I am really struggling with the concept of prolonging a valuation. Can someone please explain what 'e(E'/K)' is in the exercise below, what it means for K to be complete under a discrete valuation and help me answer the question quite thourghly?

Serre's Local Fields, question 4 Chapter 2.2:

Let K be a field complete under the discrete valuation v, and let $\Omega$ be an algebraic closure of K. a)Let S be the set of subextensions E of $\Omega$ with the property that for every finite extension E' of E, e(E'/K)=1. Show that S has maximal elements. If $K_0$ is maximal, show that v prolongs to a discrete valuation of $K_0$, and that the residue field of $K_0$ is the algebraic closure of that of K.

The hint is to use: Proposition: If A is a discrete valuation ring, and if $\bar{f}$ is irreducible, then $B_f$ is a discrete valuation ring with maximal ideal $mB_f$ and residue field $K[X]/(\bar{f})$

  • $\begingroup$ I think that book is quite self-contained; for sure it contains all definition that you need. However, $e(E'/K)=1$ means that the valuation in $E'$ of any uniformizer of $K$ is $1$, while being complete under a discrete valuation means exactly what it means, namely there is a valuation on $K$ such that $K$ is complete with respect to that valuation. $\endgroup$ – Ferra Jan 12 '15 at 10:57
  • $\begingroup$ Thank you. So does K being complete with respect to a discrete valuation mean that all Cauchy sequences of the valuations converge in K? $\endgroup$ – Camille Jan 13 '15 at 20:55
  • $\begingroup$ yes, it means that. $\endgroup$ – Ferra Jan 14 '15 at 9:29

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