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University of Paderborn Department of Mathematics Diploma Thesis ...

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4.8. SOME CONCLUSIONS AND EXPERIMENTAL RESULTS 123polynomial ring R := K[x 0 , x 1 , x 2 , x 3 , x 4 ] for k = 3, 4, 5, 6, i.e.L 3 = (x 0 , x 1 , x 5 2, x 4 2x 3 3),L 4 = (x 0 , x 1 , x 5 2, x 4 2x 4 3),L 5 = (x 0 , x 1 , x 5 2, x 4 2x 5 3),L 6 = (x 0 , x 1 , x 5 2, x 4 2x 6 3).These ideals are the unique lexicographic ideals (see Theorem 2.25) to the Hilbert polynomialsp k (z) = 4z + (k − 2), k = 3, 4, 5, 6, i.e.p R/L3 (z) = 4z + 1,p R/L4 (z) = 4z + 2,p R/L5 (z) = 4z + 3,p R/L6 (z) = 4z + 4.Note that p R/L3 (z) = 4z + 1 is the Hilbert polynomial we already considered in Example4.18 and which is also discussed by Alyson Reeves in [16]. Furthermore, note that we obtainL j+1 from L j , j = 3, 4, 5, by performing one expansion <strong>of</strong> the last monomial generator <strong>of</strong> L j .We want to compute the number <strong>of</strong> all saturated stable ideals I ⊂ R with p R/I (z) = p k (z),k = 3, 4, 5, 6, and the number <strong>of</strong> different double saturations appearing among these ideals.Using the procedure compute all ideals stated in 4.20, we obtain:number <strong>of</strong> saturated number <strong>of</strong> doublestable ideals saturationsk = 3 12 3k = 4 23 3k = 5 43 3k = 6 79 3It is interesting to see that the number <strong>of</strong> all saturated stable ideals to the given Hilbertpolynomial increases fast, whereas the number <strong>of</strong> different double saturations among theseideals remains the same.Before we try to explain these observations, we consider another example: Let L k :=(x 0 , x 3 1, x 2 1x 3 2, x 2 1x 2 2x 2 3, x 2 1x 2 2x 3 x k 4) for k = 3, 4, 5, 6 in R := K[x 0 , x 1 , x 2 , x 3 , x 4 , x 5 ], i.e.L 3 = (x 0 , x 3 1, x 2 1x 3 2, x 2 1x 2 2x 2 3, x 2 1x 2 2x 3 x 3 4),L 4 = (x 0 , x 3 1, x 2 1x 3 2, x 2 1x 2 2x 2 3, x 2 1x 2 2x 3 x 4 4),L 5 = (x 0 , x 3 1, x 2 1x 3 2, x 2 1x 2 2x 2 3, x 2 1x 2 2x 3 x 5 4),L 6 = (x 0 , x 3 1, x 2 1x 3 2, x 2 1x 2 2x 2 3, x 2 1x 2 2x 3 x 6 4).

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