Proof. Let ${(v_i)}_{i=1}^{\infty }$ be a Cauchy sequence in $V$, and let $e_1,\dots ,e_n$ be a basis for $V$.

Write $v_i={\sum }_{j=1}^n{x}_j^i{e}_j$. Then ${(x_j^i)}_{i=1}^{\infty }$ is a Cauchy sequence in $K$. Let $x_j^i\to x_j\in K$, then $v_i\to v:={\sum }_{j=1}^n{x}_j{e}_j$. □

Proof. Since equivalence defines an equivalence relation on the set of norms, it suffices to show that any norm $\parallel \bullet \parallel$ is equivalent to ${\parallel \bullet \parallel }_{\sup }$.

Let $e_1,\dots ,e_n$ be a basis for $V$, and set $D:=\underset{i}{\max }\parallel e_i\parallel >0$. Then for $x={\sum }_{i=1}^n{x}_i{e}_i$, we have

To find $C$ such that $C{\parallel \bullet \parallel }_{\sup }\le \parallel \bullet \parallel$, we induct on $n=\dim V$.

For $n=1$: $\parallel x\parallel =\parallel x_1{e}_1\parallel =\left|x_1\right|\parallel e_1\parallel$, so take $C=\parallel e_1\parallel$.

For $n>1$: set $V_i=\mathrm{span}⟨e_1,\dots ,e_{i-1},e_{i+1},\dots ,e_n⟩$. By induction, $V_i$ is complete with respect to $\parallel \bullet \parallel$, hence closed.

Then $e_i+V_i$ is closed for all $i$, and hence

is a closed subset not containing $0$. Thus there exists $c>0$ such that $B(0,C)\cap S=\varnothing$ where $B(0,C)=\{x\in V|\parallel x\parallel <C\}$.

Let $0\ne x={\sum }_{i=1}^n{x}_i{e}_i$ and suppose $\left|x_j\right|=\underset{i}{\max }\left|x_i\right|$. Then ${\parallel x\parallel }_{\sup }=|x_j|$, and $\frac{1}{x_j}\in S$. Thus $\parallel \frac{x_i}{x_j}\parallel \ge C$, and hence

$V$ is complete since it is complete with respect to ${\parallel \bullet \parallel }_{\sup }$ (see Proposition 6.4). □

Proof. Example Sheet 2. □

Proof. Let $x\in K$ be integral over $O_K$. Without loss of generality, $x\ne 0$. Let $f(X)=X^n+a_{n-1}{X}^{n-1}+\cdots +a_0\in O_K[X]$ such that $f(x)=0$. Then

If $\left|x\right|>1$, we have $\left|-a_{n-1}\frac{1}{x}-\cdots -a_0\frac{1}{x_{n-1}}\right|<1$. Thus $\left|x\right|\le 1⟹x\in O_K$. □

Proof. Let $0\ne y\in L$ and let

be the minimal (monic) polynomial of $y$.

Claim: $y$ integral over $O_K$ if and only if $f(X)\in {\mathbb{Q}}_K[X]$.

• $\Rightarrow$ Clear.
• $\Leftarrow$ Let $g(X)\in O_K[X]$ monic such that $g(y)=0$. Then $f|g$ (in $K[X]$), and hence every root of $f$ is a root of $g$. So every root of $f$ in $\overline{K}$ is integral over $O_K$, so $a_i$ are integral over $O_K$ for $i=0,\dots ,d-1$.

Hence $a_i\in O_k$ (by Lemma 6.8). By Corollary 4.5, $|a_i|\le \max (|a_0|,1)$ for $i=0,\dots ,d-1$. By property of $N_{L∕K}$, we have $N_{L∕K}(y)=\pm a_0^m$ for $m\ge 1$.

Hence

Thus and proves the Lemma. □

Proof of Theorem 6.1. We first show ${\left|\bullet \right|}_L={\left|N_{L∕K}(\bullet )\right|}^{\frac{1}{n}}$ satisfies the three axioms in the definition of absolute value.

• (i) $$\begin{array}{llll}\hfill {\left|y\right|}_L=0& ⟺{\left|N_{L∕K}(y)\right|}^{\frac{1}{n}}=0& \hfill & \\ \hfill & ⟺N_{L∕K}(y)=0& \hfill & \\ \hfill & ⟺y=0& \hfill & \end{array}$$
• (ii) $$\begin{array}{llll}\hfill {\left|y_1{y}_2\right|}_L& ={\left|N_{L∕K}(y_1,y_2)\right|}^{\frac{1}{n}}& \hfill & \\ \hfill & ={\left|N_{L∕K}(y_1)N_{L∕K}(y_2)\right|}^{\frac{1}{n}}& \hfill & \\ \hfill & ={\left|N_{L∕K}(y_1)\right|}^{\frac{1}{n}}{\left|N_{L∕K}(y_2)\right|}^{\frac{1}{n}}& \hfill & \\ \hfill & ={\left|y_1\right|}_L{\left|y_2\right|}_L& \hfill & \end{array}$$
• (iii) Set $O_L=\{y\in L|{\left|y\right|}_L\le 1\}$.

