Showing the Irreducibility of a Polynomial and "Splitting" the Field The following problem is part of my Algebra problem set:
Let $\alpha= \sqrt[3]{2}+\omega$ ,where $\omega=e ^{2\pi i /3} = -\frac{1}{2} + \frac{\sqrt{-3}}{2}$ is a primitive cube root of $1$. 
$\alpha$ is a zero of $f(x) = 9 + 9x +3x^3 + 6x^4 + 3x^5 + x^6$.  Show that $f(x)$ is irreducible.
**
Eisenstein's Criterion fails here I realize that Gauss's Lemma is the best method of proving this although I am unclear as to whether proving the irreducibility of a polynomial in a subfield is sufficient to prove that it is irreducible. **
Please Also note that we have not had any Galois Theory.
Show that $\mathbb{Q}(\alpha)$ is the smallest field which contains all three zeros of $x^3 -2 $ (this is called the splitting field of $x^3 -2$).
Thank you in advance for your help!
 A: We claim that the six degree polynomial you wrote down is irreducible by assuming that $\alpha = \sqrt[4]{2} + \omega$ is a root of it. We do this by contradiction. Suppose that it aint. 
Claim 1: If $\alpha$ is not sextic over $\mathbb{Q}$ then it must be cubic over $\mathbb{Q}$. 
As Paul said $\mathbb{Q}(\alpha)\subseteq \mathbb{Q}(\sqrt[3]{2},\omega)$ and since $[\mathbb{Q}(\sqrt[3]{2},\omega):\mathbb{Q}]=6$ this means $d=[\mathbb{Q}(\alpha):\mathbb{Q}]$ must divide $6$. Now look at the divisors of six. Clearly $d\not = 1$ as $\alpha$ is not a rational number (why not?), also $d\not = 6$ as we are assuming that $\alpha$ is not sextic over $\mathbb{Q}$. Therefore, $d=2\text{ or }3$. 
But $d$ cannot be $2$, because if it was then it would mean $\alpha$ would satisfy a quadratic polynomial $X^2+aX+b$ with $a,b\in \mathbb{Q}$. Now if $\alpha$ is a root of this polynomial then so is its complex conjugate, now $\overline{\alpha} = \overline{\omega} + \sqrt[3]{2} = \omega^2 + \sqrt[3]{2}$. Recall that the sum of the roots of $X^2 + aX+b$ add up to $-a$ and so $\alpha + \overline{\alpha} = - a \in \mathbb{Q}$ but $\alpha + \overline{\alpha} = \omega + \omega^2 + 2\sqrt[3]{2} = -1 + 2\sqrt[3]{2}$ and this number is not rational. Thus, it must be the case that $\alpha$ is cubic over $\mathbb{Q}$. 
Claim 2: $\omega \not \in \mathbb{Q}(\alpha)$ 
Mr. Paul basically showed this above. But here is another proof. If $\omega \in \mathbb{Q}(\alpha)$ then the degree of $\omega$ over $\mathbb{Q}$ would divide the degree of $\mathbb{Q}(\alpha)/\mathbb{Q}$ (why?), but the degree of $\omega$ over $\mathbb{Q}$ is two which does not divide three. 
Claim 3: $\sqrt[3]{2}\not \in \mathbb{Q}(\alpha)$ 
If it was then $\omega = \alpha - \sqrt[3]{2} \in \mathbb{Q}(\alpha)$ contradicting Claim 2. 
Claim 4: $\omega^2\sqrt[3]{2}\not \in \mathbb{Q}(\alpha)$ 
This is a bit computational. Suppose that  $\omega^2\sqrt[3]{2}\in \mathbb{Q}(\alpha)$ then: 
(i) $(\sqrt[3]{2}+\omega)\omega^2\sqrt[3]{2} = \omega^2\sqrt[3]{4} + \sqrt[3]{2} \in \mathbb{Q}(\alpha)$ 
(ii)  $(\sqrt[3]{2}+\omega)(\omega^2\sqrt[3]{2})^2 = 2\omega + \omega^2\sqrt[3]{4} \in \mathbb{Q}(\alpha)$ 
Subtract (i) from (ii) to get $2\omega - \sqrt[3]{2}\in \mathbb{Q}(\alpha)$ but then $2\omega - \sqrt[3]{2} + (\omega + \sqrt[3]{2}) = 3\omega \in \mathbb{Q}(\alpha)$. This immediately would imply that $\omega \in \mathbb{Q}(\alpha)$ which contradicts Claim 2. 
Claim 5: $\omega\sqrt[3]{2}\not \in \mathbb{Q}(\alpha)$. 
This one is a similar computation but a bit trickier than Claim 4. Begin by assuming that $\omega\sqrt[3]{2} \in \mathbb{Q}(\alpha)$. First of all $\omega^2 = -1 - \omega$ (why?). Then, $(\sqrt[3]{2}+\omega)^2 = \sqrt[3]{4} + 2\sqrt[3]{2}\omega - 1 - \omega$. This implies that $\sqrt[3]{4} - \omega \in \mathbb{Q}(\alpha)$. 
