Let $（K,v）$ be character 0 field complete with discrete normalized valuation is always non Archimedean?

Question

Let $（K,v）$ be a character $0$ field that is complete with discrete valuation is always non Archimedean?

Silverman's 'the arithmetic of elliptic curves' admits this as a fact without no explanation.

Here, I guess the definition of non Archimedean is that the metric induced by the value satisfies ultra triangle inequality.

Thank you in advance.

Answer 1

(please edit your question)

You meant: for $|.|_v$ an absolute value on a field $K$, if $|K^\times|_v$ is a discrete subgroup of $\Bbb{R^\times}$ then does $|.|_v$ satisfy the ultra-metric inequality? (so that $\log |.|_v$ is a discrete valuation)

The answer is yes.

From the discreteness $|K^\times|_v \subset r^\Bbb{Z}$ for some $r\in (0,1)$.

Take $n$ large enough so that $r < 1-r^n$ and $1+r^n<1/r$, and let $|a|_v=1,|c|_v\le 1,|b|_v\le r^n$.

• lemma: $$|a|_v-|b|_v\le |a+b|_v\le |a|_v+|b|_v$$ gives that $|a+b|_v=1$.

• If $|a+c|_v > 1$ then $|\frac{a}{a+c}|_v<1,|\frac{c}{a+c}|_v<1$ and

  $$|a+c(\frac{a}{a+c}+\frac{c}{a+c})^n|_v=|a+c|_v>1$$ which is impossible since $|a+c(\frac{a}{a+c}+\frac{c}{a+c})^n|_v$ is of the form $|a+\sum_{j=1}^{2^n} b_j|_v$ with each $|b_j|\le r^n$ and applying many times the lemma we get $|a+\sum_j b_j|_v=1$.

Whence $|a+c|_v\le 1$ which is the ultra-metric inequality.

Answer 2

(Please edit your question.)

reuns has given a nice self-contained proof for the fact you are after. Here is a way to solve it with a somewhat bigger gun:

According to this answer and/or the references therein, a sufficient condition for an absolute value $\lvert \cdot \rvert_v$ on a field $K$ to satisfy the strong triangle ("ultrametric") inequality is the original meaning of "non-archimedean", namely that $\lvert n \rvert_v \le 1$ for all $n \in \mathbb N$, where by abuse of notation we mean $\underbrace{1_K+...+1_K}_{n \text{ times}}$ by $n \in K$.

Now if $\mathrm{char}(K)=0$ i.e. $\mathbb Q \subset K$, the restriction of $\lvert \cdot \rvert_v$ to $\mathbb Q$ is an absolute value with discrete image, hence by Ostrowski's Theorem is either the trivial value or (some power of) some $p$-adic value. In both cases $\lvert \mathbb N \rvert_v \le 1$ and we conclude from the above.

(And even easier, if $\mathrm{char}(K)=p$, the restriction of the value to $\mathbb F_p$ must be trivial as any absolute value on a finite field is, and again we conclude.)