Arithmetically Cohen-Macaulay Sets of Points in P^1 x P^1 by Elena Guardo, Adam Van Tuyl

By Elena Guardo, Adam Van Tuyl

This short offers an answer to the interpolation challenge for arithmetically Cohen-Macaulay (ACM) units of issues within the multiprojective area P^1 x P^1.  It collects a few of the present threads within the literature in this subject with the purpose of delivering a self-contained, unified advent whereas additionally advancing a few new rules.  The appropriate buildings on the topic of multiprojective areas are reviewed first, through the elemental homes of issues in P^1 x P^1, the bigraded Hilbert functionality, and ACM units of issues.  The authors then convey how, utilizing a combinatorial description of ACM issues in P^1 x P^1, the bigraded Hilbert functionality may be computed and, for that reason, remedy the interpolation challenge.  In next chapters, they give thought to fats issues and double issues in P^1 x P^1 and exhibit how one can use their effects to respond to questions and difficulties of curiosity in commutative algebra.  Throughout the booklet, chapters finish with a short ancient review, citations of comparable effects, and, the place suitable, open questions that can encourage destiny study.  Graduate scholars and researchers operating in algebraic geometry and commutative algebra will locate this ebook to be a invaluable contribution to the literature.

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Example text

Since we see that that this configuration does not satisfy property ( ), it cannot be ACM. It can be also seen computing SX . We have SX = {(1, 0), (0, 1)} and the two elements are incomparable with respect to the partial order on N2 .

20. 5 satisfy property ( ). The following theorem links these concepts. 21. Let X ⊆ P1 × P1 be a finite set of reduced points. With the notation as above, the following are equivalent: (i) αX = βX∗ (or αX∗ = βX ). (ii) SX has no incomparable elements with respect to the partial order (iii) X satisfies property ( ). on Nv . Proof. 9, the set of points will resemble a Ferrers diagram of the partition αX = (α1 , . . , αh ). But this implies that for each i = 1, . . , h, si = (1, . . , 1, 0, . .

Sh } has no incomparable elements with respect to . 3 Points in P1 × P1 26 (ii) ⇒ (iii) Let A × B and A × B be two points of X with A = A and B = B . Since A, A ∈ π1 (X) = {A1 , . . , Ah }, there exist i, j ∈ {1, . . , h} such that A = Ai and A = Aj . So, consider si and sj ∈ SX . Since B, B ∈ π2 (X) = {B1 , . . , Bv }, there exist k, l ∈ {1, . . , v} such that B = Bk and B = Bl . Without loss of generality, assume that k < l. But this means the k-th coordinate of si is 1 and the l-th coordinate of sj is 1.

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