Finding a $p$-adic root of a polynomial using a convergent sequence of roots modulo $p^n$.

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I am studying $p$-adic numbers out of my own interest and I am attempting to solve the following exercise, the last part of which I have gotten stuck on. The concept I am not understanding I think is very important and powerful, especially looking ahead to what I next plan to study, and I want to make sure I fully grasp the technique before moving on...

Let $f(X) \in \mathbb{Z}[X]$ and let $p$ be a prime. Suppose $a_0$ is a root of $f(X)$ modulo $p$. Suppose that $f'(a_0)$ is not congruent to $0\mod p$. Define a sequence $\{a_n\}_{n=0}^\infty$ recursively by:

$$a_{n+1}= a_n - uf(a_n)$$

where $u \in \mathbb{Z}$ satisfies $uf'(a_0) \equiv 1 \mod p$

The first part of the exercise asks to show that the sequence is Cauchy (with respect to the $p$-adic norm; all convergence will be discussed with respect to the $p$-adic norm in this question) which I was able to do just fine. The second part asks to show that the sequence converges to a root of $f$ in $\mathbb{Q}_p$.

By repeated applications of Hansel's Lemma (unless I am incorrect), I have discovered each $a_i$ is a root of $f$ modulo $p^{i+1}$ which allows me to rewrite $a_n = a_0 -u(f(a_1) + ... + f(a_{n-1})) = c_0p - uc_1p^2 - ... -uc_{n-1}p^{n-1}$ for some $c_0, ... , c_{n-1} \in \mathbb{Z}$ which shows me the convergence of the sequence to some $p$-adic number.

However, I am not understanding why the limiting process produces a root of the polynomial in $\mathbb{Q}_p$. I do not see the relationship between the roots modulo $p^n$ in the sequence and the limit of the sequence as a root in $\mathbb{Q}_p$ (why must $f$ be zero at the limit?)...

Any hints would be greatly appreciated!

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You have to understand just what $\Bbb Q_p$ and its ring of (local) integers $\Bbb Z_p$ are. To give a number in $\Bbb Q_p$ is exactly to give a $p$-adically convergent sequence of rational numbers. Nothing more, and nothing less. So by your process, you have found an element, call it $\lambda$, of $\Bbb Q_p$. Now, you have shown that $\lambda=\lim a_n$. But a polynomial is a continuous function, so that $$ f(\lambda)=f(\lim a_n)=\lim(f(a_n))\,. $$ But since the sequence $f(a_n)$ has limit zero (again, $p$-adically), $f(\lambda)=0$.

Three comments:
First, you are not using Hensel repetitively, you are using it once, since Hensel gives the whole limiting process.
Second, you are just using a version of Newton-Raphson (which I hope you learned in Calculus), to find a root of a polynomial, since your process amounts to evaluating the polynomial $f$ at a trial value $z_n$ and corrcting $z_n$ to $z_{n+1}$ by subtracting $f(z_n)/f'(z_n)$.
Third, this is the “weak” version of Hensel. The “strong” version has to do with taking a mod-$p$ factorization of a polynomial and deducing that there is a corresponding factorization over $\Bbb Z_p$. This is much more powerful, and certainly does things that the commoner, weak, version is incapable of.