TITLE: Height of associated primes in regular rings
QUESTION [8 upvotes]: Let $I$ be an ideal in a regular ring $R$. Suppose $I$ can be generated by $n$ elements. Let $P$ be an associated primes of $I$. Is it true that the height of $P$ is bounded above by $n$?
Remark: (1) The question has an affirmative answer when $P$ is a minimal associated prime of $I$ by Krull's principal theorem.
(2) The question is false for general Noetherian rings Indeed, suppose $I = 0$ and $R$ has an embedded associated prime ideal.

REPLY [7 votes]: Just a couple of remarks:
a) You can simplify Jason's example slightly with $I=(u^2,v^2,xu-yv)$. The annihilator of $uv$ is the maximal ideal. 
b) One can in fact find an example of a three-generated $I$ with an associated prime of height $N$ for any $N$. Take $R$ to be a regular local ring of dim $N$ and maximal ideal $m$. A theorem by Bruns [1] states that the second syzygy of any ideal is the second syzygy of a three-generated ideal. So take $M = syz^2(m)$, then $M=syz^2(I)$ where $I$ has $3$ generators. Counting depth gives $depth(I)=1$, so $R/I$ has $m$ as an associated prime. 
[1] http://www.sciencedirect.com/science/article/pii/0021869376900478