# Valuations and places - decomposition and inertia group

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. I would appreciate some help to solve this paradox.

I think your understanding is good, maybe the terminology needs a little clarification:

|k':k| = |k'sep:k||k'pi:k| i.e. f = fsepfpi

|D/J| = fsep

|L:K| = efg

example: K=F(T)((X)), L=F(T1/p)((X)), valuation=X-adic, k=F(T), k'=F(T1/p), e=1, f=p, fsep=1, g=1

• thx. So, what was false is |D| = ef (despite this is what Fried and Jarden wrote). My question is : the terminology $f_{sep}$ and $f_{pi}$ is yours or is it something usual ? if it is the case, can you point to some reference for me ? thx – MikeTeX Dec 10 '14 at 19:41
• @MikeTeX : The notations were made by me for this post. The complete local case (such as the example above) has the essence of your residue field extension situation. Serre's "Local Fields" is a good reference on DVRs though I think it mainly covers separable residue field extensions. – David Lampert Dec 10 '14 at 22:00
• Here's a nice example with L/K separable degree p and k'/k purely inseparable: math.stackexchange.com/questions/57838/… – David Lampert Dec 11 '14 at 1:08
• @MikeTeX : Here's a reference: Serre, "Local Fields", p. 21 – David Lampert Dec 11 '14 at 1:35
• In the nice example cited above with y^p+xy-t over k(t)[[x]], y has an "algebraic x-adic expansion" y=t^(1/p) - t^(1/p^2)x^(1/p) + t^(1/p^3)x^((p+1)/p^2)) ... whose exponents accumulate although y is unramified. – David Lampert Dec 11 '14 at 2:14