Ideal $I$ with $\operatorname{depth}(I)=d$ in a local CM ring of dimension $d$

Question

Let $(R,m)$ be a Noetherian Cohen-Macaulay local ring, having Krull dimension $d$ (by this, necessarily $d < \infty$).

Let $I$ be an ideal of $R$ with $\operatorname{depth}(I)=d$, namely, $I$ contains a regular sequence of length $d$ (it cannot contain a longer regular sequence).

Of course $I \subseteq m$.

Question: Is it true that $I=m$?

If not, are there special cases (perhaps in regular local rings?) where $m$ is the unique ideal having maximal depth?

Example: $R=k[x,y]_{(x,y)}$, $d=2$, $m=(x,y)R$, $I=(x(x-1),y)R$; here $I=m$, because $x-1$ is invertible in $R$.

I apologize if my question is trivial.

Thank you very much!

Every m-primary ideal has the same depth as m. In particular, all powers of m do this. — user26857, Apr 27 '21 at 17:28
@user26857, please, are there cases where only powers of $m$ have depth $d$ (and in this case, all other ideals have depth $< d$). — user237522, Apr 27 '21 at 17:35
I think you mean "are there cases where only the m-primary ideals have maximal depth?". No, there aren't: maximal depth is the same with maximal height, and in this case the radical of I has to be m. — user26857, Apr 27 '21 at 17:38

score 1 · Accepted Answer · answered Apr 27 '21 at 17:44

1

If $I$ is an ideal of maximal depth (grade), then it is also an ideal of maximal height (since $R$ is Cohen-Macaulay). In this case the only prime ideal containing $I$ is $m$, and therefore $\sqrt I=m$, that is, $I$ is $m$-primary.

answered Apr 27 '21 at 17:44

user26857

53,190

Ideal $I$ with $\operatorname{depth}(I)=d$ in a local CM ring of dimension $d$

1 Answers1