Splitting field of x^5-2 over rational field and its primitive element

\mathbf{Problem:} Consturct a splitting field over \mathbb{Q} of x^5-2. Find its dimensionality over \mathbb{Q}.
\mathbf{Proof:} Let \omega is the primitive 5th root of unity in \mathbb{C}, whose minimal polynomial is x^4+x^3+x^2+x+1.
\displaystyle x^5-2 has roots \sqrt[5]{2}, \sqrt[5]{2}\omega, \sqrt[5]{2}\omega^2,\sqrt[5]{2}\omega^3,\sqrt[5]{2}\omega^4 in \mathbb{C}. Then a splitting field of x^5-2 is \mathbb{Q}(\sqrt[5]{2},\omega). And [\mathbb{Q}(\sqrt[5]{2},\omega):\mathbb{Q}]=20.
To see this, \displaystyle [\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5, because x^5-2 is irreducible by Eisenstein criterion. Furthermore f(x)=x^4+x^3+x^2+x+1 is irreducible in \mathbb{Q}(\sqrt[5]{2}). Suppose not, since \omega\not \in \mathbb{Q}(\sqrt[5]{2}), f must split into two quadratic polynomial, namely f(x)=(x^2-(\omega+\omega^4)x+1)(x^2-(\omega^2+\omega^3)x+1). So \displaystyle \omega+\omega^4\in \mathbb{Q}(\sqrt[5]{2}). While \omega+\omega^4 satisfies x^2-5x+2=0, which has roots \displaystyle\frac{5+\sqrt{17}}{2} and \displaystyle\frac{5-\sqrt{17}}{2}. So \sqrt{17}\in \mathbb{Q}(\sqrt[5]{2}). This is impossible because [\mathbb{Q}(\sqrt{17}):\mathbb{Q}]=2 does not divide 5.
We can consider this problem in another way. [\mathbb{Q}(\omega):\mathbb{Q}]=5, x^5-2 is irreducible in \mathbb{Q}(\omega). Otherwise \mathbb{Q}(\sqrt[5]{2}) can be imbedded in \mathbb{Q}(\omega). This is impossible because [\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5>4=[\mathbb{Q}(\omega):\mathbb{Q}].
\text{Q.E.D.}\hfill\square
\mathbf{Problem:} Find a primitive element of a splitting field of x^5-2 over \mathbb{Q}.
\mathbf{Proof:} The Galois group of x^5-2 over \mathbb{Q} are permutations of roots. Every permutation \eta is uniquely determined by \eta(\sqrt{2}) and \eta(\omega). Let \alpha=\sqrt{2}+\omega, the only \eta fixes \alpha is \eta=id. So \mathbb{Q}(\alpha) is the splitting field of x^5-2 and \alpha is a primitive element.

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Comments

  • MathStudent  On May 4, 2014 at 8:45 pm

    I don’t understand why you can conclude that $Q(\sqrt[5]{2})$ can be embedded in $Q(\omega)$ when $x^5 – 2$ is not irreducible over $Q(\omega)$. I know that if one root is in $Q(\omega)$ then so are the rest, but how do you show that $x^5 – 2$ cannot be factored into an irreducible cubic and an irreducible quadratic over $Q(\omega)$? Is there a standard trick that one uses in such circumstances?

    • Sun's World  On May 5, 2014 at 11:56 am

      I think all the roots of $x^5-2$ are $\sqrt[5]{2}w^i$, $i=0,1,2,3,4$. The multiplication of any two of roots are not in $Q(\omega)$. So you can not factorize to irreducible quadratic and cubic

  • MathStudent  On May 5, 2014 at 2:51 pm

    Okay, that makes sense. Thank you.

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