Showing $5^{1/3}$ is not in the field $\mathbb{Q}(2^{1/3})$

Question

I want to show $5^{1/3}$ is not in the field $\mathbb{Q}(2^{1/3})$. My reasoning was $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=3$ and $[\mathbb{Q}(5^{1/3}):\mathbb{Q}]=3$. Similarly we know that $[\mathbb{Q}(2^{1/3}, 5^{1/3}):\mathbb{Q}]=9$ so by multiplicativity of dimensions, $[\mathbb{Q}(5^{1/3}):\mathbb{Q(2^{1/3})}]=3$ which implies that $5^{1/3} \not\in \mathbb{Q}(2^{1/3})$. Is this the right approach?

Answer 1

What you wrote works as an outline, but I think your claims about the degree of each field extension would need explanation.

---

For $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]$ and $[\mathbb{Q}(5^{1/3}):\mathbb{Q}]$ you need to show that the irreducible polynomials of $2^{1/3}$ and $5^{1/3}$ over $\Bbb Q$ have degree 3.

(Since you know that $2^{1/3}$ is a root of $x^3-2 \in \Bbb Q[x]$, you just need to explain why $x^3-2$ is irreducible over $\Bbb Q$. You can use Eisenstein's criterion. Same for $5^{1/3}$.)

---

The more difficult part is explaining why $[\mathbb{Q}(2^{1/3}, 5^{1/3}):\mathbb{Q}]=9$.

I think I'd write the extensions in a different way and then use the multiplicative property of the degree:

$\Bbb Q \subseteq \Bbb Q(2^{1/3}) \subseteq [\Bbb Q(2^{1/3})](5^{1/3})=\mathbb{Q}(2^{1/3}, 5^{1/3})$, so the problem becomes showing that $5^{1/3}$ has degree $3$ over $\Bbb Q(2^{1/3})$.

$[\Bbb Q(2^{1/3}):\Bbb Q]=3$, so $\Bbb Q(2^{1/3})$ has a basis of three elements: $1, 2^{1/3}$ and $2^{2/3}$.

You can use this to show that $5^{1/3}$ is not an element of $\Bbb Q(2^{1/3})$: you could try writing it as a linear combination of the basis elements with coefficients in $\Bbb Q$ and deriving a contradiction. (This might be long and tedious.)

Answer 2

In general, if $p$ and $q$ are two different primes and $k>1$, $p^{1/k}\not\in\mathbb{Q}(q^{1/k})$.

Assuming the opposite, for any sufficiently big prime number $r\equiv 1\pmod{k}$, with $q$ being a $k$-th residue $\pmod{r}$, $p$ would be a $k$-th residue, too, but we may provide counter-examples to that situation through the Chinese theorem.