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\begin{document}

\title{Lecture 16}

\author{John Harris}

\address{Louisiana State University \\
Department of Mathematics \\
Baton Rouge, LA 70803, USA}

\date{\today}

\email{jmharris73@earthlink.net}

\maketitle



\begin{exa}
Let $K = \Q_2$, $L = \Q_2(\xi)$, where $\xi$ is a root of $X^2 + X + 1$.
\medskip

Suppose $\xi \in K$.  $\xi^3 = 1$, so $3v_2(\xi) = v_2(\xi^3) = v_2(1) = 0$,
hence $v_2(\xi) = 0$, i.e. $\xi \in \Z_2$.  But $X^2 + X + 1$ has no root in
$\Z_2$,
as it has no root in $\Z/2$.  Hence, $\xi$ does not belong to $K$ and
$L$ is quadratic over $K$.  Set $\{1, \sigma\} = \Aut_K L$.
\bigskip

Suppose $x = a + b\xi \in \mc{O}_L$.
$|\N(x)|_2 = |x|_L^2 \leq 1$, so $\N(x) \in \Z_2$.
\medskip

$|\sigma(x)|_L = \sqrt{|\N(\sigma(x))|_2} = \sqrt{|\N(x)|_2} = |x|_L$, so

$|\Tr(x)|_2 = |\Tr(x)|_L = |x + \sigma(x)|_L \leq max\{|x|_L,
|\sigma(x)|_L\} \leq 1$.
\medskip

Hence, $\Tr(x) \in \Z_2$.  Similarly, if $x \in \mc{P}_L$, then
$\Tr(x), \N(x) \in \mc{P}_L$,
and thus $\Tr(x), \N(x) \in \mc{P}_K = 2\Z_2$.
\bigskip

\noindent {Claim:} $\,$ $\mc{O}_L = \Z_2[\xi]$.
\medskip

If $x = a + b\xi \in \Z_2[\xi]$, then
$|x|_L^2 = |2a - b|_2 \leq max\{|a|_2, |b|_2\} \leq 1$, so $|x|_L \leq 1$.
\medskip

Conversely, suppose $x \in \mc{O}_L$.
Then $(\Tr(x))^2 = 4a^2 - 4ab + b^2 \in \Z_2$ and $4\N(x) = 4a^2 - 4ab +
4b^2 \in \Z_2$.
Hence $3b^2 \in \Z_2$, i.e. $|3|_2|b|_2^2 = |3b^2|_2 \leq 1$.  $|3|_2 = 1$,
so $|b|_2 \leq 1$, i.e. $b \in \Z_2$.
\medskip

$2a - b, b \in \Z_2$, so $\alpha = 2a \in \Z_2$.
$\frac{\alpha}{2}^2 - \frac{\alpha b}{2} + b^2, b^2 \in \Z_2$.
Hence, $\alpha(a - b) = \frac{\alpha^2}{2} - \alpha b \in 2\Z_2$.
Hence, $a \in \Z_2$, and so $x \in \Z_2[\xi]$.
\medskip

Hence, $\mc{O}_L = \Z_2[\xi]$.
Similarly, $\mc{P}_L = 2\Z_2[\xi]$.
\bigskip

$k_K = \Z/2 = \F{2}$.  $k_L = \mc{O}_L / \mc{P}_L = \Z_2[\xi] / 2\Z_2[\xi]$.
\medskip

Define $\varphi : \mc{O}_L \longrightarrow (\Z/2) [\overline{\xi}]$ by
$\varphi(a + b\xi) = \overline{a} + \overline{b}\overline{\xi}$.
\medskip

$\mr{Ker}(\varphi) = \mc{P}_L$.  Hence, $k_L = \F{2} [\overline{\xi}]$.
\bigskip

\begin{pro} (Generalization of Hensel's Lemma)
Let $f(X) \in \mc{O}_L[X]$, for a complete discrete valued field $L$.
If $\alpha_0 \in \mc{O}_L$ and $|f(\alpha_0)| < |f'(\alpha_0)|^2$,
then $\exists \alpha \in \mc{O}_L$ such that $f(\alpha) = 0$.
\end{pro}
\noindent {\em Proof.} $\;$  See Lang, 42.
\bigskip

\noindent {Claim:} $\,$ $\Q_2(\sqrt{5}) = \Q_2(\xi)$.
\medskip

If $\sqrt{5} \in \Q_2$, then $2v_2(\sqrt{5}) = v_2(5) = 0$, hence
$v_2(\sqrt{5}) = 0$,
i.e. $\sqrt{5} \in \Z_2$.  But $X^2 - 5 = 0$ has no solution in $\Z_2$,
since it has no solution in $\Z/8$.  Hence, $\sqrt{5}$ does not belong to
$\Q_2$ and
$\Q_2(\sqrt{5})$ is quadratic over $\Q_2$.

Thus, it suffices to show $\sqrt{5} \in \Q_2(\xi)$.
\medskip

Let $f(X) = X^2 - 5$, $\alpha_0 = 3 + 2\xi$.
$|f(\alpha_0)|_L = \sqrt{|\N(8\xi)|_2} = \sqrt{|64|_2} = \sqrt{2^{-6}} =
1/8$
and $|f'(\alpha_0)|_L^2 = |\N(6 + 4\xi)|_2 = |28|_2 = 1/4$.
Hence, $\sqrt{5} \in \Q_2(\xi)$.

Hence, $\Q_2(\sqrt{5}) = \Q_2(\xi)$.

\end{exa}

\begin{rem}
$\forall p, n$, there are only finitely many extension fields over $\Q_p$ of
degree $n$.
For example, there are only seven extensions of $\Q_2$ of degree 2:
$$\Q_2(\sqrt{-1}), \Q_2(\sqrt{\pm 2}), \Q_2(\sqrt{\pm 5}), and
\Q_2(\sqrt{\pm 10}).$$
\end{rem}

\end{document}