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 $\sigma_3^2=\sigma_4^2=\mathrm{id}$?

  • 2
    $\begingroup$ 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. $\endgroup$
    – xxxxxxxxx
    Jan 31, 2020 at 19:58
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 31, 2020 at 19:59
  • 3
    $\begingroup$ 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? $\endgroup$ Jan 31, 2020 at 20:00
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 31, 2020 at 20:21

1 Answer 1


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 $C_2 \times C_2$.

See https://kconrad.math.uconn.edu/blurbs/galoistheory/cubicquartic.pdf for details.

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 $V_4$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .