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Proceedings of the American Mathematical Society
Proceedings of the American Mathematical Society
ISSN 1088-6826(online) ISSN 0002-9939(print)

 

A note on the Cohen-Macaulay type of lines in uniform position in $ {\bf A}\sp{n+1}$


Author: William C. Brown
Journal: Proc. Amer. Math. Soc. 87 (1983), 591-595
MSC: Primary 13H10; Secondary 13H15, 14B05
MathSciNet review: 687623
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Abstract: Let $ {\mathfrak{L}_1}, \ldots ,{\mathfrak{L}_s}$ be $ s$-distinct lines in $ {\mathbf{A}}_k^{n + 1}$ passing through the origin. Assume $ s = (_n^{n + d}) - \lambda $ where $ n$, $ d \geqslant 2$. If $ {\mathfrak{L}_1}, \ldots ,{\mathfrak{L}_s}$ are in generic $ s$-position, and $ \lambda = 0$. $ 1, \ldots ,n - 1$, then the Cohen-Macaulay type, $ t({\mathfrak{L}_1}, \ldots ,{\mathfrak{L}_s})$, of $ {\mathfrak{L}_1}, \ldots ,{\mathfrak{L}_s}$ is given by the following formula: $ t({\mathfrak{L}_1}, \ldots ,{\mathfrak{L}_s}) = (_{n - 1}^{n + d - 1}) - \lambda $. This formula is known to be false for $ \lambda = n$. In this paper, we show that if $ {\mathfrak{L}_1}, \ldots ,{\mathfrak{L}_s}$ are in uniform position, and $ \lambda = n$. then $ t({\mathfrak{L}_1}, \ldots ,{\mathfrak{L}_s}) = (_{\;n - 1}^{n + d - 1}) - n$.


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DOI: http://dx.doi.org/10.1090/S0002-9939-1983-0687623-X
PII: S 0002-9939(1983)0687623-X
Keywords: Cohen-Macaulay type, generic $ s$-position, uniform position
Article copyright: © Copyright 1983 American Mathematical Society