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# value group of completion

Let $k$ be a field and $|\cdot|$ its non-archimedean valuation of rank one. Then its value group $|k\!\smallsetminus\!\{0\}|$ may be considered to be a subgroup of the multiplicative group of $\mathbb{R}$. In the completion $K$ of the valued field $k$, the extension of the valuation is defined by

$|x|\;=:\;\lim_{{n\to\infty}}|x_{n}|,$ |

when the Cauchy sequence $x_{1},\,x_{2},\,\ldots,\,x_{n},\,\ldots$ of elements of $k$ determines the element $x$ of $K$.

###### Theorem.

The non-archimedean field $k$ and its completion $K$ have the same value group.

Proof. Of course, $|k|\subseteq|K|$. Let $x=\lim_{{n\to\infty}}x_{n}$ be any non-zero element of $K$, where $x_{j}$’s form a Cauchy sequence in $k$. Then there exists a positive number $n_{0}$ such that

$|x_{n}\!-\!x|\;<\;|x|$ |

for all $n>n_{0}$. For all these values of $n$ we have

$|x_{n}|\;=\;|x\!+\!(x_{n}\!-\!x)|\;=\;|x|$ |

according to the ultrametric triangle inequality. Thus we see that $|K|\subseteq|k|$.

## Mathematics Subject Classification

13F30*no label found*13J10

*no label found*13A18

*no label found*12J20

*no label found*

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