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The definition of an arithmetical ring states that

A ring $R$ is arithmetical if the ideal lattice is distributive or equivalently $R$ is locally a valuation ring.

I was reading a paper where arithmetical rings are defined as the rings in which every finitely generated ideal is locally principal. How can we prove that the last definition make sense with the former? A proof woud be much appreciated.

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    $\begingroup$ Can you cite the paper in question? $\endgroup$
    – Sam Hopkins
    Commented Aug 20, 2023 at 22:07
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    $\begingroup$ Here is a link to the paper Check Page 102 - sciencedirect.com/science/article/pii/0021869385900961/… $\endgroup$
    – Amit Phogat
    Commented Aug 22, 2023 at 21:17
  • $\begingroup$ What is a "valuation ring" in your definition of arithmetical ring? I suspect you intend for it to mean a ring which ideals are totally ordered by inclusion, in which case your question is very easy to answer. On the other hand, if "valuation ring" is taken in the standard sense of a Manis valuation, then I believe the statement of your question is incorrect (I think that in most sense of valuation ring with zero divisors, local valuation rings need not have totally ordered ideals. On the other hand, a local ring has totally ordered ideals iff its f.g. ideals are principally generated) $\endgroup$
    – Badam Baplan
    Commented Aug 30, 2023 at 14:21
  • $\begingroup$ It's the first one. $\endgroup$
    – Amit Phogat
    Commented Sep 2, 2023 at 5:10
  • $\begingroup$ @AmitPhogat Ok. If a local ring has f.g. ideals principally generated, then let $a_1, a_2$ be elements. Then find $b_1, b_2, a_1', a_2', c$ such that $a_1 b_1 + a_2 b_2 = c$, $ca_1' = a_1, ca_2' = a_2$. Thus $c (a_1' b_1 + a_2' b_2 -1) = 0$. If $(a_1, a_2) \not= 0$, then $c \not= 0$, so $(a_1' b_1 + a_2' b_2 -1)$ is not a unit. Thus $(a_1' b_1 + a_2' b_2)$ must be a unit. Deduce that either $a_1'$ or $a_2'$ is a unit. Hence, either $cA = a_1A$ or $cA = a_2A$. Conclude that one of the $a_i$ divides the other. $\endgroup$
    – Badam Baplan
    Commented Sep 2, 2023 at 21:22

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