Let $(R, \mathfrak m)$ be a Noetherian local ring such that the residue field $k=R/\mathfrak m$ is algebraically closed.
For a fixed integer $t \ge 1$ , $V_t :=\mathfrak m^t/\mathfrak m^{t+1} $ is a finite dimensional $k$-vector space with $\dim_k V_t =\mu (\mathfrak m^t)$ (minimal no. of generators for $\mathfrak m^t$ ) . Notice that any $k$-linear subspace of $V_t$ is of the form $I/\mathfrak m^{t+1}$ for some ideal $I$ of $R$ with $\mathfrak m ^{t+1}\subseteq I \subseteq \mathfrak m^t$.
For an integer $1\le n \le \dim V_t$, consider the Grassmannian variety $Gr(n, V_t)$ with Zariski Topology (induced from $\mathbb P(\wedge^n V_t)$ via Plucker embedding ).
Consider the following subset $U :=\{ I/\mathfrak m^{t+1} \in Gr(n,V_t) | I\mathfrak m \ne \mathfrak m^{t+1} , (I:\mathfrak m)\ne \mathfrak m^t \}$ .
My questions are:
(1) How to show that $U$ is open (in Zariski Topology) subset of $Gr(n,V_t)$ ?
(2) For each integer $s > 0$, define $U_{ \le s} :=\{I/\mathfrak m^{t+1} \in U | \dim_k (I:\mathfrak m)/(\mathfrak mI: \mathfrak m) \le s \}$ , $U_{\ge s} :=\{I/\mathfrak m^{t+1} \in U | \dim_k (I:\mathfrak m)/(\mathfrak mI: \mathfrak m) \ge s\}$
and $U_ s :=\{I/\mathfrak m^{t+1} \in U | \dim_k (I:\mathfrak m)/(\mathfrak mI: \mathfrak m) = s\}$ . Which of $U_{\le s}, U_s$ and $U_{\ge s}$ is open and which is closed in $U$ ?
My thoughts: I'm not sure whether this is helpful or not but $U=U_{\ge 1}$ . Also, for $I/\mathfrak m^{t+1}\in Gr(n,V_t)$, one has that it is in $U$ if and only if $\mathfrak m(I: \mathfrak m) \ne I \mathfrak m$ . Also, just for clarification, $(I : J) :=\{r \in R | rJ \subseteq I \}$.
Please help. Thanks in advance