Unique prime ideal factorization in domains?












5












$begingroup$


This is a follow-up to this question.




Let $A$ be a domain; let $mathfrak p_1,dots,mathfrak p_k$ be distinct prime ideals of $A$ such that $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $1le ile k$, $jge1$; and let $m$ and $n$ be elements of $mathbb N^k$ such that $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$ Can we conclude that $m$ and $n$ are equal?




user26857 proved that the answer is Yes if $A$ is noetherian (see this answer). (Of course in this case the condition $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $jge1$ is equivalent to $mathfrak p_ine(0)$.)






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







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$endgroup$

















    5












    $begingroup$


    This is a follow-up to this question.




    Let $A$ be a domain; let $mathfrak p_1,dots,mathfrak p_k$ be distinct prime ideals of $A$ such that $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $1le ile k$, $jge1$; and let $m$ and $n$ be elements of $mathbb N^k$ such that $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$ Can we conclude that $m$ and $n$ are equal?




    user26857 proved that the answer is Yes if $A$ is noetherian (see this answer). (Of course in this case the condition $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $jge1$ is equivalent to $mathfrak p_ine(0)$.)






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







    share|cite|improve this question









    $endgroup$















      5












      5








      5


      2



      $begingroup$


      This is a follow-up to this question.




      Let $A$ be a domain; let $mathfrak p_1,dots,mathfrak p_k$ be distinct prime ideals of $A$ such that $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $1le ile k$, $jge1$; and let $m$ and $n$ be elements of $mathbb N^k$ such that $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$ Can we conclude that $m$ and $n$ are equal?




      user26857 proved that the answer is Yes if $A$ is noetherian (see this answer). (Of course in this case the condition $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $jge1$ is equivalent to $mathfrak p_ine(0)$.)






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







      share|cite|improve this question









      $endgroup$




      This is a follow-up to this question.




      Let $A$ be a domain; let $mathfrak p_1,dots,mathfrak p_k$ be distinct prime ideals of $A$ such that $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $1le ile k$, $jge1$; and let $m$ and $n$ be elements of $mathbb N^k$ such that $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$ Can we conclude that $m$ and $n$ are equal?




      user26857 proved that the answer is Yes if $A$ is noetherian (see this answer). (Of course in this case the condition $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $jge1$ is equivalent to $mathfrak p_ine(0)$.)






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




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






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 31 '18 at 13:21









      Pierre-Yves GaillardPierre-Yves Gaillard

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      13.5k23184






















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          $begingroup$

          The answer is no.



          Let $G$ be the abelian group $mathbb{Z}times mathbb{Z}$ with the lexicographic order, and let $Msubset G$ be the sub-monoid of elements that are greater than or equal to $(0,0)$. Define monoid ideals $I_1:={(m,n):mgeq 1text{ or }ngeq 1}subset M$ and $I_2:={(m,n):mgeq 1}subset M$. For $jinmathbb{Z}_{geq 0}$ and $Isubset M$ an ideal, write $jI:={i_1+ldots+i_j:i_1,ldots,i_jin I}subset M$. Check that for all $j$, we have $jI_1neq (j+1)I_1$ and $jI_2neq (j+1)I_2$, but $I_1+I_2=I_2$.



          We get a counterexample to your question by taking $A=k[M]$ to be the monoid ring of $M$ (where $k$ is any field), $mathfrak{p}_1=k[I_1]$, and $mathfrak{p}_2=k[I_2]$. Then $mathfrak{p}_1^jneq mathfrak{p}_1^{j+1}$ and $mathfrak{p}_2^jneq mathfrak{p}_2^{j+1}$, but $mathfrak{p}_1mathfrak{p}_2=mathfrak{p}_2$.



          An alternate way to describe this example is $A=k[x,y,yx^{-1},yx^{-2},ldots]$, $mathfrak{p}_1=(x)$, $mathfrak{p}_2=(y,yx^{-1},yx^{-2},ldots)$.






          share|cite|improve this answer









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            $begingroup$

            The answer is no.



            Let $G$ be the abelian group $mathbb{Z}times mathbb{Z}$ with the lexicographic order, and let $Msubset G$ be the sub-monoid of elements that are greater than or equal to $(0,0)$. Define monoid ideals $I_1:={(m,n):mgeq 1text{ or }ngeq 1}subset M$ and $I_2:={(m,n):mgeq 1}subset M$. For $jinmathbb{Z}_{geq 0}$ and $Isubset M$ an ideal, write $jI:={i_1+ldots+i_j:i_1,ldots,i_jin I}subset M$. Check that for all $j$, we have $jI_1neq (j+1)I_1$ and $jI_2neq (j+1)I_2$, but $I_1+I_2=I_2$.



            We get a counterexample to your question by taking $A=k[M]$ to be the monoid ring of $M$ (where $k$ is any field), $mathfrak{p}_1=k[I_1]$, and $mathfrak{p}_2=k[I_2]$. Then $mathfrak{p}_1^jneq mathfrak{p}_1^{j+1}$ and $mathfrak{p}_2^jneq mathfrak{p}_2^{j+1}$, but $mathfrak{p}_1mathfrak{p}_2=mathfrak{p}_2$.



