University of Paderborn Department of Mathematics Diploma Thesis ...

28 CHAPTER 2. STABLE IDEALSby Lemma 1.1.Since m 1 > . . . > m r−1 > m r , it follows in particular (m 1 , . . . , m r−1 ) : m r ⊂ (x 0 , . . . , x lr−1).Claim. (m 1 , . . . , m r−1 ) : m r ⊃ (x 0 , . . . , x lr−1).Proof of the Claim. Since I is stable and m r = x a r00 · . . . · x a r,lr−1l r−1· x a rlrl r, we know thatis an element of I. Since I is a monomialthe monomial ˜m := x a r00 · . . . · x a r,lr−1+1l r−1ideal, the monomial ˜m is divisible by one of the generators of I. Surely, m r does not divide· x a rlr −1l r˜m. Hence, there is m ′ ∈ {m 1 , . . . , m r−1 } such that m ′ · x c 00 · . . . · x c lrl rm ′ = x a r0−c 00 · . . . · x a r,lr−1+1−c lr−1l r−1· x a rlr −1−c lrl r.= ˜m, i.e.Because m ′ > m r in lexicographic order we may conclude that the first non-zero entry ofthe exponent vector((a r0 − c 0 ) − a r0 , . . . , (a r,lr−1 + 1 − c lr−1) − a r,lr−1, (a rlr − 1 − c lr ) − a rlr )is positive. Since c j ≥ 0, 0 ≤ j ≤ l r , it follows c lr−1 = 0. This providesgcd(m ′ , m r ) = x a r0−c 00 · . . . · x a r,lr−1l r−1· x a rlr −1−c lrl r,m ′and we getgcd(m ′ , m r ) = x l r−1, which shows that x lr−1 is contained in the set of generatorsof the ideal (m 1 , . . . , m r−1 ) : m r .Next we conclude that x lr−2 is a generator of (m 1 , . . . , m r−1 ) : m r by using the fact that˜m := x a r00 ·. . .·x a r,lr−2+1l r−2·x a r,lr−1l r−1·x a rlr −1l ris an element of I. Iteration of this process provides(m 1 , . . . , m r−1 ) : m r ⊃ (x 0 , . . . , x lr−1).Now we have (m 1 , . . . , m r−1 ) : m r = (x 0 , . . . , x lr−1) and finish the proof of the theorem: Asa consequence of the latter identity, we getWe take advantage ofR/((m 1 , . . . , m r−1 ) : m r ) ∼ = K[x lr , . . . , x n ] ∼ = K[x 0 , . . . , x n−lr ].H K[x0 ,...,x n−lr ](t) =1 (1 − t)lr=(1 − t)n−lr+1(1 − t) . n+1Since (m 1 , . . . , m r−1 ) still is a stable ideal, we know the Hilbert series of R/(m 1 , . . . , m r−1 )by induction. Together with (∗), we obtainH R/I (t) = 1 − ∑ r−1i=1 (1 − t)li · t d i(1 − t) lr− t dr ·(1 − t) n+1 (1 − t) = 1 − ∑ ri=1 (1 − t)li · t d i.n+1 (1 − t) n+1