University of Paderborn Department of Mathematics Diploma Thesis ...

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 of a surface and a curve in P 4 , where the curveis embedded into the support of 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 of a surface and a curve in P 4 , where the curve isembedded into the support of 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 of a surface and a curve in P 4 , wherethe curve is embedded into the support of 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 of theprimary decompositions of I 2 , I 3 and I 4 . As we saw above, R/(x 0 , x 4 1) defines an irreduciblesurface of degree 4. Although, (x 0 , x 4 1) does not appear in a primary decomposition of I 1(I 1 is a primary ideal itself), we saw that I 1 also defines a irreducible surface of 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 of 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 of R/I 1 respectively R/I 3 . To see, from which point on the valuesof the Hilbert functions equal those of the Hilbert polynomial (the differences have beenunderlined below), we state the first ten values of the Hilbert functions of R/I 1 , R/I 2 , R/I 3