Showing that the fifteenth cyclotomic polynomial is ireducible over $\mathbb{Q}$ I need to show that $\Phi_{15}(x) = x^8 - x^7 + x^5 - x^4 + x^3 -x + 1 = \prod_{i \in \mathbb{Z}_{15}^{\times}}(x - \omega^i)$ is irreducible over $\mathbb{Q}$, where $\omega = e^{\frac{2\pi i}{15}}$.

It is true in general that every cyclotomic polynomial is irreducible over $\mathbb{Q}$, but I am not allowed to use this result, so am looking for a way of proving this special case.

This is an exam question from a previous year, and the amount of time allowed suggests that there is a fairly quick way of doing it, which rules out any kind of major algebraic bash. 

Any help would be very much appreciated.
Let $\omega$ be a primitive $15$-th root of unity and let $G$ be the Galois group. For $\sigma \in G$, we must have $\sigma(\omega) = \omega^a$ for some $a \in (\mathbb{Z}/15 \mathbb{Z})^{\times}$. If $\sigma(\omega) = \omega^a$ and $\tau(\omega) = \omega^b$ then $\sigma(\tau(\omega)) = \sigma(\omega^b) = \sigma(\omega)^b = \omega^{ab}$, so we get an embedding of $G$ into $(\mathbb{Z}/15 \mathbb{Z})^{\times}$. The question is what subgroup it is. I see two easy things to show:


$G$ contains the element $-1 \in (\mathbb{Z}/15 \mathbb{Z})^{\times}$, because this is the action of complex conjugation.
The Chinese remainder theorem shows that $(\mathbb{Z}/15 \mathbb{Z})^{\times} \cong (\mathbb{Z}/3 \mathbb{Z})^{\times} \times (\mathbb{Z}/5 \mathbb{Z})^{\times}$. The projection of $G$ onto the two factors $(\mathbb{Z}/3 \mathbb{Z})^{\times}$ and $(\mathbb{Z}/5 \mathbb{Z})^{\times}$ gives the action of $G$ on the roots of the $3$-rd and $5$-th cyclotomic polynomial. Those polynomials are irreducible by the usual argument with Eisenstein's criterion, so $G$ acts transitively on those roots, so the projection of $G$ onto each factor is surjective.


Now, $(\mathbb{Z}/3 \mathbb{Z})^{\times} \times (\mathbb{Z}/5 \mathbb{Z})^{\times} \cong C_2 \times C_4$. Switching to additive notation, we are looking for a subgroup of $C_2 \times C_4$ which surjects onto $C_4$ and contains $(1,2)$. I leave it to you to show that the only such subgroup is all of $C_2 \times C_4$. More generally, this argument works for $3p$ for any prime $p$ which is $1 \bmod 4$.

I would be interested to see a short proof that the $21$-st cyclotomic polynomial does not have Galois group contained in the $6$ element subgroup of $(\mathbb{Z}/21 \mathbb{Z})^{\times}$ generated by $-4$.
First, observe that $\mathbb{Q}(\omega) = \mathbb{Q}(\omega^3)(\omega^5) = \mathbb{Q}(\omega^3)(\sqrt{-3})$ is a quadratic extension of $\mathbb{Q}(\omega^3)$.  Assuming you've proved that $\mathbb{Q}(\omega^3) = \mathbb{Q}(e^{2\pi i}/5)$ is Galois over $\mathbb{Q}$ (which follows from the fact that $\Phi_5$ is irreducible), then we know that $\operatorname{Gal}(\mathbb{Q}(\omega) / \mathbb{Q}(\omega^3))$ must be a normal subgroup of $\operatorname{Gal}(\mathbb{Q}(\omega) / \mathbb{Q})$ of order 2.

Now, for example, let us consider the automorphism $\sigma$ on $\mathbb{Q}(\omega^3)$ which maps $\omega^3 \mapsto \omega^6$.  Then this automorphism has an extension to an automorphism $\bar \sigma$ on $\mathbb{Q}(\omega)$; and by it being an extension, we can conclude that $\bar \sigma(\omega) \in \{ \omega^2, \omega^7, \omega^{12} \}$.  But since $\omega^{12}$ is a fifth root of unity, that image is impossible.  And in fact, from the order of $\operatorname{Gal}(\mathbb{Q}(\omega) / \mathbb{Q}(\omega^3))$, which is the kernel of the restriction homomorphism $\operatorname{Gal}(\mathbb{Q}(\omega) / \mathbb{Q}) \to \operatorname{Gal}(\mathbb{Q}(\omega^3) / \mathbb{Q})$, we know that both $\bar \sigma(\omega) = \omega^2$ and $\bar \sigma(\omega) = \omega^7$ must be possible.

Generalizing this argument, it should be straightforward to show that for every $k$ with $\gcd(k, 15) = 1$, there exists an automorphism $\bar\sigma \in \operatorname{Gal}(\mathbb{Q}(\omega) / \mathbb{Q})$ with $\bar\sigma(\omega) = \omega^k$.  Therefore, any nontrivial monic divisor of $\Phi_{15}$ must have all such $\omega^k$ as roots and therefore must be equal to $\Phi_{15}$.