Determine the splitting field $K$ over $\mathbb{Q}$ of the polynomial $x^3 - 2$

Also determine the basis over $\mathbb{Q}$ and its degree. Can I do this using only first principles?

  • 2
    $\begingroup$ Do you know what the roots of $x^3-2$ are? $\endgroup$ May 23, 2013 at 21:32
  • 1
    $\begingroup$ *\$\omega\$${}{}$ $\endgroup$
    – Git Gud
    May 23, 2013 at 21:43
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    $\begingroup$ Theory is faster than first principles. Degree is $\le 3!$. Adjoinin $\sqrt[3]{2}$ gets us degree $3$ extension, sine $x^3-2$ is irreducible over the rationals. Not enough to split, since there are non-real roots. So degree is $\le 6$, $\gt 3$, divisible by $3$. And by degree, adding $\sqrt[3]{2}$ and anything new, like another of the cube roots, gets us all. $\endgroup$ May 23, 2013 at 21:52
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    $\begingroup$ Well you've already been given the other roots: $\sqrt[3]{2}w, \sqrt[3]{2}w^2$. Why don't you plug them in there and convince yourself? $\endgroup$
    – Islands
    May 23, 2013 at 22:25
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    $\begingroup$ I can convince myself just fine. Calculators also work very well... but I'm more interested in methodology to solve other problems similar to this. Thank you for your help, the quadratic formula worked to solve for the other roots. $\endgroup$
    – Jared
    May 23, 2013 at 22:28

1 Answer 1


The roots of $x^3-2=0$ are $x_1=\sqrt[3]{2},$ $x_2= \sqrt[3]{2}\omega$, $x_3=\sqrt[3]{2}\omega^2$, where $\omega=\frac{-1+i\sqrt{3}}{2}$. It follows that the splitting field $K$ is $$K=\mathbb{Q}(x_1,x_2,x_3)=\mathbb{Q}(\sqrt[3]{2},i\sqrt{3})$$

Since $1, \sqrt[3]{2}, \sqrt[3]{2}^2$ form a basis for $\mathbb{Q}(\sqrt[3]{2})$ over $\mathbb{Q}$ and $1, i\sqrt{3}$ form a basis for $\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})$ over $\mathbb{Q}(\sqrt[3]{2})$, a basis for $\mathbb{Q}(\sqrt[3]{2},i\sqrt{3})$ over $\mathbb{Q}$ must be $1, \sqrt[3]{2}, \sqrt[3]{2}^2,i\sqrt{3}, \sqrt[3]{2}i\sqrt{3},\sqrt[3]{2}^2i\sqrt{3} $.

Edit: I have used the following facts, taken from Robert Ashs excellent book Abstract Algebra: The Basic Graduate Year:


Let $E/F$ be a field extension and suppose $\alpha\in E$ is algebraic over $F$. Then $$1, \alpha, \alpha^2, ..., \alpha^{n-1}$$ form a basis of $F(\alpha)$ over $F$, where $n=$deg Irr$(\alpha, F)$.


If $E/K/F$ are field extensions such that $\{\alpha_i\}_{i\in I}$ is a basis for $E$ over $K$ and $\{\beta_j\}_{j\in J}$ is a basis for $K$ over $F$, then $\{\alpha_i\beta_j\}_{i\in I, j\in J}$ form a basis for $E$ over $F$.

  • $\begingroup$ What would be the Galois Group of K? $\endgroup$
    – Yunus Syed
    May 8, 2016 at 18:54
  • $\begingroup$ @YunusSyed It would be $S_3$, see this post for example. $\endgroup$ May 10, 2019 at 11:35

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