How to show the following ring is not Noetherian The Next CEO of Stack OverflowHow to prove that this subring is not noetherian?If every ascending chain of primary ideals in $R$ stabilizes, is $R$ a Noetherian ring?Ascending chain “stabilizing temporarily” infinitely many timesGive an infinite sequence of principal ideals of $R$ such that the ascending chain condition does not holdDetermining if any of these three are an ideal of $mathbbR[x]$Looking for a special class of ideals such that If every ascending chain of ideals from this class stabilizes, then $R$ is a Noetherian ring.What does this field notation mean?Noetherian Rings Definition, countability?Prove that a polynomial ring over R in infinitely many variables is non-Noetherian.Infinite sums of ideals in a Noetherian ring
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How to show the following ring is not Noetherian
The Next CEO of Stack OverflowHow to prove that this subring is not noetherian?If every ascending chain of primary ideals in $R$ stabilizes, is $R$ a Noetherian ring?Ascending chain “stabilizing temporarily” infinitely many timesGive an infinite sequence of principal ideals of $R$ such that the ascending chain condition does not holdDetermining if any of these three are an ideal of $mathbbR[x]$Looking for a special class of ideals such that If every ascending chain of ideals from this class stabilizes, then $R$ is a Noetherian ring.What does this field notation mean?Noetherian Rings Definition, countability?Prove that a polynomial ring over R in infinitely many variables is non-Noetherian.Infinite sums of ideals in a Noetherian ring
$begingroup$
$R subset mathbbQ[x]$ be the subring consisting of polynomials $f = a_0 + a_1 x+cdots+a_n x^n$ s.t. $a_0in mathbbZ$. Show that $R$ is not a Noetherian ring.
Given a subgroup $Asubset mathbbQ$ (Abelian group under addition), consider the subset $Isubset R$ consisting of polynomials $a_1x+cdots+a_nx^n$ s.t. $a_1in A$.
My attempt: I've shown $I$ is an ideal and to construct an ascending chain of ideals, does this work?: Fix $a_1in A$, $<a_1x> subset<a_1x,x^2>subset....$
The main consideration is coefficient of $xin A$ which, I think, the chain above addresses and intuitively this chain does not stabilize but can I give a more rigorous statement why it doesn't? Thanks.
abstract-algebra ring-theory noetherian
edited Mar 20 at 10:21
user26857
39.5k124283
asked Mar 18 at 20:06
manifoldedmanifolded
49519
$endgroup$
add a comment |
$begingroup$
$R subset mathbbQ[x]$ be the subring consisting of polynomials $f = a_0 + a_1 x+cdots+a_n x^n$ s.t. $a_0in mathbbZ$. Show that $R$ is not a Noetherian ring.
Given a subgroup $Asubset mathbbQ$ (Abelian group under addition), consider the subset $Isubset R$ consisting of polynomials $a_1x+cdots+a_nx^n$ s.t. $a_1in A$.
My attempt: I've shown $I$ is an ideal and to construct an ascending chain of ideals, does this work?: Fix $a_1in A$, $<a_1x> subset<a_1x,x^2>subset....$
The main consideration is coefficient of $xin A$ which, I think, the chain above addresses and intuitively this chain does not stabilize but can I give a more rigorous statement why it doesn't? Thanks.
abstract-algebra ring-theory noetherian
edited Mar 20 at 10:21
user26857
39.5k124283
asked Mar 18 at 20:06
manifoldedmanifolded
49519
$endgroup$
add a comment |
$begingroup$
$R subset mathbbQ[x]$ be the subring consisting of polynomials $f = a_0 + a_1 x+cdots+a_n x^n$ s.t. $a_0in mathbbZ$. Show that $R$ is not a Noetherian ring.
Given a subgroup $Asubset mathbbQ$ (Abelian group under addition), consider the subset $Isubset R$ consisting of polynomials $a_1x+cdots+a_nx^n$ s.t. $a_1in A$.
