# Question about Galois group (acts transitively on roots)

Let $$f(x)=x^4-6x^2+4 \in \mathbb{Q}[x]$$.

The Galois group is $$\mathrm{Gal}(L/K) \simeq \mathbb{(Z/2Z)^2}$$, but I don't know how to find it.

I know that $$\mathbb{Q}(\sqrt{3+\sqrt{5}})$$ is a splitting field of $$f(x)$$ over $$\mathbb{Q}$$.

$$\mathrm{Gal}(L/K)$$ acts transitively on the roots of $$f(x)$$, so there exist $$\sigma_1, \sigma_2, \sigma_3$$ and $$\sigma_4$$ with $$\sigma_1(\sqrt{3+\sqrt{5}})=\sqrt{3+\sqrt{5}}=\mathrm{id}, \sigma_2(\sqrt{3+\sqrt{5}})=-\sqrt{3+\sqrt{5}}, \sigma_3(\sqrt{3+\sqrt{5}})=\sqrt{3-\sqrt{5}}$$ and $$\sigma_4(\sqrt{3+\sqrt{5}})=-\sqrt{3-\sqrt{5}}$$

So $$\sigma_i^2=\mathrm{id}$$ and $$\mathrm{Gal}(L/K) \simeq \mathbb{(Z/2Z)^2}$$

This is what I don't understand. I see that $$\sigma_2^2=\sigma_2 \circ \sigma_2=\sigma_1$$, but $$\sigma_3^2=\sigma_3 \circ \sigma_3 =0,7639 \neq \mathrm{id}$$ aswell as $$\sigma_4^2=\sigma_4 \circ \sigma_4 = 0,7639 \neq \mathrm{id}$$.

How to compute it to get $$\sigma_3^2=\sigma_4^2=\mathrm{id}$$?

• You have to actually work out the action of the maps on the four roots. What do you mean by $\sigma_{3} \circ \sigma_{3} = 0,7639$? I have no idea how to interpret this statement. Jan 31, 2020 at 19:58
• You have asked this type of question already often. As before, have a look again at Keith Conrad's notes, which give a very explicit description how to find these relations. See this post for the reference. $\sigma_3^2$ is an automorphism, not a number. Also, this duplicate does it in general. Jan 31, 2020 at 19:59
• I don't understand where those decimal numbers came from? Denote $\alpha=\sqrt{3+\sqrt5}$ and $\beta=\sqrt{3-\sqrt5}$. We see that $\alpha\beta=2$. Therefore, for example, $\sigma_3(\alpha)=\beta=2/\alpha$. Consequently $$\sigma_3^2(\alpha)=\sigma_3(2/\alpha)=\sigma_3(2)/\sigma_3(\alpha)=2/(2/\alpha)=\alpha.$$ In other words $\sigma_3^2=\sigma_3\circ\sigma_3$ is the identity. Do you understand how composition of functions works? Jan 31, 2020 at 20:00
• Mind you, a number of things about this field may become simpler, if you make the observation that $(\alpha+\beta)^2=10$. In other words, we also have $L=\Bbb{Q}(\sqrt5,\sqrt2)$. For example, $(1+\sqrt5)^2=6+2\sqrt5$, so $\sqrt{3+\sqrt5}=(1+\sqrt5)/\sqrt2$ etc. Jan 31, 2020 at 20:21

A quick way to compute the Galois group of a quartic like this is to use the discriminant

$$\Delta(P) = 2560X^3 - bX^2 + (ac−4d)X - (a^2d+c^2−4bd)0$$ which is a square, and the cubic resolvent

$$R_3(X) = y^3 - by^2 + (ac−4d)y - (a^2d+c^2−4bd) \\= y^3 + 6 y^2 - 16 y - 100$$

which has no rational roots in this case.

These two facts together tell us the Galois group is $$C_2 \times C_2$$.

Alternatively (inspired by the insight in the comment by Jyrki Lahtonen), write the polynomial as $$X^4 - 2 \cdot 3 X^2 + 4$$ and put $$X = \sqrt{2} Y$$ so that we have $$4 Y^4 - 4 \cdot 3 Y^2 + 4 = Y^4 - 3 Y^2 + 1 = (Y^2 - Y - 1)(Y^2 + Y - 1)$$

Both quadratics have discriminant 5. A root of the first polynomial may be written as $$Y = \frac{1 + \sqrt{5}}{2}$$ and so we have an expression $$X = \sqrt{2}\frac{1 + \sqrt{5}}{2}$$.

All four roots are got by similar expressions (conjugate the square roots) and so the splitting field is $$\mathbb Q(\sqrt{5}, \sqrt{2})$$ and therefore the Galois group is $$V_4$$.