## 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.

• 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.

• perringu  On February 8, 2018 at 4:22 am

and sure