My attempt: I've shown $I$ is an ideal and to construct an ascending chain of ideals, does this work?: Fix $a_1in A$, $<a_1x> subset<a_1x,x^2>subset....$
The main consideration is coefficient of $xin A$ which, I think, the chain above addresses and intuitively this chain does not stabilize but can I give a more rigorous statement why it doesn't? Thanks.
abstract-algebra ring-theory noetherian
edited Mar 20 at 10:21
user26857
39.5k124283
asked Mar 18 at 20:06
manifoldedmanifolded
49519
$endgroup$
$R subset mathbbQ[x]$ be the subring consisting of polynomials $f = a_0 + a_1 x+cdots+a_n x^n$ s.t. $a_0in mathbbZ$. Show that $R$ is not a Noetherian ring.
Given a subgroup $Asubset mathbbQ$ (Abelian group under addition), consider the subset $Isubset R$ consisting of polynomials $a_1x+cdots+a_nx^n$ s.t. $a_1in A$.
My attempt: I've shown $I$ is an ideal and to construct an ascending chain of ideals, does this work?: Fix $a_1in A$, $<a_1x> subset<a_1x,x^2>subset....$
The main consideration is coefficient of $xin A$ which, I think, the chain above addresses and intuitively this chain does not stabilize but can I give a more rigorous statement why it doesn't? Thanks.
abstract-algebra ring-theory noetherian
abstract-algebra ring-theory noetherian
edited Mar 20 at 10:21
user26857
39.5k124283
asked Mar 18 at 20:06
manifoldedmanifolded
49519
edited Mar 20 at 10:21
user26857
39.5k124283
asked Mar 18 at 20:06
manifoldedmanifolded
49519
edited Mar 20 at 10:21
user26857
39.5k124283
edited Mar 20 at 10:21
user26857
39.5k124283
edited Mar 20 at 10:21
user26857
39.5k124283
39.5k124283
asked Mar 18 at 20:06
manifoldedmanifolded
49519
asked Mar 18 at 20:06
manifoldedmanifolded
49519
asked Mar 18 at 20:06
manifoldedmanifolded
49519
49519
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's unclear what your chain is, or how you expect it to continue, but note that as a graded ideal, your first ideal is $newcommandZZBbbZ0oplus a_1ZZ oplus newcommandQQBbbQQQoplus QQoplus cdots$
already, since we can multiply by $fracqa_1x^i$ for $qinBbbQ$, $ige 1$. Thus your first ideal equals your second ideal.
Indeed, $I$ can easily be finitely generated. If $A=BbbZ$ for example, then $I$ is generated by $x$, so $I$ is principal, so we wouldn't expect an ascending chain of ideals in $I$.
Instead, let $I_A$ denote the ideal associated to the subgroup $A$.
Then choose
$$A_1subsetneq A_2subsetneq cdots subsetneq A_n subsetneq cdots$$
a strictly ascending chain of subgroups of $BbbQ$. Perhaps $A_n = frac12^nBbbZ$. Then note that the ideals $I_A_n$ give a strictly ascending chain of ideals in $R$. Thus $R$ is not Noetherian.
answered Mar 18 at 20:19
jgonjgon
16.1k32143
$endgroup$
$begingroup$
I see, I understand that $A_n = frac12^nmathbbZ$ is a strictly ascending chain of subgroups of $mathbbQ$ but am not sure what the elements of $I_A$ would look like and how we are guaranteed existence of an ideal associated to the subgroup $A$?
$endgroup$
– manifolded
Mar 18 at 20:33
$begingroup$
@manifolded That's what the ideal $I$ is in the question. It's the ideal associated to the subgroup $A$ from the question.
$endgroup$
– jgon
Mar 18 at 20:35
$begingroup$
Ahh I see, I was thinking of ideals associated to the same subgroup earlier which is wrong, makes sense now.
$endgroup$
– manifolded
Mar 18 at 20:37
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
It's unclear what your chain is, or how you expect it to continue, but note that as a graded ideal, your first ideal is $newcommandZZBbbZ0oplus a_1ZZ oplus newcommandQQBbbQQQoplus QQoplus cdots$
already, since we can multiply by $fracqa_1x^i$ for $qinBbbQ$, $ige 1$. Thus your first ideal equals your second ideal.
Indeed, $I$ can easily be finitely generated. If $A=BbbZ$ for example, then $I$ is generated by $x$, so $I$ is principal, so we wouldn't expect an ascending chain of ideals in $I$.
