Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Do you know what the roots of $x^3-2$ are? – Zev Chonoles May 23 '13 at 21:32
*\$\omega\$${}{}$ – Git Gud May 23 '13 at 21:43
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. – André Nicolas May 23 '13 at 21:52
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? – Islands May 23 '13 at 22:25
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. – Jared May 23 '13 at 22:28
up vote 12 down vote accepted

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 $F$.

share|cite|improve this answer
What would be the Galois Group of K? – Yunus Syed May 8 at 18:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.