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Two questions:

  1. 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 $d_i = 2^i, p_i(x)=x^2$.)

  1. 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!

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    $\begingroup$ 1 follows from the primitive element theorem, or am I missing something? $\endgroup$
    – Emil Jeřábek
    Commented Sep 14, 2015 at 18:27
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    $\begingroup$ @EmilJeřábek You can certainly get a suitable sequence by taking, say, any enumeration $\beta_1,\beta_2,\ldots$ of algebraic numbers, and then letting $\alpha_{i}$ be a generator of $\mathbb{Q}(\beta_1,\ldots,\beta_{i})$. But if you just do that, you don't necessarily get the second condition; e.g., if your enumeration starts with $\sqrt{2}$, $\sqrt{3},\ldots$, then your $\alpha_2$ could be $\sqrt{2}+\sqrt{3}$, and $\sqrt{2}$ is not the value of a rational polynomial of degree $2$ evaluated at $\sqrt{2}+\sqrt{3}$. So there's certainly work to be done beyond that. Or am I missing something? $\endgroup$
    – Arturo Magidin
    Commented Sep 14, 2015 at 18:35
  • $\begingroup$ @ArturoMagidin If $\alpha_i$ is a generator of $\mathbb Q(\beta_1,\dots,\beta_i)$ over $\mathbb Q$ for each $i$, then $\alpha_i\in\mathbb Q[\alpha_{i+1}]$, pretty much by definition. In your example, $\sqrt2+\sqrt3$ doesn’t generate $\mathbb Q(\sqrt2,\sqrt3)$. $\endgroup$
    – Emil Jeřábek
    Commented Sep 14, 2015 at 18:38
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    $\begingroup$ @EmilJeřábek: $\alpha_i\in\mathbb{Q}[\alpha_{i+1}]$ means you can express $\alpha_i$ as a rational polynomial in $\alpha_{i+1}$, but I don't see why it must be expressible as a rational polynomial of the appropriate degree in $\alpha_{i+1}$. 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 $6$. $\endgroup$
    – Arturo Magidin
    Commented Sep 14, 2015 at 18:45
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    $\begingroup$ @EmilJeřábek The degree condition $deg(p_i)=d_{i+1}/d_i$ is equivalent to writing $min(\alpha_{i+1})$ as a composition of $min(\alpha_i)$ with another polynomial. $\endgroup$
    – Mostafa - Free Palestine
    Commented Sep 14, 2015 at 18:49

1 Answer 1

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I think in question 2, what you're asking is akin to the following simpler question: if $K = \mathbb{Q}(\alpha)$ is a number field and $L/K$ is an extension of number fields such that $L = \mathbb{Q}(\beta)$, can we choose $\beta$ so that the minimal polynomial of $\beta$ over $K$ is actually defined over $\mathbb{Q}$? I think this is what the $K = \mathbb{Q}(\sqrt{2})$, $\beta = \sqrt{2} + \sqrt{3}$ example in comments is getting at.

In that case, the answer is clearly no. The reason is that $m_{\beta, K}$, the minimal polynomial of $\beta$ over $K$, has degree $[L:K]$. If the coefficients all lie in $\mathbb{Q}$, then $\mathbb{Q}(\beta)$ is actually an extension of degree $[L:K]$ and not of degree $[L:\mathbb{Q}]$.

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    $\begingroup$ The required condition $\alpha_i = p_i(\alpha_{i+1})$ (or its consequence $h_{i+1}=h_i\circ p_i$ ) gives a rational polynomial of $\alpha_{i+1}$ with value in $\mathbb{Q}[\alpha_i]$ not 0. For an example you can consider $\alpha_1 =\sqrt{2}$ and $\alpha_2 = 2^{1/4}$ so $\alpha_2^2 = \alpha_1$. $\endgroup$
    – Mostafa - Free Palestine
    Commented Sep 15, 2015 at 7:40

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