2023
DOI: 10.1307/mmj/20205951
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On Some Permanence Properties of (Derived) Splinters

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Cited by 7 publications
(5 citation statements)
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“…Then, Rm/xRm$R_{\mathfrak {m}}/xR_\mathfrak {m}$ is F ‐full (strongly F ‐injective) by definition. Hence, Rfrakturm$R_{\mathfrak {m}}$ is F ‐full (strongly F ‐injective) by [25, Theorem 4.2, Corollary 5.16], thus R is F ‐full (strongly F ‐injective) by Proposition 4.8 and [13, Theorem 5.12]. By [12, Lemma 3.4] and [13, Theorem 3.3], Rx$R_x$ is F ‐full (strongly F ‐injective), so condition (1) of G ‐deforming property definition is satisfied.…”
Section: Some G‐deforming F‐singularitiesmentioning
confidence: 99%
See 2 more Smart Citations
“…Then, Rm/xRm$R_{\mathfrak {m}}/xR_\mathfrak {m}$ is F ‐full (strongly F ‐injective) by definition. Hence, Rfrakturm$R_{\mathfrak {m}}$ is F ‐full (strongly F ‐injective) by [25, Theorem 4.2, Corollary 5.16], thus R is F ‐full (strongly F ‐injective) by Proposition 4.8 and [13, Theorem 5.12]. By [12, Lemma 3.4] and [13, Theorem 3.3], Rx$R_x$ is F ‐full (strongly F ‐injective), so condition (1) of G ‐deforming property definition is satisfied.…”
Section: Some G‐deforming F‐singularitiesmentioning
confidence: 99%
“…Hence, Rfrakturm$R_{\mathfrak {m}}$ is F ‐full (strongly F ‐injective) by [25, Theorem 4.2, Corollary 5.16], thus R is F ‐full (strongly F ‐injective) by Proposition 4.8 and [13, Theorem 5.12]. By [12, Lemma 3.4] and [13, Theorem 3.3], Rx$R_x$ is F ‐full (strongly F ‐injective), so condition (1) of G ‐deforming property definition is satisfied. Now condition (2) follows from Lemma 4.7 and [13, Theorem 3.9].$\Box$…”
Section: Some G‐deforming F‐singularitiesmentioning
confidence: 99%
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“…The study of splinters has been a celebrated endeavor with a rich history. The direct summand conjecture, now a theorem, says precisely that regular rings are splinters (refer to [DT20] for a survey, basic properties, and additional references). However we only need to use the following well known and basic facts [Ho73, Lemma 2]: Lemma 2.6.…”
Section: Preliminariesmentioning
confidence: 99%
“…Remark 6.3. Given a finite extension R → S of complete local domains, with R regular, it is not known whether there exists an R-free finitely generated S-module, but there is a countably generated such module M [28] 18 . It would be interesting to know whether there is a countably generated S-algebra which is R-free.…”
Section: Let Us Turnmentioning
confidence: 99%