$\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}$?

Question

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?

Answer 1

Here's a different method, it's not exactly nice, but I thought I'd mention it because it generalises easily to any prime root, whereas the direct argument doesn't.

Suppose not. Then, as you mention, it must be that $K=\mathbb{Q}(\sqrt[3]{2})=\mathbb{Q}(\sqrt[3]{5})$. Then we have the three embeddings of $K$ into $\mathbb{C}$, $\sigma_i$ for $i = 1,2,3$ where $\sigma_i(\sqrt[3]{2})=\sqrt[3]{2}\zeta^i$ for $\zeta=e^{2\pi i/3}$. Then $\sigma_1(\sqrt[3]{5})=\sqrt[3]{5}\zeta^j$ for some $j=1$ or $2$.

But also $K=\mathbb{Q}(\sqrt[3]{10})=\mathbb{Q}(\sqrt[3]{20})$. But depending on the value of $j$, we either have $\sigma_1(\sqrt[3]{10})=\sqrt[3]{10}$ or $\sigma_1(\sqrt[3]{20})=\sqrt[3]{20}$, contradicting the fact that $\sigma_1$ isn't the identity.

Answer 2

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 $\mathbb Q_2$.