Find $n$ elements of $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$ linearly independent We know that $\mathbb{R}$ is a linear space over $\mathbb{Q}$, denoted by $\mathbb{R_{\mathbb{Q}}}$,where the vector addition is the real numbers and scalar multiplication is the number in $\mathbb{Q}$ times the number in $\mathbb{R}$.
Then $ \forall n\in\mathbb{N^{+}}$,$\{1,\sqrt[n]{2},\sqrt[n]{2^2},\cdots,\sqrt[n]{2^{n-1}}\}$ is a linearly independent set in $\mathbb{R_{\mathbb{Q}}}.$
Further,we can get $\mathbb{R_{\mathbb{Q}}}$ is  an infinite dimensional linear space.
Let $\mathbb{Q}(\sqrt[3]{2}):=\{a+b\sqrt[3]{2}+c\sqrt[3]{2^2} \Big|a,b,c\in \mathbb{Q}\}.$
Similarly，$\mathbb{R}$ is a linear space over $\mathbb{Q}(\sqrt[3]{2})$, denoted by $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$, where the vector addition is the real numbers and scalar multiplication is the number in $\mathbb{Q}(\sqrt[3]{2})$ times the number in $\mathbb{R}$. We can proof that $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$ also is an infinite dimensional linear space. In fact, If we suppose $\dim(\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}})\in \mathbb{N^{+}},$ then  $\dim(\mathbb{R_{\mathbb{Q}}})\in\mathbb{N^{+}}$ holds.  But forall $n\in\mathbb{N^{+}},$ how can I find $n$ elements of  $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$ such that these are  linearly independent in $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$?
 A: Here’s a way of looking at your problem that makes use of ramification theory. It may be too advanced for your taste, but at least it does provide an answer to your question.
You’re looking for many real quantities, preferably algebraic, that will be linearly independent over the real field $\Bbb Q\left(2^{1/3}\right)$. As an extension of $\Bbb Q$, this field is ramified over only $2$ and $3$. So if you get lots of real fields that are unramified over $2$ and $3$, you have a solution to your problem.
A quadratic extension $\Bbb Q(\sqrt n\,)$ of $\Bbb Q$ (with $n$ squarefree) is unramified above $2$ if $n\equiv1\pmod4$. And unramified above $3$ if $n$ is prime to $3$. So take positive values of $n$ satisfying these conditions, such as $5,13,17,41,77,\cdots$ As you see with $77$, they don’t need to be prime.
Anyway, adjoin the square roots of $m$ of these numbers, and you should get a field of dimension $2^m$ over $\Bbb Q\left(2^{1/3}\right)$.
A: The following may feel like cheating: pick your favorite transcendental element of ${\mathbb R}$, e.g., $\pi$. Then $1, \pi, \pi^2, \dots, \pi^{n-1}$ are linearly independent over the field over algebraic reals and in particular over ${\mathbb Q}(\sqrt[3]{2})$. 
If you want to stick to algebraic elements, look at $1, \sqrt[n]{2}, \dots, \sqrt[n]{2^{n-1}}$ (i.e., your own example) for $n$ that are not divisible by $3$. They form a basis of ${\mathbb Q}(\sqrt[n]{2})$ over ${\mathbb Q}$, but also of ${\mathbb Q}(\sqrt[n]{2},\sqrt[3]{2})$ over ${\mathbb Q}(\sqrt[3]{2})$. 
