University of Paderborn Department of Mathematics Diploma Thesis ...

86 CHAPTER 4. ALGORITHMS FOR STABLE IDEALSwhenever the value of the integer a 0 equals 0 or the Hilbert polynomial is a constantpolynomial. Hence, in this case we have to remove the monomial x a 00 · x a 11 · . . . · x a 2+1n−2 fromthe set of generators. This monomial corresponds to x 3 1 in our example.We now use the code listed in 4.7 to compute the lexicographic ideals to the Hilbertpolynomials of Example 2.26.Example 4.8. Let R := K[x 0 , x 1 , x 2 , x 3 ], i.e. n = 3.(i) p(z) = 2z + 1.MuPAD>> compute_L_p(poly(2*z+1,[z]), 3);Output[[1, 0, 0, 0], [0, 2, 0, 0]]Indeed, the two lists [1, 0, 0, 0] and [0, 2, 0, 0] correspond to the monomials x 0 , x 2 1, i.e. L p =(x 0 , x 2 1).(ii) p(z) = 3z + 1.MuPAD>> compute_L_p(poly(3*z+1,[z]), 3);Output[[1, 0, 0, 0], [0, 4, 0, 0], [0, 3, 1, 0]]i.e. the lexicographic ideal L p is generated by x 0 , x 4 1 and x 3 1x 2 .(iii) p(z) = 2z 2 + 2z + 1 and R := K[x 0 , x 1 , x 2 , x 3 , x 4 ], i.e. n = 4.MuPAD>> compute_L_p(poly(2*z^2 + 2*z + 1,[z]), 4);Output[[1, 0, 0, 0, 0], [0, 5, 0, 0, 0], [0, 4, 3, 0, 0], [0, 4, 2, 6, 0]]i.e. L p = (x 0 , x 5 1, x 4 1x 3 2, x 4 1x 2 2x 6 3). As we can see, the lexicographic ideals computed via 4.7equal those of Example 2.26.In all of these examples, we can enlarge the value of n to obtain the specific lexicographicideal in a polynomial ring with more variables. One can see that there will be exactly one