Let $(R,m)$ be a Noetherian local ring, $M$ and $N$ finite $R$-modules, $p$ a prime ideal, and $I$ an ideal such that $IM\neq M$.

Definition: The common length of the maximal $M$-sequences in $I$ is called the grade of $I$ on $M$; denoted by $\operatorname{grade}(I,M)$.

$\operatorname{grade}(m,M)$ is denoted by $\operatorname{depth}M$. So by $\operatorname{depth}M_p$, we mean $\operatorname{grade}(pR_p,M_p)$.

Assuming $\operatorname{depth}M\ge \operatorname{depth}N$, what can one say about $\operatorname{depth}M_p$ and $\operatorname{depth}N_p$? Is there any inequality between them?

What if we impose additional assumptions on $M$ and $N$? For example, if $M=R/I$ and $N=R/J$?

Thank you.


Let $R$ be the localization of $k[x, y, z]$ at $(x, y, z)$ where $k$ is a field. Let $M = R/(x)$ and $N = R/(y)$. Then the depth of $M$ and $N$ is $2$.

  1. If $\mathfrak p = (x, y)$, then the depth of $M_{\mathfrak p}$ is 1 and the depth of $N_{\mathfrak p}$ is $1$.

  2. If $\mathfrak p = (x, z)$, then the depth of $M_{\mathfrak p}$ is $1$ and the depth of $N_{\mathfrak p}$ is $\infty$.

  3. If $\mathfrak p = (y, z)$, then the depth of $M_{\mathfrak p}$ is $\infty$ and the depth of $N_{\mathfrak p}$ is $1$.

  4. Take $M = R \oplus R/(x)$ and $N = R \oplus R/(y)$ and you can replace $\infty$ in the above by $2$.

Actually, you can get any integers subject to some constraints. Here is what I mean. We know that (a) $\dim(R) \geq \dim(M) \geq \text{depth}(M)$, (b) $\dim(R) \geq \dim(R_\mathfrak p) \geq \dim(M_\mathfrak p) \geq \text{depth}(M_\mathfrak p)$, and (c) $\dim(M) \geq \dim(M_\mathfrak p)$. I think you can make examples of $R, M, \mathfrak p$ such that the integers $\dim(R)$, $\dim(M)$, $\text{depth}(M)$, $\dim(R_\mathfrak p)$, $\dim(M_\mathfrak p)$, $\text{depth}(M_\mathfrak p)$ are an arbitrary $6$-tuple satisfying (a), (b), (c).

For example, take $6, 6, 1, 5, 4, 3$. Then you can take $R = k[x_1,\ldots, x_5, y, z]/(yz)$ localized at $(x_1, \ldots, x_5, y, z)$, take $M = R/(y) \oplus R/(z) \oplus R/(z, x_2) \oplus R/(y, z, x_2, \ldots, x_5)$, and take $\mathfrak p = (z, x_2, \ldots, x_5)$. Observe that in this case the depth of $M_\mathfrak p$ is bigger than the depth of $M$.

You can do a similar game with a pair of modules and there will be no relationship whatsoever between the depth of the modules and the depth of their localizations at $\mathfrak p$.

If you want the modules to be cyclic I am pretty sure you can do this as well. In any case, if you are only interested in depth and not dimension, then you can convert any example as above into an example with cyclic modules as follows. Step 1: we can always assume our ring $R$ is regular by writing the example ring as a quotient of a regular ring. In the example above we can write $k[x_1, \ldots, x_5, y, z]/(yz)$ as a quotient of $k[x_1, \ldots, x_5, y, z]$. Step 2. Say $M$ is our module and is generated by $n$ elements $e_1, \ldots, e_n$ subject to some relations $\sum r_{ji} e_i = 0$ for $j = 1, \ldots, m$. Then we introduce a new ring $P = R[t_1, \ldots, t_n]_{(\mathfrak m, t_1, \ldots, t_n)}$ and we consider the cyclic $P$-module $$ M_{new} = P/(t_it_{i'}, 1\leq i, i' \leq n; \sum r_{ji}t_i, 1 \leq j \leq m) $$ Then there is a short exact sequence $0 \to M \to M_{new} \to R \to 0$. A computation shows that $\text{depth}_P(M_{new}) = \text{depth}_R(M)$ provided $M$ is not zero (this is where we need that $R$ is regular so that it has depth not less than $M$). Similarly for the depth at $\mathfrak p$ and the corresponding prime of $P$.


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.