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\begin{document}

\title[Topics in Number Theory]{Topics in Number Theory}%
\author{Uroyoan R. Walker}%
\address{Mathematics Department, Louisiana State University, Baton Rouge, LA 70803}%
\email{walker@math.lsu.edu}%
\section{Norms On Finite Dimensional Vector Spaces} Let $E$ be an $n$
dimensional $k$-vector space and $e=\set {e_1,\ldots ,e_n}$ a
$k$-basis for $E$.  We can define a norm $$\supnorm{\A}=\max
_{1\leqslant i \leqslant n}{\bigl(\abs{\alpha _i}\bigr)}$$ where
$E\ni\A =\alpha _1e_1+\alpha _2e_2+\cdots +\alpha _ne_n$.  Without
loss of generality we assume here that all the absolute values we
work with satisfy the triangle inequality.\begin{thm}[4.2]Let
$\bigl( k,\abs\cdot\bigr)$ be a complete valued field.  Let $E$ be
a finite dimensional $k$ vector space. Then all (non-trivial)
norms on $E$ induce the same topology.  Thus, in particular,
$E\cong k^n$ algebraically and topologically.\end{thm} \proof (by
induction on $n$)\\ For $n=1$ it is trivial. $$\norm\A
=\norm{\alpha _1e_1}=\norm {e_1}\cdot\abs{\alpha _1}$$so in the
$1$-dimensional case $\norm{} = C\cdot\supnorm{}$ for some $C>0$.
Hence $\norm{}$ is equivalent to $\supnorm{}$, and thus gives the
same topology.\\ Now let's assume the result is true for $n-1$ and
we prove it for $n$. So $\A=\alpha _1e_1+\alpha _2e_2+\cdots
+\alpha _ne_n$.  We will prove $$\bigl(\ast\bigr)=
\left\{\begin{array}{c}\forall\,\epsilon>0, \quad\exists\,\eta>0
\mbox{ such that}\\ \mbox{ if } \norm\A <\,\eta, \mbox{ then
}\abs{\alpha _n}\,<\,\epsilon\end{array}\right.$$ Note that
$\bigl(\ast\bigr)$ says that one is able to make the last
coordinate (or any coordinate for that matter) small provided that
the vector we start with has small enough norm.  It can also be
interpreted as saying that the projection map onto the last
coordinate is continuous.  Note that the topology on $k$ comes
from $\abs{\cdot}$, while the topology on $E$ is given by
$\norm{}$. We will assume $\bigl(\ast\bigr)$ to be true and prove
the theorem, then we'll come back and prove $\bigl(\ast\bigr)$. We
proceed.
\\ $\bigl(\ast\bigr)$ implies that $$\forall\,\epsilon\,>\,0, \quad
\exists\,\eta>0 \mbox{ such that if } \norm\A <\,\eta , \mbox{
then }\supnorm{\A}\,<\,\epsilon$$ The point of all this is that
$\supnorm\A =\max\bigl(\abs{\alpha _i}\bigr)$ and for each point
we are capable of finding an $\eta$ that does the job.  So we can
find an $\eta$ that works simultaneously for all the $\alpha _i$
at once.  Thus, $$\max\bigl(\abs{\alpha _i}\bigr)\,<\,\epsilon$$
and this is equivalent to saying that the topology induced by
$\norm{}$ is stronger (finer) than the topology induced by
$\supnorm{}$.  And this is equivalent to saying that every open
$\supnorm{}$-ball is contained in an open $\norm{}$-ball.  On the
other hand,\begin{eqnarray*}\norm\A &=& \norm{\alpha _1e_1+\alpha
_2e_2+\cdots +\alpha _ne_n}\\ &\leqslant &
\abs{\alpha_1}\norm{e_1}+\cdots +\abs{\alpha _n}\norm{e_n}\\
&\leqslant & \max\bigl(\abs{\alpha _i}\bigr)
\cdot\bigl(\norm{e_1}+\cdots +\norm{e_n}\bigl)\\ &=&
C\cdot\supnorm\A \end{eqnarray*} But this implies that the
topology induced by $\supnorm{}$ is stronger (finer) than the
topology induced by $\norm{}$.  Hence both norms induce the same
