# Is there a way I can find out the degree of this extension without explicitly finding the minimal polynomial?

Suppose that $\beta$ is a real cube root of $2$ and $\omega$ is a primitive third root of unity. I've a problem that asks me to show that if $\alpha=\beta\omega$, then $\alpha+\beta$ has minimal polynomial of degree 3 over $\mathbb Q$ while $\alpha-\beta$ has minimal polynomial of degree 6. Finding the minimal polynomial seems to be a real problem, especially since the polynomial might be hard to prove to be irreducible. What way should I approach this problem? Is this just a computation exercise?

• Yes, sorry, I edited it. – adrija Nov 23 '16 at 6:15

It's not hard to show that $(\alpha+\beta)^3=-2$, either by expanding out the cube, or by observing that $$\alpha+\beta=(1+\omega)\sqrt{2}=-\omega^2\sqrt{2}$$ since $1+\omega+\omega^2=0$. And $x^3+2$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion, so this must be the minimal polynomial of $\alpha+\beta$.

One can also show that $(\alpha-\beta)^6=-108$, but it's not as clear that $x^6+108$ is irreducible over $\mathbb{Q}$. However, $$(\alpha-\beta)^3=6(\omega-\omega^2)=6+12\omega$$ so $\omega\in\mathbb{Q}(\alpha-\beta)$, and $$(\alpha-\beta)^2=2^{\frac{2}{3}}(\omega^2-2\omega+1)=2^{\frac{2}{3}}(-3\omega)$$ hence $$(\alpha-\beta)^4=2\sqrt{2}\cdot 9\omega^2$$ so that $\sqrt{2}\in \mathbb{Q}(\alpha-\beta)$ as well (since we've already shown that $\omega\in\mathbb{Q}(\alpha-\beta)$). Therefore $\mathbb{Q}(\alpha-\beta)$ contains $\mathbb{Q}(\omega,\sqrt{2})$, so has degree at least $6$ over $\mathbb{Q}$. Since it satisfies a polynomial of degree $6$, the degree of the extension must be exactly $6$, and $x^6+108$ is irreducible.

• Good answer! +1 – Learnmore Nov 23 '16 at 7:09
• Excellent strategy in the second part. I've been trying to prove the irreducibility of $x^6+108$ for some time. This will help in similar other problems too I think, so thank you. – adrija Nov 23 '16 at 7:32
• Thanks, glad to be of help. – carmichael561 Nov 23 '16 at 16:30

In general, if one has a Galois extension $K/F$, the minimal polynomial for any $a \in K$ has as its roots the elements in the orbit of $a$ under the action of $\text{Gal}(K/F)$. That is, if $S = \{ \phi(a) \ | \ \phi \in \text{Gal}(K/F) \}$, then $\displaystyle \min_a(x) = \prod_{u_k \in S} (x-u_k)$.

Proof: This is a consequence of the fact that a polynomial is irreducible $\iff$ its Galois group acts transitively on its roots. For a proof of this fact, see Theorem 2.9(b) here. Notice that $\min_a(x)$ will be an irreducible polynomial (by definition), and its Galois group will be a subgroup of $\text{Gal}(K/F)$. This latter fact is because, if $L \subseteq K$ is the splitting field of $\min_a(x)$, every $F$-automorphism of $L$ extends to an $F$-automorphism of $K$.

To apply this fact to the problem at hand, we have $K = \mathbb{Q}(\sqrt{2}, \omega)$ with $\text{Gal}(K/\mathbb{Q})$ generated by the following two automorphisms: $$i \mapsto -i$$ $$\sqrt{2} \mapsto \omega \sqrt{2}$$

Let's say we want to find the minimal polynomial of $\sqrt{2} + \omega \sqrt{2}$ over $\mathbb{Q}$. Per the argument in the first paragraph, we simply need to find the elements of the orbit of this element under the action of the Galois group, which you can check are:

• $\sqrt{2} + \omega \sqrt{2} \qquad \qquad \text{identity automorphism}$
• $\omega \sqrt{2} + \omega^2 \sqrt{2} \qquad \quad \sqrt{2} \mapsto \omega \sqrt{2} \ \ \text{ acting on original element}$
• $\sqrt{2} + \omega^2 \sqrt{2} \qquad \qquad i \mapsto -i \ \ \text{ acting on original element}$

And this list is exhaustive. The minimal polynomial of $\sqrt{2} + \omega \sqrt{2}$ over $\mathbb{Q}$ therefore has $3$ roots and is of degree $3$. Repeating this process for $- \sqrt{2} + \omega \sqrt{2}$ will demonstrate that its minimal polynomial is of degree $6$.

At least for degree 3, it's not too hard to show that a polynomial is irreducible--simply use the rational root theorem. Since all irreducible polynomials of degree 2 or 3 have no roots over the base field, we can easily show it's irreducible.

My next thought if you want to avoid computations is to try and use the degree formula (I'm not 100% on this route working, though). With this route, you would show the other polynomial has degree six. by showing $\mathbb{Q}[\alpha + \beta][\alpha - \beta] = \mathbb{Q}[\alpha + \beta, \alpha - \beta] = \mathbb{Q}[\alpha, \beta]$ has degree 6 over $\mathbb{Q}$.

Let $K = \mathbb{Q}[\alpha + \beta][\alpha - \beta]$. Then by the degree formula we would obtain that $6 = [K:\mathbb{Q}] = [K:\mathbb{Q}[\alpha - \beta]][\mathbb{Q}[\alpha - \beta]:\mathbb{Q}]$. So essentially if you could show $[K:\mathbb{Q}[\alpha - \beta]] = 1$, or equivalently, $\alpha + \beta \in \mathbb{Q}[\alpha - \beta]$, that would show your claim.