110 CHAPTER 4. ALGORITHMS FOR STABLE IDEALSthese one can find seven saturated stable ideals, whose double saturation differs from thedouble saturation <strong>of</strong> the ideal L p .The set and the number <strong>of</strong> different double saturations, which appear among all saturatedstable ideals to the Hilbert polynomial p(z), is given byMuPAD>> M:= compute_all_ideals(poly(2*z^2 + 2*z + 1, [z]), 4, All):>> M[2];Output{[[2, 0, 0, 0], [1, 1, 0, 0], [0, 3, 0, 0]],[[1, 0, 0, 0], [0, 5, 0, 0], [0, 4, 2, 0]],[[2, 0, 0, 0], [1, 1, 0, 0], [1, 0, 2, 0], [0, 4, 0, 0]],[[2, 0, 0, 0], [1, 1, 0, 0], [1, 0, 1, 0], [0, 5, 0, 0], [0, 4, 1, 0]]}>> M[3];Output4As we can see, there are four different double saturations:I 1 :=(x 2 0, x 0 x 1 , x 3 1),I 2 :=(x 0 , x 5 1, x 4 1x 2 2),I 3 :=(x 2 0, x 0 x 1 , x 0 x 2 2, x 4 1),I 4 :=(x 2 0, x 0 x 1 , x 0 x 2 , x 5 1, x 4 1x 2 ).The ideal I 2 is the double saturation <strong>of</strong> the unique lexicographic ideal L p associated top(z) = 2z 2 + 2z + 1, since L p = (x 0 , x 5 1, x 4 1x 3 2, x 4 1x 2 2x 6 3) and we obtain sat x3 ,x 4(L p ) by settingx 4 := 1 as well as x 3 := 1 in the set <strong>of</strong> minimal generators <strong>of</strong> L p . This providessat x3 ,x 4(L p ) = I 2 .To obtain an impression <strong>of</strong> the geometric objects corresponding to R/I 1 , R/I 2 , R/I 3 andR/I 4 , we consider the primary decompositions <strong>of</strong> the ideals I 1 , I 2 , I 3 and I 4 :I 1 – The ideal I 1 is a primary ideal to the prime ideal Rad I 1 = (x 0 , x 1 ). Hence, R/I 1defines an irreducible surface in 4-dimensional projective space P 4 .I 2 – This ideal is not a primary ideal. Its primary decomposition is given byI 2 = (x 0 , x 4} {{ 1) ∩ (x} 0 , x 5 1, x 2} {{ 2)}=:q 2,1=:q 2,2.
4.5. COMPUTING STABLE IDEALS TO A GIVEN HILBERT POLYNOMIAL 111The associated prime ideals are Rad q 2,1 = (x 0 , x 1 ) and Rad q 2,2 = (x 0 , x 1 , x 2 ). Hence,R/q 2,1 describes an irreducible surface and R/q 2,2 a curve support on a line in P 4 .Thus, R/I 2 corresponds to the union <strong>of</strong> a surface and a curve in P 4 , where the curveis embedded into the support <strong>of</strong> the surface.I 3 – Again, as the ideal I 2 , I 3 is not a primary ideal. Its primary decomposition isI 3 = (x 0 , x 4} {{ 1) ∩ (x 2} 0, x 0 x 1 , x 4 1, x 2} {{ 2)}=:q 3,1=:q 3,2,where Rad q 3,1 = (x 0 , x 1 ) and Rad q 3,2 = (x 0 , x 1 , x 2 ). It follows that R/q 3,1 againdefines an irreducible surface and R/q 3,2 describes a curve support on a line. Hence,R/I 3 corresponds to the union <strong>of</strong> a surface and a curve in P 4 , where the curve isembedded into the support <strong>of</strong> the surface.I 4 – The ideal I 4 is also not a primary ideal. The primary decomposition is given byI 4 = (x 0 , x 4} {{ 1) ∩ (x 2} 0, x 0 x 1 , x 5 1, x 2 )} {{ }=:q 4,1=:q 4,2.We obtain Rad q 4,1 = (x 0 , x 1 ) and Rad q 4,2 = (x 0 , x 1 , x 2 ). As above, R/q 4,1 definesan irreducible surface. R/q 4,2 defines a curve support on a line, such that – as in theupper cases – R/I 4 corresponds to the union <strong>of</strong> a surface and a curve in P 4 , wherethe curve is embedded into the support <strong>of</strong> the surface.Note that we have q 2,1 = q 3,1 = q 4,1 = (x 0 , x 4 1), i.e. the ideal (x 0 , x 4 1) appears is any <strong>of</strong> theprimary decompositions <strong>of</strong> I 2 , I 3 and I 4 . As we saw above, R/(x 0 , x 4 1) defines an irreduciblesurface <strong>of</strong> degree 4. Although, (x 0 , x 4 1) does not appear in a primary decomposition <strong>of</strong> I 1(I 1 is a primary ideal itself), we saw that I 1 also defines a irreducible surface <strong>of</strong> degree 4.In order to distinguish between the geometric objects defined by R/I 1 , R/I 2 , R/I 3 andR/I 4 , we additionally compute their Hilbert polynomials and the first 10 values <strong>of</strong> theirHilbert functions: We obtain (e.g. by using procedure compute Hilbert polynomial presentedin 4.3):p R/I1 (z) = 2z 2 + 2z + 1, p R/I2 (z) = 2z 2 + 2z − 5,p R/I3 (z) = 2z 2 + 2z + 1, p R/I4 (z) = 2z 2 + 2z − 1.As expected, the four Hilbert polynomials only differ within their constant terms. Wehave p R/I1 (z) = p R/I3 (z), whereas p R/I2 (z) and p R/I4 (z) differ from each other and fromthe Hilbert polynomial <strong>of</strong> R/I 1 respectively R/I 3 . To see, from which point on the values<strong>of</strong> the Hilbert functions equal those <strong>of</strong> the Hilbert polynomial (the differences have beenunderlined below), we state the first ten values <strong>of</strong> the Hilbert functions <strong>of</strong> R/I 1 , R/I 2 , R/I 3