TITLE: $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}$?
QUESTION [9 upvotes]: Is there an easy way to see that
$$\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}?$$
I know that $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})$ is a subfield of $\mathbb{Q}(\sqrt[3]{5})$ which is a field of degree 3. Since only 1 and 3 divides 3, $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})$ must be either $\mathbb{Q}$ or $\mathbb{Q}(\sqrt[3]{5})$. Hence, if $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})\neq\mathbb{Q}$ then $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}(\sqrt[3]{5})$, so $\mathbb{Q}(\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt[3]{2})$. But $\mathbb{Q}(\sqrt[3]{2})$ is spanned by $\{1,2^{1/3},2^{2/3}\}$, so
$$5^{1/3}=a+b2^{1/3}+c2^{2/3},$$
for some $a,b,c\in\mathbb{Q}$. Cubing both sides would - I think - lead to a contradiction, but this is a very tedious computation. Is there a simpler approach?

REPLY [6 votes]: It’s probably a more advanced method than you want, but $p$-adic theory rides to the rescue here. If the two fields $\mathbb Q(2^{1/3})$ and $\mathbb Q(5^{1/3})$ were the same, then we’d have $\mathbb Q_2(2^{1/3})=\mathbb Q_2(5^{1/3})$. But $\mathbb Q_2$ already has a cube root of $5$ in it, as one easily sees from Hensel (factoring $X^3-5\equiv X^3-1\pmod2$ ) or from the Binomial expansion of $(1+X)^{1/3}$, which has only $3$’s in its coefficients’ denominators. Thus $\mathbb Q_2(5^{1/3})=\mathbb Q_2$. But of course $2^{1/3}\notin\mathbb Q_2$, in fact $2$ is irreducible in the integers $\mathbb Z_2$ of $\mathbb Q_2$.