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!

  • 3
    $\begingroup$ 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". $\endgroup$ – Dietrich Burde Dec 19 '18 at 11:47
  • 1
    $\begingroup$ For the first part, see math.stackexchange.com/a/2809524/589 $\endgroup$ – lhf Dec 19 '18 at 11:49
  • $\begingroup$ Is it possible that the Galois group of a polynomial of degree 3 has cardinality $6$ ? $\endgroup$ – NewMath Dec 19 '18 at 11:56
  • $\begingroup$ @NewMath Yes, $S_3$ has $6$ elements. Please click on the link given by lhf (and read his answer). $\endgroup$ – Dietrich Burde Dec 19 '18 at 12:00
  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Dec 19 '18 at 12:24

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