Proof.

• (i) $⟹$ (ii) Let $U\ni 0$ be a compact neighbourhood of $0$ ($0\in U\subseteq Z$ with $U$ open, $Z$ compact). Then there exists $x\in O_K$ such that $x{O}_K\subseteq U$. Since $x{O}_K$ is closed, $x{O}_K$ is compact. Hence $O_K$ is compact ($x{O}_K\stackrel{\cdot x^{-1}}{\to }{O}_K$ is a homeomorphism).
• (ii) $⟹$ (i) $O_K$ compact implies $a+O_K$ is compact for all $a\in K$. So $K$ is locally compact.
• (ii) $⟹$ (iii) Let $x\in m$, and $A_x\subseteq O_K$ be a set of coset representatives for $O_K∕x{O}_K$. Then $O_K={\bigcup }_{y\in A}y+x{O}_K$ is a disjoint open cover. So $A_x$ is finite by compactness of $O_K$. So $O_K∕x{O}_K$ is finite, hence $O_K∕m{O}_K$ is finite.o

  Suppose $v$ is not discrete. Then let $x_1,x_2,\dots$ such that

  $$v(x_1)>v(x_2)>\cdots >0.$$

  Then $x{O}_K⊊x_2{O}_K⊊x_3{O}_K⊊\cdots ⊊O_K$. But $O_K∕x{O}_K$ is finite so can only have finitely many subgroups, contradiction.

• (iii) $⟹$ (ii) Since $O_K$ is a metric space, it suffices to prove $O_K$ is sequentially compact.

  Let ${(x_n)}_{n=1}^{\infty }$ be a sequence in $O_K$, and fix $\pi \in O_K$ a uniformiser. Since ${\pi }^i{O}_K∕{\pi }^{i+1}{O}_K\cong k$, $O_K∕{\pi }^i{O}_K$ is finite for all $i$ ($O_K\supseteq \pi O_K\supseteq \cdots \supseteq {\pi }^i{O}_K$). Since $O_K∕\pi O_K$ is finite, there exists $a_1\in O_K∕\pi O_K$ and a subsequence ${(x_n)}_{n=1}^{\infty }$ such that $x_{1n}\equiv amod\pi$ for all $n$.

  Since $O_K∕{\pi }^2{O}_K$ is finite, there exists $a_2\in O_K∕{\pi }^2{O}_K$ and a subsequence ${(x_{2n})}_{n=1}^{\infty }$ of ${(x_{1n})}_{n=1}^{\infty }$ such that $x_{2n}\equiv a_2{(mod\pi )}^2{O}_K$. Continuing, this, we obtain sequences ${(x_{in})}_{n=1}^{\infty }$ for $i=1,2,\dots$ such that

  • (1) ${(x_{(i+1)n})}_{n=1}^{\infty }$ is a subsequence of ${(x_{in})}_{n=1}^{\infty }$
  • (2) For any $i$, there exists $a_i\in O_K∕{\pi }^i{O}_K$ such that $x_{in}\equiv a_{i}mod{\pi }^i$ for all $n$.

  Then necessarily $a_i\equiv a_{i+1}mod{\pi }^i$ for all $i$.

  Now choose $y_i=x_{ii}$. This defines a subsequence of ${(x_n)}_{n=1}^{\infty }$. Moreover, $y_i\equiv a_i\equiv a_{i+1}\equiv y_{i+1}mod{\pi }^i$. Thus $y_i$ is Cauchy, hence converges by completeness. □

Proof. One checks that the sets

is a basis of open sets in both topologies.

For $\left|\bullet \right|$: clear.

For profinite topology: $O_K∕O_K∕{\pi }^n{O}_K$ is continuous if and only if $a+{\pi }^n{O}_K$ is open for all $a\in O_K$. □

Proof. Theorem 6.1 implies that $L$ is complete and discretely valued. Suffices to show $k_L:=O_L∕m_L$ is finite. Let ${\alpha }_1,\dots ,{\alpha }_n$ be a basis for $L$ as a $K$ vector space.

$\parallel \bullet {\parallel }_{\sup }$ (sup norm) equivalent to $|\bullet {|}_L$ implies that there exists $r>0$ such that

Take $a\in K$ such that $|a|\ge r$, then

Then $O_L$ is finitely generated as a module over $O_K$, hence $k_L$ is finitely generated over $k$. □

Proof. $K$ complete discretely valued, $\mathrm{characteristic}K>0$. Moreover, $k\cong {𝔽}_{p^n}$ is finite, hence perfect.

