Two questions:

- Does there exist a sequence $\alpha_1,\alpha_2,...$ of algebraic numbers with degrees $d_1,d_2,...$ s.t. for each $i$, $d_i|d_{i+1}$ and $\alpha_i= p_i(\alpha_{i+1})$ with $p_i$ a polynomial of degree $d_{i+1}/d_i$ with rational coefficients and $\bar{\mathbb{Q}}= \cup_i\mathbb{Q}[\alpha_i]$?

(For an example of $\alpha_i$'s satisfying all of these conditions except $\bar{\mathbb{Q}}= \cup_i\mathbb{Q}[\alpha_i]$, one can consider $2^{1/2},2^{1/4},2^{1/8},...$ and in this case $d_i = 2^i, p_i(x)=x^2$.)

- If the answer to the previous question is yes, then for irreducible polynomials $h_i=$ (minimal polynomial$/\mathbb{Q}$ of $\alpha_i$) we have $h_{i+1}(x) =h_i(p_i(x))$ and for every polynomial $f\in \mathbb{Q}[x]$ there is $i\in \mathbb N$ and a polynomial $g\in\mathbb{Q}[x]$ s.t. $h_i(x) | f(g(x))$. (it's because $f$ has a root that can be expressed as a polynomial $g$ of some $\alpha_i$.) Does there exist an explicit construction for a sequence $h_1,h_2,...$ satisfying these conditions?

Thanks!

Imissing something? $\endgroup$appropriate degreein $\alpha_{i+1}$. As for the example, there are only three proper nontrivial subfields of $\mathbb{Q}(\sqrt{2},\sqrt{3})$, namely $\mathbb{Q}(\sqrt{n})$, $n=2,3,6$. But $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ is none of them; it is trivially not equal when $n=2,3$; and no number $a+b(\sqrt{2}+\sqrt{3})$, $a,b\in\mathbb{Q}$, has a square equal to $6$. $\endgroup$1more comment