18
$\begingroup$

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!

$\endgroup$
6
  • 3
    $\begingroup$ 1 follows from the primitive element theorem, or am I missing something? $\endgroup$ Sep 14 '15 at 18:27
  • 4
    $\begingroup$ @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? $\endgroup$ Sep 14 '15 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 $\mathbb Q(\sqrt2,\sqrt3)$. $\endgroup$ Sep 14 '15 at 18:38
  • 1
    $\begingroup$ @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 $6$. $\endgroup$ Sep 14 '15 at 18:45
  • 1
    $\begingroup$ @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
    Sep 14 '15 at 18:49
-1
$\begingroup$

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 $[L:\mathbb{Q}]$.

$\endgroup$
1
  • 1
    $\begingroup$ 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
    Sep 15 '15 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.