$ it follows that ,
$x_0=\dfrac{-\alpha-i\alpha\sqrt3}2\quad\lor\quad x_0=\dfrac{-\alpha+i\alpha\sqrt3}2\;.$
Without loss of generality, we can suppose that
$x_0=\dfrac{-\alpha+i\alpha\sqrt3}2\in\mathbb Q(\alpha)\;.$
Moreover ,
$i\sqrt3=\dfrac{2x_0+\alpha}{\alpha}\in\mathbb Q(\alpha)\;.$
Consequently, there exist $\;p,q,r\in\mathbb Q\;$ such that
$i\sqrt3=p+q\alpha+r\alpha^2\quad,\quad$ hence $\;,\quad\alpha\not\in\mathbb R\;.$
Since $\;\alpha^3=2\;$ and $\;\alpha\not\in\mathbb R\;,\;$ it follows that ,
$\alpha^2+\sqrt[3]2\alpha+\sqrt[3]4=0\quad,$
$\alpha=\dfrac{-1-i\sqrt3}2\sqrt[3]2\quad\lor\quad\alpha=\dfrac{-1+i\sqrt3}2\sqrt[3]2\;.$
Without loss of generality, we can suppose that
$\alpha=\dfrac{-1+i\sqrt3}2\sqrt[3]2\;.$
Moreover ,
$\sqrt[3]2=\dfrac{2\alpha}{-1+i\sqrt3}\in\mathbb Q(\alpha)\;.$
Consequently, there exist $\;s,t,v\in\mathbb Q\;$ such that
$\sqrt[3]2=s+t\alpha+v\alpha^2\;.$
Since $\;\alpha^2=-\sqrt[3]2\alpha-\sqrt[3]4\;,\;$ it follows that
$\sqrt[3]2=s+t\alpha+v\left(-\sqrt[3]2\alpha-\sqrt[3]4\right)\;\;,$
$\sqrt[3]2=s-v\sqrt[3]4+\alpha\left(t-v\sqrt[3]2\right)\;.$
Since $\;\sqrt[3]2\not\in\mathbb Q\;,\;$ it results that $\;t-v\sqrt[3]2\ne0\;,\,$ consequently ,
$\alpha=\dfrac{\sqrt[3]2-s+v\sqrt[3]4}{t-v\sqrt[3]2}\;\;,$
but it is impossible because $\;\alpha\not\in\mathbb R\;.$
Hence, the polynomial $\;x^2+\alpha x+\alpha^2\,$ cannot be reducible over $\,\mathbb Q(\alpha)\;.$
A: Let
$P(x)=x^2+\alpha x+\alpha^2.$
In$$\Bbb Q(\alpha)\simeq\Bbb Q(\sqrt[3]2)\subset\Bbb R,$$
$-3$ is not a square, whereas
$$P(z)=0\iff\left(1+\frac{2z}\alpha\right)^2=-3.$$
Therefore, the quadratic polynomial $P(x)$ has no root in $\Bbb Q(\alpha),$ hence is irreducible over it.