topology on $E$.  So the theorem is true provided that
$\bigl(\ast\bigr)$ is true.  Hence let's go ahead and prove it,
we'll do it by contradiction.  Suppose $\bigl(\ast\bigr)$ does not
hold.  Then $\exists\,\epsilon\,>\,0$ such that
$\forall\,\eta\,>\,0$ we can find $\A$ with $\norm\A\,<\,\eta$ but
$\abs{\alpha _n}\,\geqslant\,\epsilon$.  Define $\B =\A /\alpha _n
= \frac{\alpha _1}{\alpha _n}e_1+\cdots +\frac{\alpha
_{n-1}}{\alpha _n}e_{n-1}+e_n$.  Now,  $$\norm\B
=\frac{\norm\A}{\abs{\alpha _n}}\leqslant \frac{\eta}{\epsilon}$$
call this new bound $\frac{\eta}{\epsilon},\,\eta '$.  We can
reformulate this as:  $\exists\,\epsilon>0$ such that
$\forall\,\eta>0$ we can find a vector in the form $\A =\alpha
_1e_1+\cdots +\alpha _{n-1}e_{n-1}+e_n$ such that
$\norm\A\,<\,\eta$.  Now, we can find a sequence of these such
that $\norm{\beta _\nu}\,<\,\frac{1}{\nu}$ where $\beta _\nu =
\beta _1^{(\nu)}e_1+\cdots +\beta _{n-1}^{(\nu)}e_{n-1}+e_n$. Note
that $\beta _\nu -\beta _\mu\in\mbox{Span}\set{e_1,\cdots
,e_{n-1}}$.  Notice as well that $\norm{\beta _\nu -\beta
_\mu}\,\leqslant\,\norm{\beta _\nu}+\norm{\beta _\mu}\,<\,1/\nu
+1/\mu$.\\ By induction, the topology induced on
$\mbox{Span}\set{e_1,\cdots ,e_{n-1}}$ by the $\norm{}$-norm is
equivalent to the product topology.  Since the vectors form a
Cauchy sequence, each coordinate $\bigl(\beta
_i^{(\nu)}\bigl)_\nu$, forms a Cauchy sequence.  By completeness
of $k$, there is a limit
$\beta_i=\displaystyle{\lim_{\nu\to\infty}{\beta _i^{(\nu)}}}$ for
each $i=1, 2, \ldots , n-1$.  Call $\gamma$ the element obtained
by putting this limits as coefficients of the $e_i$'s.  So
$$\gamma=\beta_1e_1+\beta_2e_2+\cdots+\beta_{n-1}e_{n-1}+e_n$$ Now
$$\norm{\gamma -\beta_\nu} \leqslant \abs{\beta_1-\beta_1^{(\nu)}}
\norm{e_1}+\cdots +\abs{\beta_{n-1}-\beta_{n-1}^{(\nu)}}
\norm{e_{n-1}}<\epsilon_1$$ for any $\epsilon_1>0$ for $\nu>0$
sufficiently large.  This follows since the $\beta_i$'s are the
limits of the $\beta_i^{(\nu)}$.  We also have,
$$\norm\gamma\leqslant\norm{\gamma
-\beta_\nu}+\norm{\beta_\nu}\,<\,\epsilon_1+\frac{1}{\nu}$$ so
$\norm\gamma=0$.  Hence $\gamma =0\quad\rightarrow\leftarrow$ This
contradicts the linear independence of the $e_i$'s.  This finishes
the proof of theorem 4.2.\endproof As a little remark note that if
a vector converges to zero, then its individual coordinates go to
zero as well.\begin{thm}[4.3]Let $\bigl( k,\abs\cdot\bigr)$ be a
complete valued field.  Let $L$ be a finite field extension of $k$
of degree $n$.  Then, there is a unique extension of $\abs\cdot$
to an absolute value $\norm\cdot$ on $L$, namely $\norm
x=\abs{N_{L/k}(x)}^{1/n}$\end{thm}\proof \underline{Uniqueness}:
If $\norm\cdot_1$ and $\norm\cdot_2$ are two (non-trivial)
absolute values on $L$ extending $\abs\cdot$ from $k$.  Each of
these defines a normed vector space structure on $L$.  By the
above theorem, they define the same topology.  On the other hand,
by proposition 3.1, two absolute values define the same topology
if and only if there is a $\lambda>0$ such that $\norm x_1=\norm
x_2^\lambda$.  But if we take $x\in k$, the ground field, we have
$\abs x=\abs x^\lambda$, forcing $\lambda$ to be $1$.  Thus,
$\norm\cdot_1=\norm\cdot_2$.

\end{document}