If we consider the polynomial $f(x)=x^3-3x-1 \in \mathbb{Q}[x]$ which has 3 real roots $\{x_1,x_2,x_3\}$. I read that its galois group is isomorphic to $A_3$ and not to $S_3$. I don't really understand this. If I consider the map (Here $E$ is the splitting field of $f$ over $\mathbb{Q})$:
\begin{align*} \sigma \colon E \to E\\ x_1 \mapsto x_2\\ x_3 \mapsto x_3\\ \mathbb{Q} \mapsto \mathbb{Q} \end{align*}

This would correspond to a transposition $(12)$ which is not in the alternating group.
But isn't this a $\mathbb{Q}$-automorphism? I actually don't really grasp why the Galois group isn't in general isomorphic to Sn (for n distinct roots).

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    $\begingroup$ What do you mean by $E$? $\endgroup$ – Servaes Nov 29 '18 at 15:24
  • $\begingroup$ $E$ is the splitting field of $f$ over $\mathbb{Q}$. Thanks for the comment, I edited. $\endgroup$ – roi_saumon Nov 29 '18 at 15:41
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    $\begingroup$ There may be hidden relations among the roots. Your cubic is a case in point. If $r=x_1$ is one of the zeros, then the other zeros are $x_2=2-r^2$ and $x_3=r^2-r-2$. So the splitting field is $\Bbb{Q}(r)$. If you know $\sigma(x_1)=\sigma(r)$ then $$\sigma(x_2)=\sigma(2-r^2)=2-\sigma(r)^2$$ is uniquely determined. In other words, you cannot choose $\sigma(x_1)$ and $\sigma(x_2)$ independently from each other in the case of this polynomial. Looking at the discriminant as in Servaes' answer is a more common constraint, I simply wanted to shed more light to this. $\endgroup$ – Jyrki Lahtonen Nov 30 '18 at 4:10
  • $\begingroup$ @JyrkiLahtonen Oh, I see, but how did you find this hidden relation? $\endgroup$ – roi_saumon Nov 30 '18 at 11:09
  • $\begingroup$ The polynomial is kinda "famous". The zeros are $x_1=-2\cos(2\pi/9)$, $x_2=-2\cos(4\pi/9)$ and $x_3=-2\cos(8\pi/9)$. The implication $r$ is a root $\implies$ $2-r^2$ is a root, is simple the double angle cosine formula $\cos2\alpha=2\cos^2\alpha-1$ in disguise. More generally, adjoining a single zero $r$ may allow us to factor the polynomial more than just splitting off a factor $x-r$ of $f(x)$, and such factorizations reduce the extension degree of the splitting field (and hence also the number of automorphisms). $\endgroup$ – Jyrki Lahtonen Nov 30 '18 at 11:36

TL;DR The element $(x_1-x_2)(x_1-x_3)(x_2-x_3)$ is contained in $\Bbb{Q}$ but not fixed by $\sigma$.

The following theorem tells us when the Galois group of a polynomial over $\Bbb{Q}$ is contained in $A_n$:

Let $f\in\Bbb{Q}[x]$ be irreducible of degree $n$, and let $E$ be a splitting field of $f$. Identify $\operatorname{Gal}(E/\Bbb{Q})$ with a subgroup of $S_n$ by enumerating the roots of $f$ in $E$. If $\Delta(f)\in\Bbb{Q}$ is a square in $\Bbb{Q}$ then $\operatorname{Gal}(E/\Bbb{Q})$ is contained in $A_n$.

The proof shows in particular why your map is not a field automorphism of $E$:

Let $x_1,\ldots,x_n\in E$ be the roots of $f$ and let $S_n$ act on $E$ by its action on the indices of the roots, i.e. $\sigma(x_i):=x_{\sigma(i)}$ for all $i$. Consider the element $$\delta:=\prod_{1\leq i<j\leq n}(x_i-x_j)\in E.$$ Note that for all $\sigma\in S_n$ we have $\sigma(\delta)=\operatorname{sgn}(\sigma)\delta$ and that $\delta^2=\Delta(f)$.

If $\Delta(f)$ is a square in $\Bbb{Q}$ then it $\delta\in\Bbb{Q}$. It follows that $\sigma(\delta)=\delta$ for all $\sigma\in\operatorname{Gal}(E\Bbb{Q})$ and hence that $\operatorname{sgn}(\sigma)=1$ for all $\sigma\in\operatorname{Gal}(E\Bbb{Q})$. This means precisely that $\operatorname{Gal}(E\Bbb{Q})$ is contained in $A_n$.

In this particular case we see that $\Delta(x^3-3x-1)=81$, so the element $$\delta=(x_1-x_2)(x_1-x_3)(x_2-x_3)=\pm9,$$ is not fixed by your map $\sigma$, so $\sigma$ cannot be a field automorphism of $E$.


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