Extension of a complete discrete valuation ring

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, eventually not discrete. I have in mind an example which proves that this extension may produce a not discrete valuation ring, but 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.

• Being height $1$ is equivalent to the value group being a subgroup of $\mathbf R$. Since you say you're looking for an example, try $R = {\mathbf Z}_p$ and let the algebraic extension of $K = {\mathbf Q}_p$ be the algebraic closure of $K$, so $L = {\mathbf C}_p$. Do you agree that the valuation on $L$ extending that on $K$ has rank $1$? Commented Apr 3, 2015 at 22:43
• First of all thank you for your suggestion, but sorry, maybe I didn't explain well my question. I know that this is a good example of an extension not producing a dvr and I agree that on $\mathbb{C}_{p}$ the valuation has rank $1$. But what I'm trying to prove is the general fact that an extension constructed in this way is again a complete rank $1$ valuation ring.
– rime
Commented Apr 4, 2015 at 7:08
• That example contains essentially all the seeds of the general statement. Do you know how to extend an absolute value from a complete field to any finite extension, and then further to the algebraic closure? Do you see anything about $K = \mathbf Q_p$ that doesn't extend to the general case? Commented Apr 4, 2015 at 12:06