Infinite sums of ideals in a Noetherian ring
$begingroup$
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
asked Dec 3 '18 at 11:51
38917803891780
816
$endgroup$
add a comment |
$begingroup$
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
asked Dec 3 '18 at 11:51
38917803891780
816
$endgroup$
$begingroup$
Note that countability of $R$ does not imply that it has only countably many ideals.
$endgroup$
– Lubin
Dec 3 '18 at 21:27
add a comment |
$begingroup$
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
asked Dec 3 '18 at 11:51
38917803891780
816
$endgroup$
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
asked Dec 3 '18 at 11:51
38917803891780
816
asked Dec 3 '18 at 11:51
38917803891780
816
asked Dec 3 '18 at 11:51
38917803891780
816
asked Dec 3 '18 at 11:51
38917803891780
816
asked Dec 3 '18 at 11:51
38917803891780
816
816
$begingroup$
Note that countability of $R$ does not imply that it has only countably many ideals.
$endgroup$
– Lubin
Dec 3 '18 at 21:27
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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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 $Box$
answered Dec 3 '18 at 14:18
Rafay AsharyRafay Ashary
83618
$endgroup$
add a comment |
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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 $Box$
answered Dec 3 '18 at 14:18
Rafay AsharyRafay Ashary
83618
$endgroup$
add a comment |
$begingroup$
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 $Box$
answered Dec 3 '18 at 14:18
Rafay AsharyRafay Ashary
83618
$endgroup$
add a comment |
$begingroup$
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 $Box$
answered Dec 3 '18 at 14:18
Rafay AsharyRafay Ashary
83618
$endgroup$
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 $Box$
answered Dec 3 '18 at 14:18
Rafay AsharyRafay Ashary
83618
answered Dec 3 '18 at 14:18
Rafay AsharyRafay Ashary
83618
answered Dec 3 '18 at 14:18
Rafay AsharyRafay Ashary
83618
answered Dec 3 '18 at 14:18
Rafay AsharyRafay Ashary
83618
83618
add a comment |
add a comment |
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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