Calculating the degree of a splitting field of $f(x)=x^4-4x^3-4x^2+16x-8$ over $\mathbb{Q}$.

Question

I have shown that the above polynomial is irreducible in $\mathbb{Q}[x]$ and that $\alpha=1+\sqrt{2}+\sqrt{3}$ is a root. As $f(x+1)$ is an even function, $f$ is symmetric in the line $x=1$, so $2-\alpha$ is also a root. Graphing $f$, one sees that $f$ has two real roots larger than $1$. One of these is $\alpha$ so if we call the other one $\beta$, by the symmetry we've already noted $2-\beta$ would also a root. We could then factor $f$ into linear factors in the following way: \begin{equation*} f(x)=(x-\alpha)(x+\alpha-2)(x-\beta)(x+\beta-2) \end{equation*} Therefore a splitting field of $f$ over $\mathbb{Q}$ is given by $\mathbb{Q}(\alpha,\beta)$. By the tower law, \begin{equation*} [\mathbb{Q}(\alpha,\beta):\mathbb{Q}]=[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)]\cdot [\mathbb{Q}(\alpha):\mathbb{Q}] \end{equation*} and we know $[\mathbb{Q}(\alpha):\mathbb{Q}]=4$ ($f$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$). It remains to calculate $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)]$. I know that $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)]\leq [\mathbb{Q}(\beta):\mathbb{Q}]=4$ so $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] \leq 16$. This is a small improvement on the general bound that the splitting field of a degree $4$ polynomial is less than or equal to $4!=24$. I'd appreciate any advice on how to continue, or feedback on my approach so far.

Answer 1

In this case, I think the best thing to do is to actually compute the roots! Your observation that $f(x+1)$ is even is very useful: it means that $f(x+1)$ has no odd-degree terms, so that $f(x+1)$ will be a polynomial in $x^2$. Indeed, we find $f(x+1) = x^4 - 10x^2 + 1 = (x^2)^2 -10(x^2) + 1$. Thus, $f(x+1) = 0$ if and only if $$x = \pm \sqrt{5 \pm 2 \sqrt{6}}.$$ Equivalently, the roots of $f$ are $$1 \pm \sqrt{5 \pm 2 \sqrt{6}}$$ $\alpha$ is the root $1 + \sqrt{5 + 2\sqrt{6}} = 1 + \sqrt{2} + \sqrt{3}$. Then we must have $\beta = 1 + \sqrt{5 - 2 \sqrt{6}} = 1 - \sqrt{2} + \sqrt{3}$.

Now we notice that $\mathbb{Q}(\alpha, \beta) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$, and also that $$\sqrt{2} = \frac{\alpha - \beta}{2} \in \mathbb{Q}(\alpha, \beta)$$ $$\sqrt{3} = \frac{\alpha + \beta - 2}{2} \in \mathbb{Q}(\alpha, \beta)$$ Thus, $\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\sqrt{2},\sqrt{3})$. Now we have $$[\mathbb{Q}(\alpha,\beta) : \mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}) : \mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}) : \mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] \\ = 2 [\mathbb{Q}(\sqrt{2},\sqrt{3}) : \mathbb{Q}(\sqrt{2})],$$ so we only need to compute $[\mathbb{Q}(\sqrt{2},\sqrt{3}) : \mathbb{Q}(\sqrt{2})]$. This is just the degree of the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}(\sqrt{2})$. Since $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$ (exercise!), $x^2 - 3$ does not have any roots in $\mathbb{Q}(\sqrt{2})$, so $x^2 - 3$ is the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}(\sqrt{2})$. Thus, $[\mathbb{Q}(\sqrt{2},\sqrt{3}) : \mathbb{Q}(\sqrt{2})] = 2$, and we get $$[\mathbb{Q}(\alpha,\beta) : \mathbb{Q}] = 4.$$