Instead, let $I_A$ denote the ideal associated to the subgroup $A$.
Then choose
$$A_1subsetneq A_2subsetneq cdots subsetneq A_n subsetneq cdots$$
a strictly ascending chain of subgroups of $BbbQ$. Perhaps $A_n = frac12^nBbbZ$. Then note that the ideals $I_A_n$ give a strictly ascending chain of ideals in $R$. Thus $R$ is not Noetherian.
answered Mar 18 at 20:19
jgonjgon
16.1k32143
$endgroup$
$begingroup$
I see, I understand that $A_n = frac12^nmathbbZ$ is a strictly ascending chain of subgroups of $mathbbQ$ but am not sure what the elements of $I_A$ would look like and how we are guaranteed existence of an ideal associated to the subgroup $A$?
$endgroup$
– manifolded
Mar 18 at 20:33
$begingroup$
@manifolded That's what the ideal $I$ is in the question. It's the ideal associated to the subgroup $A$ from the question.
$endgroup$
– jgon
Mar 18 at 20:35
$begingroup$
Ahh I see, I was thinking of ideals associated to the same subgroup earlier which is wrong, makes sense now.
$endgroup$
– manifolded
Mar 18 at 20:37
add a comment |
$begingroup$
It's unclear what your chain is, or how you expect it to continue, but note that as a graded ideal, your first ideal is $newcommandZZBbbZ0oplus a_1ZZ oplus newcommandQQBbbQQQoplus QQoplus cdots$
already, since we can multiply by $fracqa_1x^i$ for $qinBbbQ$, $ige 1$. Thus your first ideal equals your second ideal.
Indeed, $I$ can easily be finitely generated. If $A=BbbZ$ for example, then $I$ is generated by $x$, so $I$ is principal, so we wouldn't expect an ascending chain of ideals in $I$.
Instead, let $I_A$ denote the ideal associated to the subgroup $A$.
Then choose
$$A_1subsetneq A_2subsetneq cdots subsetneq A_n subsetneq cdots$$
a strictly ascending chain of subgroups of $BbbQ$. Perhaps $A_n = frac12^nBbbZ$. Then note that the ideals $I_A_n$ give a strictly ascending chain of ideals in $R$. Thus $R$ is not Noetherian.
answered Mar 18 at 20:19
jgonjgon
16.1k32143
$endgroup$
$begingroup$
I see, I understand that $A_n = frac12^nmathbbZ$ is a strictly ascending chain of subgroups of $mathbbQ$ but am not sure what the elements of $I_A$ would look like and how we are guaranteed existence of an ideal associated to the subgroup $A$?
$endgroup$
– manifolded
Mar 18 at 20:33
$begingroup$
@manifolded That's what the ideal $I$ is in the question. It's the ideal associated to the subgroup $A$ from the question.
$endgroup$
– jgon
Mar 18 at 20:35
$begingroup$
Ahh I see, I was thinking of ideals associated to the same subgroup earlier which is wrong, makes sense now.
$endgroup$
– manifolded
Mar 18 at 20:37
add a comment |
$begingroup$
It's unclear what your chain is, or how you expect it to continue, but note that as a graded ideal, your first ideal is $newcommandZZBbbZ0oplus a_1ZZ oplus newcommandQQBbbQQQoplus QQoplus cdots$
already, since we can multiply by $fracqa_1x^i$ for $qinBbbQ$, $ige 1$. Thus your first ideal equals your second ideal.
Indeed, $I$ can easily be finitely generated. If $A=BbbZ$ for example, then $I$ is generated by $x$, so $I$ is principal, so we wouldn't expect an ascending chain of ideals in $I$.
Instead, let $I_A$ denote the ideal associated to the subgroup $A$.
