# Finding Galois group of a separable polynomial

Let $$E$$ be the splitting field of $$x^4-10x^2-20$$ over $$\mathbb{Q}$$. Find $$Gal(E/\mathbb{Q})$$.

Since this polynomial is irreducible by Eisenstein's criterion, we know that the Galois group acts transitively on the set of four distinct roots of this polynomial. However, after that, I am very lost as to how should I proceed. One approach which I thought of is to check the transitive subgroups of $$S_4$$, but that seems to make things more complicated. I would appreciate hints on this problem.

Let $$\alpha_1=\sqrt{5+3\sqrt{5}},\alpha_2=\sqrt{5-3\sqrt{5}}$$, then the splitting field of $$f$$ is $$E=\mathbf{Q}(\alpha_1,\alpha_2)$$. As $$f$$ is irreducible, we know that $$\alpha_1$$ has degree $$4$$ over $$\mathbf{Q}$$. In particular, $$\mathbf{Q}(\alpha_1)$$ and $$\mathbf{Q}(\alpha_2)$$ are quadratic extensions of $$\mathbf{Q}(\sqrt{5})$$. They are equal if and only if $$\alpha_1\alpha_2=\sqrt{-20}\in \mathbf{Q}(\sqrt{5})$$, which is not the case (see comment in this answer). Now, $$\alpha_2$$ is a zero of $$X^2+\alpha_1^2-10\in \mathbf{Q}(\alpha_1)[X]$$, so $$|\operatorname{Gal}(E/\mathbf Q)|=[E:\mathbf{Q}]=[\Omega:\mathbf{Q}(\alpha_1)][\mathbf{Q}(\alpha_1):\mathbf{Q}]=2\cdot 4=8.$$

As $$D_8$$ is the only transitive subgroup of order 8 of $$S_4$$, we conclude $$\operatorname{Gal}(E/\mathbf{Q})\cong D_8$$.

The resolvent cubic of the quartic is $$C(x) = x^3+20x^2+180x=x(x^2+20x+180)$$Note that the discriminant of the quadratic is $$20^2-4\cdot180=-320$$, so $$C(x)$$ factors into a linear and quadratic term.

The discriminant of the cubic is $$D=20^2180^2-4\cdot180^3=160^2\cdot5$$

The original quartic is reducible in $$\mathbb Q(\sqrt D)=\mathbb Q(\sqrt5)$$ since it can be written as $$(x^2-5)^2-45=(x^2-5+3\sqrt5)(x^2-5-3\sqrt5)$$

So, the Galois group is $$D_8$$.

• Thanks! We didn't cover discriminant in lecture at all but your answer of the Galois group being cyclic actually allowed me to prove it the following steps: Since exists a $\sigma$ s.t. $\sigma(\sqrt{5+3\sqrt5}) = \sqrt{5-3\sqrt5}$ , we know that $\sigma(\sqrt5) = -\sqrt{5}$. This completely determines how $\sigma$ on the set of roots. Combine this with the fact that $[E:\mathbb{Q}] = 4$ (by checking $E = \mathbb{Q}(\sqrt{5+3\sqrt5})$), we obtain that $Gal(E/F) \cong Z/4Z$ as every element has order four. May 24 '20 at 21:50
• But I think it is the dihedral group of order $8$. To get $E$ you have to adjoin square roots of both $5+3\sqrt{5}$ and $5-3\sqrt{5}$ to ${\mathbb Q}(\sqrt{5})$, so $|E:{\mathbb Q}| = 8$. Note also that complex conjugation swaps two of those but fixes the other two. May 24 '20 at 21:54
• @DerekHolt Oops I think you are right. May 24 '20 at 22:03
• @DerekHolt Yes, my bad, I'll edit May 24 '20 at 22:03
• I must admit, I was very confused by your answer: it seems to directly contradict the criteria you linked in the comments above! But there is just a small arithmetic mistake---the discriminant of the resolvent cubic is $D = 20^2 \cdot 180^2 - 4\cdot180^3= -5 \cdot 2^6 \cdot 180^2$, so $\mathbb{Q}(\sqrt{D}) = \mathbb{Q}(\sqrt{-5})$. Note the minus sign! And $f$ is indeed irreducible over $\mathbb{Q}(\sqrt{-5})$, so the Galois group is $D_8$ according to your link. Jun 20 '20 at 5:58