Hello and thanks in advance for any responses. I'm stuck on the following problem: "Show that the polynomial $X^3 – 2$ is irreducible over $\mathbb{Q}(\omega)$, where $\omega = e^{\frac{2\pi i}{3}}$, and that it is the minimal polynomial of $\xi = \sqrt[3]{2}$ over $\mathbb{Q}(\omega)$ with $[\mathbb{Q}(\xi, \omega): \mathbb{Q}] = 6$
$\textbf{My Attempt:}$
Since $deg(X^3 – 2) =3$ and its roots are $\xi, \xi\omega, \xi\omega^2 \in \mathbb{C}\setminus \mathbb{Q}(\omega)$ then $X^3 – 2$ is irreducible over $Q(\omega)$ (is this enough to prove irreducible over any field if we have degree 2 or 3 and none of the roots in it?). Now, by the Tower Theorem we know $$[\mathbb{Q}(\xi, \omega) : \mathbb{Q}] = [\mathbb{Q}(\xi, \omega) : \mathbb{Q}(\omega)][\mathbb{Q}(\omega): \mathbb{Q}]$$ I'm tempted to say that $[\mathbb{Q}(\xi, \omega) : \mathbb{Q}(\omega)] = 3$ because $\xi, \xi\omega, \xi\omega^2 \in \mathbb{Q}(\xi, \xi\omega, \xi\omega^2)=\mathbb{Q}(\xi, \omega)$ which are roots of irreducible $X^3 – 2 \in \mathbb{Q}(\omega)$ and therefore the minimal polynomial of $\xi$ over $\mathbb{Q}(\omega)$ is exactly $m_{\xi}(X) = X^3 -2$ and therefore $[\mathbb{Q}(\xi, \omega) : \mathbb{Q}(\omega)] = deg(m_{\xi}(X)) = 3$.
Now if so far the process is correct, we're only left to show that $[\mathbb{Q}(\omega): \mathbb{Q}] = 2$.
Well, we know that $\omega = e^{\frac{2\pi i}{3}} = \frac{\sqrt{3} – 1}{2}$ so let's calculate a polynomial which has $\omega$ has a root over $\mathbb{Q}$. After some algebraic manipulations I've figured that $f(X) =4X^4 – 8X^2 + 1 \in \mathbb{Q}[X]$ is such that $f(\omega) = 0$, therefore $\omega \in \mathbb{Q}(\omega)$ is an algebraic element over $\mathbb{Q}$ and therefore $m_{\omega}(X) | f(X)$ where $m_{\omega}(X)$ is its minimal polynomial over $\mathbb{Q}$. This means that $deg(m_{\omega}(X)) \in \{1, 2, 4\}$. Well, obviously $deg(m_{\omega}(X)) \neq 1$ otherwise it would mean $\omega \in \mathbb{Q}$, which is false. I don 't really know how to handle the case where possibly $deg(m_{\omega}(X)) = 2$, therefore I thought that maybe if I could prove $f(X)$ is reducible over $\mathbb{Q}$ I would know that $m_{\omega}(X) \neq f(X)$ which would imply $deg(m_{\omega}(X)) \neq 4$ leaving us with $deg(m_{\omega}(X)) = 2$.
But how do I show $f(X) = 4X^4 – 8X^2 +1$ reducibility over $\mathbb{Q}$? Is it in first place? For example, rational root test didn't work. In doubt I went and checked some tests for its irreducibility over $\mathbb{Q}$ such as Eisenstein Criterium, reciprocal polynomial $X^4f(\frac{1}{X})$ and irreducibility modulo $p$ with $p = 3$ and $p = 5$. With $p = 5$ I tried to get a contradiction arguing that $f(X) = g(X)h(X)$ where $deg(g) = deg(h) = 2$ and ended up with a contradiction (admittedly with some calculation errors through the way). So if its irreducible over $\mathbb{Z}_5$ it is over $\mathbb{Z}$ and by Gauss Lemma it is over $\mathbb{Q}$.
But this can't be, otherwise we have that $m_{\omega}(X) = f(X)$ and therefore $[\mathbb{Q}(\omega): \mathbb{Q}] = 4$ which won't lead us to the desire result of $[\mathbb{Q}(\xi, \omega): \mathbb{Q}] = 6$.
Again, thanks for any suggestions in advance.
Best Answer
Actually there is a simpler argument. Look at the two towers : $\mathbf{Q} \subset \mathbf{Q}(\sqrt[3]{2}) \subset \mathbf{Q}(\sqrt[3]{2}, \zeta_3)$ and $\mathbf{Q} \subset \mathbf{Q}(\zeta_3) \subset \mathbf{Q}(\sqrt[3]{2}, \zeta_3)$, where $\zeta_3$ is a primitive third unit root. We know that $[\mathbf{Q}(\sqrt[3]{2}) : \mathbf{Q}] = 3$ and $[\mathbf{Q}(\zeta_3) : \mathbf{Q}] =2$. Furthermore, the extension $\mathbf{Q}(\sqrt[3]{2}) \subset \mathbf{Q}(\sqrt[3]{2}, \zeta_3)$ is at most of degree $2$, whereas $\mathbf{Q}(\zeta_3) \subset \mathbf{Q}(\sqrt[3]{2}, \zeta_3)$ is at most of degree $3$. Now, just compare different possibilities. You can see that $\mathbf{Q}(\sqrt[3]{2}) \subset \mathbf{Q}(\sqrt[3]{2}, \zeta_3)$ has to be of degree $2$ and the other has to be of degree $3$.