2 LaTeX cleanup

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

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

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 $$\lim_n \ell(A/(\mathfrak{m}^n ;\ell(A/(\mathfrak{m}^n + sA))/ \ell(A/(\mathfrak{m}^n) = 0.$$

1

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 $$\lim_n \ell(A/(\mathfrak{m}^n + sA))/ \ell(A/(\mathfrak{m}^n) = 0.$$