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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):

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 $ht(\mathfrak{p}) = s$?

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    $\begingroup$ What about $R = \mathbb{C}[[t,s]]$ and $I_k = \langle t(t-s)(t-2s)\cdot \dots \cdot (t-(k-1)s) \rangle$? $\endgroup$
    – Jason Starr
    Commented Oct 29, 2013 at 21:01

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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 $\langle t(t-s)(t-2s)\cdots (t-(k-1)s) \rangle$.

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  • $\begingroup$ Thank you! If someone is interested in the details: $I_{k}$ is by the Weistrass division theorem the intersection $\cap_{l=0}^{k-1} ker \ \epsilon_{l}$, where $\epsilon_{l}$ is the evaluation map $F(s,t) \mapsto F((l-1)s,s)$. And since $ht( ker \ \epsilon_{l}) = 1$ for all $l$, all ideals $I_{k}$ are radical equidimensional ideals of height one. Also by the Weierstrass division theorem, a power series $F(x,y)$ where $y$ is a single variable and $x=(x_{1},\dots, x_{n})$ has only finitely many implict solutions $F(x,y(x))=0$, which yields $\cap I_{l} = \{0\}$. $\endgroup$
    – Sebastian
    Commented Oct 30, 2013 at 13:17

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