This is a similar question from the book "Valued Fields by Antonio J. Engler and Alexander Prestel, Springer, 2005 " page 82, Exercise 3.5.4.(b).


Let $(K\_{1}, V\_{1})\subseteq (K\_{2}, V\_{2})$ be finite extension of valued fields. Assume that $[K\_{2} : K\_{1}] = n$. Let $G\_{1}$ and $G\_{2}$ be value groups of $V\_{1}$ and $V\_{2}$ respectively. Prove that $[G\_{2} : G\_{1}] = n$. Thanks



Finite extension of valuation
As remarked before, such a statement will not hold in general. One has the usual formula $n=\sum\_{w|v\_1\ on\ K\_2} d(w/v\_1) e(w/v\_1) f(w/v)$ where $n$ is the degree, $d(w/v\_1)$ is the defect of $w/v\_i$, $e(w\_i/v)$ is the ramification degree of $w/v\_1$ and $f(w/v\_1)$ is the residue field extension degree of $w/v\_1$. Hence the statement is true if and only if there is a unique extension of $v\_1$ to $K\_2$ and there is no defect or ramification.


