2001
DOI: 10.1081/agb-100106812
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Going Down: Ascent/Descent

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Cited by 2 publications
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“…Now, let T be any overring of R. Then T is an overring of R , and so T is a going-down domain. As T ⊆ T satisfies going-down, [32,Lemma B] gives that T must be a going-down domain.…”
Section: Proof It Is Clear That (I) ⇒ (Ii) Conversely Suppose (Ii)mentioning
confidence: 99%
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“…Now, let T be any overring of R. Then T is an overring of R , and so T is a going-down domain. As T ⊆ T satisfies going-down, [32,Lemma B] gives that T must be a going-down domain.…”
Section: Proof It Is Clear That (I) ⇒ (Ii) Conversely Suppose (Ii)mentioning
confidence: 99%
“…(iv) ⇒ (v): Assume (iv). By[32, Lemma D], it suffices to prove that if Q ∈ Spec(R ), with n := ht R (Q) (< ∞), and q := Q ∩ R (∈ Max(R)), with m := ht R (q) (< ∞), then n = m. In view of the conditions in (iv), we can assume, without loss of generality, that Q is a non-maximal prime ideal of R and that q is unibranched in R . As the assertion is trivial if Q (or q) is (0), we can also assume that n = 0 and m = 0.…”
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confidence: 99%
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“…Let R be a universally going-down weak Baer ring and an absolutely flat ring morphism f : R → S. Then S is a universally going-down weak Baer ring. McAdam proved a result (see [23]) that gives in our context an ascent criterion: if R → S is an absolutely flat ring morphism between weak Baer rings and R is a going-down ring, then S is a going-down ring if and only if S is treed. We deduce from this criterion the next result.…”
mentioning
confidence: 99%