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Let $\newcommand{\frakm}{\mathfrak{m}}(R,\frakm)$ be a reduced Noetherian local ring of dimension $d$ and $f\in\frakm^\alpha\setminus\frakm^{\alpha+1}$ a parameter of $R$, i.e. $\dim R/(f)=d-1$. Let $e(R)$ denote the Hilbert-Samuel multiplicity of $R$ with respect to the maximal ideal. By Theorem 3.2 of Flenner and Vogel's On multiplicities of local rings, we have $$ e(R/(f))=\alpha\cdot e(R) \tag{$*$}\label{468646_star}$$ if and only if $\DeclareMathOperator{\gr}{gr}\dim\gr_{\frakm}(R)/(f^*)=d-1$ where $f^*=f+\frakm^{\alpha+1}$ is the initial form of $f$ in the associated graded ring.

A element $f\in\frakm^\alpha\setminus\frakm^{\alpha+1}$ is superficial of order $\alpha$ if (among many equivalent definitions) there exists an integer $c$ such that $(\frakm^{n+\alpha}:(f))\cap \frakm^c=\frakm^n$ for all $n\geq c$. It's well-known that superficial elements satisfy equality \eqref{468646_star} when $d>1$, so in this case $f$ superficial implies $\dim\gr_{\frakm}(R)/(f^*)=d-1$.

My question is, what is an example of a reduced Noetherian local ring $R$ of dimension $d>1$ and parameter $f$ such that $\dim\gr_{\frakm}(R)/(f^*)=d-1$ but $f$ is not superficial? Are there any conditions on $R$ where the set of superficial elements are exactly the ones where $\dim\gr_{\frakm}(R)/(f^*)=d-1$?

If relevant, I'd prefer not to assume the residue field is infinite, and I am particularly interested in the case $\alpha=1$.

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$\newcommand{\kk}{\mathbb{k}} \newcommand{\mmm}{\mathfrak{m}} \newcommand{\ppp}{\mathfrak{p}} \newcommand{\qqq}{\mathfrak{q}} \DeclareMathOperator{\gr}{gr}$Assume $G := \gr_\mmm(R)$ satisfies the following two conditions:

  1. $G$ has pure dimension $d$.
  2. The zero ideal of $G$ has no embedded prime which does not contain the maximal ideal $G_+ := \oplus_{i \geq 1}\mmm^i/\mmm^{i+1}$.

Theorem. If $G$ satisfies the above conditions, then the answer is negative, i.e. if $f \in R$ is such that $\dim(\gr_\mmm(R)/\langle f^* \rangle) = d-1$, then $f$ is superficial.

Before proving the theorem, some examples (please doublecheck that the claims in the examples are correct - I have not checked these carefully):

  • An example where Condition 1 fails: If $R$ is the localization at the origin of $\kk[x,y,z]/\langle xy, xz \rangle$, where $\kk$ is a field, then $\gr_\mmm(R) \cong \kk[x^*,y^*,z^*]/\langle x^*y^*, x^*z^* \rangle$. In particular, $d := \dim(R) = 2$, $\dim(\gr_\mmm(R)/\langle y^* \rangle) = 1$, but $x^{n-1} \in (\mmm^{n+1}: y) \setminus \mmm^n$ for each $n \geq 2$, i.e. $y$ is not supercial.
  • Edit: this example is incorrect (see the comments). Let $R$ be the localization at the origin of $\kk[x,y,z,w]/\langle x^2 - z^3 , xy - w^3 \rangle$. Then $\gr_\mmm(R) \cong \kk[x^*,y^*,z^*,w^*]/\langle (x^*)^2, x^*y^* \rangle$. Consequently, $\dim(\gr_\mmm(R)/\langle y^* \rangle) = 1$, but $xz^{n-2} \in (\mmm^{n+1}: y) \setminus \mmm^n$ for each $n \geq 2$, so that $y$ is not superficial.
  • An example where Condition 1 is satisfied but Condition 2 fails: Let $R$ be the localization at the origin of $\kk[x,y,z,w]/\langle xz-y^3,yz-x^4,z^2-x^3y^2 \rangle$. Then $\gr_\mmm(R) \cong \kk[x^*,y^*,z^*,w^*]/\langle (z^*)^2, y^*z^*, x^*z^*, (y^*)^4 \rangle$. Consequently, $\dim(\gr_\mmm(R)/\langle x^* \rangle) = 1$, but $zw^{n-2} \in (\mmm^{n+1}: x) \setminus \mmm^n$ for each $n \geq 2$, so that $x$ is not superficial.