            An alternate way to describe this example is $A=k[x,y,yx^{-1},yx^{-2},ldots]$, $mathfrak{p}_1=(x)$, $mathfrak{p}_2=(y,yx^{-1},yx^{-2},ldots)$.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              The answer is no.



              Let $G$ be the abelian group $mathbb{Z}times mathbb{Z}$ with the lexicographic order, and let $Msubset G$ be the sub-monoid of elements that are greater than or equal to $(0,0)$. Define monoid ideals $I_1:={(m,n):mgeq 1text{ or }ngeq 1}subset M$ and $I_2:={(m,n):mgeq 1}subset M$. For $jinmathbb{Z}_{geq 0}$ and $Isubset M$ an ideal, write $jI:={i_1+ldots+i_j:i_1,ldots,i_jin I}subset M$. Check that for all $j$, we have $jI_1neq (j+1)I_1$ and $jI_2neq (j+1)I_2$, but $I_1+I_2=I_2$.



              We get a counterexample to your question by taking $A=k[M]$ to be the monoid ring of $M$ (where $k$ is any field), $mathfrak{p}_1=k[I_1]$, and $mathfrak{p}_2=k[I_2]$. Then $mathfrak{p}_1^jneq mathfrak{p}_1^{j+1}$ and $mathfrak{p}_2^jneq mathfrak{p}_2^{j+1}$, but $mathfrak{p}_1mathfrak{p}_2=mathfrak{p}_2$.



              An alternate way to describe this example is $A=k[x,y,yx^{-1},yx^{-2},ldots]$, $mathfrak{p}_1=(x)$, $mathfrak{p}_2=(y,yx^{-1},yx^{-2},ldots)$.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                The answer is no.



                Let $G$ be the abelian group $mathbb{Z}times mathbb{Z}$ with the lexicographic order, and let $Msubset G$ be the sub-monoid of elements that are greater than or equal to $(0,0)$. Define monoid ideals $I_1:={(m,n):mgeq 1text{ or }ngeq 1}subset M$ and $I_2:={(m,n):mgeq 1}subset M$. For $jinmathbb{Z}_{geq 0}$ and $Isubset M$ an ideal, write $jI:={i_1+ldots+i_j:i_1,ldots,i_jin I}subset M$. Check that for all $j$, we have $jI_1neq (j+1)I_1$ and $jI_2neq (j+1)I_2$, but $I_1+I_2=I_2$.



                We get a counterexample to your question by taking $A=k[M]$ to be the monoid ring of $M$ (where $k$ is any field), $mathfrak{p}_1=k[I_1]$, and $mathfrak{p}_2=k[I_2]$. Then $mathfrak{p}_1^jneq mathfrak{p}_1^{j+1}$ and $mathfrak{p}_2^jneq mathfrak{p}_2^{j+1}$, but $mathfrak{p}_1mathfrak{p}_2=mathfrak{p}_2$.



                An alternate way to describe this example is $A=k[x,y,yx^{-1},yx^{-2},ldots]$, $mathfrak{p}_1=(x)$, $mathfrak{p}_2=(y,yx^{-1},yx^{-2},ldots)$.






                share|cite|improve this answer









                $endgroup$



                The answer is no.



                Let $G$ be the abelian group $mathbb{Z}times mathbb{Z}$ with the lexicographic order, and let $Msubset G$ be the sub-monoid of elements that are greater than or equal to $(0,0)$. Define monoid ideals $I_1:={(m,n):mgeq 1text{ or }ngeq 1}subset M$ and $I_2:={(m,n):mgeq 1}subset M$. For $jinmathbb{Z}_{geq 0}$ and $Isubset M$ an ideal, write $jI:={i_1+ldots+i_j:i_1,ldots,i_jin I}subset M$. Check that for all $j$, we have $jI_1neq (j+1)I_1$ and $jI_2neq (j+1)I_2$, but $I_1+I_2=I_2$.



                We get a counterexample to your question by taking $A=k[M]$ to be the monoid ring of $M$ (where $k$ is any field), $mathfrak{p}_1=k[I_1]$, and $mathfrak{p}_2=k[I_2]$. Then $mathfrak{p}_1^jneq mathfrak{p}_1^{j+1}$ and $mathfrak{p}_2^jneq mathfrak{p}_2^{j+1}$, but $mathfrak{p}_1mathfrak{p}_2=mathfrak{p}_2$.



                An alternate way to describe this example is $A=k[x,y,yx^{-1},yx^{-2},ldots]$, $mathfrak{p}_1=(x)$, $mathfrak{p}_2=(y,yx^{-1},yx^{-2},ldots)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 3 at 0:46









                Julian RosenJulian Rosen

                12.1k12349




                12.1k12349






























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