14
$\begingroup$

$\mathbf{1.\space Proposition}$

Let $\gamma$ be a solution to the equation:

$$ \sum_{i=0}^n \rm a_i\rm x^i=0, \rm a_i\in\Bbb Z, a_n=1. $$

Then, there exists a polynomial $p\in \Bbb Z[x]$ such that: $$ p=\sum _{i=0}^m\rm s_ix^i $$

Where $\rm s_i=(-1)^{k_i}\tau_i^2 \space\space\forall \rm i\in \{0,..,m\}, k_i\in \Bbb Z, \tau_i\in \Bbb Z$, for which $\gamma$ is a root.

$\mathbf{1.1\space Generalization}$.

There exist a monic polynomial satisfying the above conditions.

$\mathbf{1.2\space Generalization}$

Same conditions as $\rm1$, but $\rm s_i=(-1)^{k_i}\tau_i^r$, for all natural $\rm r$.

$\mathbf{1.3\space Generalization}$

There exists a monic polynomial satisfying $\rm1.2$.

$\mathbf{Important \space implication \space of \space 1.1 }$

If $\rm 1.1$ is true, it means that the condition of $\gamma$ being the root of a monic polynomial $p\in \Bbb Z[x]$ is actually equivalent to the 'stronger' condition of being a solution to a $\pm$quadratic coefficiented polynomial $q\in \Bbb Z[x]$.

$\endgroup$
  • 3
    $\begingroup$ Can you find such an equation for $\sqrt3$ ? $\endgroup$ – lhf Sep 30 '15 at 16:55
  • 2
    $\begingroup$ $(x^4-9)$ works for $\sqrt{3}$. $\endgroup$ – Aravind Sep 30 '15 at 17:50
  • 2
    $\begingroup$ You can always do it for $\sqrt[n]{m}$ since the polynomial $x^{2n}-m^2=(x^n-m)(x^n+m)$ will work...but for more complicated things, I have no idea. $\endgroup$ – Ben Sheller Sep 30 '15 at 19:23
  • 2
    $\begingroup$ Could you write $1+\sqrt5$ in this manner ? $\endgroup$ – Lucian Sep 30 '15 at 19:34
  • 2
    $\begingroup$ @BenS.: Then the answer to the OP's question is yes. Try $x^4-4x^3+4x^2-9$. $\endgroup$ – Lucian Sep 30 '15 at 20:48
1
$\begingroup$

Proposition: Let $\gamma$ be an algebraic integer, $r$ an odd positive integer. Then there exists a non-zero polynomial $f\in\mathbb{Z}[x]$ with $r$-power coefficients such that $f(\gamma)=0$.

This is a consequence of Birch's theorem on odd-degree forms over number fields in many variables. We will need the following special case of Birch's theorem.

Theorem (Birch): Let $k$ and $r$ be postive integers, with $r$ odd. Then there exists a positive integer $n$ such that, if $f_1,\ldots,f_k$ are homogeneous polynomials with integer coefficients in $n$ variables, there exists $\vec{x}=(x_1,\ldots,x_n)\neq (0,\ldots,0)\in\mathbb{Z}^n$ such that $f_1(\vec{x})=\ldots=f_k(\vec{x})=0$.

Proof of proposition: We are given $\gamma$ and odd $r$. Let $k=\deg(\gamma)$, and let $n$ be as in the conclusion of Birch's Theorem. Choose an integral basis $\alpha_1,\ldots,\alpha_k$ of $\mathbb{Z}[\gamma]$. For each $m\geq 0$, there are unique integers $a_{m,1},\ldots,a_{m,k}$ such that $\gamma^m=a_{m,1}\alpha_1+\ldots+a_{m,k}\alpha_k$. For $i=1,\ldots,k$, define $$ f_i(x_1,\ldots,x_n)=a_{1,i}x_1^r+\ldots+a_{n,i}x_n^r. $$ The $f_i$ are $k$ homogeneous polynomial of degree $r$ in $n$ variables, so by Birch's theorem, they have a common non-zero solution $(x_1,\ldots,x_n)$. Now $$ x_1^r\gamma+x_2^r\gamma^2+\ldots+x_n^r\gamma^n=0, $$ as desired.

$\endgroup$

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.