In the proof Proposition 8.4 we need to show that if Z is fifteen points in the plane defined by the (5x5)-minors of a general (5x6) matrix of linear forms, then there are no curves of degree 8, singular along Z.
i1 : kk=ZZ/101
o1 = kk
o1 : QuotientRing
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i2 : S=kk[x0,x1,x2]
o2 = S
o2 : PolynomialRing
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i3 : M=random(S^5,S^{6:-1});--a random (5x6) matrix linear forms in P2
5 6
o3 : Matrix S <-- S
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i4 : m5=minors(5,M);--the ideal of 15 points
o4 : Ideal of S
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i5 : m52=saturate m5^2;--the square of the ideal of 15 points, saturated, with no forms of degree 8.
o5 : Ideal of S
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i6 : betti res m52
0 1 2
o6 = total: 1 10 9
0: 1 . .
1: . . .
2: . . .
3: . . .
4: . . .
5: . . .
6: . . .
7: . . .
8: . 10 9
o6 : BettiTally
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In Y, a smooth 4-fold in $\mathbb{P}^7$ defined as the rank 2 locus of (3x5) matrix M of linear forms, let S1 be the 3-fold scroll defined as the rank 1 locus of the submatrix M12 of M consisting of its two first rows, and let S2 be the 3-fold scroll defined as the rank 1 locus of the submatrix M23 of M consisting of its two last rows Then we show that every cubic hypersurface that contains both S1 and S2, contains all of Y. Hence this is the case for any divisor on Y equivalent to S1+S2.
i7 : kk=ZZ/101
o7 = kk
o7 : QuotientRing
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i8 : S=kk[x0,x1,x2,x3,x4,x5,x6,x7];
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i9 : M=random(S^3,S^{5:-1});--a random (3x5) matrix linear forms in P7
3 5
o9 : Matrix S <-- S
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i10 : m3=minors(3,M);--the ideal of a 4-fold Y,
o10 : Ideal of S
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i11 : M12=submatrix (M, {0,1},{0,1,2,3,4});
2 5
o11 : Matrix S <-- S
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i12 : M23=submatrix (M, {1,2},{0,1,2,3,4});
2 5
o12 : Matrix S <-- S
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i13 : S1=minors (2,M12);-- a divisor in Y of degree 5
o13 : Ideal of S
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i14 : S2=minors (2,M23);-- another divisor equivalent to S1
o14 : Ideal of S
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i15 : S12=saturate intersect(S1,S2);--the divisor S1+S2 on Y
o15 : Ideal of S
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i16 : betti res S12 --the ten cubics are the same as the ones in the ideal of Y
0 1 2 3 4 5 6 7
o16 = total: 1 60 240 426 420 240 75 10
0: 1 . . . . . . .
1: . . . . . . . .
2: . 10 15 6 . . . .
3: . 50 225 420 420 240 75 10
o16 : BettiTally
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