Extension of a complete discrete valuation ring

Question

My question came when I was reading the famous Tate's paper on $p$-divisible groups. At the beginning of chapter $(2.4)$ he cites this fact as obvious. If you take a complete discrete valuation ring $R$, you extend its fraction field $K$ algebraically (the extension is allowed to be infinite), you take the completion $L$ of the extended field and you consider inside it the ring of integers over $R$, call it $S$, then $S$ is a complete height $1$ valuation ring. I really don't know how to prove that the extended ring is a complete rank 1 valuation ring, especially since the field extension can be infinite. If you have any suggestion or reference, I'll be really thankful.

Answer 1

The valuation on the infinite extension has rank 1, it just may not be discrete.

Suppose $v$ is the valuation of $K$, and $L$ is a possibly infinite algebraic extension of $K$, with $w$ the extension of $K$ to $L$. Let $L'$ be the completion of $L$ with respect to $w$ and $w'$ the extension of $w$ to $L$. Let also $\Gamma_v=v(K)$, $\Gamma_w=w(L)$, $\Gamma_w'=w'(L')$ be the value groups.

We know $v$ has rank 1, because $K$ is a dvr. By definition that means $\dim_\mathbb{Q} \Gamma_v \otimes \mathbb{Q} = 1$. We want to see that $\dim_\mathbb{Q} \Gamma_w' \otimes \mathbb{Q} = 1$.

First note that $\Gamma_w = \Gamma_w'$. Suppose you have $x\in L'$ and $x = \lim_{n\rightarrow \infty} x_n$, with $x_n \in L$. Then since $|\cdot|_{w'}$ is continuous $|x|_{w'} = \lim_{n\rightarrow \infty} |x_n|_{w'}$. Now if $|x|_{w'}$ is non-zero, we can take $n$ large enough that $|x - {x_n}|_{w'}< |x_n|_{w'}$. Then since the valuation is non-archimedian, $|x|_{w'} = \max \{|x_n|_{w'},|x-x_n|_{w'}\} = |x_n|_{w'}=|x_n|_{w}$. This shows $\Gamma_{w}' = \Gamma_w$.

Then it's enough to show that $\dim_\mathbb{Q} \Gamma_w \otimes\mathbb{Q}=1$. In fact $\Gamma_w \otimes \mathbb{Q} = \Gamma_v \otimes \mathbb{Q}$.

To see this, let $a\in L$. Then $a$ belongs to a finite extension $K'$ of $K$, with valuation $v'$ extending $v$. Then the value group $\Gamma_v$ has finite index in $\Gamma_{v'}$, equal to the ramification index of $K'/K$. That shows $\Gamma_v' \otimes \mathbb{Q} = \Gamma_v \otimes \mathbb{Q}$, and $w(a)=v'(a)\in \Gamma_v \otimes \mathbb{Q}$.