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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?

Martin Sleziak
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Jared
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    Do you know what the roots of $x^3-2$ are? – Zev Chonoles May 23 '13 at 21:32
  • $\sqrt{2}$? I feel like $i$ must be involved, but I don't clearly see how. – Jared May 23 '13 at 21:38
  • No, it's not the square root of $2$... – Zev Chonoles May 23 '13 at 21:39
  • The roots are $$\sqrt[3]2;,;\sqrt[3]2,w;,;\sqrt[3]2,w^2;,;;w:=e^{\frac{2\pi i}3}$$ – DonAntonio May 23 '13 at 21:40
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    *$\omega$${}{}$ – Git Gud May 23 '13 at 21:43
  • Hint: the splitting field is not $\mathbb{Q}(\sqrt[3]{2})$, because this is contained in $\mathbb{R}$, and $\sqrt[3]{2} \omega$ isn't – Cocopuffs May 23 '13 at 21:51
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    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
  • I'm sorry. Can you explain to me where $ \omega$ comes from? I can't see where roots of unity happen in this polynomial. – Jared May 23 '13 at 22:04
  • Since $\sqrt[3]{2}$ is a root, we can factor out $x-\sqrt[3]{2}$: $x^3 - 2 = (x-\sqrt[3]{2})(x^2+\sqrt[3]{2}x+\sqrt[3]{2}^2)$ – Islands May 23 '13 at 22:14
  • And then use the quadratic equation to find the complex roots in the second polynomial? – Jared May 23 '13 at 22:22
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    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
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    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
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    Not entirely unrelated question: http://math.stackexchange.com/questions/294250/dimension-of-a-splitting-field-of-a-cubic-polynomial-over-mathbbq? – Gerry Myerson May 24 '13 at 12:56

1 Answers1

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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:

3.1.7

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$(\alpha, F)$.

3.1.8

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 $F$.

Kortlek
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