Componentwise Linearity Under Square-Free Gr\"obner Degenerations
Using the recent results on square-free Gr\"obner degenerations by Conca and Varbaro, we proved that if a homogeneous ideal $I$ of a polynomial ring is such that its initial ideal $\mathrm{in}_
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creator | Yu, Hongmiao |
description | Using the recent results on square-free Gr\"obner degenerations by Conca and
Varbaro, we proved that if a homogeneous ideal $I$ of a polynomial ring is such
that its initial ideal $\mathrm{in}_ |
doi_str_mv | 10.48550/arxiv.2303.05190 |
format | Article |
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Varbaro, we proved that if a homogeneous ideal $I$ of a polynomial ring is such
that its initial ideal $\mathrm{in}_<(I)$ is square-free and $\beta_0(I) =
\beta_0(\mathrm{in}_<(I))$, then $I$ is a componentwise linear ideal if and
only if $\mathrm{in}_<(I)$ is a componentwise linear ideal. In particular, if
furthermore one of $I$ and $\mathrm{in}_<(I)$ is componentwise linear, then
their graded Betti numbers coincide.</description><identifier>DOI: 10.48550/arxiv.2303.05190</identifier><language>eng</language><subject>Mathematics - Commutative Algebra</subject><creationdate>2023-03</creationdate><rights>http://arxiv.org/licenses/nonexclusive-distrib/1.0</rights><oa>free_for_read</oa><woscitedreferencessubscribed>false</woscitedreferencessubscribed></display><links><openurl>$$Topenurl_article</openurl><openurlfulltext>$$Topenurlfull_article</openurlfulltext><thumbnail>$$Tsyndetics_thumb_exl</thumbnail><link.rule.ids>228,230,777,882</link.rule.ids><linktorsrc>$$Uhttps://arxiv.org/abs/2303.05190$$EView_record_in_Cornell_University$$FView_record_in_$$GCornell_University$$Hfree_for_read</linktorsrc><backlink>$$Uhttps://doi.org/10.48550/arXiv.2303.05190$$DView paper in arXiv$$Hfree_for_read</backlink></links><search><creatorcontrib>Yu, Hongmiao</creatorcontrib><title>Componentwise Linearity Under Square-Free Gr\"obner Degenerations</title><description>Using the recent results on square-free Gr\"obner degenerations by Conca and
Varbaro, we proved that if a homogeneous ideal $I$ of a polynomial ring is such
that its initial ideal $\mathrm{in}_<(I)$ is square-free and $\beta_0(I) =
\beta_0(\mathrm{in}_<(I))$, then $I$ is a componentwise linear ideal if and
only if $\mathrm{in}_<(I)$ is a componentwise linear ideal. In particular, if
furthermore one of $I$ and $\mathrm{in}_<(I)$ is componentwise linear, then
their graded Betti numbers coincide.</description><subject>Mathematics - Commutative Algebra</subject><fulltext>true</fulltext><rsrctype>article</rsrctype><creationdate>2023</creationdate><recordtype>article</recordtype><sourceid>GOX</sourceid><recordid>eNpjYJA0NNAzsTA1NdBPLKrILNMzMjYw1jMwNbQ04GRwdM7PLcjPS80rKc8sTlXwycxLTSzKLKlUCM1LSS1SCC4sTSxK1XUrSk1VcC-KUcpPygOKuqSmpwLpxJLM_LxiHgbWtMSc4lReKM3NIO_mGuLsoQu2LL6gKDM3sagyHmRpPNhSY8IqAEfZNyk</recordid><startdate>20230309</startdate><enddate>20230309</enddate><creator>Yu, Hongmiao</creator><scope>AKZ</scope><scope>GOX</scope></search><sort><creationdate>20230309</creationdate><title>Componentwise Linearity Under Square-Free Gr\"obner Degenerations</title><author>Yu, Hongmiao</author></sort><facets><frbrtype>5</frbrtype><frbrgroupid>cdi_FETCH-arxiv_primary_2303_051903</frbrgroupid><rsrctype>articles</rsrctype><prefilter>articles</prefilter><language>eng</language><creationdate>2023</creationdate><topic>Mathematics - Commutative Algebra</topic><toplevel>online_resources</toplevel><creatorcontrib>Yu, Hongmiao</creatorcontrib><collection>arXiv Mathematics</collection><collection>arXiv.org</collection></facets><delivery><delcategory>Remote Search Resource</delcategory><fulltext>fulltext_linktorsrc</fulltext></delivery><addata><au>Yu, Hongmiao</au><format>journal</format><genre>article</genre><ristype>JOUR</ristype><atitle>Componentwise Linearity Under Square-Free Gr\"obner Degenerations</atitle><date>2023-03-09</date><risdate>2023</risdate><abstract>Using the recent results on square-free Gr\"obner degenerations by Conca and
Varbaro, we proved that if a homogeneous ideal $I$ of a polynomial ring is such
that its initial ideal $\mathrm{in}_<(I)$ is square-free and $\beta_0(I) =
\beta_0(\mathrm{in}_<(I))$, then $I$ is a componentwise linear ideal if and
only if $\mathrm{in}_<(I)$ is a componentwise linear ideal. In particular, if
furthermore one of $I$ and $\mathrm{in}_<(I)$ is componentwise linear, then
their graded Betti numbers coincide.</abstract><doi>10.48550/arxiv.2303.05190</doi><oa>free_for_read</oa></addata></record> |
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subjects | Mathematics - Commutative Algebra |
title | Componentwise Linearity Under Square-Free Gr\"obner Degenerations |
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