By L. D. Olson

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If pa (S) ≤ 0 then δS − d−2 ≤ −g − d + 4, which is negative unless d = 4, in which 2 case (ii) is trivial. 5) h1 (HS , OHS (i)), i=1 where pg (S) := h (S, OS (KS )) is the geometric genus of S. 1) holds by (ii). e. 1) with strict inequality. When δS = d−2 2 the previous argument yields e ≤ 0, so pa (S) ≤ 0, and S is a rational scroll by (ii). 5) we deduce pa (S) ≤ h1 (HS , OHS (1)). By Cliﬀord Theorem we have h0 (HS , OHS (1)) ≤ 1 + d2 . Therefore pa (S) ≤ h1 (HS , OHS (1)) = h0 (HS , OHS (1)) − (1 − g + d) ≤ g − d2 .

Hence, it suﬃces to show that k[u1 , u2 , . . , un ] is integral over k[u1 + u2 + · · · + un ] under the same hypothesis. We prove the statement (with η1 = · · · = ηn = 1) by induction on n and assume n ≥ 2 since the assertion is trivial for n = 1. We ﬁrst show the base case 1 α2 n = 2. For simplicity on the notation, in this case we write uα 1 u2 = 0. This α1 α2 +1 = 0 so we may assume that α2 is odd. By a straightforward implies u1 u2 computation, 1 +α2 uα 1 We conclude 1 +α2 uα − 1 α2 α2 α1 α2 1 2 = uα + uα 1 (u2 + u1 ) = u1 {[(u1 + u2 ) − u1 ] 1 } α1 α2 α2 α2 −1 = u1 [(u1 + u2 ) − 1 (u1 + u2 ) u1 + · · · α2 −1 2 (u + u )u ].

Introduction and Motivation Let R be a commutative ring with identity and let J ⊆ I be ideals of R. I is integral over J, if for every element u in I, there exist elements ai in J i such that un + a1 un−1 + a2 un−2 + · · · + an−1 u + an = 0. J is a reduction of I, if I m+1 = JI m for some positive integer m. f. 1]). The reduction of ideals was ﬁrst introduced by Northcott and Rees [NR] and Rees [R] extended the notion to modules. Since then the reduction of ideals and modules have been discussed extensively.