Then choose
$$A_1subsetneq A_2subsetneq cdots subsetneq A_n subsetneq cdots$$
a strictly ascending chain of subgroups of $BbbQ$. Perhaps $A_n = frac12^nBbbZ$. Then note that the ideals $I_A_n$ give a strictly ascending chain of ideals in $R$. Thus $R$ is not Noetherian.
answered Mar 18 at 20:19
jgonjgon
16.1k32143
$endgroup$
It's unclear what your chain is, or how you expect it to continue, but note that as a graded ideal, your first ideal is $newcommandZZBbbZ0oplus a_1ZZ oplus newcommandQQBbbQQQoplus QQoplus cdots$
already, since we can multiply by $fracqa_1x^i$ for $qinBbbQ$, $ige 1$. Thus your first ideal equals your second ideal.
Indeed, $I$ can easily be finitely generated. If $A=BbbZ$ for example, then $I$ is generated by $x$, so $I$ is principal, so we wouldn't expect an ascending chain of ideals in $I$.
Instead, let $I_A$ denote the ideal associated to the subgroup $A$.
Then choose
$$A_1subsetneq A_2subsetneq cdots subsetneq A_n subsetneq cdots$$
a strictly ascending chain of subgroups of $BbbQ$. Perhaps $A_n = frac12^nBbbZ$. Then note that the ideals $I_A_n$ give a strictly ascending chain of ideals in $R$. Thus $R$ is not Noetherian.
answered Mar 18 at 20:19
jgonjgon
16.1k32143
answered Mar 18 at 20:19
jgonjgon
16.1k32143
answered Mar 18 at 20:19
jgonjgon
16.1k32143
answered Mar 18 at 20:19
jgonjgon
16.1k32143
16.1k32143
$begingroup$
I see, I understand that $A_n = frac12^nmathbbZ$ is a strictly ascending chain of subgroups of $mathbbQ$ but am not sure what the elements of $I_A$ would look like and how we are guaranteed existence of an ideal associated to the subgroup $A$?
$endgroup$
– manifolded
Mar 18 at 20:33
$begingroup$
@manifolded That's what the ideal $I$ is in the question. It's the ideal associated to the subgroup $A$ from the question.
$endgroup$
– jgon
Mar 18 at 20:35
$begingroup$
Ahh I see, I was thinking of ideals associated to the same subgroup earlier which is wrong, makes sense now.
$endgroup$
– manifolded
Mar 18 at 20:37
add a comment |
$begingroup$
I see, I understand that $A_n = frac12^nmathbbZ$ is a strictly ascending chain of subgroups of $mathbbQ$ but am not sure what the elements of $I_A$ would look like and how we are guaranteed existence of an ideal associated to the subgroup $A$?
$endgroup$
– manifolded
Mar 18 at 20:33
$begingroup$
@manifolded That's what the ideal $I$ is in the question. It's the ideal associated to the subgroup $A$ from the question.
$endgroup$
– jgon
Mar 18 at 20:35
$begingroup$
Ahh I see, I was thinking of ideals associated to the same subgroup earlier which is wrong, makes sense now.
$endgroup$
– manifolded
Mar 18 at 20:37
$begingroup$
I see, I understand that $A_n = frac12^nmathbbZ$ is a strictly ascending chain of subgroups of $mathbbQ$ but am not sure what the elements of $I_A$ would look like and how we are guaranteed existence of an ideal associated to the subgroup $A$?
$endgroup$
– manifolded
Mar 18 at 20:33
$begingroup$
I see, I understand that $A_n = frac12^nmathbbZ$ is a strictly ascending chain of subgroups of $mathbbQ$ but am not sure what the elements of $I_A$ would look like and how we are guaranteed existence of an ideal associated to the subgroup $A$?
$endgroup$
– manifolded
Mar 18 at 20:33
$begingroup$
@manifolded That's what the ideal $I$ is in the question. It's the ideal associated to the subgroup $A$ from the question.
$endgroup$
– jgon
Mar 18 at 20:35
$begingroup$
@manifolded That's what the ideal $I$ is in the question. It's the ideal associated to the subgroup $A$ from the question.
$endgroup$
– jgon
Mar 18 at 20:35
$begingroup$
Ahh I see, I was thinking of ideals associated to the same subgroup earlier which is wrong, makes sense now.
$endgroup$
– manifolded
Mar 18 at 20:37
$begingroup$
Ahh I see, I was thinking of ideals associated to the same subgroup earlier which is wrong, makes sense now.
$endgroup$
– manifolded
Mar 18 at 20:37
add a comment |
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