University of Paderborn Department of Mathematics Diploma Thesis ...
University of Paderborn Department of Mathematics Diploma Thesis ...
University of Paderborn Department of Mathematics Diploma Thesis ...
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4.2. COMPUTING HILBERT POLYNOMIALS OF STABLE IDEALS 81Example 4.4. Let R := K[x 0 , x 1 , x 2 ], i.e. n = 2.(i) The Hilbert polynomial <strong>of</strong> R/I 1 for I 1 = (x 2 0, x 0 x 1 , x 2 1) is computed byMuPAD>> compute_Hilbert_polynomial([[2,0,0],[1,1,0],[0,2,0]], 2);Outputpoly(3, [z])Thus, we obtain p R/I (z) = 3.(ii) The Hilbert polynomial <strong>of</strong> R/I 2 for I 2 = (x 2 0, x 0 x 1 , x 0 x 2 , x 3 1) isMuPAD>> compute_Hilbert_polynomial([[2,0,0],[1,1,0],[1,0,1],[0,3,0]], 2);Outputpoly(3, [z])i.e. p R/I2 (z) = p R/I1 (z) = 3, as expected.(iii) Now, as in Example 4.2 (iii), let R := K[x 0 , x 1 , x 2 , x 3 ]. The Hilbert polynomial <strong>of</strong>R/I 3 , where I 3 = (x 0 , x 2 1, x 1 x 2 2, x 4 2), is obtained byMuPAD>> compute_Hilbert_polynomial([[1,0,0,0],[0,2,0,0],[0,1,2,0],>> [0,0,4,0]], 3);Outputpoly(6, [z])i.e. p R/I3 (z) = 6.(iv) Finally, let R := K[x 0 , x 1 , x 2 , x 3 , x 4 ] and I 4 := (x 0 , x 5 1, x 4 1x 3 2, x 4 1x 2 2x 6 3). In this case,the Hilbert polynomial <strong>of</strong> R/I 4 is not a constant term. Its degree is 2, as we can easily seeby using the procedure stated in 4.3.
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