Why isn't the the galois group of a polynomial with n distinct roots isomorphic to Sn?

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).

• What do you mean by $E$? – Servaes Nov 29 '18 at 15:24
• $E$ is the splitting field of $f$ over $\mathbb{Q}$. Thanks for the comment, I edited. – roi_saumon Nov 29 '18 at 15:41
• 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. – Jyrki Lahtonen Nov 30 '18 at 4:10
• @JyrkiLahtonen Oh, I see, but how did you find this hidden relation? – roi_saumon Nov 30 '18 at 11:09
• 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). – 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 $$\delta:=\prod_{1\leq i Note that for all $$\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 $$E$$.