2013
DOI: 10.1016/j.jalgebra.2013.06.029
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Subintegrality, invertible modules and polynomial extensions

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Cited by 5 publications
(6 citation statements)
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“…The following result is due to Sadhu and Singh (Theorem 1.5 of [5]) which we use frequently throughout this paper. Proof.…”
Section: The Map I(a B) → I(a[x X −1 ] B[x X −1 ])mentioning
confidence: 99%
See 1 more Smart Citation
“…The following result is due to Sadhu and Singh (Theorem 1.5 of [5]) which we use frequently throughout this paper. Proof.…”
Section: The Map I(a B) → I(a[x X −1 ] B[x X −1 ])mentioning
confidence: 99%
“…Proof. By Lemma 2.2 of [5], it is enough to show that ϕ (A, D, B) is surjective for every subring D of C containing A such that D is finitely generated as an A-algebra. Let such a ring D be given.…”
Section: Proofmentioning
confidence: 99%
“…) is the direct sum of I(A, B), n terms of the form LI(A, B) and 2 i n i terms of the form N i I(A, B), 1 ≤ i ≤ n. Since we know from [8] that NI(A, B) = 0 is equivalent to A being seminormal in B (Definition 6.5) we can further conclude: Corollary 1.3. For A ⊂ B, the following are equivalent:…”
Section: Corollary 12 For Every Ring Extension a ⊂ B I(a[t ] B[t ]mentioning
confidence: 85%
“…Since we know from [8] that NI(A, B) = 0 is equivalent to A being seminormal in B (Definition 6.5) we can further conclude: It is immediate from our Main Theorem that LI(A, B) is a torsionfree group (we give a simple proof in Corollary 3.6); it is free abelian of finite rank when A is pseudo-geometric and finite dimensional (Proposition 3.7).…”
Section: Corollary 12 For Every Ring Extension a ⊂ B I(a[t ] B[t mentioning
confidence: 88%
“…Note that NK 0 (f ) ∼ = NPic (f). By Theorem 1.5 of [15], NPic (f) = 0 if and only if f is seminormal. Therefore, NK 0 (f ) = 0 and hence f is not K 0 -regular.…”
Section: On Regularitymentioning
confidence: 98%