How to generate an algebraically independent set over rational number field?

Question

Algebraic independence. In abstract algebra, a subset ${\displaystyle S}$ of a field ${\displaystyle }$L is algebraically independent over a subfield ${\displaystyle }$K if the elements of ${\displaystyle }$S do not satisfy any non-trivial polynomial equation with coefficients in ${\displaystyle }$K

Over $\mathbb{Q}$, for any $n\in \mathbb{N}$, is there an algorithm to generate an algebraically independent set whose cardinality is $n$?

Answer 1

One such algorithm would be

Lindemann-Weierstrass Theorem: If $\alpha_1,\ldots,\alpha_n$ are algebraic numbers which are linearly independent over $\mathbb{Q}$, then $e^{\alpha_1},\ldots,e^{\alpha_n}$ are algebraically independent over $\mathbb{Q}$.

which reduces the problem to finding lineraly independent algebraic numbers. And this can be done for example by taking $\{1, \sqrt[m]{2}^2, \sqrt[m]{2}^3,\ldots,\sqrt[m]{2}^{m-1}\}$ which you can verify here.