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..


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 $n$?!

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.