By Grauert G. (ed.)
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Additional info for Complex Analysis and Algebraic Geometry
14. Suppose that the k-algebra O is a complete discrete k-valuation ring with residue class map η : O F. Assume further that F is a finite separable extension of k. Given any local parameter t, there is a unique isometric ˆ isomorphism µˆ : F[[X]] O such that µ(X) = t. Proof. 12). Define µˆ : F[[X]] → O via µˆ ∑ ai X i i := ∑ µ(ai )t i . i This map is clearly well-defined and injective, and is uniquely determined by µ and t. To show that it is surjective, put F := im(µ). Then O = F + P, and F ∩ P = 0.
Proof. 13) we have D(i) y ( f ) = 0 for 0 < i < p. 4 Residues In this section, we discuss Tate’s elegant theory of abstract residues, closely following . For a variation based on topological ideas, see the appendix of . Let V be a (not necessarily finite-dimensional) vector space over a field k. Recall that a k-linear map y : V → V has finite rank if y(V ) is finite-dimensional. We can generalize this notion by calling y finitepotent if yn (V ) is finite-dimensional for some positive integer n.
Now consider the polynomial D( f ) ∈ A[X]. We have D( f ) ≡ f1 mod M where M is the maximal graded ideal of A. It follows that v is a root of D( f ) modulo M and that D( f ) (v) is invertible modulo M. By Hensel’s Lemma, there is a unique root v1 of D( f ) in A congruent to v modulo M. To each such root there corresponds a unique extension D1 of D to K1 , defined by D1 (u) = v1 . 5. Let K1 /K be a finite extension of fields. Then K1 /K is separable if and only if every derivation of K into a K1 -module M extends uniquely to K1 .