# $\alpha \in \Omega_{\mathbb Q}^{x^3-2}$(splitting field) is such that $\alpha^5 \in \mathbb Q$ then $\alpha \in \mathbb Q$

Let $f(x)=x^3-2$ and let $E$ be it's splitting field. Prove that if $\alpha \in E$ is such that $\alpha^5 \in \mathbb Q$ then $\alpha \in \mathbb Q$. Let $\zeta_n$ denote a n-root of unity.

This is what I did. I think that there exist a simpler way.

Note $E=\mathbb Q(\sqrt[3]{2},\zeta_3)=\mathbb Q(\sqrt[3]{2},\sqrt{-3})$ it's an extension of degree $6$ over $\mathbb Q$. By towers law the degree $[\mathbb Q(\alpha):\mathbb Q]$ divides 6. And clearly $[\mathbb Q(\alpha):\mathbb Q]\le 5$ thus $[\mathbb Q(\alpha):\mathbb Q]=1,2,3$. Let $r=\alpha^5 \in \mathbb Q$ and consider $p(x)=x^5-r \in \mathbb Q[x]$. The factors of $f$ over it's splitting field are $(x-\zeta_5^i \sqrt[5]{r})$ (for some $i=0,1..4)$.

Assume that the minimal polynomial over $f$ has degree $2$ then:

$$(x-\zeta_5^i \sqrt[5]{r})(x-\zeta_5^j \sqrt[5]{r})= x^2-x\sqrt[5]{r}(\zeta_5^i+\zeta_5^j)+\sqrt[5]{r^2}\zeta_5^{i+j}\in \mathbb Q[x]$$

But in this case $\zeta_5^i, \zeta_5^j$ are conjugate thus $\zeta_5^i \cdot \zeta_5^j = 1$

Therefore:

$$x^2-2\sqrt[5]{r}Re(\zeta_5^i)x+\sqrt[5]{r^2} \in \mathbb Q[x]$$

We conclude:

• $\sqrt[5]{r^2}\in \mathbb Q$
• $2\sqrt[5]{r}Re(\zeta_5^i) \in \mathbb Q$

Then $Re(\zeta_5^i)\in \mathbb Q$ and this is a contradiction.

if the minimal polynomial has degree $3$ the argument it's similar.

• Once you know for a rational $r$ that $r^{2/5}$ is in $\mathbb{Q}$ you also know $r^{1/5} \in \mathbb{Q}$. Sep 10, 2014 at 5:10

If $a$ is a rational number such that the polynomial $X^5 - a$ is reducible over $\mathbb{Q}$ then $a$ is the fifth power of a rational number. Now, we have the following decomposition over $\mathbb{C}$ $$X^5 - a= \prod_{\eta^5 = 1} ( X- \eta \cdot a^{1/5})$$ Assume that $X^5 - a$ is reducible over $\mathbb{Q}$. Then the product of some $k<5$ of the factors on the right will be a polynomial with coefficients in $\mathbb{Q}$. In particular, the free term in that product will be in $\mathbb{Q}$. Taking absolute values we obtain $$(a^{1/5})^k \in \mathbb{Q}$$ and hence $a \in \mathbb{Q}$. Indeed, look at the expansion of $a$ in a product of prime numbers with some integral (positive or negative) exponents. $(a^{k/5})= (a^k)^{1/5}$ is in $\mathbb{Q}$ means that all the exponents for $a^k$ are divisible by $5$, that is, all exponents of $a$, multiplied by $k$, are divisible by $5$. Since $1\le k < 5$, it follows that all the exponents of $a$ are divisible by $5$.
This argument works for any $p$ prime instead of $5$.
We are done now: The degree of the extension $\mathbb{Q}(\alpha)/ \mathbb{Q}$ is $1$ or $5$ and since it divides $6 = \deg E/ \mathbb{Q}$ it must be $1$, hence $\alpha \in E$.
• Clearly $(\sqrt[5]{a})^k\in \mathbb Q$ but why $a\in \mathbb Q?$ Sep 10, 2014 at 5:17
• What do you mean by a product of primes over $\mathbb Q$? Sep 10, 2014 at 7:09
• for example $\frac{15}{28} = 2^{-2}3^1 5^1 7^{-1}$ Sep 10, 2014 at 8:23