if $u=3^{1/5}$ then $\mathbb Q(u)=\mathbb Q(u²)$

Question

if $u=3^{1/5}$ then $\mathbb Q(u)=\mathbb Q(u²)$

As $u=3^{1/5}$ then elements of $\mathbb Q(u)$ are of the type $a+b\,3^{1/5}$ and of $\mathbb Q(u²)$ are $a+b\,9^{1/5}$

Then bases for $\mathbb Q(u)$ are $\{ 1, 3^{1/5} \} $ and for $\mathbb Q(u²)$ are $\{ 1, 9^{1/5}\}$

How they can be equal then ?

Answer 1

... As $u=3^{1/5}$ then elements of $\mathbb Q(u)$ are of the type $a+b\,3^{1/5}$ and of $\mathbb Q(u^2)$ are $a+b\,9^{1/5}$ ...

This statement is incorrect. The elements of $\mathbb Q(u)$ are actually of the type $a+b\,3^{1/5}+c\,3^{2/5}+d\,3^{3/5}+e\,3^{4/5}$; those of $\mathbb Q(u^2)$ are of type $a+b\,9^{1/5}+c\,9^{2/5}+d\,9^{3/5}+e\,9^{4/5}$. Now we note that $9^{1/5} = \,3^{2/5}$, $9^{2/5} = \,3^{4/5}$, $9^{3/5} = \,3^{6/5} = 3 \times \,3^{1/5}$, and $9^{4/5} = \,3^{8/5} = 3 \times \,3^{3/5}$, and we easily see that these two forms both denote the same set of values.

Answer 2

Yes, they are! Since $\Bbb Q(u)$ is a field, then $u^2$ belongs to $\Bbb Q(u)$, hence all elements of the form $a+bu^2$ belong to $\Bbb Q(u)$; hence $\Bbb Q(u^2)\subseteq \Bbb Q(u)$.

For the other inclusion, observe that $(u^2)^3\in \Bbb Q(u^2)$, because $\Bbb Q(u^2)$ is a field. But $(u^2)^3=u^6=3u$, hence $\Bbb Q(u^2)$ contains all elements of the form $a+bu$, hence $\Bbb Q(u)\subseteq \Bbb Q(u^2)$.

Then $\Bbb Q(u)=\Bbb Q(u^2)$.