## 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). $\square$

Comment #680 by Keenan Kidwell on

In 00L5, maybe it should be assumed that $M$ is non-zero, or else the statement should be changed to $\mathrm{supp}(M)\subseteq\{\mathfrak{m}\}$.

Comment #5975 by Juan on

What is the definition of $d(M)$?

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