By Michael Jollenbeck, Volkmar Welker
Quantity 197, quantity 923 (end of volume).
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Example text
Clearly, w | lcm(B). We define w := v xt,min ({i1 , M (u0 )}) xMj1(u ) 0 The critical vertices in homological degree l1 + 1 ≥ 1 (resp. κ + l1 ≥ 1) are now given by Type I: ⎧ ⎫ rt ⎪ ⎪ v xt,max ({i1 })up0 u, ⎨ ⎬ prt B1+l1 = xi u0 ⎪ u, j = 1, . . , l1 ⎪ ⎩ vj xt,min ({mi1 (u0 )}) xm 1 (u0 ) ⎭ i1 [p] with 1 ≤ i1 < M (u0 ) and v1 , . . , vl1 ∈ I Lκ ≥ 1) are now given by Type I: B1+l1 = ⎧ ⎪ ⎨ rt v xt,max ({i1 })up0 ⎪ ⎩ v1j xt,min ({mi1 (u0 )}) u, xi1 u0 xmi (u0 ) 1 prt ⎫ ⎪ ⎬ u, j = 1, . . , l1 ⎪ ⎭ , 3. CELLULAR MINIMAL RESOLUTION FOR A CLASS OF p-BOREL FIXED IDEALS 47 [p] 1 ≤ i1 < M (u0 ) and v11 , . . +lκ = 2j t,min 1 0 xM (u0 ) ⎪ ⎪ . ⎪ ⎪ .. ⎪ ⎪ ⎪ ⎪ ⎪ xiκ u0 ⎩ v x κj t,min ({i1 , M (u0 )}) xM (u ) rt p prt prt 0 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ u, j = 1, . . , l1 , ⎪ ⎪ ⎪ ⎪ ⎬ u, j = 1, . . , l2 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u, j = 1, . . , lκ ⎭ [p] I Step 6. In the second step we have to remove a monorr mial vi upi u from B. In order to have a cycle we have to add this element again. rt Since v0 up0 u is the smallest monomial of B, the construction of the monomials w in Step 1,. . , Step 6 implies that it can never be added in the cycle. Hence all sets rt occurring in the cycle have smallest element v0 up0 u. The construction of each monomial w in Step 1,. . , Step 6 depends either on the monomial u0 or on the monomials u0 and uj for some j.