Infinite sums of ideals in a Noetherian ring












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$begingroup$


Let R be a Noetherian ring, and $mathcal{I}$ a set of ideals of R. Let $J = sum_{Iin mathcal{I}} I$ be the set of all finite sums $i_1 + cdots + i_n$ where $i_k in I_k in mathcal{I},$ for all $k leq n.$



Suppose $mathcal{I}$ is countable, and enumerate the ideals in $mathcal{I}$. Let $S_n = sum_{i=1}^n I_k,$ where $I_k in mathcal{I}.$



Since R is Noetherian, the chain of ideals $$S_1 subset S_2subset S_3 subset cdots $$
satisfies the ascending chain condition, i.e. there exits $nin mathbb{N}$ such that $S_n = S_{n+1} = cdots$ .



Question 1: Does this imply that $J = S_n$?



Question 2: If $mathcal{I}$ is an uncountable set of ideals, can we still say that J is the sum of finitely many ideals in $mathcal{I}$? Can you prove this using the ascending chain condition?






abstract-algebra ring-theory







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$endgroup$












  • $begingroup$
    Note that countability of $R$ does not imply that it has only countably many ideals.
    $endgroup$
    – Lubin
    Dec 3 '18 at 21:27
















0












$begingroup$


Let R be a Noetherian ring, and $mathcal{I}$ a set of ideals of R. Let $J = sum_{Iin mathcal{I}} I$ be the set of all finite sums $i_1 + cdots + i_n$ where $i_k in I_k in mathcal{I},$ for all $k leq n.$



Suppose $mathcal{I}$ is countable, and enumerate the ideals in $mathcal{I}$. Let $S_n = sum_{i=1}^n I_k,$ where $I_k in mathcal{I}.$



Since R is Noetherian, the chain of ideals $$S_1 subset S_2subset S_3 subset cdots $$
satisfies the ascending chain condition, i.e. there exits $nin mathbb{N}$ such that $S_n = S_{n+1} = cdots$ .



Question 1: Does this imply that $J = S_n$?



Question 2: If $mathcal{I}$ is an uncountable set of ideals, can we still say that J is the sum of finitely many ideals in $mathcal{I}$? Can you prove this using the ascending chain condition?






abstract-algebra ring-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Note that countability of $R$ does not imply that it has only countably many ideals.
    $endgroup$
    – Lubin
    Dec 3 '18 at 21:27














0












0








0





$begingroup$


Let R be a Noetherian ring, and $mathcal{I}$ a set of ideals of R. Let $J = sum_{Iin mathcal{I}} I$ be the set of all finite sums $i_1 + cdots + i_n$ where $i_k in I_k in mathcal{I},$ for all $k leq n.$



Suppose $mathcal{I}$ is countable, and enumerate the ideals in $mathcal{I}$. Let $S_n = sum_{i=1}^n I_k,$ where $I_k in mathcal{I}.$



Since R is Noetherian, the chain of ideals $$S_1 subset S_2subset S_3 subset cdots $$
satisfies the ascending chain condition, i.e. there exits $nin mathbb{N}$ such that $S_n = S_{n+1} = cdots$ .



Question 1: Does this imply that $J = S_n$?



Question 2: If $mathcal{I}$ is an uncountable set of ideals, can we still say that J is the sum of finitely many ideals in $mathcal{I}$? Can you prove this using the ascending chain condition?






abstract-algebra ring-theory







share|cite|improve this question









$endgroup$




Let R be a Noetherian ring, and $mathcal{I}$ a set of ideals of R. Let $J = sum_{Iin mathcal{I}} I$ be the set of all finite sums $i_1 + cdots + i_n$ where $i_k in I_k in mathcal{I},$ for all $k leq n.$



Suppose $mathcal{I}$ is countable, and enumerate the ideals in $mathcal{I}$. Let $S_n = sum_{i=1}^n I_k,$ where $I_k in mathcal{I}.$



Since R is Noetherian, the chain of ideals $$S_1 subset S_2subset S_3 subset cdots $$
satisfies the ascending chain condition, i.e. there exits $nin mathbb{N}$ such that $S_n = S_{n+1} = cdots$ .



Question 1: Does this imply that $J = S_n$?



Question 2: If $mathcal{I}$ is an uncountable set of ideals, can we still say that J is the sum of finitely many ideals in $mathcal{I}$? Can you prove this using the ascending chain condition?






abstract-algebra ring-theory




abstract-algebra ring-theory






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share|cite|improve this question











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asked Dec 3 '18 at 11:51









38917803891780

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  • $begingroup$
    Note that countability of $R$ does not imply that it has only countably many ideals.
    $endgroup$
    – Lubin
    Dec 3 '18 at 21:27


















  • $begingroup$
    Note that countability of $R$ does not imply that it has only countably many ideals.
    $endgroup$
    – Lubin
    Dec 3 '18 at 21:27
















$begingroup$
Note that countability of $R$ does not imply that it has only countably many ideals.
$endgroup$
– Lubin
Dec 3 '18 at 21:27




$begingroup$
Note that countability of $R$ does not imply that it has only countably many ideals.
$endgroup$
– Lubin
Dec 3 '18 at 21:27










1 Answer
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$begingroup$

Question 1: Observe that $forall jin J$, $j=i_{k_1}+i_{k_2}+cdots+i_{k_n}in S_{k_n}$ (for some $k_1<k_2<dots<k_n$ that may depend on $j$). So $Jsubseteq bigcup_{k=1}^infty S_k=S_n$. But also $Jsupseteq S_n$ for obvious reasons. It follows that $J=S_n$ $Box$