Then: 
(iii) $(\sqrt[3]{4}-\omega)\omega\sqrt[3]{2} = 2\omega - \omega^2\sqrt[3]{2}$
(iv) $(\sqrt[3]{4}-\omega)(\omega\sqrt[3]{2})^2 = 2\omega^2\sqrt[3]{2} - \sqrt[3]{4}$
Add 2(iii) + (iv) to get that $4\omega - \sqrt[3]{4}\in \mathbb{Q}(\alpha)$ be we shown just above that $\sqrt[3]{4} - \omega \in \mathbb{Q}(\alpha)$ so when we add them we get $3\omega \in \mathbb{Q}(\alpha)$ which is a contradiction. 
Claim 6: $\sqrt[3]{2}$ is cubic over $\mathbb{Q}(\alpha)$. 
We know from Claim 3 that $\sqrt[3]{2}$ is not in the field and $\sqrt[3]{2}$ solves the polynomial $X^3-2$ so the degree of $\sqrt[3]{2}$ is either two or three. It cannot be two. Because if it degree two then $X^3-2$ must be reducible over $\mathbb{Q}(\alpha)$, in particular one of its roots: $\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2}$ must be contained in $\mathbb{Q}(\alpha)$ but Claim 3,4,5 show that is impossible. Thus $\sqrt[3]{2}$ must be cubic. 
Claim 7: CONTRADICTION ARISES from assuming $\alpha$ was cubic. 
Let $K=\mathbb{Q}(\alpha)$. By the above claims $K(\omega)/K$ is quadratic field and $K(\sqrt[3]{2})/K$ is a cubic field. Thus, $K(\omega,\sqrt[3]{2})/K$ is a sextic field which is a contradiction (do you see it?)* 
*)Fact from Field Theory: If $F(\alpha)/F$ is degree $n$ and $F(\beta)/F$ is degree $m$ with $\gcd(n,m)=1$ then $F(\alpha,\beta)/F$ is degree $n\cdot m$.  
A: Hint for the second part: Note that the splitting field of $x^3-2$ is exactly ${\mathbb Q}(\sqrt[3]{2},\omega)$. If you can show that $[{\mathbb Q}(\sqrt[3]{2},\omega) : {\mathbb Q}] = 6$, you should be done (why?).
For convinience consider: $ [{\mathbb Q}(\sqrt[3]{2},\omega) : {\mathbb Q}] = [{\mathbb Q}(\sqrt[3]{2}):{\mathbb Q}] \times [{\mathbb Q}(\sqrt[3]{2},\omega) : {\mathbb Q}(\sqrt[3]{2})] $ and note that $\omega\not\in {\mathbb Q}(\sqrt[3]{2})$ (why?). :)
EDIT: Without Galois Theory, the first part looks very messy.
Here goes an idea: We are done, in view of the previous hint, if we can show ${\mathbb Q}(\alpha) = {\mathbb Q}(\sqrt[3]{2},\omega)$ (in fact, this should prove the second part directly).
Note that if $\omega\in {\mathbb Q}(\alpha)$, the so is $\sqrt[3]{2}$, and similarly if $\sqrt[3]{2}\in {\mathbb Q}(\alpha)$ then $\omega\in {\mathbb Q}(\alpha)$. Thus we will assume that $\omega\not\in {\mathbb Q}(\alpha)$ and $\sqrt[3]{2}\not\in {\mathbb Q}(\alpha)$, else we would be done.
We have, since ${\mathbb Q}(\alpha,\omega) = {\mathbb Q}(\sqrt[3]{2},\omega)$ (why?) :
$$6 = [{\mathbb Q}(\alpha,\omega):{\mathbb Q}] =[{\mathbb Q}(\alpha,\omega):{\mathbb Q}(\alpha)]\times [{\mathbb Q}(\alpha):{\mathbb Q}]$$
Now use this and $\omega\not \in {\mathbb Q}(\alpha)$ to conclude that $[{\mathbb Q}(\alpha):{\mathbb Q}]=3$. Use this to prove that $2=[{\mathbb Q}(\alpha,\sqrt[3]{2}):{\mathbb Q}(\alpha)]$.
Now the minimal polynomial $m(x)$ for $\sqrt[3]{2}$ over ${\mathbb Q}(\alpha)[x]$ must divide $x^3-2 \in {\mathbb Q}(\alpha)[x]$ and $\deg m(x) = 2$.
Dicuss then according to what the 2 roots of $m(x)$ should be, bearing in mind that $m(x) | (x^3 - 2)$.
Finally, prove that if a possible choices produced a polynomial in ${\mathbb Q}(\alpha)[x]$, then either $\omega\in {\mathbb Q}(\alpha)$ or $\sqrt[3]{2}\in {\mathbb Q}(\alpha)$  contradicting our assumption.
