Questions tagged [prufer-domain]
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7 questions
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Decomposing semihereditary rings
Let $R$ be a commutative semihereditary ring (i.e. every finitely generated ideal of $R$ is projective https://en.wikipedia.org/wiki/Hereditary_ring). Then is $R$ a finite direct product of Prufer ...
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Integral domain $R$ with fraction field $K$ such that for every $u \in K$, the subring $R[u]$ of $K$ is flat $R$-module
Let $R$ be an integral domain with fraction field $K$. If for every $u \in K$, the subring $R[u]$ of $K$ is a flat $R$-module, then is it true that $R$ is a Prufer domain ?
If $R$ were moreover a ...
user111524
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GCD and LCM of elements in Prufer domain
Let $R$ be a Prufer domain. If $0 \ne a \in R$ is such that $Ra \cap Rb$ is principal ideal for every $b \in R$, then is it true that $Ra+Rb$ is also principal for every $b\in R$ ?
Over Prufer ...
user111524
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What kind of arithmetic information does the ring of integers in an infinite extension carry?
The fact that the ring of integers in a finite extension of $\Bbb Q$ is a Dedekind domain and purely algebraic properties of Dedekind domains are absolutely essential for algebraic number theory. So ...
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Indecomposable quotient of Prüfer domains
Let $D$ be a Prüfer domain. I am looking for equivalent condition on an ideal $I$ of $D $ under which $D/I $ is an indecomposable ring.
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On Prufer domains
Are there any Prufer domains that have an infinity of prime ideals but only one of those primes is not finitely generated?
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Is the ring of all cyclotomic integers a Bezout domain?
My previous question about the theorem (apparently due to Dedekind -- thanks, Arturo Magidin!) that the ring $\overline{\mathbb{Z}}$ of all algebraic integers is a Bezout domain got me thinking about ...
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