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The question asks the following:

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.

Bernard
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Bihu Duo
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2 Answers2

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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 $D_8$.

Rushabh Mehta
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  • 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.

     – Bihu Duo May 24 '20 at 21:50
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    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. – Derek Holt May 24 '20 at 21:54
  • @DerekHolt Oops I think you are right. – Bihu Duo May 24 '20 at 22:03
  • @DerekHolt Yes, my bad, I'll edit – Rushabh Mehta 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. – Viktor Vaughn Jun 20 '20 at 05:58
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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 $\operatorname{Gal}(E/\mathbf{Q})\cong D_8$.

rae306
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