Finitely many $\alpha$ s.t. all conjugates $\leq N$ ($N \geq 1$)

Question

Let $K$ be a number field, and let $N \geq 1$. Then there are only finitely many $\alpha \in O_K$ such that all conjugates of $\alpha$ have complex absolute value $\leq N$.

The solution goes as follows:

Let $s_1, \cdots , s_n$ be the elementary symmetric polynomials in $n$ variables. Let $σ_1,\cdots , σ_n$ be the complex embeddings and denote $α_i = σ_i(α)$ for $i = 1, \cdots , n$. If $|α_i| ≤ N$ for all $i$, then $|s_k(α_1, \cdots , α_n)| ≤ 2^nN^n$ for any $1 ≤ k ≤ n$. In particular, there are only finitely many integral polynomials $x_n − s_1x^{n−1} + \cdots + (−1)^ns_n$ satisfying this bound. It implies that there are only finitely many $α$ with conjugates of bounded complex absolute value.

My bound would be $(nN)^n$, arguing as follows :

Arguing for $1 \leq k \leq n$, each $s_k$:

• sums at $ \leq n^n$ different terms.
• each summed term is the product of at most $n$ Terms $\leq N$ and thus each such term is $\leq N^n$.

Combining the two using absolute values everywhere gives $|s_k| \leq (nN)^n$.

How do attain the $2$?

Answer 1

Thanks to Anne Bauval:

The $2^n$ stems from the fact that: $$ \binom{n}{k} \leq \sum_{i=0}^n \binom{n}{i} = 2^n$$

Since $s_k$ consists of $\binom{n}{k}$ different summands, each summand being the product of at most $n$-terms $\leq N$, we obtain: \begin{align*} |s_k| \leq \binom{n}{k} N^n \leq \sum_{i=0}^n \binom{n}{i} N^n \leq 2^nN^n \end{align*}