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\begin{document}



Take $b=c^n$. Then:
\[
|c|^n \leq M(\frac{n\cdot log(c)}{log(a)}+1)\cdot max(1, |a|^{\frac{n\cdot log(c)}{log(a)}})
\] 
Now take the $n^{th}$ root in both sides and let $n$ go to $\infty$. We get:
\[
|c| \leq max(1, |a|^{\frac{log(c)}{log(a)}}) \ \ \ \ \ \ \  (*).
\] 
\par
Now consider the following two cases:
{\it Case} 1: There exists $c>1$ with $|c|>1$.
Then (*) implies $|a|>1$, for all $a>1$, and (*) becomes:
\[
|c| \leq |a|^{\frac{log(c)}{log(a)}}
\] 
hence
\[
|c|^{\frac{1}{log(c)}} \leq |a|^{\frac{1}{log(a)}}.
\] 
\par
Now reverse the roles of $a$ and $c$, and get:
\[
|c|^{\frac{1}{log(c)}} \geq |a|^{\frac{1}{log(a)}}.
\] \par
The last two inequalities imply:
\[
|c|^{\frac{1}{log(c)}} = |a|^{\frac{1}{log(a)}}
\] 
which means that 
\[
|\ | = |\ |_{\R}^{\lambda}
\]
with 
\[
\lambda = \frac{log|a|}{log(a)}= \frac{log|c|}{log(c)}.
\]

\par
{\it Case} 2: $c\leq1$ for all $c\in {\Z}$.\par
Note that in this case $|\ |$ is archimedian. Let
\[
P = \{x\in {\Z}: |x|<1 \}.
\]
Then $P$ is a prime ideal in $\Z$: for any $x, y\in P$ and $z\in {\Z}$ we have:
\[
|x+y|\leq max{|x|, |y|}<1
\]
\[
|z\cdot x| = |z|\cdot |x| <1
\]
because $x\in P$.\par
Now let $x, y\in {\Z}$ such that $x\cdot y\in P$. Then $|x\cdot y|<1$ and, since $|x|\leq 1$ and $|y|\leq 1$, necessarily $|x|<1$ or $|y|<1$, which means that P is prime.
Moreover, $P\neq \Z$ because $1\notin P$, and $P\neq (0)$: since $|\ |$ is non-trivial, there is $u\in {\Q}$, $u\neq 0$, with $|u|<1$. Write $\displaystyle u=\frac{x}{y}$ with $x, y\in \Z$.
Then 
\[
|x| = |y| \cdot |u| < 1
\]
hence $x\in P$ (and $x\neq 0$). \par
Consequently, $P =  p\Z$, for some prime integer $p$. Then $|\ |$ is equivalent $|\ |_p$ iff
\[
\{x\in {\Q}: |x| < 1\} = \{x\in {\Q}: |x|_p < 1 \}
\]
which is obvious.

\section{4. Completion}
\par
\begin{Def} If $(K, |\ |)$ is a valued field, we say that it is complete if every Cauchy sequence converges to a limit in $K$.
\end{Def}
\par
\begin{exa}
\par
1. ($\R, |\ |_{\R}$) and ($\Q_p, |\ |_p$) are both locally compact, hence complete.
\par
2. ($\Q, |\ |_{\R}$) and ($\Q, |\ |_p$) are not complete.
\end{exa}
\par
\begin{Def} Let $(K, |\ |)$ be a valued field. A completion of $(K, |\ |)$ is a valued field $(\hat{K}, \norm{\ } )$ such that:
\par
1. $K$ is a subfield of $\hat{K}$.
\par
2. The restriction to $K$ of $\norm{\ }$ is $|\ |$.
\par
3. $K$ is dense in $\hat{K}$.
\end{Def}
\par
\begin{thm}
Given a valued field $(K, |\ |)$ there exists a completion which is unique up to an isomorphism.
\end{thm}
\par 
\begin{exa}
\par
1. The completion of $(\Q, |\ |_{\R})$ is $(\R, |\ |_{\R})$;
\par
2. The completion of $(\Q, |\ |_{p})$ is $(\Q_p, |\ |_p)$.
\end{exa}
\par
{\it Sketch of proof}:
\par 1. Let
\[
F=\{(a_n)_{n\geq 0}: a_n\in K, (a_n) - Cauchy\ sequence\}
\]
Then $F$ is ring with the componentwise addition and multiplication of sequences:
\[
(a_n)+(b_n)=(a_n+b_n)
\]
\[
(a_n)\cdot (b_n)=(a_n\cdot b_n)
\]\par
Define $\displaystyle ||(a_n)||=lim_{n\rightarrow \infty}|a_n|$ which exists because the absolute function is continuous with values in the
complete field $\R$.
\par
2. Let
\[
N=\{(a_n)\in F:\ a_n\rightarrow 0\}
\]
Then $N$ is an ideal of $F$ (any sum of sequences that go to 0 goes to 0, and a product of any sequence that goes to 0 with any Cauchy sequence goes to 0).
Moreover, $N$ is a maximal ideal of $F$ because if a Cauchy sequence $(a_n)_n$ does not converge to 0 one can find an $\epsilon_0>0$ which is a lower bound for 
the absolute values of all the terms in the sequence (beyond a certain rank $n_0$), hence the sequence 
\vskip .01cm
\[
b_1=b_2=...=b_{n_0-1}=0;
\]
\[
b_n=\frac{1}{a_n}, n\geq n_0
\]
is Cauchy. Moreover, $(a_n)\cdot (b_n)=1,\ \forall n\geq n_0$, hence $(a_n)_n$ is invertible.
\par
3. Define $\hat{K}=F/N$ and note that the map $a\rightarrow (a,a,...)$ is a field homomorphism from $K$ to $\hat{K}$.
\par
4. Show that $\norm{\ }$ is well defined on $\hat{K}$, which means that
\[
\norm{(a_n)+(b_n)}=\norm{(a_n)},\ \forall (b_n)\in N
\]
\par
5. One verifies that $(\hat{K}, \norm{\ })$ is complete.
\par
6. One verifies that $K$ is dense in $\hat{K}$.
\par
\begin{Def}
Let $(K,|\ |)$ be a valued field. By a normed vector space over $(K,|\ |)$ we mean a $K$-vector space $E$ and a function
$e\rightarrow \norm{e}:E\rightarrow \R$ such that:
\par 
1. $\norm{e}\geq 0$ and $\norm{e}=0$ iff $e=0$
\par 
2. $\norm{e_1+e_2}\leq \norm{e_1}+\norm{e_2}$
\par 
3. $\norm{\alpha\cdot e}=|\alpha|\cdot \norm{e},\ \forall \alpha\in K$.
\end{Def}
\par
\begin{exa}
Let $E=L$ be a field extension of $K$. Let $\norm{\ }$ be an absolute value on $L$ that satisfies the triangle inequality.
Let $|\ |$ be the restriction of $\norm{\ }$ to the subfield $K$ of $L$.
\end{exa}
\end{document}