# Galois group of $x^3+x+1$ over $\mathbb{Q}$ is isomorphic to $S_3$

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

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

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!

• 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
• For the first part, see 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 $6$ ? – NewMath 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