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Let n be a positive integer, and let $P_0\subsetneq P_1\subsetneq ...\subsetneq P_n$ be a chain of prime ideals in a Noetherian ring R. Moreover, let $a\in P_n$. Prove:

1.There is a chain of prime ideals $P_0'\subsetneq P_1'\subsetneq ...\subsetneq P'_{n-1}\subsetneq P_n$, s.t. $a\in P'_1$.

2.There is in general no such chain with $a\in P'_0$.

Use this to prove Krull height theorem, i.e. any minimal prime ideal containing n fixed elements in a Noetherian ring R has cxdimension at most n.

Part 2 seems easy, e.g. we can take R to be a PID and $a\neq0$, which then forces $P_0'=0$. But I have no clue how to do 1 and use this to prove Krull height theorem. I find it difficult constructing a chain of prime ideals of length exactly n, since in general R can have two maximal chains of prime ideals of different length..

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We may assume $P_0 = 0$, and more, $R$ is local with the unique maximal ideal $P_n$. Since the title is 'application of Krull Principle ideal theorem', we will use the theorem to show the statements.

For $n=2$:

If $n=2$, if $a = 0$, namely $a\in P_0$, done; if $a\neq 0$, by Krull Principle ideal theorem, for every prime ideal $P$ minimal over $a$, $ht(P)\leq 1$, so $ht(P)=1$, thus $0\subsetneq P \subsetneq P_2$, we are done too.

For $n\geq 2$:

If $a\in P_{n-2}$, by induction, we can find a chain of length $n-2$ starting with $0$ and ending with $P_{n-2}$ such that $a$ is in the first link.

If $a\notin P_{n-2}$, consider $R/P_{n-2}$, by result of $n=2$, we can find a $P$ such that $P_{n-2}\subsetneq P \subsetneq P_n$ with $a \in P$, use induction again, we are done!


If a prime $P$ is minimal over $(a_1,\ldots, a_n)$, if $P$ was of height $\geq n+1$. We can find $P_0\subsetneq P_1\cdots\subsetneq P_{n+1}=P$, by above, we can find another chain of length $n+1$ with $a_1\in P_1$. Now consider $R/P_1$, check if $P/P_1$ is minimal over $(a_2,\ldots,a_n)$, so by induction $ht(P/P_1)\leq n-1$, but here we have a chain $P_1/P_1\subset P_2/P_1\cdots\subset P_{n+1}/P_1=P/P_1$ of length $n$?!

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  • $\begingroup$ principle $\ne$ principal $\endgroup$ – user26857 Mar 9 '15 at 11:43

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