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\begin{document}

\setcounter{section}{7}
%
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%Feb. 25, Friday, note by Changheon Kang
%----------------------------------------
%
\begin{pro}
\label{P:I.7.1} Let $(K,v)$ be a complete valued field with respect to
a discrete valuation $v$. Let $L/K$ be a finite extension with
$[L:K]=n$. (Then by Theorem {T:I.4.3} $v$ extends uniquely to a
discrete valuation $w$ of $L$ and $L$ is complete in $w$.) Let
$$e=e(w|v) \mbox{ and } f=f(w|v).$$
Then $n=ef$.

\end{pro}

\begin{proof}
In fact, let $\Pi$ be a prime element of $L$
and let $B_1,\cdots,B_f \in \ok{L}$ be such that
$$\bar{B}_1,\cdots,\bar{B}_f \in \ok{L}/\mfr{p}_L = k_L$$
is a $k_K$-basis for $k_L$ as a vector space of $k_K$.
Then it's enough to show that
$$\msr{B} =\{B_i\Pi^j : 1\leq i\leq f, 0\leq j\leq e-1\}$$
is a basis for $\ok{L}$ as an $\ok{K}$-module.

(i) Show that $\msr{B}$ is free over $\ok{K}$, that is $\msr{B}$ is
$K$-linear independent.
Suppose $$\sum_j\sum_i a_{ij}B_i\Pi^j=0,\;\mbox{ for some }\;
a_{ij} \in K.$$
Since $a_{ij}\in K$, we can write each $a_{ij}$ as a unit times a power of
a prime element $\pi$ of $K$.
Multiplying out the equation above by a proper power of $\pi$
we can assume that
$$a_{ij} \in \ok{K} \;\mbox{ and }\; \mr{sup} |a_{ij}|=1.$$

Let $I,\;J$ be such that $|a_{IJ}|=1$.
Choose the first $J$ that
$$|a_{ij}|<1 \;\mbox{ for }\; j \leq J, \mbox{ all } 1\leq i\leq f.$$

We claim that $|\sum_i a_{iJ}B_i | =1.$
Note that $a_{iJ} \in \ok{K}$, and $B_i \in \ok{L}$.
So $|\sum_i a_{iJ}B_i | \leq 1.$ Suppose it's not equal to 1.
Then  $|\sum_i a_{iJ}B_i | < 1$, i.e.
$$\sum_i a_{iJ}B_i \in \mfr{p}_L.$$
Taking $\bmod \; \mfr{p}_L$ :
$$\sum_{i=1}^{f} \bar{a}_{iJ}\bar{B}_i =0.$$
And since $\bar{a}_{IJ} \neq 0$ (because $|a_{IJ}|=1$),
this equation is a nontrivial linear combination of $\bar{B}_i$
which is a contradiction of the linear independence of $\bar{B}_i$'s.
Hence
$$\left|\sum_i a_{iJ}B_i \right| =1.$$

