2

Let $\alpha \in \mathbb{C}$ be an algebraic integer, which means that it has a monic polynomial $f=X^d + a_1X^{d-1} + \dots + a_d\in \mathbb{Z}[X]$ such that $f(\alpha)=0$. Over $\bar{\mathbb{Q}}$ this factorizes as $f=\prod_{i=1}^d (X-\beta_i)$, and we call $\beta_i$ the conjugates of $\alpha$.

Now the question is, for fixed $C,d\in \mathbb{Z}_{\geq 1}$, how many algebraic integers $\alpha \in \mathbb{C}$ are there given that $\max\{|\beta_i| : 1\leq i \leq \deg(\alpha)\}\leq C$ and $\deg(\alpha)\leq d$? My idea was that it was the number of irreducible monic polynomials of degree at most $d$ over $\mathbb{Z}[X]$, but I'm not sure.

MarvinsSister
  • 126

1 Answers1

4

A (possibly trivial) bound can be obtained by realizing that given $$ f(X) = X^d + a_1X^{d-1} + \dots + a_d = \prod_{i = 1}^d(X - \beta_i)$$ with $|\beta_i| \leq C$ we can find upper bounds for $|a_i|$. Indeed, by comparing coefficients and alluding to Vieta's formulas, we find $$|a_i| \leq \binom d i C^i.$$

Now, the number of degree $d$ polynomials with $i$-th coefficient in $\Bbb Z$ bounded by $\binom di C^i$ is trivially less than $$ \prod_{i=1}^d \left(2 \binom di C^i + 1 \right) \leq 3^d C^{\frac{d(d+1)}2} \prod_{i=1}^d \binom di. $$

As we count the number of zeroes of polynomials, not the number of polynomials, the number of algebraic integers of degree $\leq d$ and absolute value $\leq C$ is at most $$ d 3^d C^{\frac{d(d+1)}2} \prod_{i=1}^d \binom di.$$

The last product can be bounded more explicitely. Using This stackexchange post, we have the bound $$ \prod_{i=1}^d \binom di \leq \left(\frac{2^d-2}{d-1}\right)^{d-1}.$$

LurchiDerLurch
  • 625