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.



Height of associated primes in regular rings
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 $4$.


