Dyson's Lemma and a Theorem of Esnault and Viehweg

Two ingredients are needed to prove Lemma 4.1: first an analysis of fk−1 I(s) and secondan analysis of L|V k . In the case where X = P the reader can find an explicit statement andproof of both steps in [EV1] Lemma 2.9.Lemma 4.3 (cf. [EV1] Lemma 2.9i) Let T i = {j : P j ≠ π j (ζ i )} with notation as in Lemma4.1 and write⎧⎫⎨τ i = max⎩ t i − ∑ ⎬a j d j , 0⎭ , 1 ≤ i ≤ M.j∈T iLet ζ ′ i = π {k+1,...,m}(ζ i ) and let d ′ = (d k+1 , . . .,d m ). Thenf −1k [I(s)] = ⋂I ζ ′i ,d ′ ,τ ii∈SProof: This can be shown by direct computation as in [EV1] Lemma 2.9 (i) by consideringgenerators of I ζi ,d,t i.Lemma 4.4 Using the notation of Lemma 4.1, write Z(s) = Z(s 1 ) + β 1 V 1 such that V 1 ⊄supp[Z(s 1 )] and inductivelyAlso writeZ(s i |V i ) = Z(s i+1 ) + β i+1 V i+1 with V i+1 ⊄ supp[Z(s i+1 )].t ′ i = ind a (ζ ′ i, s k |V k ).Then τ i > t ′ i for at most one i. Moreover, in this case if e = τ i − t ′ i > 0 then for all j ≠ ieither τ j = 0 or t ′ j − τ j ≥ e.Proof: One can directly verify, following the method of Lemma 2.2, that⎧ ⎫⎨t ′ j ≥ max ⎩ t j − ∑ a i d i + ∑a i β i − ∑ ⎬a i β i , 0 , 1 ≤ j ≤ M. (4.4.1)⎭i∈T j i∈T j i∉T jFor all inidices j for which T j = {1, . . .,k} it is clear that t ′ j ≥ τ j and that if τ j ≠ 0 thent ′ j − τ j =m∑i=k+1a i β i ≥ e,the last inequality following from (4.4.1) and Lemma 4.3. So suppose T j ≠ {1, . . .,k} andτ j > 0. For simplicity of notation, we assume that j = 2 and that τ 1 − t ′ 1 = e > 0. Then⎛⎞2∑ 2∑ 2∑t ′ k ≥ τ k + ⎝ ∑a i β i − ∑a i β i⎠k=1 k=1 k=1 i∈T k i∉T k18