Computations and Combinatorics in Commutative Algebra by Anna M. Bigatti Philippe Gimenez & Eduardo Sáenz-de-Cabezón

Author:Anna M. Bigatti, Philippe Gimenez & Eduardo Sáenz-de-Cabezón
Language: eng
Format: epub
Publisher: Springer International Publishing, Cham

X (2, 2, 1, …, 1) to

X (1, 2, 1, …, 1) to

and X (2, 1, 1, …, 1) to

so that X (1, 1, …, 1) X (2, 2, 1, …, 1) − X (1, 2, 1, …, 1) X (2, 1, 1, …, 1) is mapped by φ to 0. Thus I G  ⊆ P.

As φ is a homogeneous monomial map of positive degree, P is generated by binomials and does not contain any variables. It follows that I G : (∏ a X a ) ∞  ⊆  P.

Now let f ∈ P. The proof below that f ∈ I G : (∏ a X a ) ∞ is fairly elementary, only long in notation. Since P is the kernel of a homogeneous monomial map, we may assume that for some n-tuples a 1, …, a m , b 1, …, b m . To show that f ∈ I G : (∏ a X a ) ∞ , it suffices to prove that any monomial multiple of f is in I G : (∏ a X a ) ∞ . Fix a non-edge (i, j). Suppose that in a k neither the ith nor the jth component is 1. Let c k be the n-tuple whose ith and jth components are 1 and whose other components agree with the components of a k . Both and lie in the same submatrix of [X a ] a that gives I ij , so that reduces modulo I ij and hence modulo I G to where a k ′ and c k ′ each have entry 1 either in the ith or the jth components. Let U be the product of all such Then modulo I G , Thus U f reduces with respect to I G to a binomial in which the subscripts of all the variables appearing in the first monomial have at least one of i, j components equal to 1, and in the second monomial the number of non-1ith and jth components in the subscripts does not increase. By repeating this for the second monomial as well, we may assume that for each variable appearing in f, the ith or the jth component in the subscript is 1. If we next similarly clean positions i ′ , j ′ in this way, we do not at the same time lose the cleaned property of positions i and j: as factors of the multipliers U keep the clean (i, j) property. By repeating this cleaning, in finitely many rounds we get a binomial f in P such that for each non-edge (i, j) and for each variable appearing in f, the ith or the jth component of the subscript of that variable is 1.

With the assumption that for each non-edge (i, j), the ith or the jth component of a k and of b k is 1, we claim that f = 0 ∈ I G . If a i  = b j for some i, j ∈ [m], then the binomial has the same property of many components being 1, and it suffices to prove that Thus without loss of generality we may assume that m > 0 and that a i  ≠ b j for all i, j ∈ [m]. Let K j (resp. L j ) be the set of all i ∈ [n] such that the ith component in a j (resp. b j ) is not 1. By possibly reindexing we may assume that K 1 is maximal among all such sets.

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