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I know the Galois group of $x^3-x-1$ over $\mathbb{Q}$ is $S_3$. But to find the Galois group over $\mathbb{Q}(i\sqrt{23})$ we need to find a splitting field. To this end, the only idea I know of is writing the polynomial in the form $(x^2-a)(x-b)$. But we don't have $x^2$ in the polynomial. So, I don't know how split the polynomial. I cannot even guess the roots. Let's assume we know how to split it. Then I hope it has a factor like $(x^2-\sqrt{23})$ so that I can consider it over $\mathbb{Q}(\sqrt[4]{23},i)$ or something similar as a subfield. Then find the two degrees of extension ( I mean the deg of extesion of a splitting field over $\mathbb{Q}(\sqrt[4]{23},i)$ and that of the latter over $\mathbb{Q}(i\sqrt{23})$) and finally use them to find the order of the Galois group. I don't know if my approach is right or there are better tricks...

Tim
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    Look at the discriminant? – Eric Apr 27 '24 at 20:21
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    Yeah, seconding @Eric's comment, unless you have some reason to believe (or hope) that a cubic is somehow special (related to cyclotomic fields?), and especially if it has no $x^2$ term, just check whether its disciminant is a square in the ground field... Not always a trivial matter... and, often, answerable specific versions of this question depend upon some algebraic number theory... – paul garrett Apr 27 '24 at 20:26
  • @paulgarrett The discriminant is 31, so it is $S_3$ over $\mathbb{Q}(i\sqrt{23})$ as well? – Tim Apr 27 '24 at 20:30
  • @Eric I see! Since it's separable, as the discriminant is negative, and irreducible, don't know how to show this by the way, the Galois group is $S_3$, as the disc is not a square, right? – Tim Apr 27 '24 at 20:51
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    @Tim, well, the salient question is about whether the disc $31$ (assuming you've computed it correctly, etc, ... I'm not gonna try :) ) is a square in the field $\mathbb Q(\sqrt{-23})$. Luckily, the norm form on this field is positive-definite, so you can figure out a way to check only finitely-many things. (I seem to recall that the ring of algebraic integers in that field is not a PID, so be careful...) – paul garrett Apr 27 '24 at 21:14
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    "Secretly", since $31$ does not ramify in the quadratic extension $\mathbb k(\sqrt{31})$ of $k=\mathbb Q(\sqrt{-23})$ (proof? well, it's a "standard consequence"...), even the ideal cannot be a square... but such lines of reasoning depend on what context you want... – paul garrett Apr 27 '24 at 21:17

2 Answers2

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The discriminant mentioned in the comments is incorrect, it should be $-4\cdot (-1)^3 - 27\cdot (-1)^2 = -23$ which is a square in $\mathbb{Q}(\sqrt{-23})$. The polynomial is irreducible since the discriminant is nonzero and also separable since it is over a perfect field, and therefore the Galois group is $A_3$ over $\mathbb{Q}(\sqrt{-23})$.

Victor Lu
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We can obtain the form of the transformations in the Galois group by writing the homographic relation between two roots of a cubic like here. In our case we get:

if $u$ is a root of $x^3-x-1$, then the other two roots are Möbius transforms of $u$ by the matrices

$$\left(\begin{matrix} -\frac{1}{2} \pm \frac{9}{ \sqrt{-23}} & \pm\frac{1}{\sqrt{-23}}\\ \mp\frac{3}{\sqrt{-23}} & -\frac{1}{2} \mp\frac{9}{2 \sqrt{-23}} \end{matrix}\right)$$

One can check that the above matrices are of order $3$ and invariate the cubic form $x^3-x y^2 - y^3$.

orangeskid
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