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Let $(R, \mathfrak m,k)$ be a Noetherian local ring such that the residue field $k$ is infinite. Let $n=\mu(\mathfrak m)$. Then $n=\dim_k(\mathfrak m/\mathfrak m^2)$ . By fixing $x_1,...,x_n \in \mathfrak m$ such that $\bar x_1,...,\bar x_n\in \mathfrak m/\mathfrak m^2$ gives a $k$-vector space basis, we can identify $\mathfrak m/\mathfrak m^2$ with $\mathbb A^n(k)$ by sending $\bar x_i \to e_i$. So we can transfer the classical Zariski topology of $\mathbb A^n(k)$ to $\mathfrak m/\mathfrak m^2$.

I have two similar kind of questions :

(1) If $I$ is $\mathfrak m$-primary i.e. $\sqrt I=\mathfrak m$ and the set $\{\bar x\in \mathfrak m/\mathfrak m^2: x\in \mathfrak m \setminus \mathfrak m^2 \space \text{and} \space (\mathfrak mI:x)=I \}$ is non-empty, then how to show that the set is Zariski Open ?

(2) If $I=\overline I$ is $\mathfrak m$-primary i.e. $\sqrt I=\mathfrak m$ and the set $\{\bar x\in \mathfrak m/\mathfrak m^2: x\in \mathfrak m \setminus \mathfrak m^2 \space \text{and} \space (\overline{\mathfrak mI}:x)=I \}$ is non-empty, then how to show that the set is Zariski Open ?

Here for an ideal $I$, by $\overline I$ we denote the integral closure of $I$ https://en.m.wikipedia.org/wiki/Integral_closure_of_an_ideal

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  • $\begingroup$ You always have $(\mathfrak mI \colon x) \supseteq I$, right? So the question is when is there nothing else. You would like to have access to $R/J$ for any $\mathfrak m$-primary ideal $J$ as an affine space over $k$ (or a successive fibration of affine spaces), and use some sort of structure constants to show that multiplication $m \colon R/I \times \mathfrak m/\mathfrak m I \to R/\mathfrak m I$ is a morphism of varieties. This is certainly ok if $R$ is a $k$-algebra, but probably also works in other settings. Then study the map $m$ geometrically. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 21, 2020 at 2:32
  • $\begingroup$ @R. van Dobben de Bruyn: I'm having some issues seeing as to how the multiplication $R/I \times \mathfrak m/\mathfrak mI\to R/\mathfrak m I$ being a morphism of schemes would be helpful (and what scheme structure exactly are you talking about on $ \mathfrak m/\mathfrak mI$ ?) ... a priori, I don't see any connection between the usual Zariski topology of Spec $(R/I)$ and the Zariski topology I've defined on $\mathfrak m/\mathfrak m^2$ ... $\endgroup$
    – uno
    Commented Apr 29, 2020 at 16:31
  • $\begingroup$ Sorry, I didn't mean to suggest that this is a full answer, but it is certainly a strategy in case $R$ is a $k$-algebra. Then $R/I$ is a finite dimensional $k$-algebra, so you can view it as a ring object in affine $k$-schemes (in particular, addition and multiplication are given by polynomials; to make this explicit use the structure constants of $R/I$). For example this immediately shows that for a single $f\not\in I$, the set of $x$ with $f\in(\mathfrak mI:x)$ is open, because $x \mapsto fx \in R/\mathfrak mI$ is continuous. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 29, 2020 at 17:50

1 Answer 1

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Case 1 was treated carefully in Appendix A of J. Watanabe's paper "$m$-full ideals". Case 2 can be treated the same way, as sketched below.

Let $J$ be either $mI$ or $\overline{mI}$ and $A=R/J$. Since $J:x \supset I$, we have that $l(A/xA)= l(0:_Ax) \geq l(I/J)$. Thus we need to show that the set of $x$ such that the length $A/xA$ is minimal possible is Zariski open. Then the idea is to consider the "universal generic element". Let $S=A[Y_1,\dots, Y_n]$ and $A'=S_{mS}$. Consider the element $X= \sum Y_ix_i$ in $A'$. Prove that $l(A/xA)\geq l(A'/XA')$ with equality happens if and only if the ideal $(Y_1-a_1,\dots Y_n-a_n)$ does not contain certain radical ideal in $k[Y_1,\dots, Y_n]$ where $x = (a_1,\dots, a_n) \in m/m^2=\mathbb A^n_k$

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