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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 $\epsilon>0$, 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 $n$, $$\lim_{m\to \infty} \mbox{length}(A/(I_{n,m}+As)) /\mbox{length}(A/I_{n,m}) = 1$$

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Here is a heuristic argument. Let $B=A/sA$ and $d=dim(A)$. Then the dimension of $B$ is $d-1$ as $A$ is a domain. Ideals in $n$ power of $m$ in $B$ should have colength of size $n^{d-1}$, while those in $A$ has size $n^d$ (cf. Hilbert-Samuel multiplicity). This seems a little messy to write down and I don't have time right now, but may be it helps. –  Hailong Dao Mar 30 '12 at 14:12
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The above works for example if $I=m^n$. By the way, I think the parameter ideal is red herring, you may only need $I$ to be $m$-primary. –  Hailong Dao Mar 30 '12 at 14:17
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Thank you for your comment! Maybe " $I$ is parameter ideal" can be replaced by "number of generators of $I$ is bounded". I added an example that shows that some restriction on $I$ is necessary. –  Nico Bellic Mar 30 '12 at 22:08
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1 Answer

Here is a counterexample for you question.

Let $A = k[[s, x]]$, $\dim A = 2$

For each pair $n, m$, $n < m$, we consider the parameter ideal $$\mathfrak{q}_{n, m} = (s^n+x^m, sx^{n-1})$$

We have $\mathfrak{q}_{n, m} + sA = (s, x^m)$. Hence $$\ell(A/(\mathfrak{q}_{n, m} + sA)) = m$$

On the other hand, we can check that $s^{n+1}$ and $x^{m+n-1}$ is contained in $\mathfrak{q}_{n, m}$. Thus $$\ell(A/\mathfrak{q}_{n, m}) \leq \ell(A/(s^{n+1}, sx^{n-1},x^{m+n-1})) = m + n^2-1.$$ Therefore $$\lim_{m \to \infty} \ell(A/(\mathfrak{q}_{n, m} + sA))/ \ell(A/\mathfrak{q}_{n, m}) = 1$$

Remark: (i) It should be noted that, I contruct this example based thinking the minimal reduction of the ideal $I_{n,m}$ of your question.

(ii) Your question is true in the case $I = \mathfrak{m}^n$, it means $$\lim_n \;\ell(A/(\mathfrak{m}^n + sA))/ \ell(A/(\mathfrak{m}^n) = 0.$$

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Thanks a lot! I accept the answer. –  Nico Bellic Apr 20 '12 at 3:46
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