 Solution

Use the simple-root form of Hensel lemma. Let $R$ be a complete discrete valuation ring, let $\pi$ be a uniformizer, and let $F\in R[T]$. If $\bar{a}\in R/(\pi )$ is a root of $\bar{F}$ with ${\bar{F}}^{\prime }(\bar{a})\ne 0$, then there is exactly one root $a\in R$ reducing to $\bar{a}$.

Choose a lift $a_1$ of $\bar{a}$. Suppose $F(a_r)\in {\pi }^{r}R$. To construct a lift satisfying the next congruence, write $a_{r+1}=a_r+{\pi }^r{t}_r$. The polynomial expansion givesSince $F^{\prime }(a_r)$ is a unit, its residue is invertible in the residue field. There is a unique residue class of $t_r$ making this zero modulo ${\pi }^{r+1}$. Thus $a_{r+1}\equiv a_r(\mathrm{mod}{\pi }^r)$ and $F(a_{r+1})\equiv 0(\mathrm{mod}{\pi }^{r+1})$. This produces a Cauchy sequence in the complete discrete valuation ring. Its limit $a$ satisfies $F(a)=0$ by continuity of polynomial evaluation, and $a\equiv a_1(\mathrm{mod}\pi )$.

For uniqueness, suppose $a,b$ are two such roots. Polynomial division of their difference givesBecause $b-a\in \pi R$, the second factor is a unit. Hence $b-a=0$. The simple-root hypothesis is used both for constructing successive lifts and for this uniqueness argument.