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Valuations and places - decomposition and inertia group

  1. Dec 9, 2014 #1
    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: Dec 9, 2014
  2. jcsd
  3. Dec 11, 2014 #2
    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.
     
  4. Dec 13, 2014 #3
    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.
     
  5. Dec 13, 2014 #4

    marcus

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    Dearly Missed

    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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