# Let $f(x)=x^4-x^2-1 \in \Bbb Q[x]$, $K$ is the splitting field of $f$ over $\Bbb Q$.

Let $$f(x)=x^4-x^2-1 \in \Bbb Q[x]$$, $$K$$ is the splitting field of $$f$$ over $$\Bbb Q$$.

I’m to find out $$\text{Gal}(K/\Bbb Q)$$, how do I do?

If the four roots of $$f(x)$$ in $$\Bbb C$$ are $$x_1, x_2, x_3, x_4$$, then $$K = \Bbb Q(x_1, x_2, x_3, x_4)$$. Also, every element in $$\text{Gal}(K/\Bbb Q)$$ permutes the four roots. But how can I determine $$\text{Gal}(K/\Bbb Q)$$? Is there a universal method?

• For quartic polynomials such as this there is a method for determining the Galois group using the so-called "cubic resolvent", have you heard of this? – Dave Mar 6 at 16:13
• First find out what the solutions look like. HINT: the equation can be rewritten as $(x^2)^2 - (x^2) - 1 = 0$ – Justin Stevenson Mar 6 at 16:13
• @Dave I can determine the four roots, but can’t determine the Galois group. – Mingwei Zhang Mar 7 at 1:08

If my computation is right, the solutions are $$\alpha = \sqrt{\frac{1+\sqrt{5}}{2}}, -\alpha, i\alpha^{-1}, -i\alpha^{-1}$$. Then do the standard adjoining of roots to $$\mathbb{Q}$$ to determine the splitting field.

Adjoining $$\alpha$$ to $$\mathbb{Q}$$ we get the field $$\mathbb{Q}(\alpha)$$. Now obviously it contains $$-\alpha$$, while on the other side the minimum polynomial of $$i\alpha^{-1}$$ over $$\mathbb{Q}(\alpha)$$ is $$x^2 + \alpha^{-2} = x^2 + (\alpha^2 -1)$$. So we get that $$\mathbb{Q}(\alpha, i\alpha^{-1}) = \mathbb{Q}(\alpha,i)$$ is the splitting field of $$f$$ over $$\mathbb{Q}$$ and its extension degree over $$\mathbb{Q}$$ is $$8$$.

As $$\alpha$$ and $$i$$ are independent over $$\mathbb{Q}$$ we have that all the automorphisms are: $$Id : \begin{array}{lr} \alpha \to \alpha\\ i \to i \end{array}\quad \quad \sigma : \begin{array}{lr} \alpha \to i\alpha\\ i \to i \end{array}\quad \quad \sigma^2 : \begin{array}{lr} \alpha \to -\alpha\\ i \to i \end{array}\quad \quad \sigma^3 : \begin{array}{lr} \alpha \to -i\alpha\\ i \to i \end{array}\quad \quad$$ $$\tau : \begin{array}{lr} \alpha \to \alpha\\ i \to -i \end{array}\quad \quad \sigma\tau : \begin{array}{lr} \alpha \to i\alpha\\ i \to -i \end{array}\quad \quad \sigma^2\tau : \begin{array}{lr} \alpha \to -\alpha\\ i \to -i \end{array}\quad \quad \sigma^3\tau : \begin{array}{lr} \alpha \to -i\alpha\\ i \to -i \end{array}\quad \quad$$

Now they satisfy the relations $$\sigma^4 = \tau^2 = Id$$ and $$\tau\sigma\tau = \sigma^3$$. However this is exactly the description of the dihedral group of order $$8$$. Thus we have that $$\text{Gal}(K/\mathbb{Q}) \cong D_4$$

On the other hand, the Galois group of every quartic can be computed without explicitly computing the automorphisms. As mentioned in the comments, you can use the resultant. You can use Keith Conrad's notes, which elaborate on the method quite well.

• $i\alpha$ is not a root of $x^4 - x^2 - 1$. $(i\alpha)^4 - (i\alpha)^2 - 1 = \alpha^4 + \alpha^2 - 1 = \alpha^4 - \alpha^2 + 1 + 2\alpha^2 = 2\alpha^2 = 1 + \sqrt{5}$. – André 3000 Mar 8 at 20:43
• @André3000 Oops. I made a sign mistake, the complex roots are $\pm i\sqrt{\frac{\sqrt{5}-1}{2}} = i\alpha^{-1}$. Anyway the conclusion wouldn't change much. – Stefan4024 Mar 8 at 20:59