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Bounty Ended with no winning answer by Nico Bellic
Bounty Started worth 50 reputation by Nico Bellic
narrowed the question to the special case B= dim A_m.
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Nico Bellic
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Let $A$ be a noetherian domain, $\mathfrak{m}$ а maximal ideal, $s$ a non-zero element of $\mathfrak{m}$, $d= \dim A_\mathfrak{m}$. Is the following claim true?

Claim: For any $B>0$ and $\epsilon>0$, there exists a positive integer $n$ s.t. for any ideal $I$ satisfying

  1. $ I \subset\mathfrak{m^n}$

  2. $\sqrt I = \mathfrak{m}$

  3. $I$ can be generated by (at most) $B$$d$ elements,

the following holds: $$ \mbox{length}(A/(I+As)) /\mbox{length}(A/I) < \epsilon$$

Note: The following example shows that the claim can be false if one drops the requirement that that the number of generators of $I$ be bounded.

Example: $A:= k[x,s]$, and let $\mathfrak{m}$ denote the ideal $(x,s)$. Let $I_{n,m}$ be an ideal of $A$ given by $$ I_{n,m}= s\mathfrak{m}^{n-1} + \mathfrak{m}^m$$.

We can calculate that for any $n$, $$\lim_{m\to \infty} \mbox{length}(A/(I_{n,m}+As)) /\mbox{length}(A/I_{n,m}) = 1$$

Let $A$ be a noetherian domain, $\mathfrak{m}$ а maximal ideal, $s$ a non-zero element of $\mathfrak{m}$. Is the following claim true?

Claim: For any $B>0$ and $\epsilon>0$, there exists a positive integer $n$ s.t. for any ideal $I$ satisfying

  1. $ I \subset\mathfrak{m^n}$

  2. $\sqrt I = \mathfrak{m}$

  3. $I$ can be generated by (at most) $B$ elements,

the following holds: $$ \mbox{length}(A/(I+As)) /\mbox{length}(A/I) < \epsilon$$

Note: The following example shows that the claim can be false if one drops the requirement that that the number of generators of $I$ be bounded.

Example: $A:= k[x,s]$, and let $\mathfrak{m}$ denote the ideal $(x,s)$. Let $I_{n,m}$ be an ideal of $A$ given by $$ I_{n,m}= s\mathfrak{m}^{n-1} + \mathfrak{m}^m$$.

We can calculate that for any $n$, $$\lim_{m\to \infty} \mbox{length}(A/(I_{n,m}+As)) /\mbox{length}(A/I_{n,m}) = 1$$

Let $A$ be a noetherian domain, $\mathfrak{m}$ а maximal ideal, $s$ a non-zero element of $\mathfrak{m}$, $d= \dim A_\mathfrak{m}$. Is the following claim true?

Claim: For any $\epsilon>0$, there exists a positive integer $n$ s.t. for any ideal $I$ satisfying

  1. $ I \subset\mathfrak{m^n}$

  2. $\sqrt I = \mathfrak{m}$

  3. $I$ can be generated by $d$ elements,

the following holds: $$ \mbox{length}(A/(I+As)) /\mbox{length}(A/I) < \epsilon$$

Note: The following example shows that the claim can be false if one drops the requirement that that the number of generators of $I$ be bounded.

Example: $A:= k[x,s]$, and let $\mathfrak{m}$ denote the ideal $(x,s)$. Let $I_{n,m}$ be an ideal of $A$ given by $$ I_{n,m}= s\mathfrak{m}^{n-1} + \mathfrak{m}^m$$.

We can calculate that for any $n$, $$\lim_{m\to \infty} \mbox{length}(A/(I_{n,m}+As)) /\mbox{length}(A/I_{n,m}) = 1$$

rephrased the question so that it makes sense for any (not just local) noetherian ring
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Nico Bellic
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Let $A$ be a noetherian local domain, $\mathfrak{m}$ theа maximal ideal, $s$ a non-zero, non-unit element of $A$$\mathfrak{m}$. Is the following claim true?

Claim: For any $B>0$ and $\epsilon>0$, there exists a positive integer $n$ s.t. for any parameter ideal $I$ contained in $\mathfrak{m^n}$, $$ \mbox{length}(A/(I+As)) /\mbox{length}(A/I) < \epsilon$$satisfying

  1. $ I \subset\mathfrak{m^n}$

  2. $\sqrt I = \mathfrak{m}$

  3. $I$ can be generated by (at most) $B$ elements,

($I$ is a parameter ideal if it can be generated by $\dim A$ elements and $\sqrt I = \mathfrak{m}$.) the following holds: $$ \mbox{length}(A/(I+As)) /\mbox{length}(A/I) < \epsilon$$

Note: regarding the requirement that $I$ be a parameter ideal, theThe following example shows that the claim iscan be false if one puts no restriction ondrops the requirement that that the number of generators of $I$ other than it be $\mathfrak{m}$-primarybounded.

Example: $A:= k[x,s]$, and let $\mathfrak{m}$ denote the ideal $(x,y)$$(x,s)$. Let $I_{n,m}$ be an ideal of $A$ given by $$ I_{n,m}= s\mathfrak{m}^{n-1} + \mathfrak{m}^m$$.

We can calculate that for any $n$, $$\lim_{m\to \infty} \mbox{length}(A/(I_{n,m}+As)) /\mbox{length}(A/I_{n,m}) = 1$$

Let $A$ be a noetherian local domain, $\mathfrak{m}$ the maximal ideal, $s$ a non-zero, non-unit element of $A$. Is the following claim true?

