= 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 $\bar F'(\bar a)\ne0$, then \b[there is exactly one root $a\in R$ reducing to $\bar a$.]
Choose a lift $a_1$ of $\bar a$. Suppose $F(a_r)\in\pi^rR$. To construct a lift satisfying the next congruence, write $a_{r+1}=a_r+\pi^r t_r$. The <polynomial> expansion gives
$$
F(a_r+\pi^r t_r)\equiv F(a_r)+\pi^r t_rF'(a_r)\pmod{\pi^{r+1}}.
$$
Since $F'(a_r)$ 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 $a_{r+1}\equiv a_r\pmod{\pi^r}$ and $F(a_{r+1})\equiv0\pmod{\pi^{r+1}}$. This produces a <Cauchy sequence> in the <complete discrete valuation ring>. Its limit $a$ satisfies $F(a)=0$ by <continuity> of <polynomial> evaluation, and $a\equiv a_1\pmod\pi$.
For uniqueness, suppose $a,b$ are two such roots. Polynomial division of their difference gives
$$
F(b)-F(a)=(b-a)\bigl(F'(a)+(b-a)H(a,b)\bigr),\qquad H\in R[S,T].
$$
Because $b-a\in\pi R$, 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.
Back to article page
