3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ In the ring $R=k[x,y,u,v,w]$, for the ideal $I=\langle xu^3,yv^3,x^2u^2w-yuvw^2,y^2v^2w-xuvw^2\rangle$, for the congruence class of the element $xyu^2v^2w$ in $R/I$, what is the annihilator of this congruence class? $\endgroup$ – Jason Starr Aug 7 '17 at 19:32
3
$\begingroup$

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 $4$.

$\endgroup$
  • $\begingroup$ Nice! Do you have any example for monomial ideal? Could you explain more the sentence: 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 $n$. $\endgroup$ – Pham Hung Quy Aug 8 '17 at 3:02
  • 1
    $\begingroup$ "for an ideal $I$ generated by $n$ elements ..." In fact, for every local Cohen-Macaulay ring $(R,\mathfrak{m})$, for every sequence $(a_1,\dots,a_n)$ of elements in $\mathfrak{m}$, the sequence is a regular sequence if and only if every minimal prime over $I=\langle a_1,\dots,a_n \rangle$ has height $\geq n$, and then $R/I$ is also Cohen-Macaulay, cf. Theorem 17.4, p. 135, Matsumura, Commutative ring theory. Then, by the unmixedness theorem, every associated prime of $I$ is minimal of height $n$. Thus, if $I$ has embedded primes, then some minimal prime has height $<n$. $\endgroup$ – Jason Starr Aug 8 '17 at 11:51
  • $\begingroup$ @JasonStarr Thank you for a helpful answer. Could you explain briefly why every associated prime $\mathfrak{p}$ of $J$ being of height $2$ implies that every associated prime of $I$ has height $1$ or $2$? $\endgroup$ – O-Ren Ishii Mar 30 '18 at 19:21
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$

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.