By Hager A.W.

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A point of Jc(IQ) corresponds to an ideal class of the integral closure of lQ[x] in F(x, y) := lQ[x, y] IUd which is a free module of rank 2 over lQ[x] with integral basis (1, Ll c ) (Lla is the discriminant of F). t. this basis. 42 G. Frey, M. 1. Splitting of SJ]ew(N) 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 1 0 0 1 1 0 1 0 1 1 1 2 2 1 0 2 1 2 2 3 2 1 3 3 3 1 2 4 3 3 3 5 3 4 3 1 0 0 1 1 0 1 0 1 1 1 0 2 1 0 2 1 0 2 1 2 1 1 1 3 1 2 2 3 1 3 1 3 1 1 1(1) 46 47 48 1(1) 49 1(1) 50 51 1(1) 52 53 54 1(1) 1(1) 55 1(1) 56 57 58 2(1) 1(1) 59 60 1(2) 61 62 1(1) 63 2(1) 64 65 1(1) 2(1) 66 1(1) 67 1(1) 68 69 1(1) 1(1), 2(1) 70 1(1) 71 1(2) 72 73 1(2) 1(1), 2(1) 74 1(1) 75 3(1) 76 1(1) 77 1(1), 2(1) 78 1(1) 79 1(1) 80 5 4 3 1 2 5 5 4 4 5 5 5 6 5 7 4 7 5 3 5 9 5 7 7 9 6 5 5 8 5 8 7 11 6 7 1 4 1 1 2 3 1 4 2 3 2 3 2 5 0 4 3 3 1 5 3 5 2 3 1 6 1 5 4 3 1 5 1 6 2 1(1) 4(1) 1(1) 1(1) 1(2) 1(1), 2(1) 1(1) 1(1), 3(1) 1(2) 1(1), 2(1) 1(2) 1(3) 1(2) 5(1) 1(1), 3(1) 1(1), 2(1) 1(1), 2(1) 1(1) 1(1), 2(2) 1(3) 1(1), 2(2) 2(1) 1(1), 2(1) 1(1) 3(2) 1(1) 1(1), 2(2) 2(2) 1(3) 1(1) 1(3), 2(1) 1(1) 1(1), 5(1) 1(2) Explanation of the fourth column: The notation a(b) means that SJ]ew(N) has b a-dimensional'll'N[GQ]-invariant subspaces.

S. 1. en does not depend on the level N and the weight k. 2. The 'claBsical' method of computing the Hecke operator on M is to use the isomorphism (17) to compute the Hecke operator on H1 (Xo(N), cusp, Z) and to map the result back to the space M. To do this explicitly one has to calculate a continued fraction expansion of some rational numbers (see [6] or [28]). ' 3. The same formula holds if we replaces M by M+ corresponding to the +l-eigenspace under complex conjugation (see below). 9. Take N = 31.

For the same reason as in characteristic two, we restrict our attention to the case whtlre v«(3) = -k with some kEN, k = 61 - m, m = {O, 1, ... , 5}. 7) where /3 := 11'61 (3 has v(/3) = m. 8 Proposition. , if m E {I, 2, 4, 5}. The conductors f(EI K) and Kodaira types are f = 6l + 1 = k + 2, type = II for m = 1, f = 6l = k + 2, type = IV for m = 2, f = 6l- 2 = k + 2, type = IV* for m = 4 and f = 6l- 3 = k + 2, type = II* for m = 5. Proof. 1 = 11'61/3. From this, the first two cases are straightforward.

### Alpha-cut-complete Boolean algebras by Hager A.W.

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