Now, fix $j$. To show
$|\sum_i a_{ij}B_i\Pi^j|\leq|\Pi|^J$ for any $j$ we claim
$$\left|\sum_i a_{ij}B_i\Pi^j \right|=
\left\{ \begin{array}{ll}
\leq |\pi|=|\Pi|^e< |\Pi|^J & \mbox{if } j<J \\
= |\Pi|^J          & \mbox{if } j=J \\
\leq |\Pi|^{J+1}<|\Pi|^J & \mbox{if } j\geq J+1
\end{array} \right.$$
\newline
\underline{Case I}: If $j<J$ then
$$\left|\sum_i a_{ij}B_i\Pi^j \right| \leq \max_i(|a_{ij}||B_i||\Pi|^j).$$
Note $|B_i|\leq 1\mbox{ and }|\Pi|^j\leq 1\Rightarrow|B_i||\Pi|^j\leq 1$.
Since $J$ is chosen so that $|a_{ij}|<1$ for $j<J$ (and $a_{ij}\in\ok{K}$),
we have $|a_{ij}|\leq|\pi|=|\Pi|^e< |\Pi|^J$.
Hence, $\max(|a_{ij}||B_i||\Pi|^j)\leq |\pi|=|\Pi|^e< |\Pi|^J$.
\newline
\underline{Case II}: If $j=J$ then
$$\left|\sum_i a_{iJ}B_i\Pi^J \right|
=\underbrace{\left|\sum_i a_{iJ}B_i\right|}_{=1}\left|\Pi \right|^J
=\left|\Pi \right|^J.$$
\newline
\underline{Case III}: If $j\geq J+1$ then
$$\left|\sum_i a_{ij}B_i\Pi^j \right|
=\underbrace{\left|\sum_i a_{ij}B_i\right|}_{\leq 1}\left|\Pi \right|^j
\leq\left|\Pi \right|^j\leq \left|\Pi \right|^{J+1}<|\Pi|^J.$$
Thus we have
$$\left|\Pi\right|^J=\left|\sum_i a_{iJ}B_i\Pi^J \right|
\leq \left|\sum_j\sum_i a_{ij}B_i\Pi^j \right|
\leq \max_j \left( \left|\sum_i a_{ij}B_i\Pi^j \right| \right)
\leq\left|\Pi\right|^J,$$
and hence
$$\left|\sum_j\sum_i a_{ij}B_i\Pi^j \right|
=\left|\Pi\right|^J\neq \infty$$
which is a contradiction that $|0|=\infty$.

(ii) Now, we want to show that $\msr{B}$ is an $\ok{K}$-module generator
of $\ok{L}$.
\newline
Let $A\in\ok{L}$ and $\bar{A}\in k_L$ then we can write $\bar{A}$ as
$$\bar{A}=\sum_{i=1}^f \bar{a}_{i0}\bar{B}_i, \;\;\bar{a}_{i0}\in k_K.$$
Then
$$A-\sum_{i=1}^f a_{i0} B_i \in \mfr{p}_L = \langle \Pi \rangle$$
that is
$$A-\sum_{i=1}^f a_{i0} B_i
=\Pi A_1 \;\;\mbox{ for some }\;\; A_1 \in \ok{L}.$$
Similarly for $A_1$, we have
$$A_1-\sum_{i=1}^f a_{i1} B_i
=\Pi A_2 \;\;\mbox{ for some }\;\; A_2 \in \ok{L}.$$
Combining these two gives
$$A-\sum_{i=1}^f a_{i0} B_i
- \Pi\sum_{i=1}^f a_{i1} B_i \in \langle \Pi^2 \rangle.$$
Continuing this process up to $e-1$, we get
$$A-\sum_{j=0}^{e-1}\left(\sum_{i=1}^f a_{ij} B_i\right) \Pi^j
= \Pi^e A_e
=\pi A^{(1)} \;\mbox{ for some }\; A_e=A^{(1)} \in \ok{L}.$$
Doing the similar process with $A^{(1)}$, we can write
$$A^{(1)}=\sum_{i,j} a_{ij}^{(1)}B_i\Pi^j + \pi A^{(2)} \;\mbox{ for some }\;
A^{(2)} \in \ok{L}.$$
Hence, (putting $\;a_{ij}^{(0)}=a_{ij}$) we have
$$A-\sum_{i,j} a_{ij}^{(0)}B_i\Pi^j
=\pi\sum_{i,j} a_{ij}^{(1)}B_i\Pi^j + \pi^2 A^{(2)}.$$
Then repeating the process:
$$A-C_0-\pi C_1-\pi^2C_2-\cdots-\pi^sC_s\in\langle\pi^{s+1}\rangle$$
where
$$C_k=\sum_{i,j}a_{ij}^{(k)}B_i\Pi^j.$$
As $s$ goes to $\infty$, we have
$$A=\sum_{j=0}^{e-1}\sum_{i=1}^f \alpha_{ij}B_i\Pi^j$$
with
$$\alpha_{ij}=\sum_{k=0}^\infty a_{ij}^{(k)}\pi^k\in\ok{K}.$$
\end{proof}

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\end{document}