By Theorem 5.6, $K\cong {𝔽}_{p^n}((t))$. □

Proof.

• $\Rightarrow$ Since $|-1|=1$, $|-n|=|n|$, it suffices to show that $|n|$ bounded for $n\ge 1$. Then note that

  $$|n|=|1+1+\cdots +1|\le 1.$$

• $\Leftarrow$ Suppose $|n|\le B$ for all $n\in \mathbb{Z}$. Let $x,y\in K$ with $|x|\le |y|$. Then we have $$\begin{array}{llll}\hfill |x+y{|}^m& =\left|\sum _{i=0}^m\left(\genfrac{}{}{0ex}{}{m}{i}\right)x^i{y}^{m-i}\right|& \hfill & \\ \hfill & \le \sum _{i=0}^m\left|\left(\genfrac{}{}{0ex}{}{m}{i}\right)x^i{y}^{m-i}\right|& \hfill & \\ \hfill & \le |y{|}^{m}B(m+1)& \hfill & \end{array}$$

  Taking $m$-th roots gives

  $$|x+y|\le |y|{[B(m+1)]}^{\frac{1}{m}}.$$

  The right hand side tends to $|y|$ as $m\to \infty$, hence

  $$|x+y|\le |y|=\max (|x|,|y|)\square .$$

Proof. Case: $\left|\bullet \right|$ is archimedean. We fix $b>1$ an integer such that $|b|>1$ (exists by Lemma 7.8). Let $a>1$ be an integer and write $b^n$ in base $a$:

with $0\le c_i<a$, $c_m\ne 0$. Let $B=\underset{0\le c<a-1}{\max }(|c|)$, and then we have

Then $\left|a\right|>1$ and

Switching roles of $a$ and $b$, we also obtain

Then ($\ast$) and ($\ast \ast$) gives (using ):

Hence $\left|a\right|=a^{\lambda }$ for all $a\in {\mathbb{Z}}_{>1}$, hence $\left|x\right|=|x{|}_{\infty }^{\lambda }$ for all $x\in \mathbb{Q}$.

Case 2: $\left|\bullet \right|$ is non-archimedean. As in Lemma 7.8, we have $|n|\le 1$ for all $n\in \mathbb{Z}$. Since $\left|\bullet \right|$ is non-trivial, there exists $n\in {\mathbb{Z}}_{>1}$ such that $\left|n\right|<1$. Write $n=p_1^{e_1}\cdots p_r^{e_r}$ decomposition into prime factors. Then $\left|p\right|<1$, for some $p\in \{p_1,\dots ,p_r\}$. Suppose $\left|q\right|<1$ for some prime $q$, $q\ne p$. Write $1=rp+sq$ with $r,s\in \mathbb{Z}$. Then

contradiction. Thus $\left|p\right|=\alpha <1$ and $\left|q\right|=1$ for all primes $q\ne p$. Hence $\left|\bullet \right|$ is equivalent to ${\left|\bullet \right|}_p$. □

Proof. $K$ mixed characteristic implies that $\mathrm{characteristic}K=0$, hence $\mathbb{Q}\subseteq K$. $K$ non-archimedean implies that $\left|\bullet \right|{|}_{\mathbb{Q}}={\left|\bullet \right|}_p$ for some prime $p$. Since $K$ is complete, ${\mathbb{Q}}_p\subseteq K$. Suffices to show that $O_K$ is finite as a ${\mathbb{Z}}_p$-module.

Let $\pi \in O_K$ be a uniformiser, $v$ a normalised valuation and set $v(p)=e$. Then $O_K∕p{O}_K\cong O_K∕{\pi }^e{O}_K$ is finite since ${\pi }^i{O}_K∕{\pi }^{i+1}{O}_K\cong O_K∕\pi O_K$ is finite. Since ${𝔽}_p\cong \mathbb{Z}∕p\mathbb{Z}\hookrightarrow O_K∕p{O}_K$ we have $O_K∕p{O}_K$ a finite dimensional vector space over ${𝔽}_p$.

Let $x_1,\dots ,x_n\in O_K$ be coset representatives for ${𝔽}_p$-basis of $O_K∕p{O}_K$. Then

is a set of coset representatives for $O_K∕p{O}_K$. Let $y\in O_K$. Proposition 3.4(ii) tells us that

Hence $O_K$ is finite over ${\mathbb{Z}}_p$. □