Irreducible cubic polynomial in $\mathbb{Q}[x]$ has no roots in $\mathbb{Q}(\sqrt{2}, 5^{1/4})$

Question

I am working on the following problem:

Consider the field extension $F = \mathbb{Q}(\sqrt{2}, 5^{1/4})$ of $\mathbb{Q}$. Let $f(x)$ be an irreducible cubic polynomial in $\mathbb{Q}[x]$. Prove that $F$ contains no roots of $f(x)$.

I know that, since $f(x)$ an irreducible degree $3$ polynomial in $\mathbb{Q}[x]$, $f(x)$ has no roots in $\mathbb{Q}$. In the specified extension $F$ of $\mathbb{Q}$, we're only adjoining $\sqrt{2}$ and $5^{1/4}$. Thus, it's left to show that $f(x)$ can't have a root that is a power of $\sqrt{2}$ or a power of $5^{1/4}$.

I think I can see how to partially argue this. I believe $f(x)$ can't have a root that is an odd power of $5^{1/4}$, right off the bat, because the corresponding minimal polynomial for such an element would have to be degree $4$, already contradicting the degree of $f(x)$. Thus, the problem has now reduced down to showing that $f(x)$ can't have a root that is a power of $\sqrt{2}$ or a power of $\sqrt{5}$. But I'm not sure how to argue these remaining two points. Does it have to do with the minimal polynomial of such an element being of degree $2$, which would need to be a factor of $f(x)$ ? And this contradicts $f(x)$ being irreducible?

Any help would be appreciated. Thanks!

Answer 1

It's a simple argument in field theory.

If $F$ does contain a root $r$ of $f$, then we would have a field embedding $K = \mathbb Q[x]/f(x) \hookrightarrow F$, sending $x$ to $r$.

We then have a problem with the dimensions: $\dim_\mathbb Q F = \dim_\mathbb Q K \cdot \dim_K F$. It should be clear that $\dim_\mathbb Q F = 8$, but $\dim_\mathbb Q K = 3$, so that's not possible.