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We have the following commutative diagram: 0 / Ix ,K / Vx ,K / Dx ,K / Vx ,K /Γ /0 0  / Ix,K / Vx,K  / Dx,K / Vx,K  / Autk (kx ) v / 0. 10 gives us a splitting of the second sequence provided that we can show Hom(Autkx (kx ), Ix,K / Vx,K ) = 0. 6). Suppose we have such a non-trivial morphism. Then as ∆x /∆v is torsion, we can find an element of prime order l coprime to pv in ∆x /∆v not mapping to zero (note that the order of l divids ord(Ix,K / Vx,K )). This gives us a surjective morphism Autkx (kx ) → Z/lZ.

IM0 /L0 (diff(L0 /K0 )) = diff(M0 /L1 ); ii. iM0 /L0 (qdiff(L0 /K0 )) = qdiff(M0 /L1 ). Proof. Let q = [M0 : K0 ]i . 14 it follows that L0 = M0q , and K0 = Lq1 . The statement then follows easily. 20. Let K ⊆ L ⊆ M be a tower of function fields over a field k. If M/K is separable, or k is perfect, one has i. diff(M/K) = diff(M/L) + iM/L (diff(L/K)); ii. qdiff(M/K) = qdiff(M/L) + iM/L (qdiff(L/K)) . Proof. Assume first that M/K is separable. The case for the different follows from [Ser79, Proposition 8, Chapter III].

11 respectively L ∩ L = K). Contradiction. 5. 16 it is not enough to require that L ∩ L = K. Here is an example where all three statements are false. Consider the extension L = Q(α) of Q where α is a root of x(x − 1)2 + 2. Well-known techniques show that there are two primes above (2), namely p = (2, α) and q = (2, α − 1). One has (2) = pq2 . It follows that Q(α)/Q is not Galois. Hence the Galois closure M of this extension has group S3 . Let α be another root in this Galois closure and let L = Q(α).