# Did I find the splitting field of $x^3-3x+1$ over $\Bbb Q$?

I want to find the splitting field of $$x^3-3x+1$$ over $$\Bbb Q$$. I think I've got it but I'm not sure …

Here's what I did :

Using the formula for cubic roots I said that

$$x=\dfrac{-b\pm\sqrt{b^2-3ac}}{3a}=\dfrac{3\pm\sqrt6}{3}=\dfrac {1\pm\sqrt{2}}{\sqrt3}.$$

The field $$\Bbb Q(\sqrt2/\sqrt{3})$$ contains both roots so the splitting field is a subfield of $$\Bbb Q(\sqrt2/\sqrt{3})$$. The polynomial $$x^2-\frac23$$ is irreducible over $$\Bbb Q$$. So $$|\Bbb Q(\sqrt{2/3}):\Bbb Q|=2.$$

So $$\Bbb Q (\sqrt2/\sqrt{3})$$ is the splitting field of $$x^3-3x+1$$ with degree $$2$$.

Is this correct ?

• are you looking for the splitting field of $x^2-3x+1$ over the rationals? or $x^3-3x+1$?. Your method works for the quadratic, but is clearly wrong for the cubic. – Alexandros Mar 9 at 15:11
• @Alexandros I'm looking for the cubic, I thought that you could use that formula based off Wikipedia. what method must be used instead ? – Voltron Mar 9 at 15:16
• I think $x=\frac{-b\pm\sqrt{b^2-3ac}}{3a}$ is just something you made up by substituting some $3$ where in another formula $2$ and $4$ appear. Which, on a side note, would raise the question of why not $\frac{-b\pm\sqrt[3]{b^3-9ac}}{3a}$. – Saucy O'Path Mar 9 at 15:17
• I would use Vieta's substitution: mathworld.wolfram.com/VietasSubstitution.html – Dr. Mathva Mar 9 at 15:17
• That is the formula for finding the critical points of the cubic.The fact that your splitting field is a degree 2 extension means that your answer must be wrong, as you'd expect a degree 3 extension. Maybe try a substitution of the form $x=z+\frac{1}{z}$, and see what you get. one of the roots is 2cos(2π/9). Any idea what the conjugates of that are? – Alexandros Mar 9 at 15:22