Valuations and places - decomposition and inertia group

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Discussion Overview

The discussion revolves around the theory of valuations, places, and valuation rings in the context of finite Galois extensions of fields. Participants explore the implications of separability in the relationship between decomposition groups and inertia groups, particularly focusing on the validity of certain formulas and definitions in the literature.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses discomfort with the theory, particularly regarding the relationship between ramification index, residual degree, and the decomposition and inertia groups when the residual field extension is not separable.
  • The participant notes that while the literature suggests a formula [L:K] = e f g holds under certain conditions, it appears to break down when k'/k is not separable.
  • Another participant mentions having asked the same question elsewhere and suggests that it is false that D = ef if k'/k is not separable.
  • A subsequent post corrects the previous statement, asserting that it is always true that D = ef, but that J = e may not hold if k'/k is not separable.
  • A later reply shifts the focus away from the technical discussion, expressing gratitude for a reminder of a film, indicating a potential shift in the conversation's tone.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the implications of separability for the relationships discussed, with conflicting views on the validity of certain equations and definitions in the context of non-separable extensions.

Contextual Notes

The discussion highlights limitations in the definitions and assumptions regarding the relationships between decomposition groups, inertia groups, and the nature of field extensions, particularly in the case of non-separable extensions.

coquelicot
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Hello,
I feel very uncomfortable with some aspects of the theory of valuations, places, and valuation rings.
Here is one of my problems :
Assume that L/K is a finite Galois extension of fields, and that F is a place from K to its residual field k, whose associated valuation ring is discrete.
F extends to a place F' of L, from L to a finite algebraic extension k' of k, which we suppose to be the residual field of F'.
According to the litterature (Bourbaki, "commutative algebra", chap VI, Fried and Jarden, "Field arithmetic"), if e(F'/F) is the ramification index of F' over K, f(F'/F) is the residual degree of F' over K, and g is the number of non equivalent places extending F to L, there holds [L:K] = e f g.
Also, if D is the decomposition group of F' over K and J is its inertia group over K, then e(F'/F) = |J| and |D| = e(F'/F)f(F'/F).
Everything seems to be nice. Well, let see : I agree with everything as long as k'/k is separable, which is always the case if K is a number field (since then k is a finite field). But suppose that k'/k is not separable.
The decomposition group D of F' over F is the set of all automorphisms s of Gal(L/K) that leaves the valuation ring associated to F' globally invariant: this means that F's is equivalent to F' for all s in D (two places A and B are equivalent if there exists an automorphism s' of k'/k such that B = s'A, hence the place As is equivalent to A by the definition of the decomposition group). Since the natural homomorphism from D to Aut(k'/k) is surjective (Frobenius theorem), it is clear that Aut(k'/k) is isomorphic to D/J (J is the kernel of the said epimorphism). So, |D|/|J| = |Aut(k'/k)|, and there holds |D| = |J|.|Aut(k'/k)| = |J|.[k' : k]sep. Consequently, |D| is different from e(F'/F)f(F'/F) since f(F'/F) = [k' : k] and e(F'/F) = |J|, in contradiction with the literature.
Everything would have been OK if f(F'/F) had been defined to be equal to [k' : k]sep, but then the formula [L:K] = e f g would not hold anymore.

Every help will be greatly appreciated.
 
Last edited:
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Well, I have asked the same question in math overflow, and get the answer here. In fact, it is false that D = ef if k'/k is not separable.
 
coquelicot said:
Well, I have asked the same question in math overflow, and get the answer here. In fact, it is false that D = ef if k'/k is not separable.

Sorry, this is a mistake. It is always true that D = ef, but it is false that J = e if k'/k is not separable.
 
I trust that you now have a satisfactory reply, so I will not try to respond but simply wish to thank you for having reminded me
of a lovely film called "king of hearts" in which one of the characters has the name Coquelicot.
 

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