# Generalized Hensel lifting in a Dedekind domain

The following theorem is known as generalized Hensel lifting(see here). Can we prove this without using $P$-adic completion?

Theorem Let $A$ be a Dedekind domain. Let $P$ be a non-zero prime ideal of $A$. Let $f(x) \in A[x]$ be a polynomial. Let $a \in A$. Suppose $P^r||f'(a)$. Let $m = 2r + 1$. Let $k \geq m$ be an integer. Suppose $f(a) \equiv 0$ (mod $P^k$). Then there exists $b \in A$ such that $b \equiv a$ (mod $P^{k-r}$), $P^r||f'(b)$ and $f(b) \equiv 0$ (mod $P^{k + 1}$).

This can be an immediate corollary of the above theorem.

Motivation This theorem can be used in class field theory. I'm interested in proving CFT without $p$-adic numbers(see here).

-
There is no need to use completions. Even your link to the Wikipedia page does not use completions. Since the rings $A/P^n$ are unchanged by localization, in your problem you can replace $A$ with its localization $A_P$, where the ideal $PA_P$ becomes principal. Let $\pi$ be a generator of this ideal (e.g., pick any element of $A - P$ as $\pi$). Then you want to find $b = a + \pi^{k-r}c$ with $c$ to be determined such that $f'(b) \equiv 0 \bmod \pi^r$ and $f(b) \equiv 0 \bmod \pi^{k+1}$. Use a first-term Taylor expansion for $f(x)$. – KCd Jul 21 '12 at 11:22
Dear KCd, Thanks! "Even your link to the Wikipedia page does not use completions." I meant the "Generalizations" section of the Wikipedia page, where they assume that $A$ is complete with respect to an ideal $P$. Regards. – Makoto Kato Jul 21 '12 at 11:35
That section of the Wikipedia page is not using completeness to get a result like you want (an approximate root mod $P^k$ leading to an approximate root mod $P^{k+1}$). Completeness is needed there to be sure the approximation process converges to an actual root in the ring $A$. To define the sequence of approximate roots doesn't need any completeness, just like Newton's method for solving real equations does not need completeness in order to define the Newton recursion; completeness becomes essential when you want to show the recursion converges. – KCd Jul 23 '12 at 10:10