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\documentclass{book}

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
%\usepackage{head}

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\begin{document}

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\setcounter{secnumdepth}{1}

\newtheorem{thm}{Theorem}[section]
\newtheorem{prop}[thm]{Proposition}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{defi}[thm]{Definition}
\newtheorem{ex}[thm]{Example}
\newtheorem{rem}[thm]{Remark}

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%\newtheorem{alem}[athm]{Lemma}
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%\newtheorem{adefi}[athm]{Definition}
%\newtheorem{aex}[athm]{Example} %\newtheorem{arem}[athm]{Remark}

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\pagestyle{plain}

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\newcommand{\nequiv}{\equiv\!\!\!\!\!\!/\;}

\begin{lem}
\label{L:I.2.1}
Let $\;\cdots \morph{\varphi_2} X_2 \morph{\varphi_1} X_1$
be a projective system of finite nonempty sets.

Then $\prlim X \neq \emptyset$.
\end{lem}
\begin{proof}
Suppose that, $\forall i$, $\varphi_i$ is surjective.
Let $x_1 \in X_1$, and $\forall i$, let $x_{i+1} \in \varphi_i^{-1} x_i$.
Then $\prlim X \neq \emptyset$.
\medskip

Consider $\cdots \subseteq \phi_2 X_{n+2} \subseteq \phi_1 X_{n+1} \subseteq X_n$,
where $\phi_i = \varphi_{n+i-1} \cdots \varphi_n$.  This chain must terminate,
since the sets are finite, in a nonempty set $Y_n \subseteq X_n$.
\medskip

Let $\eta_n = \varphi_n |_{Y_{n+1}}$.
Then $\cdots \morph{\eta_2} Y_2 \morph{\eta_1} Y_1$ is a projective system, and
$\forall n$, $\eta_n$ is surjective.
Hence, $\prlim X \supseteq \prlim Y \neq \emptyset$.
\end{proof}
\bigskip

\begin{prop}
\label{P:I.2.1}
Let ${f_\alpha} \subset \ZZ_p[X_1, \ldots, X_n]$.  Then the following are equivalent:
\newline\indent(a) The $f_\alpha$'s have a common zero in $(\ZZ_p)^n$.
\newline\indent(b) $\forall \nu \geq 1$, the $f_\alpha$'s have a common zero in %%@
$(\ZZ/p^\nu)^n$.
\end{prop}
\begin{proof}
If $\exists (x_1, \ldots, x_n) \in (\ZZ_p)^n$ such that
$f_\alpha(x_1, \ldots, x_n) = 0, \forall \alpha$, then
$f_\alpha(\overline{x}_1, \ldots, \overline{x}_n) = 0$ in $(\ZZ/p^\nu)^n$.
\medskip

Conversely, $\forall \nu$, let $X_\nu = \{(x_1, \ldots, x_n) \in (\ZZ/p^\nu)^n :
f_\alpha(x_1, \ldots, x_n) = 0, \forall \alpha\}$.
\medskip

$X_\nu$ is nonempty and finite.  Furthermore,
$\cdots \morph{\eta_2} X_2 \morph{\eta_1} X_1$, where
$\eta_\nu = (\varphi_\nu, \ldots, \varphi_\nu)$ and
$\varphi_\nu: [z]_{\nu+1} \longmapsto [z]_{\nu}$, forms a projective system.
Hence, by Lemma \ref{L:I.2.1}, $\prlim X \neq \emptyset$, and these are the
solutions in $\ZZ_p$.
\end{proof}
\bigskip

\begin{thm}
\label{T:I.2.1} (Hensel's Lemma)
Let $f \in \ZZ_p[X_1, \ldots, X_m]$, $x = (x_1, \ldots, x_m) \in (\ZZ_p)^m$, and
$n, k \in \ZZ$ such that $0 \leq 2k < n$.

If $f(x) \equiv 0$ mod $p^n$ and $v_p(\frac{\partial f}{\partial X_j}(x)) = k$ for some
$j$, then $\exists y \in (\ZZ_p)^m$ with $y \equiv x$ mod $p^{n-k}$ such that $f(y) = %%@
0$.
\end{thm}

Hence, when $n = 1$ and $k = 0$, i.e. $f(x) \equiv 0$ mod $p$ and $f'(x) \nequiv 0$ mod $p$,
the approximate solution $x$ can be lifted to an exact solution $y$ with
$y \equiv x$ mod $p$.

\begin{ex}
2 is a solution of $f(X) =  X^2 + 1$ in $\ZZ/5$ and $f'(2) = 4 \nequiv 0$ mod 5,
so $f$ has a 5-adic solution $y$ with $y \equiv 2$ mod 5.
\end{ex}

\begin{proof}
It is sufficient to prove the result for $m = 1$, since a solution $y \equiv x_j$ mod $p$
for $g(X) = f(x_1, \ldots, x_{j-1}, X, x_{j+1}, \ldots, x_n)$ yields a solution
$(x_1, \ldots, x_{j-1}, y, x_{j+1}, \ldots, x_n) \equiv x$ mod $p$ for $f$.
\end{proof}

\end{document}