I'd like to calculate the Galois group of the polynomial $f = X^4 + 4X^2 + 2$ over $\mathbb Q$.

My thoughts so far:

  1. By Eisenstein, $f$ is irreducible over $\mathbb Q$. So $\mathrm{Gal}(f)$ must be a transitive subgroup of $S_4$, i.e. $\mathrm{Gal}(f) = S_4, \ A_4, \ D_8, \ V_4$ or $C_4$.

  2. $X^4 + 4X^2 + 2 = 0 \Leftrightarrow X^2 = -2 \pm \sqrt{2}$. Write $\alpha_1 = \sqrt{-2 + \sqrt{2}}, \ \alpha_2 = -\sqrt{-2+\sqrt{2}}, \ \alpha_3 = \sqrt{-2-\sqrt{2}}, \ \alpha_4 = - \sqrt{-2-\sqrt{2}}$ for the roots of $f$. Then $\mathbb Q(\sqrt{2}, \alpha_1, \alpha_2) = \mathbb Q(\sqrt{2}, \alpha_1) $ is a degree $2$ extension of $\mathbb Q(\sqrt{2})$, and likewise for $\mathbb Q(\sqrt{2}, \alpha_3, \alpha_4) = \mathbb Q(\sqrt{2}, \alpha_3)$. So, by the tower law, the splitting field of $f$ is at most a degree $8$ extension of $\mathbb Q$, so $\mathrm{Gal}(f) = D_8, \ V_4$ or $C_4$.

  3. If I could show that $\mathbb Q(\sqrt{2}, \alpha_1) \neq \mathbb Q(\sqrt{2}, \alpha_3)$, then I'd have that $\mathrm{Gal}(f) = D_8$. At a glance this looks to be true, but I don't know how to prove it.

  4. $\mathrm{Gal}(\mathbb Q (\sqrt{2}) / \mathbb Q) = C_2 \lhd D_8, \ V_4$ and $C_4$, so this doesn't rule anything out.

Any comments on my thoughts 1-4, or hints / explanations would be greatly appreciated.

  • 3
    $\begingroup$ actually, $\alpha_1$ and $\alpha_3$ generate the same field - just multiply them to see why. It shouldn't be hard from here. $\endgroup$ – KotelKanim Jan 29 '12 at 18:14
  • 1
    $\begingroup$ $$\alpha_1\alpha_3=(i\sqrt{2-\sqrt2})(i\sqrt{2+\sqrt2})=-\sqrt2,$$ so $\alpha_3\in\mathbf{Q}(\sqrt2,\alpha_1)$. $\endgroup$ – Jyrki Lahtonen Jan 29 '12 at 18:19
  • $\begingroup$ Great, thanks. In fact, $\alpha_i \alpha_j \in \mathbb Q (\sqrt{2})$ for all $i$ and $j$. Which means the Galois group is $C_4$. Did I approach the entire question in the 'right' way? $\endgroup$ – Matt Jan 29 '12 at 18:29
  • $\begingroup$ You approached it in a reasonable way. I don't know that there is any 'right' way in general. Sometimes, some kinds of information are more accessible than usual, and you can use that as a foothold for the problem, but just as there is no right way solve a jigsaw puzzle, most mathematical problems will have a few reasonable approaches. $\endgroup$ – Aaron Jan 29 '12 at 19:14
  • $\begingroup$ I see that since $\mathbb{Q}(\alpha_1)$ contains all the other roots, that $[\mathbb{Q}(\alpha_i : i =1, 2, 3, 4) : \mathbb{Q}] = 4$ and so the Galois group is $C_2 \times C_2$ or $C_4$. Why does the fact that $\alpha_i\alpha_j \in \mathbb{Q}(\sqrt{2})$ mean that the group is $C_4$? $\endgroup$ – D_S Jan 4 '15 at 21:59

As discussed in the comments, it is usually easiest to piece something like this together by hacking around than to follow a methodical approach. But I thought you might be interested in seeing a methodical approach written up.

In general, let's understand the Galois group $G$ of the splitting field of $x^4+bx^2+c$. Let the roots of $x^4+b x^2+c$ be $\pm \alpha$ and $\pm \beta$. We will assume that the polynomial doesn't have repeated roots. This is equivalent to $(b^2-4c)c \neq 0$.

Any Galois symmetry must either take the pair $\{ \alpha, -\alpha \}$ to itself, or to $\{ - \beta, \beta \}$, because these are the two two-element subsets of the roots which sum to $0$. So the group is a subgroup of the dihedral group $D_8$. I like to think of $D_8$ as the symmetries of a square, with $\pm \alpha$ and $\pm \beta$ at diagonally opposite corners of the square.

