Math8410Ex1.24Jim - Exercise: Endler, Valuation Theory,...

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Unformatted text preview: Exercise: Endler, Valuation Theory, (3.9). Let (K, v ) be a non-Archimedean valued field of residue characteristic p > 0. [Remark: The word ”residue” was omitted when this problem was first posted, which makes the problem almost trivial.] Suppose that x and y both lie in the valuation ring. Show that v (x − y ) > 0 implies v (xp − y p ) > v (x − y ). [stankewicz] Following a hint, let’s recall that x = y + (x − y ) so xp − y p = = (y + (x − y ))p − y p p p −1 p yp + y (x − y ) + · · · + y (x − y )p−1 + (x − y )p − y p 1 p−1 p −1 = (x − y ) (x − y )p−1 + i=1 p−1 = (x − y ) (x − y )p−1 + p i=1 p p−i y (x − y )i−1 i p − 1 p −i y (x − y )i−1 i p −1 Define c to be i=1 p−1 y p−i (x − y )i−1 and note that c is in the valuation i ring (this holds for any commutative ring since we’ve only really used the bip−1 nomial theorem and the distributive law so far) because x, y and p−1 i=1 i are. Thus we have shown that xp − y p = (x − y ) (x − y )p−1 + pc . Thus by the nonarchimedean assumption v ( xp − y p ) = v (x − y ) (x − y )p−1 + pc = v (x − y ) + v (x − y )p−1 + pc ≥ v (x − y ) + min {(p − 1)v (x − y ), v (pc)} Now recall that v (pc) = v (p) + v (c) ≥ v (p) since c lies in the valuation ring. Moreover, the residue characteristic is p which implies among other things that p lies in the maximal ideal of of the valuation ring and thus has positive valuation. Therefore v (pc) > 0. Moreover, v (x − y ) > 0 by assumption, so min {(p − 1)v (x − y ), v (pc)} > 0. Combining all this together, we find that v (xp − y p ) ≥ v (x − y ) + min {(p − 1)v (x − y ), v (pc)} > v (x − y ) as claimed. 1 ...
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This note was uploaded on 10/26/2011 for the course MATH 8410 taught by Professor Staff during the Fall '11 term at University of Georgia Athens.

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