# Minimal length of quotient by parameter ideals

Consider a commutative noetherian local ring $R$ of dimension $d$ and define

$$c_R\colon=\min_{(x_1,\ldots,x_d)} \{\mathrm{length}\ R/(x_1,\ldots,x_d)R\mid (x_1,\ldots,x_d)\ \mathrm{is\ a\ system\ of parameters\ of\ }R\}.$$

1) Does $c_R$ always give information about the ring $R$? For instance, clearly $c_R=1$ if and only if $R$ is regular. When is $c_R=2$? Does $R$ have to be any particular type of ring for $c_R$ to be $2$?

2) In a given local ring $R$ can one characterize those system of parameters $(x_1,\ldots,x_d)$ for which $c_R=\mathrm{length}\ R/(x_1,\ldots,x_d)R$? For instance, one could say that a necessary condition for $c_R=\mathrm{length}\ R/(x_1,\ldots,x_d)R$ is that $(x_1,\ldots,x_d)$ has to be "maximal", that is, not contained in any larger parameter ideal. Is this condition also sufficient?

Here is answer of (1), $c_R=2$. I assume that $R$ is complete (as $\mathfrak{m}$-adic topology) and the residue field $A/\mathfrak{m}$ is infinite (if necessary).

Let $v(R) = \dim_{R/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2$ be the embedded dimension of $R$. Since $c_R = 2$ we have $R$ is not a regular local ring, so $v(R) \ge d+1$. Let $\mathfrak{q} = (x_1, ..., x_d)$ be a parameter ideal such that $\ell(R/\mathfrak{q}) = c_R=2$. We have $\ell(\mathfrak{m}/\mathfrak{q})= 1$, so $v(R) = d+1$. Furthermore $\mathfrak{m}^2 \subseteq \mathfrak{q}$ and $\mathfrak{m} = (x_1, ..., x_d, y)$ for some $y$.

First we consider the case $(R, \mathfrak{m})$ is unmixed i.e. $\dim R/\mathfrak{p} = d$ for all $\mathfrak{p} \in \mathrm{Ass}R$.

We note that the multiplicity $e(R) = e_R(\mathfrak{m}) \ge 2$, since $R$ is not regular.

Let $\mathfrak{q} = (x_1, ..., x_d)$ be a parameter ideal such that $\ell(R/\mathfrak{q}) = c_R=2$. It is well known that the multiplicity $e_R(\mathfrak{q}) \le \ell(R/\mathfrak{q})$. Thus we have two cases:

Case 1: $e_R(\mathfrak{q}) = 1$, so the multiplicity of $R$, $e(R) =1$. It is a contradiction.

Thus we have

Case 2: $e_R(\mathfrak{q}) = 2$, so $e(R) = 2$. Moreover $e_R(\mathfrak{q}) = \ell(R/\mathfrak{q})$, hence $(R, \mathfrak{m})$ is Cohen-Macaulay.

By Abhyankar's inequality we have $v(R) \le e(R) + d -1 = d+1$. Thus $R$ is a Cohen-Macaulay local ring of maximal embedded dimension. Furthermore, we have $\mathfrak{q}$ is a minimial reduction of $\mathfrak{m}$ and $\mathfrak{m}^2 = \mathfrak{q}\mathfrak{m}$ (by Sally?). The converse is true. Thus we have

Conclusion. Suppose $(R, \mathfrak{m})$ is unmixed. Then $c_R = 2$ if and only if $R$ is a Cohen-Macaulay (in fact, Gorenstein) of maximal embedded dimension with $v(R) = d+1$.

The case $R$ is not unmixed. Let $U$ be the largest ideal of $R$ of dimension least than $d$ (as $R$-module). Let $S = R/U$, we have $S$ in unmixed. It is clear that $c_S \le c_R = 2$. As above we have $S$ is Cohen-Macalay. So $x_1, ..., x_d$ is an regular sequence on $S$, we have $U \cap \mathfrak{q} = \mathfrak{q}U$. Therefore we have the following short exact sequence $$0 \to U/\mathfrak{q}U \to R/\mathfrak{q} \to S/\mathfrak{q}S \to 0.$$ Since $\ell(R/\mathfrak{q}) = 2$ we have $\ell(U/\mathfrak{q}U) = \ell(S/\mathfrak{q}S) = 1$. Therefore $R$ is a regular local ring and $U = (a)$ is a principal ideal. Moreover we have $\mathfrak{m} = (x_1, ..., x_d, a)$.

Examples. Take $S = K[[X_1, ..., X_d]]$ and $T$ is a non-zero quotient ring of $S$. Let $R = S \ltimes T$ be the idealization (or trivial extension). We can check that the maximal ideal of $R$ generated by $d+1$ elements $(X_1, 0), ..., (X_d, 0), (0, 1)$. If $T = S$ we have $R$ is unmixed. If $T \neq S$ we have $R$ is not unmixed.

Edit: I give a counter-example for (2). Let $R$ be the localization of $\mathbb{C}[X,Y]/(Y^2) = \mathbb{C}[x, y]$ at the maximal ideal $\mathfrak{m} = (x,y)$. We have $R$ is a Cohen-Macaulay local ring of dimension one, and $e(R) = 2$, $v(R) = 2$, $c_R=2$. Thus $R$ is maximal embedded dimension.

Consider the parameter ideal $\mathfrak{q} = (x^2+y)$. Since $\mathfrak{q} \nsubseteq \mathfrak{m}^2$ we have there is no parameter ideal containing $\mathfrak{q}$. Therefore $\mathfrak{q}$ is a maximal parameter ideal. One can check that $$\ell(R/\mathfrak{q}) = \ell(\mathbb{C}[X,Y]/(X^2+Y, Y^2)) = 4.$$