Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.