Characterize elements in $\mathbb Q(\sqrt2,\sqrt[3]5)$

Question

I'm studying algebra by Herstein's Topics in Algebra, 2ed, and stuck at Ex.5.1.6(b):

In $\mathbb Q(\sqrt2,\sqrt[3]5)$ characterize all the elements $w$ such that $\mathbb Q(w)\ne \mathbb Q(\sqrt2,\sqrt[3]5)$.

My approach is that since $\mathbb Q(\sqrt2,\sqrt[3]5)=6$, $\mathbb Q(w)\ne \mathbb Q(\sqrt2,\sqrt[3]5)$ if and only if $w$ is not of degree 6, hence $\mathbb Q(w)\ne \mathbb Q(\sqrt2,\sqrt[3]5)$ if and only if $w$ is of degree 1,2, or 3.

I get a hint from the answer to question Finding all reals such that two field extensions are equal. that $w$ is of degree 2 (resp. 3) if and only if it is in form of $a+b\sqrt2$ (resp. $a+b\sqrt[3]5+c(\sqrt[3]5)^2$). It is easy to show the if part is valid, but how to prove that $w$ is of degree 2 (resp. 3) only if it is in form of $a+b\sqrt2$ (resp. $a+b\sqrt[3]5+c(\sqrt[3]5)^2$)? Thanks.

Answer 1

Your hypothesis is correct. Let $w=\sqrt{2}+a\sqrt[3]{5}+b\sqrt[3]{25}$ and assume that $w$ has degree $3$, then we have $\sqrt{2} \in \mathbb{Q}(\sqrt[3]{5}, w)$ and $$[ \mathbb{Q}(\sqrt[3]{5}, w): \mathbb{Q}]=[ \mathbb{Q}(\sqrt[3]{5}, w): \mathbb{Q}(w)][ \mathbb{Q}(w): \mathbb{Q}]=3 \ \text{or} \ 9$$ but this is impossible since $[ \mathbb{Q}(\sqrt[3]{5}, w): \mathbb{Q}]=[ \mathbb{Q}(\sqrt[3]{5}, w): \mathbb{Q}(\sqrt{2})][ \mathbb{Q}(\sqrt{2}): \mathbb{Q}]$ is even.

The other case can be solved similarly.