Are unique prime ideal factorization domains noetherian?












6














Let $A$ be a domain satisfying the following condition:



If $mathfrak p_1,dots,mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $mathbb N^k$, then we have
$$
mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}nemathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}.
$$




Is $A$ necessarily noetherian?




This question is motivated by these outstanding answers of user26857 and Julian Rosen.



user26857's answer shows that noetherian domains do satisfy the above condition, whereas Julian's answer shows that many non-noetherian domains do not satisfy it.






commutative-algebra maximal-and-prime-ideals noetherian integral-domain







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  • 1




    It's true for $1$-dimensional valuation domains. Is it true for valuation domains in general?
    – Badam Baplan
    yesterday










  • I think there are 1-dimensional valuation domains such that $m^2=m$.
    – user26857
    yesterday










  • Yes exactly. A $1$-dimensional valuation domain is Noetherian iff its max ideal is not idempotent iff it satisfies the property in question.
    – Badam Baplan
    yesterday
















6














Let $A$ be a domain satisfying the following condition:



If $mathfrak p_1,dots,mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $mathbb N^k$, then we have
$$
mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}nemathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}.
$$




Is $A$ necessarily noetherian?




This question is motivated by these outstanding answers of user26857 and Julian Rosen.



user26857's answer shows that noetherian domains do satisfy the above condition, whereas Julian's answer shows that many non-noetherian domains do not satisfy it.






commutative-algebra maximal-and-prime-ideals noetherian integral-domain







share|cite|improve this question




















  • 1




    It's true for $1$-dimensional valuation domains. Is it true for valuation domains in general?
    – Badam Baplan
    yesterday










  • I think there are 1-dimensional valuation domains such that $m^2=m$.
    – user26857
    yesterday










  • Yes exactly. A $1$-dimensional valuation domain is Noetherian iff its max ideal is not idempotent iff it satisfies the property in question.
    – Badam Baplan
    yesterday














6












6








6


1





Let $A$ be a domain satisfying the following condition:



If $mathfrak p_1,dots,mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $mathbb N^k$, then we have
$$
mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}nemathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}.
$$




Is $A$ necessarily noetherian?




This question is motivated by these outstanding answers of user26857 and Julian Rosen.



user26857's answer shows that noetherian domains do satisfy the above condition, whereas Julian's answer shows that many non-noetherian domains do not satisfy it.






commutative-algebra maximal-and-prime-ideals noetherian integral-domain







share|cite|improve this question















Let $A$ be a domain satisfying the following condition:



If $mathfrak p_1,dots,mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $mathbb N^k$, then we have
$$
mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}nemathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}.
$$




Is $A$ necessarily noetherian?




This question is motivated by these outstanding answers of user26857 and Julian Rosen.



user26857's answer shows that noetherian domains do satisfy the above condition, whereas Julian's answer shows that many non-noetherian domains do not satisfy it.






commutative-algebra maximal-and-prime-ideals noetherian integral-domain




commutative-algebra maximal-and-prime-ideals noetherian integral-domain






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday

























asked yesterday









Pierre-Yves Gaillard

13.2k23181




13.2k23181








  • 1




    It's true for $1$-dimensional valuation domains. Is it true for valuation domains in general?
    – Badam Baplan
    yesterday










  • I think there are 1-dimensional valuation domains such that $m^2=m$.
    – user26857
    yesterday










  • Yes exactly. A $1$-dimensional valuation domain is Noetherian iff its max ideal is not idempotent iff it satisfies the property in question.
    – Badam Baplan
    yesterday














  • 1




    It's true for $1$-dimensional valuation domains. Is it true for valuation domains in general?
    – Badam Baplan
    yesterday










  • I think there are 1-dimensional valuation domains such that $m^2=m$.
    – user26857
    yesterday










  • Yes exactly. A $1$-dimensional valuation domain is Noetherian iff its max ideal is not idempotent iff it satisfies the property in question.
    – Badam Baplan
    yesterday








1




1




It's true for $1$-dimensional valuation domains. Is it true for valuation domains in general?
– Badam Baplan
yesterday




It's true for $1$-dimensional valuation domains. Is it true for valuation domains in general?
– Badam Baplan
yesterday












I think there are 1-dimensional valuation domains such that $m^2=m$.
– user26857
yesterday




I think there are 1-dimensional valuation domains such that $m^2=m$.
– user26857
yesterday












Yes exactly. A $1$-dimensional valuation domain is Noetherian iff its max ideal is not idempotent iff it satisfies the property in question.
– Badam Baplan
yesterday




Yes exactly. A $1$-dimensional valuation domain is Noetherian iff its max ideal is not idempotent iff it satisfies the property in question.
– Badam Baplan
yesterday










1 Answer
1






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oldest

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Let's call domains having this property $UPIF$-domains, as in your first post.