Claim: For any $\epsilon>0$, there exists a positive integer $n$ s.t. for any parameter ideal $I$ contained in $\mathfrak{m^n}$, $$ \mbox{length}(A/(I+As)) /\mbox{length}(A/I) < \epsilon$$

($I$ is a parameter ideal if it can be generated by $\dim A$ elements and $\sqrt I = \mathfrak{m}$.)

Note: regarding the requirement that $I$ be a parameter ideal, the following example shows that the claim is false if one puts no restriction on $I$ other than it be $\mathfrak{m}$-primary.

Example: $A:= k[x,s]$, and let $\mathfrak{m}$ denote the ideal $(x,y)$. Let $I_{n,m}$ be an ideal of $A$ given by $$ I_{n,m}= s\mathfrak{m}^{n-1} + \mathfrak{m}^m$$.

We can calculate that for any $n$, $$\lim_{m\to \infty} \mbox{length}(A/(I_{n,m}+As)) /\mbox{length}(A/I_{n,m}) = 1$$

Let $A$ be a noetherian domain, $\mathfrak{m}$ а maximal ideal, $s$ a non-zero element of $\mathfrak{m}$. Is the following claim true?

Claim: For any $B>0$ and $\epsilon>0$, there exists a positive integer $n$ s.t. for any ideal $I$ satisfying

  1. $ I \subset\mathfrak{m^n}$

  2. $\sqrt I = \mathfrak{m}$

  3. $I$ can be generated by (at most) $B$ elements,

the following holds: $$ \mbox{length}(A/(I+As)) /\mbox{length}(A/I) < \epsilon$$

Note: The following example shows that the claim can be false if one drops the requirement that that the number of generators of $I$ be bounded.

Example: $A:= k[x,s]$, and let $\mathfrak{m}$ denote the ideal $(x,s)$. Let $I_{n,m}$ be an ideal of $A$ given by $$ I_{n,m}= s\mathfrak{m}^{n-1} + \mathfrak{m}^m$$.

We can calculate that for any $n$, $$\lim_{m\to \infty} \mbox{length}(A/(I_{n,m}+As)) /\mbox{length}(A/I_{n,m}) = 1$$

Add an example showing that some restriction on $I$ is necessary
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Nico Bellic
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Let $A$ be a noetherian local domain, $\mathfrak{m}$ the maximal ideal, $s$ a non-zero, non-unit element of $A$. Is the following claim true?

Claim: For any $\epsilon>0$, there exists a positive integer $n$ s.t. for any parameter ideal $I$ contained in $\mathfrak{m^n}$, $$ \mbox{length}(A/(I+As)) /\mbox{length}(A/I) < \epsilon$$

($I$ is a parameter ideal if it can be generated by $\dim A$ elements and $\sqrt I = \mathfrak{m}$.)

Note: regarding the requirement that $I$ be a parameter ideal, the following example shows that the claim is false if one puts no restriction on $I$ other than it be $\mathfrak{m}$-primary.

Example: $A:= k[x,s]$, and let $\mathfrak{m}$ denote the ideal $(x,y)$. Let $I_{n,m}$ be an ideal of $A$ given by $$ I_{n,m}= s\mathfrak{m}^{n-1} + \mathfrak{m}^m$$.

We can calculate that for any $n$, $$\lim_{m\to \infty} \mbox{length}(A/(I_{n,m}+As)) /\mbox{length}(A/I_{n,m}) = 1$$

Let $A$ be a noetherian local domain, $\mathfrak{m}$ the maximal ideal, $s$ a non-zero, non-unit element of $A$. Is the following claim true?

Claim: For any $\epsilon>0$, there exists a positive integer $n$ s.t. for any parameter ideal $I$ contained in $\mathfrak{m^n}$, $$ \mbox{length}(A/(I+As)) /\mbox{length}(A/I) < \epsilon$$

($I$ is a parameter ideal if it can be generated by $\dim A$ elements and $\sqrt I = \mathfrak{m}$.)

Let $A$ be a noetherian local domain, $\mathfrak{m}$ the maximal ideal, $s$ a non-zero, non-unit element of $A$. Is the following claim true?

Claim: For any $\epsilon>0$, there exists a positive integer $n$ s.t. for any parameter ideal $I$ contained in $\mathfrak{m^n}$, $$ \mbox{length}(A/(I+As)) /\mbox{length}(A/I) < \epsilon$$

($I$ is a parameter ideal if it can be generated by $\dim A$ elements and $\sqrt I = \mathfrak{m}$.)

Note: regarding the requirement that $I$ be a parameter ideal, the following example shows that the claim is false if one puts no restriction on $I$ other than it be $\mathfrak{m}$-primary.

Example: $A:= k[x,s]$, and let $\mathfrak{m}$ denote the ideal $(x,y)$. Let $I_{n,m}$ be an ideal of $A$ given by $$ I_{n,m}= s\mathfrak{m}^{n-1} + \mathfrak{m}^m$$.

We can calculate that for any $n$, $$\lim_{m\to \infty} \mbox{length}(A/(I_{n,m}+As)) /\mbox{length}(A/I_{n,m}) = 1$$

polished up and put def'n of parameter ideal
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Nico Bellic
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Source Link
Nico Bellic
  • 635
  • 4
  • 12
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