You act as if there are two four element subgroups of $D_8$, but there are really three: $C_4$, the copy of $V_4$ generated by reflections over lines parallel to the sides of the square and the copy of $V_4$ generated by reflections over the diagonals of the square. The last $V_4$ doesn't act transitively, so you rule it out at an earlier stage, but I'd rather keep it around.

Reflections over lines parallel to the sides of the square: Consider the element $\gamma: =\alpha \beta$ in the splitting field. If the full $D_8$ acts on the roots, then the orbit of $\gamma$ is $\pm \gamma$ and the stabilizer of $\gamma$ is this $V_4$. In general, if the group is $G$, then the stabilizer of $\alpha \beta$ is $G \cap V_4$. So $G$ is contained in this $V_4$ if and only if $\alpha \beta$ is fixed by the full Galois action, if and only if $\alpha \beta$ is rational.

Now, $(\alpha \beta)^2 = c$. So we get that $G$ is contained in this $V_4$ if and only if $c$ is square.

Reflections over the diagonals of the square: The element $\alpha^2$ is stabilized by this copy of $V_4$; so is the element $\beta^2$. So $G$ is contained in this $V_4$ if and only if $\alpha^2$ and $\beta^2$ are rational. Now, $\alpha^2$ and $\beta^2$ are the roots of $x^2+bx+c$, and the roots of this quadratic are rational if and only if $b^2-4c$ is square.

So $G$ is contained in this $V_4$ if and only if $b^2-4c$ is a square.

The group $C_4$: Again, I think of $C_4$ as a subgroup of the symmetries of a square -- specifically, the rotational symmetries. I am gong to find an element $\delta$ whose stabilizer is $C_4$; this will play a role analogous to $\gamma$ in the first section and $\alpha^2$ in the second.

I found $\gamma$ and $\alpha^2$ just by guessing, but $\delta$ took me a little thought. I'd like a polynomial in $\alpha$ and $\beta$ which has odd degree in each, so that it is not fixed under reflection over any of the diagonals of the squares. We saw above that $\alpha \beta$ doesn't work -- it's stabilizer is $V_4$. Let's try a linear combination of $\alpha \beta^3$ and $\alpha^3 \beta$. A $90^{\circ}$ rotation of the square takes $\alpha \mapsto \beta$ and $\beta \mapsto - \alpha$, so it negates and switches the preceding monomials. In short, we take $$\delta = \alpha \beta^3 - \alpha^3 \beta.$$

If all of $D_8$ acts, then the orbit of $\delta$ is $\pm \delta$. So, as above, the Galois group is contained in $C_4$ if and only if $\delta$ is rational.

Now, $$\delta^2 = (\alpha^2 \beta^2) (\alpha^2 - \beta^2)^2 = c \cdot (b^2-4c).$$ (Remember that $\alpha^2$ and $\beta^2$ are the roots of $x^4+bx^2+c$.)

So $G \subseteq C_4$ if and only if $c(b^2-4c)$ is square.

In your case, we have $c \cdot (b^2 - 4c) = 2 \cdot (4^2-4 \cdot 2) = 16$, so your Galois group is contained in $C_4$.

By the way, notice that the intersection of any two of these groups is contained in the third. Correspondingly, if any two of $c$, $b^2-4c$ and $c \cdot (b^2-4c)$ are square, so is the third.

  • 1
    $\begingroup$ This is a really great answer, thanks. I've learnt Galois theory from a more algebraic perspective; it's useful to see this consideration of the ideas. I'm going to need to spend some time going through this to make sure I understand it all - the fact that I don't immediately see why the statement "any Galois symmetry must take... because these are the two two-element subsets of the roots which sum to zero" is true suggests I'm lacking an understanding of how Galois groups relate to the symmetry behind polynomials, which I shall attempt to rectify. $\endgroup$ – Matt Jan 30 '12 at 19:11
  • $\begingroup$ David, how do you know that $\alpha \beta$ fixed by $G$ iff $\alpha \beta$ rational? $\endgroup$ – Justine Jun 3 '15 at 17:31
  • $\begingroup$ @Michael by definition, $G$ is the group of symmetries of the splitting field fixing the rationals. So, if $\alpha \beta$ is rational, then $G$ is required to fix $\alpha \beta$. Conversely, the field fixed by $G$ is just the ground field; this is the Galois correspondence. $\endgroup$ – David E Speyer Jun 3 '15 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.