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Let E be splitting field of $x^3+x+1 \in \mathbb{Q}[X]$ over $\mathbb{Q}$.

Proof that Gal{E/$\mathbb{Q}$}$\cong S_3$. Specify all extension fields L with $\mathbb{Q} \subset L \subset E$ and the degree [L : $\mathbb{Q}$] of the extensions.

Set $f(x):= x^3+x+1 \in \mathbb{Q}[X]$. It's easy to see that f is irreducible over $\mathbb{Q}$.

At first, I wanted to determine E. With Cardano formula I got the complex roots

$$x_2 = \sqrt[3]{- \frac{1}{2} + \sqrt{\frac{31}{108}}} + i \sqrt[3]{\frac{1}{2} + \sqrt{\frac{31}{108}}}\quad\text{and}\quad x_3 = \sqrt[3]{- \frac{1}{2} + \sqrt{\frac{31}{108}}} - i \sqrt[3]{\frac{1}{2} + \sqrt{\frac{31}{108}}}$$

Since f has degree 3, the roots $x_2, x_3$ must be simple. So it exists another real root $x_1$ of f. Therefore we can write $f(x)= (x-x_1)(x-x_2)(x-x_3)$ in $E[x] \Rightarrow f$ is separable and E is splitting field of $f \Rightarrow E/\mathbb{Q}$ is galois. We got that: $[E:K] = |Gal(E/K)| \Leftrightarrow E/K$ is galois. So $|Gal(E/ \mathbb{Q})| = [E:\mathbb{Q}]$.

Now i don't know how to go on. Can i somehow get the degree of E/$\mathbb{Q}$ by means of $x_1, x_2, x_3$? And how can i easily calculate $x_1$ or isn't that necessary to solve the problem?

Thanks a lot!

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Zorro_C
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    See this duplicate and the article of Keith Conrad here, where he determines the Galois group for all cubic (and quartic) polynomials. Then it is clear how to "go on". – Dietrich Burde Dec 19 '18 at 11:47
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    For the first part, see https://math.stackexchange.com/a/2809524/589 – lhf Dec 19 '18 at 11:49
  • Is it possible that the Galois group of a polynomial of degree 3 has cardinality $6$ ? – user621345 Dec 19 '18 at 11:56
  • @NewMath Yes, $S_3$ has $6$ elements. Please click on the link given by lhf (and read his answer). – Dietrich Burde Dec 19 '18 at 12:00
  • Vieta's formulas make it particularly easy to find the $n$th zero of a degree $n$ polynomial if you already know $n-1$ of them. – Jyrki Lahtonen Dec 19 '18 at 12:24
  • I think the first thing to see is that if $K$ is the splitting field of $f$ of degree $n$ and $[K:Q] = n!$ then $Gal(K/Q) = S_n$. If $f = x^3+x+1$ is irreducible modulo $2$ then it is irreducible over $Q$ and $3$ divides $[K:Q]$. And if $f$ has a complex root then $2$ divides $[K:Q]$. – reuns Dec 19 '18 at 12:31
  • @reuns Can you explain why [K:Q] = n! and 2 divides [K:Q] when complex roots exists? – Zorro_C Dec 19 '18 at 15:30
  • Think to $K$ as a subfield of $\mathbb{C}$. If it is not contained in $\mathbb{R}$ then you know a subgroup of $Gal(K/Q)$ of order $2$ : the one generated by the complex conjugaison (exchanging the two complex roots of $f$) – reuns Dec 19 '18 at 15:49
  • @reuns Ok thanks!! – Zorro_C Dec 20 '18 at 12:39
  • Hint: -31 is not a square in $\Bbb Q$ – Brozovic Sep 17 '20 at 15:07

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