# If $\beta^n + \alpha_1 \beta^{n-1} + \dots + \alpha_n = 0$, $|\alpha_i | \leq 1$, then $|\beta| \leq 1$, with $|\cdot|$ ultrametric absolute value.

Let $$|\cdot|$$ be an ultrametric absolute value (i.e it is a function from a field $$\mathcal{K}\rightarrow [0,\infty)$$ that satisfies that $$|\alpha \beta| = |\alpha||\beta|$$ and $$|\alpha + \beta| \leq \max \{ |\alpha|, |\beta|\}$$, and $$|\alpha| = 0$$ if and only if $$\alpha = 0$$).

I want to show that if $$\beta$$ is a root of the polynomial $$X^n + \alpha_1 X^{n-1} + \cdots + \alpha_n = 0$$ where $$|\alpha_i| \leq 1$$ for all $$i$$ ,then $$|\beta| \leq 1$$.

I've tried several things with the ultrametric inequality like showing that $$|\beta||\beta^{n-1}+ \alpha_1 \beta^{n-2}+ \cdots +\alpha_{n-1}| = |\alpha_n| \leq 1,$$ and I've tried to bound the factor $$|\beta^{n-1}+ \alpha_1 \beta^{n-2} + \cdots +\alpha_{n-1}|$$ using what we know about the $$\alpha_i$$ and the ultrametric inequality, but I'm just going in circles. Any hints or solutions would be greatly appreciated.

## 1 Answer

Since $$-\beta^n = \alpha_1 \beta^{n-1} + \dots + \alpha_n$$, we get $$|\beta|^n = |\alpha_1 \beta^{n-1} + \dots + \alpha_n|$$. Suppose $$|\beta| > 1$$. Then, $$|\beta|^n = |\alpha_1 \beta^{n-1} + \dots + \alpha_n| \leq \max \{ |\alpha_1||\beta|^{n-1},\dots,|\alpha_n| \} \leq \max \{ |\beta|^{n-1},\dots,1 \} = |\beta|^{n-1}.$$ Hence, $$|\beta|^n \leq |\beta|^{n-1} \implies |\beta| \leq 1$$, a contradiction.

Thus, $$|\beta| \leq 1$$.