Question 2: Consider the following (non-deterministic) algorithm $mathcal A$. At step $0$ we have the ideal $J_0=(0)subseteq J$. At step $t+1$ we choose an arbitrary $Iin mathcal I$ such that $Inotsubseteq J_t$, and add it to $J_tsubseteq J$ to obtain $J_{t+1}=I+J_tsubseteq J$. If such an $I$ does not exist, the process terminates and outputs $J_t$, which by the above considerations would necessarily equal $J$. Suppose that $mathcal A$ does not necessarily terminate. Then there would necessarily exist an infinite strictly ascending chain of ideals $$(0)=J_0subsetneq J_1subsetneq J_2subsetneq dots subsetneq J$$
A contradiction to the assumption that $R$ is Noetherian $Box$






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    $begingroup$

    Question 1: Observe that $forall jin J$, $j=i_{k_1}+i_{k_2}+cdots+i_{k_n}in S_{k_n}$ (for some $k_1<k_2<dots<k_n$ that may depend on $j$). So $Jsubseteq bigcup_{k=1}^infty S_k=S_n$. But also $Jsupseteq S_n$ for obvious reasons. It follows that $J=S_n$ $Box$



    Question 2: Consider the following (non-deterministic) algorithm $mathcal A$. At step $0$ we have the ideal $J_0=(0)subseteq J$. At step $t+1$ we choose an arbitrary $Iin mathcal I$ such that $Inotsubseteq J_t$, and add it to $J_tsubseteq J$ to obtain $J_{t+1}=I+J_tsubseteq J$. If such an $I$ does not exist, the process terminates and outputs $J_t$, which by the above considerations would necessarily equal $J$. Suppose that $mathcal A$ does not necessarily terminate. Then there would necessarily exist an infinite strictly ascending chain of ideals $$(0)=J_0subsetneq J_1subsetneq J_2subsetneq dots subsetneq J$$
    A contradiction to the assumption that $R$ is Noetherian $Box$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Question 1: Observe that $forall jin J$, $j=i_{k_1}+i_{k_2}+cdots+i_{k_n}in S_{k_n}$ (for some $k_1<k_2<dots<k_n$ that may depend on $j$). So $Jsubseteq bigcup_{k=1}^infty S_k=S_n$. But also $Jsupseteq S_n$ for obvious reasons. It follows that $J=S_n$ $Box$



      Question 2: Consider the following (non-deterministic) algorithm $mathcal A$. At step $0$ we have the ideal $J_0=(0)subseteq J$. At step $t+1$ we choose an arbitrary $Iin mathcal I$ such that $Inotsubseteq J_t$, and add it to $J_tsubseteq J$ to obtain $J_{t+1}=I+J_tsubseteq J$. If such an $I$ does not exist, the process terminates and outputs $J_t$, which by the above considerations would necessarily equal $J$. Suppose that $mathcal A$ does not necessarily terminate. Then there would necessarily exist an infinite strictly ascending chain of ideals $$(0)=J_0subsetneq J_1subsetneq J_2subsetneq dots subsetneq J$$
      A contradiction to the assumption that $R$ is Noetherian $Box$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Question 1: Observe that $forall jin J$, $j=i_{k_1}+i_{k_2}+cdots+i_{k_n}in S_{k_n}$ (for some $k_1<k_2<dots<k_n$ that may depend on $j$). So $Jsubseteq bigcup_{k=1}^infty S_k=S_n$. But also $Jsupseteq S_n$ for obvious reasons. It follows that $J=S_n$ $Box$



        Question 2: Consider the following (non-deterministic) algorithm $mathcal A$. At step $0$ we have the ideal $J_0=(0)subseteq J$. At step $t+1$ we choose an arbitrary $Iin mathcal I$ such that $Inotsubseteq J_t$, and add it to $J_tsubseteq J$ to obtain $J_{t+1}=I+J_tsubseteq J$. If such an $I$ does not exist, the process terminates and outputs $J_t$, which by the above considerations would necessarily equal $J$. Suppose that $mathcal A$ does not necessarily terminate. Then there would necessarily exist an infinite strictly ascending chain of ideals $$(0)=J_0subsetneq J_1subsetneq J_2subsetneq dots subsetneq J$$
        A contradiction to the assumption that $R$ is Noetherian $Box$






        share|cite|improve this answer









        $endgroup$



        Question 1: Observe that $forall jin J$, $j=i_{k_1}+i_{k_2}+cdots+i_{k_n}in S_{k_n}$ (for some $k_1<k_2<dots<k_n$ that may depend on $j$). So $Jsubseteq bigcup_{k=1}^infty S_k=S_n$. But also $Jsupseteq S_n$ for obvious reasons. It follows that $J=S_n$ $Box$



        Question 2: Consider the following (non-deterministic) algorithm $mathcal A$. At step $0$ we have the ideal $J_0=(0)subseteq J$. At step $t+1$ we choose an arbitrary $Iin mathcal I$ such that $Inotsubseteq J_t$, and add it to $J_tsubseteq J$ to obtain $J_{t+1}=I+J_tsubseteq J$. If such an $I$ does not exist, the process terminates and outputs $J_t$, which by the above considerations would necessarily equal $J$. Suppose that $mathcal A$ does not necessarily terminate. Then there would necessarily exist an infinite strictly ascending chain of ideals $$(0)=J_0subsetneq J_1subsetneq J_2subsetneq dots subsetneq J$$
        A contradiction to the assumption that $R$ is Noetherian $Box$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 14:18









        Rafay AsharyRafay Ashary

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