University of Paderborn Department of Mathematics Diploma Thesis ...

36 CHAPTER 2. STABLE IDEALSis stable, since its generators fulfill the condition of Theorem 2.7. Furthermore, L(a 0 , . . . , a r )is saturated by Theorem 2.13, because the last variable x n does not appear in any of itsgenerators.We have to show that L(a 0 , . . . , a r ) is a lexicographic ideal. The monomials in{x 0 , x 1 , . . . , x n−r−2 ,x ar+1n−r−1,x arn−r−1 · x a r−1+1n−r ,x arn−r−1 · x a r−1n−r · x a r−2+1n−r+1 , . . . ,x arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 2n−3 · x a 1+1n−2 ,x arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 1n−2 · x a 0n−1}are ordered in lexicographic order. The vector space [L(a 0 , . . . , a r )] 1 is generated byx 0 , . . . , x n−r−2 , which are the n − r − 1 greatest monomials in M of degree 1. It followsthat the vector spaces [L(a 0 , . . . , a r )] j are also lexicographic subsets of [R] j for 2 ≤ j ≤ a r :The only monomial generator of degree a r + 1 is x ar+1n−r−1. Since L(a 0 , . . . , a r ) is stable,( ) s xj· x ar+1n−r−1 ∈ L(a 0 , . . . , a r )x n−r−1for all j < n−r−1 and s < a r +1. Thus, and because of x 0 , . . . , x n−r−2 ∈ L(a 0 , . . . , a r ), thevector space [L(a 0 , . . . , a r )] ar+1 is a lexicographic subset of [R] ar+1. Analogous conclusionsshow that [L(a 0 , . . . , a r )] j is a lexicographic subset for all j > a r + 1, since from one to thenext monomial generator, we always take into account the next variable in lexicographicorder.To show that p R/L(a0 ,...,a r)(z) = p(z), we first write m r = a r , m r−1 = a r + a r−1 , . . . ,m 0 = a r + a r−1 + . . . + a 0 . Thus, we write p(z) in the formp(z) =r∑i=0[( ) z + i−i + 1( z + i −∑ rj=i a ji + 1Now we prove the assertion by induction on r. ( ( )z z − a0The case r = 0 provides the Hilbert polynomial p(z) = − . The ideal L(a 0 )1)1is generated by the monomials x 0 , . . . , x n−2 , x a 0n−1. Thus,R/L(a 0 ) = K[x 0 , . . . , x n ]/(x 0 , . . . , x n−2 , x a 0n−1) ∼ = K[x n−1 , x n ]/(x a 0n−1)( ( )z z − a0and therefore p R/L(a0 )(z) = a 0 = − = p(z).1)1Now let r ≥ 1. We will make use of some results, which also occurred in the proof ofTheorem 2.17 and in Lemma 2.16. Finally, the exact sequence used to prove Lemma 2.16)].