0
$\begingroup$

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

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ What if $K_1=\mathbf{Q}_2$ and $K_2=K_1(\sqrt5)$ ? $\endgroup$
    – Chandan Singh Dalawat
    Commented Jan 18, 2013 at 6:44

1 Answer 1

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .