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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.

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    $\begingroup$ 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
    Commented Jan 29, 2012 at 18:14
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    $\begingroup$ $$\alpha_1\alpha_3=(i\sqrt{2-\sqrt2})(i\sqrt{2+\sqrt2})=-\sqrt2,$$ so $\alpha_3\in\mathbf{Q}(\sqrt2,\alpha_1)$. $\endgroup$ Commented Jan 29, 2012 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
    Commented Jan 29, 2012 at 18:29
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    $\begingroup$ 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
    Commented Jan 4, 2015 at 21:59
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    $\begingroup$ Related: math.stackexchange.com/questions/204709 $\endgroup$
    – Watson
    Commented Dec 26, 2016 at 13:02

2 Answers 2

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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.

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    $\begingroup$ 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
    Commented Jan 30, 2012 at 19:11
  • $\begingroup$ David, how do you know that $\alpha \beta$ fixed by $G$ iff $\alpha \beta$ rational? $\endgroup$
    – Justine
    Commented Jun 3, 2015 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. $\endgroup$ Commented Jun 3, 2015 at 19:08
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Let $K=\mathbb{Q}(\alpha_1)$. Of course, $\alpha_2 \in K$. We have $\alpha_1^2=-2+\sqrt{2} \in K \implies \sqrt{2} \in K \implies \alpha_3=-\sqrt{2}/\alpha_1 \in K$ and $\alpha_4=-\alpha_3 \in K$. So, $K$ is the splitting field of the irreducible polynomial $f$ over $\mathbb{Q}$.

So, $[K:\mathbb{Q}]=\deg(f)=4=|Gal(f)| \implies Gal(f) \cong D_2$ or $Gal(f) \cong C_4$.

Now show that $\sigma \in Gal(f)$ having $\sigma(\alpha_1)=\alpha_3$ is an element of order $4$, but all elements of $D_2$ have order $2$. So we conclude that $Gal(f) \cong C_4$(or alternatively using the result that you mentioned: $Gal(f)$ is isomorphic to a transitive subgroup of $S_4$ gives $Gal(f) \cong C_4$ directly as well).

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