Newton's criterion in Q_p

Put $h_n:=f(a_n)\mathrm{/}{f}^{\mathrm{\prime }}(a_n)$ whenever $a_n$ is defined. Since $\mathrm{\mid }{h}_0{\mathrm{\mid }}_p<\mathrm{\mid }{f}^{\mathrm{\prime }}(a_0){\mathrm{\mid }}_p\le 1$, we have $h_0\in p{\mathbb{Z}}_p$, so $a_1=a_0-h_0$ lies in ${\mathbb{Z}}_p$. For any $x\in {\mathbb{Z}}_p$, Taylor expansion gives $f(x-h)=f(x)-h{f}^{\mathrm{\prime }}(x)+h^2{g}_x(h)$ with $g_x(h)\in {\mathbb{Z}}_p$. Hence $f(a_{n+1})=h_n^2{g}_{a_n}(h_n),$ so $\mathrm{\mid }f(a_{n+1}){\mathrm{\mid }}_p\le \mathrm{\mid }{h}_n{\mathrm{\mid }}_p^2$.

The derivative also satisfies $f^{\mathrm{\prime }}(a_{n+1})=f^{\mathrm{\prime }}(a_n)+h_n{u}_n$ for some $u_n\in {\mathbb{Z}}_p$. Because $\mathrm{\mid }{h}_n{\mathrm{\mid }}_p<\mathrm{\mid }{f}^{\mathrm{\prime }}(a_n){\mathrm{\mid }}_p$, the ultrametric inequality gives $\mathrm{\mid }{f}^{\mathrm{\prime }}(a_{n+1}){\mathrm{\mid }}_p=\mathrm{\mid }{f}^{\mathrm{\prime }}(a_n){\mathrm{\mid }}_p\mathrm{.}$ By induction, every iterate is defined, each $f^{\mathrm{\prime }}(a_n)$ has the same nonzero absolute value, and $\mathrm{\mid }{a}_{n+1}-a_n{\mathrm{\mid }}_p=\mathrm{\mid }{h}_n{\mathrm{\mid }}_p\le \frac{\mathrm{\mid }f(a_n){\mathrm{\mid }}_p}{\mathrm{\mid }{f}^{\mathrm{\prime }}(a_0){\mathrm{\mid }}_p}\le {\left(\frac{\mathrm{\mid }f(a_0){\mathrm{\mid }}_p}{\mathrm{\mid }{f}^{\mathrm{\prime }}(a_0){\mathrm{\mid }}_p^2}\right)}^{2^n-1}\mathrm{\mid }{h}_0{\mathrm{\mid }}_p\mathrm{.}$ So the differences tend to $0$ quadratically.

The series of successive differences is therefore Cauchy, so $(a_n)$ converges in ${\mathbb{Z}}_p$ by [L2]; call its limit $a$. Continuity of polynomial evaluation gives $f(a)=0$, and the ultrametric inequality applied to $a-a_0={\sum }_{n\ge 0}(a_{n+1}-a_n)$ shows $\mathrm{\mid }a-a_0{\mathrm{\mid }}_p\le \mathrm{\mid }{h}_0{\mathrm{\mid }}_p=\mathrm{\mid }f(a_0)\mathrm{/}{f}^{\mathrm{\prime }}(a_0){\mathrm{\mid }}_p$.

If $b$ is another root in the closed ball of radius $\mathrm{\mid }{h}_0{\mathrm{\mid }}_p$ about $a_0$, then $0=f(b)-f(a)=(b-a)v$ for the usual divided-difference element $v\in {\mathbb{Z}}_p$. Because $a-a_0$ and $b-a_0$ both have absolute value at most $\mathrm{\mid }{h}_0{\mathrm{\mid }}_p$, each term of $v-f^{\mathrm{\prime }}(a_0)$ contains one factor from $(a-a_0){\mathbb{Z}}_p$ or $(b-a_0){\mathbb{Z}}_p$. Thus $\mathrm{\mid }v-f^{\mathrm{\prime }}(a_0){\mathrm{\mid }}_p\le \mathrm{\mid }{h}_0{\mathrm{\mid }}_p<\mathrm{\mid }{f}^{\mathrm{\prime }}(a_0){\mathrm{\mid }}_p\mathrm{.}$ The ultrametric inequality therefore gives $\mathrm{\mid }v{\mathrm{\mid }}_p=\mathrm{\mid }{f}^{\mathrm{\prime }}(a_0){\mathrm{\mid }}_p\mathrm{\ne }0$, so $v$ is nonzero and hence $b-a=0$. Therefore $b=a$.