Lengths over a local ring - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T09:01:10Z http://mathoverflow.net/feeds/question/92604 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92604/lengths-over-a-local-ring Lengths over a local ring Nico Bellic 2012-03-29T19:56:17Z 2012-04-17T19:07:11Z <p>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?</p> <p>Claim: For any $\epsilon>0$, there exists a positive integer $n$ s.t. for any ideal $I$ satisfying</p> <p>1) $I \subset\mathfrak{m^n}$</p> <p>2) $\sqrt I = \mathfrak{m}$</p> <p>3) $I$ can be generated by $d$ elements,</p> <p>the following holds: $$\mbox{length}(A/(I+As)) /\mbox{length}(A/I) &lt; \epsilon$$</p> <p>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. </p> <p>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 $$I_{n,m}= s\mathfrak{m}^{n-1} + \mathfrak{m}^m$$.</p> <p>We can calculate that for any $n$, $$\lim_{m\to \infty} \mbox{length}(A/(I_{n,m}+As)) /\mbox{length}(A/I_{n,m}) = 1$$ </p> http://mathoverflow.net/questions/92604/lengths-over-a-local-ring/94313#94313 Answer by Pham Hung Quy for Lengths over a local ring Pham Hung Quy 2012-04-17T18:56:14Z 2012-04-17T19:07:11Z <p>Here is a counterexample for you question.</p> <p>Let $A = k[[s, x]]$, $\dim A = 2$</p> <p>For each pair $n, m$, $n &lt; m$, we consider the parameter ideal $$\mathfrak{q}_{n, m} = (s^n+x^m, sx^{n-1})$$</p> <p>We have <code>$\mathfrak{q}_{n, m} + sA = (s, x^m)$</code>. Hence $$\ell(A/(\mathfrak{q}_{n, m} + sA)) = m$$</p> <p>On the other hand, we can check that <code>$s^{n+1}$</code> and <code>$x^{m+n-1}$</code> is contained in <code>$\mathfrak{q}_{n, m}$</code>. Thus <code>$$\ell(A/\mathfrak{q}_{n, m}) \leq \ell(A/(s^{n+1}, sx^{n-1},x^{m+n-1})) = m + n^2-1.$$</code> Therefore <code>$$\lim_{m \to \infty} \ell(A/(\mathfrak{q}_{n, m} + sA))/ \ell(A/\mathfrak{q}_{n, m}) = 1$$</code></p> <p>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.</p> <p>(ii) Your question is true in the case $I = \mathfrak{m}^n$, it means <code>$$\lim_n \;\ell(A/(\mathfrak{m}^n + sA))/ \ell(A/(\mathfrak{m}^n) = 0.$$</code> </p>