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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 ?

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  • $\begingroup$ 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. $\endgroup$ – Alexandros Mar 9 at 15:11
  • $\begingroup$ @Alexandros I'm looking for the cubic, I thought that you could use that formula based off Wikipedia. what method must be used instead ? $\endgroup$ – Voltron Mar 9 at 15:16
  • $\begingroup$ 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 $\frac{-b\pm\sqrt[3]{b^3-9ac}}{3a}$. $\endgroup$ – Saucy O'Path Mar 9 at 15:17
  • $\begingroup$ I would use Vieta's substitution: mathworld.wolfram.com/VietasSubstitution.html $\endgroup$ – Dr. Mathva Mar 9 at 15:17
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    $\begingroup$ 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? $\endgroup$ – Alexandros Mar 9 at 15:22
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As the polynomial is of degree 3 and having no roots in field of rationals so the polynomial is irreducible over rationals, so degree of splitting field can not be less than 3 over rationals.To find the roots of the cubic , we can use cardan method and so the splitting field.

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  • $\begingroup$ card on? Did you mean Cardano (or maybe Cardan)? Stupid spellcheckers. $\endgroup$ – Peter Shor Mar 10 at 11:10
  • $\begingroup$ Yes , Cardan Method to solve cubic equations. $\endgroup$ – Priyanka Mar 10 at 12:12

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