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.
For a regular ring $R$ and an ideal $I$ generated by $n$ elements, the embedded primes of $I$ can, indeed, have height strictly larger than $n.$ For instance, let $R$ be $k[x,y,u,v].$ Let $I$ be the ideal generated by $3$ elements, $$ I = \langle xu^2,yv^2,x^2u-y^2v\rangle.$$ The congruence class of the element $xyuv$ is annihilated by $\langle x,y,u,v\rangle.$ Thus the maximal ideal $\langle x,y,u,v\rangle$ is an embedded prime of $I$. The height of the maximal ideal is $4$, which is strictly larger than $3$.
Here is the result that is true: for an ideal $I$ generated by $n$ elements in a regular local ring $R$, every embedded prime is contained in a minimal prime of height strictly less than $n.$ For every regular local ring, for every nonzero ideal $I$ generated by $n$ elements, there exists an element $f$ of $R$ and a nonzero ideal $J$ generated by $n$ elements such that $I$ equals $fJ$ and the ideal $J$ has no minimal primes of height $1$. In particular, for $n=2,$ there can be an embedded prime only if there is a minimal prime of height $1$. Since every associated prime of $J$ has height $2$, every associated prime of $I$ has height $1$ or $2$. Thus, $n=3$ is the minimal integer such that there exists an ideal $I$ generated by $n$ elements with an embedded prime of height strictly larger than $n$.
Original example. For instance, let $R$ be $k[x,y,u,v,w].$ Let $I$ be the ideal generated by $4$ elements, $$ I = \langle xu^3,yv^3,x^2u^2w-yuvw^2,y^2v^2w-xuvw^2 \rangle. $$
Denote by $S$ the $k$-subalgebra $k[x,y,u,v]$ of $R.$ The $S$-submodule of $R/I$ generated by $\overline{1}$ and $\overline{w}$ is $$ \left(S/\langle xu^3,yv^3\rangle\cdot 1 \right) \oplus \left( S/\langle xu^3,yv^3,x^2u^2v^2,y^2u^2v^2\rangle\cdot \overline{w}\right). $$ Thus, the image of $xyu^2v^2w$ in $R/I$ is nonzero. Yet the annihilator equals all of $\langle x,y,u,v,w\rangle.$ Thus, the maximal ideal $\langle x,y,u,v,w\rangle$ is an embedded prime of $I.$ This maximal ideal has height $5,$ which is strictly greater than $4.$
For a regular ring $R$, for an ideal $I$ generated by $n$ elements, it is true that every embedded prime contains a minimal prime of height strictly less than $n$. For instance, for the ideal above, the minimal primes are $$\langle x,y\rangle,\ \langle x,v \rangle,\ \langle y,u \rangle,\ \langle u,v \rangle,$$ and these each have height $2$, which is strictly less than $4$.
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
After Jason Starr's beautiful and explicit example, another answer is not necessary, but I jot this down to hopefully demystify (at least for myself) this calculation.
Let $E$ be a vector bundle on $\mathbb{P}^n$ of rank $<n$ and $H^1_*(E)\neq 0$. (For example, you can find such when $n=3,4$ and for many other $n$) Then, after twisting suffficiently, you can find an exact sequence $0\to E\to F\to \mathcal{I}\to 0$, where $F$ is a direct sum of line bundles with rank of $F=m\leq n$ equal to rank of $E+1$ and $\mathcal{I}$ defines a codimension 2 subvariety. Then, we have $J$, the image of $H^0_*(F)$ in $I=H^0_*(\mathcal{I})$ and $J\neq I$ by our assumption on $H^1_*(E)$. Notice that $J$ is generated by $m$ equations and letting $R=H^0_*(\mathbb{P}^n)$, we have $0\neq I/J\subset R/J$. Since $I/J$ is of finite length, we see that the maximal ideal is associated to $R/J$ and $m<\dim R=n+1$. This gives you such an example as you desire.
