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\begin{document}

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%note on Jan.31 by Changheon Kang, kang_c@math.lsu.edu
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%
\begin{lem}
\label{L:I.2.2} Let $f(X) \in \Z_p[X]$, and $a \in \Z_p$. Then $$
\frac{f^{(n)}(a)}{n!} \in \Z_p$$
\end{lem}

\begin{proof}
By the linearity, it's enough to check for $f(X)=X^N$.
In this case, we have
$$
\frac{f^{(n)}(a)}{n!} = \frac{N(N-1) \cdots (N-n+1)}{n!} a^{N-n}
                      = \left(
                          \begin{array}{c}
                          N \\ n
                          \end{array}
                          \right)
                          a^{N-n} \in \Z_p
$$
since $\left(
       \begin{array}{c}
       N \\ n
       \end{array}
       \right)$
is in $\Z$ and $\Z \subset \Z_p$.
\end{proof}

\begin{pro}
\label{P:I.2.2} Let $f(X) \in \Z_p[X]$ and let $n,k \in \Z$ with
$0 \leq 2k < n$.

Suppose $x \in \Z_p$ is such that
\begin{eqnarray*}
f(x) & \equiv& 0 \pmod{ p^n}, \\
\mbox{and } \;\;v_p(f'(x)) &=&k.
\end{eqnarray*}

Then there exists $y \in \Z_p$ such that
\begin{eqnarray*}
y & \equiv & x \pmod{ p^{n-k}}, \\
f(y) & \equiv & 0 \pmod{ p^{n+1}},\\
\mbox{and } \;\;v_p(f'(y))&=&k.
\end{eqnarray*}
\end{pro}

\begin{proof}
Let $y=x+p^{n-k}z$ with $z \in \Z_p$.
The Taylor expansion for $f$ gives
\begin{eqnarray*}
f(y) &=& \sum_{n\geq 0} \frac{f^{(n)}(x)}{n!} (y-x)^n \\
     &=& f(x) + f'(x)p^{n-k}z+ap^{2(n-k)},  \;\mbox{ where }\; a \in \Z_p.
\end{eqnarray*}
By hypothesis
\begin{eqnarray*}
f(x) &=& p^nb, \;\;\;\; b \in \Z_p \\
\mbox{and }\;\;f'(x) &=& p^kc, \;\;\;\; c \in \Z_p^\times.
\end{eqnarray*}
Thus $$f(y)=p^n(b+cz)+ap^{2(n-k)}.$$
Since $c \in \Z_p^\times$ is a unit, we can find $z \in \Z_p$ so that
$b+cz \equiv 0 \pmod{ p},$
and hence $p^n(b+cz) \equiv 0 \pmod{ p^{n+1}} \;\mbox{ for any }\; n \geq 1.$
Since $2(n-k) \geq n+1$, we have $ap^{2(n-k)} \equiv 0 \pmod{ p^{n+1}}$
which gives $f(y) \equiv 0 \pmod{ p^{n+1}}$.

Now, differentiating the Taylor expansion for $f(x)$ gives
\begin{eqnarray*}
f'(y) &=& f'(x) +f''(x)p^{n-k}z+ a'p^{2(n-k)}, \;\mbox{ where }\; a' \in \Z_p
\\
      & \equiv & f'(x) \pmod{ p^{n-k}}.
\end{eqnarray*}
Since $n-k \geq k$, we have
$v_p(f'(y))=v_p(f'(x))=k.$
\end{proof}

\begin{thm}[Hensel's lemma]
\label{T:I.2.1} Let $f \in \Z_p[X_1,\ldots,X_m]$, and let $n, \; k \in \Z$
with $0 \leq 2k <n$.
Suppose $x=(x_1, \ldots,x_m) \in \Z_p^m$ is such that
$$f(x) \equiv 0 \pmod{ p^n} \;\mbox{ and }\; v_p\left(\frac{\partial
f}{\partial X_j}(x)\right)=k.$$ Then there exists $y \in \Z_p^m$
such that $$y \equiv x \pmod{p^{n-k}} \;\mbox{ and }\; f(y)=0.\;\;\;\;\;$$
\end{thm}

