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If $R$ is a noetherian local domain which is catenary, and $a_1,...,a_n$ are elements of the maximal ideal of $R$ with $\operatorname{height}(a_1,...,a_n)=n$, could we conclude that $\operatorname{height}(a_1,...,a_i)=i$ for each $i$ with $1≤i≤n$?

Being a domain is a necessary condition, as could be seen for $$R=K[X,Y,Z]/{(X,Y)∩(Z)}=K[x,y,z],$$ where $\operatorname{height}(x,y+z)=2$ and $\operatorname{height}(x)=0$. Thanks in advance.

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  • $\begingroup$ why height(x,y+z)=2 and height(x)=0? $\endgroup$ – user147308 Jun 15 '14 at 14:05
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Proposition: $\DeclareMathOperator{\ht}{ht}$If $R$ is a catenary local domain, and $I$ is an $R$-ideal, then $\ht I + \dim R/I = \dim R$.

Proposition: If $(R, m)$ is Noetherian local, a sequence of elements $a_1, \ldots, a_d \in m$ ($d = \dim R$) form a system of parameters iff $\dim R/(a_1, \ldots, a_i) = \dim R - i$ for each $i = 1, \ldots, d$.

Thus in a Noetherian catenary local domain $R$, elements $a_1, \ldots a_n \in m$ with $n \le \dim R$ satisfy $\ht (a_1, \ldots, a_i) = i$ for each $i = 1, \ldots, n$ iff $a_1, \ldots, a_n$ form part of a system of parameters.

Edit: I had initially overlooked the assumption that $\ht(a_1, \ldots, a_n) = n$. This does imply that $a_1, \ldots, a_n$ is part of a system of parameters, as follows: by the first proposition above, $\dim R/(a_1, \ldots, a_n) = \dim R - n =: d - n$. Choose a s.o.p. of $R/(a_1, \ldots, a_n)$, necessarily of length $d - n$, say $\overline{b_1}, \ldots, \overline{b_{d-n}}$. Then $(\overline{b_1}, \ldots, \overline{b_{d-n}})$ is $\overline{m}$-primary, so $(a_1, \ldots, a_n, b_1, \ldots, b_{d-n})$ is $m$-primary, so $a_1, \ldots, a_n, b_1, \ldots, b_{d-n}$ is an s.o.p. of $R$.

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  • $\begingroup$ @user26857: Yes, I had wanted to remove the space as well, but it somehow messed up the boldface "Proposition". I think I found a fix though, just by inserting the \DeclareMathOperator suitably before the first use $\endgroup$ – zcn Jun 17 '14 at 18:02

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