2013
DOI: 10.2478/auom-2013-0043
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Stanley depth of squarefree Veronese ideals

Abstract: We compute the Stanley depth for the quotient ring of a square free Veronese ideal and we give some bounds for the Stanley depth of a square free Veronese ideal. In particular, it follows that both satisfy the Stanley's conjecture.

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Cited by 8 publications
(15 citation statements)
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“…I(P n,3 ) = I(P n,3 ) ∩ R n ⊕ ⊕ n−1 l=1 y n−l (I(P n,3 ) ∩ R l : y n−l )R l ⊕ y n (I(P n,3 ) : y n )S n, 3 . Therefore where J 2 := (x n−2 , z n−2 , x n−1 , z n−1 , x n−3 , y n−3 , z n−3 ), using the same arguments as in case (1) we have sdepth((I(P n,3 ) ∩ R 2 : y n−2 )R 2 ) > ⌈ n 3 ⌉.…”
Section: Definitions and Notationmentioning
confidence: 99%
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“…I(P n,3 ) = I(P n,3 ) ∩ R n ⊕ ⊕ n−1 l=1 y n−l (I(P n,3 ) ∩ R l : y n−l )R l ⊕ y n (I(P n,3 ) : y n )S n, 3 . Therefore where J 2 := (x n−2 , z n−2 , x n−1 , z n−1 , x n−3 , y n−3 , z n−3 ), using the same arguments as in case (1) we have sdepth((I(P n,3 ) ∩ R 2 : y n−2 )R 2 ) > ⌈ n 3 ⌉.…”
Section: Definitions and Notationmentioning
confidence: 99%
“…Let n ≥ 3, if n ≡ 0, 2 (mod 3), then depth(S n,3 /I(C n,3 )) = sdepth(S n,3 /I(C n,3 )) = ⌈ n−1 3 ⌉, and if n ≡ 1 (mod 3), then ⌈ n−1 3 ⌉ ≤ depth(S n,3 /I(C n,3 )), sdepth(S n,3 /I(C n,3 )) ≤ ⌈ n 3 ⌉. Proof.We first prove the result for depth.…”
mentioning
confidence: 99%
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“…Remark 1.3. The (multigraded) Stanley depth of I n,d has been conjectured (see [6] and [10]) to be equal to d + n d+1 / n d . M. Keller et al [10] and M. Ge et al [8] partially confirmed this conjecture but it is still open.…”
Section: Introductionmentioning
confidence: 99%