# Upper bound on number of algebraic integers of degree $\leq d$

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.

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 $$\prod_{i=1}^d \binom di \leq \left(\frac{2^d-2}{d-1}\right)^{d-1}.$$