\begin{proof}

We can reduce to one-variable case: replace $f(X_1, \ldots,X_m)$
\newline by $f(x_1, \ldots,x_{j-1},X_j,x_{j+1}, \ldots,x_m)$.
By Proposition~\ref{P:I.2.2} we can construct a sequence $x_0,x_1,\ldots \in
\Z_p$ as follows:
\begin{eqnarray*}
 x_0 &=& x, \\
 \mbox{and }\;\; x_{q+1} &\equiv& x_q \pmod{p^{n+q-k}} \;\mbox{ for }\; q
\geq 0.
\end{eqnarray*}
Then $$f(x_q) \equiv 0 \pmod{p^{n+q}}.\mbox{\hspace{16 mm}}$$
Let $y = \lim_{q \rightarrow \infty} x_q \in \Z_p$ (note that $\{ x_q \}$ is
a Cauchy sequence).
Then $y \equiv x \pmod{p^{n-k}}$, and $f(y) = \lim_{q \rightarrow \infty}
f(x_q) = 0$ since $f$ is continuous.
\end{proof}

\begin{exa} For an odd prime $p$,
$$\sqrt{-1} \in \Q_p \; \Longleftrightarrow \; p \equiv 1 \pmod{4}.$$
\end{exa}

$( \Rightarrow )$ Suppose that $\sqrt{-1} \in \Q_p$, then $\sqrt{-1} \in\Z_p^\times$ since $x^2=-1 \Rightarrow 2v_p(x)=0$.
Reducing $\bmod \; p$, we have $\sqrt{-1} \in \Z/p \Z=\F{p}
\Longleftrightarrow p \equiv 1 \pmod{4}$.

$( \Leftarrow )$ Conversely, suppose that $p \equiv 1 \pmod{4}$.
Let $x \in \Z_p$ be such that $$f(x):=x^2+1 \equiv 0 \pmod{p}.$$
Then $$v_p(f'(x))=v_p(2)+v_p(x)=0.$$
By Hensel's lemma($n=1, \; k=0$), there is a solution to $f$ in $\Z_p \subset
\Q_p$.
Since $\pm \sqrt{-1}$ are the solutions to $f$, $\sqrt{-1} \in \Q_p$.

\begin{exa}
$x^4-17=2y^2$ has solutions over $\Q_p$ for every prime $p$ and over $\R$,
but no solution over $\Q$.
\end{exa}

For example, let $f(x,y)=2y^2-x^4+17$ and let $p=2$. Then
$$ f(5,0) \equiv 0 \pmod{32} \;\mbox{ and }\; v_2\left(\frac{\partialf}{\partial x}(5,0)\right)=2.$$
By Hensel's lemma ($n=5, \; k=2$), there exists a solution over $\Z_2 \subset
\Q_2$.

To show that $f(x,y)$ has no solution over $\Q$
suppose that there is a solution $(x,y)$ over $\Q$. Let $x=a/c$ as a fraction
in its lowest terms.
Then (putting $b=yc^2 \in \Z$)$$a^4 -17c^4 =2b^2, \;\;\;
gcd(a,c)=gcd(b,c)=gcd(a,b)=1.$$
Putting $A=a^2, \;\; C=c^2$ we have
$$ A^2-17C^2=2b^2. \;\;\; (*)$$
This equation is solvable everywhere locally, so globally, and in fact
$$5^2-17 \cdot 1^2=2 \cdot 2^2.$$

Now, from $(*)$, it's easy to show $$(5A+17C+4b)(5A+17C-4b)=17(A+5C)^2.$$
If there is a common odd prime divisor of the two factors on the left hand
side, it divides
$5A+17C$ and $A+5C$, so divides $8A$ and $8C$: a contradiction. The twofactors on the left hand side
have the same sign, which for $A=a^2,\; C=c^2$ must be positive. Hence for
integers $u, \; v$
there is one of two possibilities:

$$
\begin{array}{cccc}
                  &  & \mbox{First Case} & \mbox{Second Case} \\
 5a^2+17c^2 \pm 4b& =       & 17u^2 & 34u^2 \\
 5a^2+17c^2 \mp 4b& =       & v^2   & 2v^2  \\
         a^2+5c^2 & =       & uv    & 2uv
\end{array}
$$

In the first case, we have
\begin{eqnarray*}
10a^2+34c^2 &=& 17u^2+v^2 \\
   a^2+5c^2 &=& uv.
\end{eqnarray*}
We claim that this is impossible in $\Q_{17}$. Write $| \sim | = | \sim|_{17}$.
By homogeneity, $\max(|a|,|c|,|u|,|v|)=1$.
Since $10$ is a quadratic non-residue mod 17, we have $|a|<1, \;\; |v|<1$.
Then the second equation gives $|c|<1$.
Finally, the first equation gives $|u|<1$: a contradiction.

The second case gives
\begin{eqnarray*}
 5a^2+17c^2&=&17u^2+v^2 \\
   a^2+5c^2&=&2uv.
\end{eqnarray*}
The proof that this is impossible in $\Q_{17}$ is similar.

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\end{document}