0
$\begingroup$

I was reading the section on Completions in Neukirch's Algebraic Number Theory.

Neukirch uses the term multiplicative valuation for what other authors seem to call absolute value. He uses the term exponential valuation for what other authors call valuation. Anyway, I will stick to his conventions.

On page 126 Neukirch mentions the following:

Let $v$ be an exponential valuation of a field $K$. Then $v$ is canonically continued to an exponential valutation $v^{\wedge}$ of the completion $K^{\wedge}$, by setting $v^{\wedge}(a) = \lim_{n \to \infty} v(a_n)$, where $a = \lim_{n \to \infty} a_n \in K^{\wedge}$, $a_n \in K$.

Perhaps this is obvious, but why does $\lim_{n \to \infty} v(a_n)$ exist?

1 Answers 1

2

If $a_n\rightarrow 0$ then clearly $v(a_n)\rightarrow \infty$, so let's exclude this case. This means that $a_n$ is a Cauchy sequence in $K$ with absolute value bounded above by $C$. Pick $N$ large enough so that for all $n\geq N$ $v(a_n-a_N)>C$. Now suppose $v(a_n)\neq v(a_N)$ for $n\geq N$. Since $v$ is a valuation it follows that $v(a_n-a_N)=\min(v(a_n),v(a_N)). A contradiction!

So we've shown that for all $n\geq N$ $v(a_n)=v(a_N)$ i.e. the sequence $v(a_n)$ becomes constant after $N$.