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How to show the following ring is not Noetherian

How 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 $mathbb{R}[x]$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

0

$begingroup$

$R subset mathbb{Q}[x]$ be the subring consisting of polynomials $f = a_0 + a_1 x+cdots+a_n x^n$ s.t. $a_0in mathbb{Z}$. Show that $R$ is not a Noetherian ring.

Given a subgroup $Asubset mathbb{Q}$ (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

share|cite|improve this question

edited Mar 20 at 10:21

user26857

39.5k124283

asked Mar 18 at 20:06

manifoldedmanifolded

49519

$endgroup$

add a comment | 

0

$begingroup$

$R subset mathbb{Q}[x]$ be the subring consisting of polynomials $f = a_0 + a_1 x+cdots+a_n x^n$ s.t. $a_0in mathbb{Z}$. Show that $R$ is not a Noetherian ring.

Given a subgroup $Asubset mathbb{Q}$ (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

share|cite|improve this question

edited Mar 20 at 10:21

user26857

39.5k124283

asked Mar 18 at 20:06

manifoldedmanifolded

49519

$endgroup$

add a comment | 

$begingroup$

$R subset mathbb{Q}[x]$ be the subring consisting of polynomials $f = a_0 + a_1 x+cdots+a_n x^n$ s.t. $a_0in mathbb{Z}$. Show that $R$ is not a Noetherian ring.

Given a subgroup $Asubset mathbb{Q}$ (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

share|cite|improve this question

edited Mar 20 at 10:21

user26857

39.5k124283

asked Mar 18 at 20:06

manifoldedmanifolded

49519

$endgroup$

share|cite|improve this question

edited Mar 20 at 10:21

user26857

39.5k124283

asked Mar 18 at 20:06

manifoldedmanifolded

49519

share|cite|improve this question

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

asked Mar 18 at 20:06

manifoldedmanifolded

49519

asked Mar 18 at 20:06

manifoldedmanifolded

49519

1 Answer
1

active

oldest

votes

1

$begingroup$

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 $newcommandZZ{Bbb{Z}}0oplus a_1ZZ oplus newcommandQQ{Bbb{Q}}QQoplus QQoplus cdots$
already, since we can multiply by $frac{q}{a_1}x^i$ for $qinBbb{Q}$, $ige 1$. Thus your first ideal equals your second ideal.

Indeed, $I$ can easily be finitely generated. If $A=Bbb{Z}$ 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 $Bbb{Q}$. Perhaps $A_n = frac{1}{2^n}Bbb{Z}$. Then note that the ideals $I_{A_n}$ give a strictly ascending chain of ideals in $R$. Thus $R$ is not Noetherian.

share|cite|improve this answer

answered Mar 18 at 20:19

jgonjgon

16.4k32143

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, I understand that $A_n = frac{1}{2^n}mathbb{Z}$ is a strictly ascending chain of subgroups of $mathbb{Q}$ 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

$begingroup$

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 $newcommandZZ{Bbb{Z}}0oplus a_1ZZ oplus newcommandQQ{Bbb{Q}}QQoplus QQoplus cdots$
already, since we can multiply by $frac{q}{a_1}x^i$ for $qinBbb{Q}$, $ige 1$. Thus your first ideal equals your second ideal.

Indeed, $I$ can easily be finitely generated. If $A=Bbb{Z}$ 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 $Bbb{Q}$. Perhaps $A_n = frac{1}{2^n}Bbb{Z}$. Then note that the ideals $I_{A_n}$ give a strictly ascending chain of ideals in $R$. Thus $R$ is not Noetherian.

share|cite|improve this answer

answered Mar 18 at 20:19

jgonjgon

16.4k32143

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, I understand that $A_n = frac{1}{2^n}mathbb{Z}$ is a strictly ascending chain of subgroups of $mathbb{Q}$ 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 | 

1

$begingroup$

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 $newcommandZZ{Bbb{Z}}0oplus a_1ZZ oplus newcommandQQ{Bbb{Q}}QQoplus QQoplus cdots$
already, since we can multiply by $frac{q}{a_1}x^i$ for $qinBbb{Q}$, $ige 1$. Thus your first ideal equals your second ideal.

Indeed, $I$ can easily be finitely generated. If $A=Bbb{Z}$ 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 $Bbb{Q}$. Perhaps $A_n = frac{1}{2^n}Bbb{Z}$. Then note that the ideals $I_{A_n}$ give a strictly ascending chain of ideals in $R$. Thus $R$ is not Noetherian.

share|cite|improve this answer

answered Mar 18 at 20:19

jgonjgon

16.4k32143

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, I understand that $A_n = frac{1}{2^n}mathbb{Z}$ is a strictly ascending chain of subgroups of $mathbb{Q}$ 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 $newcommandZZ{Bbb{Z}}0oplus a_1ZZ oplus newcommandQQ{Bbb{Q}}QQoplus QQoplus cdots$
already, since we can multiply by $frac{q}{a_1}x^i$ for $qinBbb{Q}$, $ige 1$. Thus your first ideal equals your second ideal.

Indeed, $I$ can easily be finitely generated. If $A=Bbb{Z}$ 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 $Bbb{Q}$. Perhaps $A_n = frac{1}{2^n}Bbb{Z}$. Then note that the ideals $I_{A_n}$ give a strictly ascending chain of ideals in $R$. Thus $R$ is not Noetherian.

share|cite|improve this answer

answered Mar 18 at 20:19

jgonjgon

16.4k32143

$endgroup$

share|cite|improve this answer

answered Mar 18 at 20:19

jgonjgon

16.4k32143

answered Mar 18 at 20:19

jgonjgon

16.4k32143

answered Mar 18 at 20:19

jgonjgon

16.4k32143