  Claim: $O_L$ is the integral closure of $O_K$ inside $L$.

  Assuming this, we prove (iii). Let $x,y\in L$, and without loss of generality assume ${\left|x\right|}_L\le {\left|y\right|}_L$. Then ${\left|\frac{x}{y}\right|}_L$ hence $\frac{x}{y}\in O_L$. Since $1\in O_L$ and $O_L$is a ring, we have $1+\frac{x}{y}\in O_L$ and hence ${\left|1+\frac{x}{y}\right|}_L\le 1$. Hence ${\left|x+y\right|}_L\le {\left|y\right|}_L=\max ({\left|x\right|}_L,{\left|y\right|}_L)$ thus (iii) is satisfied.

So we have proved that $|\bullet {|}_L$ is an absolute value on $L$.

Since $N_{L∕K}(x)=x^n$ for $x\in K$, $|x{|}_L$ extends $|\bullet |$ on $K$.

If ${\left|\bullet \right|}_L^{\prime }$ is another absolute value on $L$ extending $\left|\bullet \right|$, then $|\bullet {|}_L,|\bullet {|}_L^{\prime }$ are norms on $L$.

Theorem 6.5 tells us that $|\bullet {|}_L^{\prime },|\bullet {|}_L$ induce the same topology on $L$. Hence $|\bullet {|}_L^{\prime }=|\bullet {|}_L^c$ for some $c>0$ (by Proposition 1.4) since $|\bullet {|}_L^{\prime }$ extends $|\bullet |$, we have $c=1$.

Now we show that $L$ is complete with respect to $|\bullet {|}_L$: this is immediate by Theorem 6.5. □

Proof.

• (i) $v$ a valuation on $K$, $v_L$ valuation on $L$ such that $v_L$ extends $v$. Let $n=[L:K]$, and let $y\in L^{\times }$. Then $|y{|}_L=|N_{L∕K}(y){|}^{\frac{1}{n}}$ hence $v_L(y)=\frac{1}{n}v(N_{L∕K}(y))$, hence $v_L(L^{\times })\le \frac{1}{n}v(K^{\times })$, so $v_L$ is discrete.
• (ii) Lemma 6.9.

Proof. Let $x\in \overline{K}$, then $x\in L$ for some $L∕K$ finite. Define $|x{|}_{\overline{K}}=|x{|}_L$. Well-defined, i.e. independent of $L$ by the uniqueness in Theorem 6.1.

The axioms for $|\bullet {|}_{\overline{K}}$ to be an absolute value can be checked over finite extensions.

Uniqueness: clear. □

Proof. We’ll choose $\alpha \in O_L$ such that:

• there exists $\beta \in O_L[\alpha ]$ a uniformiser for $O_L$

• $O_K[\alpha ]\to k_L$ surjective

$k_L∕k$ separable tells us that there exists $\overline{\alpha }\in k_L$ such that $k_L=k(\overline{\alpha })$.

Let $\alpha \in O_L$ a lift of $\overline{\alpha }$, and $g(X)\in O_K[X]$ a monic lift of the minimal polynomial of $\overline{\alpha }$.

Fix ${\pi }_L\in O_L$ a uniformiser. Then $\overline{g}(X)\in k[X]$ irreducible and separable, hence $g(\alpha )\equiv 0mod{\pi }_L$ and $g^{\prime }(\alpha )⁄\equiv 0mod{\pi }_L$.

If $g(\alpha )\equiv 0mod{\pi }_L^2$, then

Thus

($v_L$ normalised valuation on $L$).

Thus either $v_L(g(\alpha ))=1$ or $v_L(g(\alpha +{\pi }_L))=1$. Upon possibly replacing $\alpha$ by $\alpha +{\pi }_L$, we may assume $v_L(g(\alpha ))=1$.

Set $\beta =g(\alpha )\in O_K[\alpha ]$ a uniformiser. Then $O_K[\alpha ]\subseteq L$ is the image of a continuous map:

where $n=[K(\alpha ):K]$. Since $O_K$ is compact, $O_K[\alpha ]\subseteq L$ is compact, hence closed. Since $k_L=k(\overline{\alpha })$, $O_K[\alpha ]$ contians a set of coset representatives for $k_L=O_L∕\beta O_L$.

Let $y\in O_L$. Then Proposition 3.4 gives us

Then $y_m={\sum }_{i=0}^m{\lambda }_i{\beta }^i\in O_K[\alpha ]$. Hence $y\in O_K[\alpha ]$, since $O_k[\alpha ]$ is closed. □