The Stacks project

10.62 Support and dimension of modules

Some basic results on the support and dimension of modules.

Lemma 10.62.1. Let $R$ be a Noetherian ring, and let $M$ be a finite $R$-module. There exists a filtration by $R$-submodules

\[ 0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M \]

such that each quotient $M_ i/M_{i-1}$ is isomorphic to $R/\mathfrak p_ i$ for some prime ideal $\mathfrak p_ i$ of $R$.

First proof. By Lemma 10.5.4 it suffices to do the case $M = R/I$ for some ideal $I$. Consider the set $S$ of ideals $J$ such that the lemma does not hold for the module $R/J$, and order it by inclusion. To arrive at a contradiction, assume that $S$ is not empty. Because $R$ is Noetherian, $S$ has a maximal element $J$. By definition of $S$, the ideal $J$ cannot be prime. Pick $a, b\in R$ such that $ab \in J$, but neither $a \in J$ nor $b\in J$. Consider the filtration $0 \subset aR/(J \cap aR) \subset R/J$. Note that both the submodule $aR/(J \cap aR)$ and the quotient module $(R/J)/(aR/(J \cap aR))$ are cyclic modules; write them as $R/J'$ and $R/J''$ so we have a short exact sequence $0 \to R/J' \to R/J \to R/J'' \to 0$. The inclusion $J \subset J'$ is strict as $b \in J'$ and the inclusion $J \subset J''$ is strict as $a \in J''$. Hence by maximality of $J$, both $R/J'$ and $R/J''$ have a filtration as above and hence so does $R/J$. Contradiction. $\square$

Second proof. For an $R$-module $M$ we say $P(M)$ holds if there exists a filtration as in the statement of the lemma. Observe that $P$ is stable under extensions and holds for $0$. By Lemma 10.5.4 it suffices to prove $P(R/I)$ holds for every ideal $I$. If not then because $R$ is Noetherian, there is a maximal counter example $J$. By Example 10.28.7 and Proposition 10.28.8 the ideal $J$ is prime which is a contradiction. $\square$

Lemma 10.62.2. Let $R$, $M$, $M_ i$, $\mathfrak p_ i$ as in Lemma 10.62.1. Then $\text{Supp}(M) = \bigcup V(\mathfrak p_ i)$ and in particular $\mathfrak p_ i \in \text{Supp}(M)$.

Lemma 10.62.3. Suppose that $R$ is a Noetherian local ring with maximal ideal $\mathfrak m$. Let $M$ be a nonzero finite $R$-module. Then $\text{Supp}(M) = \{ \mathfrak m\} $ if and only if $M$ has finite length over $R$.

Proof. Assume that $\text{Supp}(M) = \{ \mathfrak m\} $. It suffices to show that all the primes $\mathfrak p_ i$ in the filtration of Lemma 10.62.1 are the maximal ideal. This is clear by Lemma 10.62.2.

Suppose that $M$ has finite length over $R$. Then $\mathfrak m^ n M = 0$ by Lemma 10.52.4. Since some element of $\mathfrak m$ maps to a unit in $R_{\mathfrak p}$ for any prime $\mathfrak p \not= \mathfrak m$ in $R$ we see $M_{\mathfrak p} = 0$. $\square$

Lemma 10.62.4. Let $R$ be a Noetherian ring. Let $I \subset R$ be an ideal. Let $M$ be a finite $R$-module. Then $I^ nM = 0$ for some $n \geq 0$ if and only if $\text{Supp}(M) \subset V(I)$.

Proof. Indeed, $I^ nM = 0$ is equivalent to $I^ n \subset \text{Ann}(M)$. Since $R$ is Noetherian, this is equivalent to $I \subset \sqrt{\text{Ann}(M)}$, see Lemma 10.32.5. This in turn is equivalent to $V(I) \supset V(\text{Ann}(M))$, see Lemma 10.17.2. By Lemma 10.40.5 this is equivalent to $V(I) \supset \text{Supp}(M)$. $\square$

Lemma 10.62.5. Let $R$, $M$, $M_ i$, $\mathfrak p_ i$ as in Lemma 10.62.1. The minimal elements of the set $\{ \mathfrak p_ i\} $ are the minimal elements of $\text{Supp}(M)$. The number of times a minimal prime $\mathfrak p$ occurs is

\[ \# \{ i \mid \mathfrak p_ i = \mathfrak p\} = \text{length}_{R_\mathfrak p} M_{\mathfrak p}. \]

Proof. The first statement follows because $\text{Supp}(M) = \bigcup V(\mathfrak p_ i)$, see Lemma 10.62.2. Let $\mathfrak p \in \text{Supp}(M)$ be minimal. The support of $M_{\mathfrak p}$ is the set consisting of the maximal ideal $\mathfrak p R_{\mathfrak p}$. Hence by Lemma 10.62.3 the length of $M_{\mathfrak p}$ is finite and $> 0$. Next we note that $M_{\mathfrak p}$ has a filtration with subquotients $ (R/\mathfrak p_ i)_{\mathfrak p} = R_{\mathfrak p}/{\mathfrak p_ i}R_{\mathfrak p} $. These are zero if $\mathfrak p_ i \not\subset \mathfrak p$ and equal to $\kappa (\mathfrak p)$ if $\mathfrak p_ i \subset \mathfrak p$ because by minimality of $\mathfrak p$ we have $\mathfrak p_ i = \mathfrak p$ in this case. The result follows since $\kappa (\mathfrak p)$ has length $1$. $\square$

Lemma 10.62.6. Let $R$ be a Noetherian local ring. Let $M$ be a finite $R$-module. Then $d(M) = \dim (\text{Supp}(M))$ where $d(M)$ is as in Definition 10.59.8.

Proof. Let $M_ i, \mathfrak p_ i$ be as in Lemma 10.62.1. By Lemma 10.59.10 we obtain the equality $d(M) = \max \{ d(R/\mathfrak p_ i) \} $. By Proposition 10.60.9 we have $d(R/\mathfrak p_ i) = \dim (R/\mathfrak p_ i)$. Trivially $\dim (R/\mathfrak p_ i) = \dim V(\mathfrak p_ i)$. Since all minimal primes of $\text{Supp}(M)$ occur among the $\mathfrak p_ i$ (Lemma 10.62.5) we win. $\square$

Lemma 10.62.7. Let $R$ be a Noetherian ring. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of finite $R$-modules. Then $\max \{ \dim (\text{Supp}(M')), \dim (\text{Supp}(M''))\} = \dim (\text{Supp}(M))$.

Proof. If $R$ is local, this follows immediately from Lemmas 10.62.6 and 10.59.10. A more elementary argument, which works also if $R$ is not local, is to use that $\text{Supp}(M')$, $\text{Supp}(M'')$, and $\text{Supp}(M)$ are closed (Lemma 10.40.5) and that $\text{Supp}(M) = \text{Supp}(M') \cup \text{Supp}(M'')$ (Lemma 10.40.9). $\square$


Comments (3)

Comment #680 by Keenan Kidwell on

In 00L5, maybe it should be assumed that is non-zero, or else the statement should be changed to .

Comment #5975 by Juan on

What is the definition of ?


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00KY. Beware of the difference between the letter 'O' and the digit '0'.