# 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 $$\text{Gal}(K/\mathbb{Q}) \cong D_4$$
• $i\alpha$ 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