I have been wondering about the following (and allready posted a similar question, see [Dimension of ring completion wrt to a decreasing chain of ideals](https://mathoverflow.net/questions/143048/dimension-of-ring-completion-wrt-to-a-decreasing-chain-of-ideals)): 


Let $R$ be the ring of formal power series in $n$ indeterminates over $\mathbb{C}$, and let $(I\_{k})\_{k\in \mathbb{N}}$ be a strictly decreasing chain of unmixed radical ideals, which **all** have the same height $s$. Further assume that $\bigcap I\_{n} =: \mathfrak{p}$ is prime and that $I\_{1}$ is prime.


Then obviously $ht(\mathfrak{p}) \le s$.


Is it true that $ht(\mathfrak{p}) = s$?



Chain of Ideals of same height
No, it is not true that $\mathfrak{p}$ must have height $s$. For one counterexample, let $R$ be $\mathbb{C}[[s,t]]$, and let $I\_k$ be $\langle t(t-s)(t-2s)\cdots (t-(k-1)s) \rangle$. 