Now we prove the theorem following arguments from the proof of the existence of superficial elements in Volume 2 of Zariski-Samuel (Chapter VIII, Section 8, Lemma 5).

Let $\ppp_1, \ldots, \ppp_k$ be the prime ideals of $G := \gr_\mmm(R)$ associated to the zero ideal. We may assume that $\ppp_j$ contains the maximal ideal $G_+ := \oplus_{i \geq 1}\mmm^i/\mmm^{i+1}$ if and only if $j > k'$, where $k' \leq k$. Since $G$ satisfies the two conditions mentioned in the beginning, each of $\ppp_1, \ldots, \ppp_{k'}$ is a minimal prime ideal of dimension $d$ associated to zero. In particular, $f^* \not\in \ppp_j$ for $j = 1, \ldots, k'$.

Let $\qqq_j$ be a primary component of the zero ideal of $G$ corresponding to $\ppp_j$. For each $j > k'$, $\qqq_j$ contains a power of $G_+$. Pick $c$ such that $$G_+^c \subseteq \bigcap_{j > k'} \qqq_j$$ Pick $n \geq c$ and $g \in (\mmm^{n+\alpha}: \langle f \rangle) \cap \mmm^c$. It suffices to show that $g \in \mmm^n$. Indeed, if $g \not\in \mmm^n$, then $g^*f^* = 0 \in G$, where $g^*$ is the initial form of $g$. Since $f^* \not\in \ppp_j$, $j = 1, \ldots, k'$, it follows that $g^* \in \qqq_j$, $j = 1, \ldots, k'$. On the other hand, $g^* \in G_+^c$ (since $g \in \mmm^c$). Consequently, $$ g^* \in \bigcap_{j=1}^k \qqq_j = 0$$ But then $g = 0 \in \mmm^n$. This contradiction proves the theorem.

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  • $\begingroup$ Thanks! Can you elaborate on your dimension calculations in the examples? For example in the first one, the associated graded corresponds to the union of the $yz$-plane and the $x$-axis, so quotienting by $x$ should leave the $yz$-plane which has dimension 2, right? A quick check in Macaulay2 also says $\dim(\operatorname{gr}_{\mathfrak{m}}(R)/(x^*))=2$, but perhaps I am misunderstanding $\endgroup$
    – mbert
    Commented Apr 9 at 20:55
  • $\begingroup$ I am also not sure about your calculation of the initial form ideal in the second example, since in the associated graded we should have $(x^*)^2=(z^*)^3=0\in \mathfrak{m}^2/\mathfrak{m}^3$ $\endgroup$
    – mbert
    Commented Apr 9 at 21:05
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    $\begingroup$ You are interested in $f$ such that $\dim(G)/\langle f^* \rangle = d-1$. This translates to $f$ not being any of the prime ideals corresponding to $d$ dimensional irreducible components of $G$. If $G$ is not pure dimensional, then this does not prevent the possibility that $f^*$ could be inside one of those $\mathfrak{p}_j$ which are prime ideals corresponding to lower dimensional irreducible components of $G$. This was the case in the first example. $\endgroup$
    – pinaki
    Commented Apr 11 at 13:19
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    $\begingroup$ I used the TangentCone package of Macaulay2 to check your initial form ideal for the second example. According to it, the initial form ideal in this case is not generated by the initial forms of the generators of the original ideal (it gives $y^*(z^*)^3-x^*(w^*)^3$ as another generator) and the dimension of $\operatorname{gr}_{\mathfrak{m}}(R)/(y^*)$ is 2 $\endgroup$
    – mbert
    Commented Apr 11 at 23:26
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    $\begingroup$ Yeah - you are right. I am glad that you are carefully checking the claims. I was trying to find an example where the first condition on $G$ is satisfied, but not the second one. It turns out that the first example from the Macaulay2 webpage on the "tangentCone" function gives such an example :) I will update the answer. $\endgroup$
    – pinaki
    Commented Apr 12 at 4:30

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