Observe that a domain which is locally $UPIF$ is $UPIF$.



Indeed, given $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$
we can localize at each of the $mathfrak{p_i}$ one at a time and use the assumption that $D_{mathfrak{p_i}}$ is $UPIF$ to deduce that $m_i = n_i$.



From your linked Questions/Answers, we thus have that a locally Noetherian domain is locally UPIF, hence




A locally Noetherian domain is $UPIF$.




However, there are locally Noetherian domains that are not Noetherian.



A classic reference for such examples is section 2 of this paper of Heinzer and Ohm.



For a more recent reference, I'd suggest Loper's article on Almost Dedekind Domains Which Are Not Dedekind. In section 3, he discusses five distinct techniques for producing examples of non-Noetherian domains which are locally discrete valuation rings.






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    Let's call domains having this property $UPIF$-domains, as in your first post.



    Observe that a domain which is locally $UPIF$ is $UPIF$.



    Indeed, given $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$
    we can localize at each of the $mathfrak{p_i}$ one at a time and use the assumption that $D_{mathfrak{p_i}}$ is $UPIF$ to deduce that $m_i = n_i$.



    From your linked Questions/Answers, we thus have that a locally Noetherian domain is locally UPIF, hence




    A locally Noetherian domain is $UPIF$.




    However, there are locally Noetherian domains that are not Noetherian.



    A classic reference for such examples is section 2 of this paper of Heinzer and Ohm.



    For a more recent reference, I'd suggest Loper's article on Almost Dedekind Domains Which Are Not Dedekind. In section 3, he discusses five distinct techniques for producing examples of non-Noetherian domains which are locally discrete valuation rings.






    share|cite|improve this answer




























      1














      Let's call domains having this property $UPIF$-domains, as in your first post.



      Observe that a domain which is locally $UPIF$ is $UPIF$.



      Indeed, given $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$
      we can localize at each of the $mathfrak{p_i}$ one at a time and use the assumption that $D_{mathfrak{p_i}}$ is $UPIF$ to deduce that $m_i = n_i$.



      From your linked Questions/Answers, we thus have that a locally Noetherian domain is locally UPIF, hence




      A locally Noetherian domain is $UPIF$.




      However, there are locally Noetherian domains that are not Noetherian.



      A classic reference for such examples is section 2 of this paper of Heinzer and Ohm.



      For a more recent reference, I'd suggest Loper's article on Almost Dedekind Domains Which Are Not Dedekind. In section 3, he discusses five distinct techniques for producing examples of non-Noetherian domains which are locally discrete valuation rings.






      share|cite|improve this answer


























        1












        1








        1






        Let's call domains having this property $UPIF$-domains, as in your first post.



        Observe that a domain which is locally $UPIF$ is $UPIF$.



        Indeed, given $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$
        we can localize at each of the $mathfrak{p_i}$ one at a time and use the assumption that $D_{mathfrak{p_i}}$ is $UPIF$ to deduce that $m_i = n_i$.



        From your linked Questions/Answers, we thus have that a locally Noetherian domain is locally UPIF, hence




        A locally Noetherian domain is $UPIF$.




        However, there are locally Noetherian domains that are not Noetherian.



        A classic reference for such examples is section 2 of this paper of Heinzer and Ohm.



        For a more recent reference, I'd suggest Loper's article on Almost Dedekind Domains Which Are Not Dedekind. In section 3, he discusses five distinct techniques for producing examples of non-Noetherian domains which are locally discrete valuation rings.






        share|cite|improve this answer














        Let's call domains having this property $UPIF$-domains, as in your first post.



        Observe that a domain which is locally $UPIF$ is $UPIF$.



        Indeed, given $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$
        we can localize at each of the $mathfrak{p_i}$ one at a time and use the assumption that $D_{mathfrak{p_i}}$ is $UPIF$ to deduce that $m_i = n_i$.



        From your linked Questions/Answers, we thus have that a locally Noetherian domain is locally UPIF, hence




        A locally Noetherian domain is $UPIF$.




        However, there are locally Noetherian domains that are not Noetherian.



        A classic reference for such examples is section 2 of this paper of Heinzer and Ohm.



        For a more recent reference, I'd suggest Loper's article on Almost Dedekind Domains Which Are Not Dedekind. In section 3, he discusses five distinct techniques for producing examples of non-Noetherian domains which are locally discrete valuation rings.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 5 hours ago

























        answered 7 hours ago









        Badam Baplan

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