By Alain Escassut

The behaviour of the analytic components on an infraconnected set D in ok an algebraically closed whole ultrametric box is principally defined by means of the round filters and the monotonous filters on D, particularly the T-filters: zeros of the weather, Mittag-Leffler sequence, factorization, Motzkin factorization, greatest precept, injectivity, algebraic houses of the algebra of the analytic components on D, difficulties of analytic extension. this is often utilized to the differential equation y'=hy (y,h analytic components on D), analytic interpolation, p-adic crew duality on meromorphic items and to the p-adic Fourier rework

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D — 1. Thus, we have polynomials V n , Wn satisfying v(Vng -f Wnh — x - n) > 0, v(Vn) > 0, v(W n ) > 0, deg(yn) < deg(h), deg(Wn) < deg(g). We put rf-i An = v(Vng + Wnh - x n ) , (0 < n < d - 1). Now let Q = ^ a n x n , let V = n=0 d-1 y2 d-1 a n^ni Wn — S2 anWn and let A(#, h) = n=0 n=0 v(Vg + Wh-Q) > v y X(g,h). min A n . Clearly we have ~ ~ min (v(a n ) + An) > ' ~ 0 min v(an), v(W) > v(Q), deg(V) < 0

This ends the proof. 4: Let L be complete and let ft be an algebraic closure of L provided with the unique absolute value \ . \ that extends the one of L. Then UQ is equal to the integral closure ofU^. Besides \ft\ = { tfr\r G \L\). 5: Let B be a subfield of L. The residue class field of B is a subfield ofC. 6: Let L be complete, and let ft be an algebraic closure of L provided with the unique absolute value extending the one of L. Then the residue class field of ft is an algebraic closure of C.

1 and a0 i M\. 5: (Eisenstein) Let L have a discrete valuation . Let P(x) 2_^ajXJ € L[x] be an Eisenstein polynomial = Then P is irreducible in L[x]. j=o Proof : We suppose P not irreducible. Then P splits in L[x] in the form m S(x)T(x) n a xJ with S(x) = J2 i > T ^) = j=0 1L,PJXJ and <*m = Pn = 1. Since S,T j=0 are monic we have ||5|| > 1, ||T|| > 1 and since ||5||||T||||5T|| - ||P|| = 1, we have | | 5 | | = ||T|| = 1. Hence, both 5 , T belong to UL[x\. First we notice that if ao belongs to ML then f30 does not, because a 0 ^ (ML)2- Hence we may assume ceo G ML and f30 £ ML- Besides we have OL3 £ ML for every j = 0 , .

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