
## 10.61 Support and dimension of modules

Lemma 10.61.1. 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 $R$.

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 $aR/(J \cap aR)$ is a quotient of $R/(J + bR)$ and the second quotient equals $R/(aR + J)$. Hence by maximality of $J$, each of these has a filtration as above and hence so does $R/J$. Contradiction. $\square$

Lemma 10.61.2. Let $R$, $M$, $M_ i$, $\mathfrak p_ i$ as in Lemma 10.61.1. Then $\text{Supp}(M) = \bigcup V(\mathfrak p_ i)$ and in particular $\mathfrak p_ i \in \text{Supp}(M)$.

Lemma 10.61.3. 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 10.61.1 are the maximal ideal. This is clear by Lemma 10.61.2.

Suppose that $M$ has finite length over $R$. Then $\mathfrak m^ n M = 0$ by Lemma 10.51.4. 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 10.61.4. 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. It is clear that $I^ nM = 0$ for some $n \geq 0$ implies $\text{Supp}(M) \subset V(I)$. Suppose that $\text{Supp}(M) \subset V(I)$. Choose a filtration $0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M$ as in Lemma 10.61.1. Each of the primes $\mathfrak p_ i$ is contained in $V(I)$ by Lemma 10.61.2. Hence $I \subset \mathfrak p_ i$ and $I$ annihilates $M_ i/M_{i - 1}$. Hence $I^ n$ annihilates $M$. $\square$

Lemma 10.61.5. Let $R$, $M$, $M_ i$, $\mathfrak p_ i$ as in Lemma 10.61.1. 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 10.61.2. 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 10.61.3 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 10.61.6. Let $R$ be a Noetherian local ring. Let $M$ be a finite $R$-module. Then $d(M) = \dim (\text{Supp}(M))$.

Proof. Let $M_ i, \mathfrak p_ i$ be as in Lemma 10.61.1. By Lemma 10.58.10 we obtain the equality $d(M) = \max \{ d(R/\mathfrak p_ i) \}$. By Proposition 10.59.8 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 10.61.5) we win. $\square$

Lemma 10.61.7. 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 10.61.6 and 10.58.10. 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 10.39.5) and that $\text{Supp}(M) = \text{Supp}(M') \cup \text{Supp}(M'')$ (Lemma 10.39.9). $\square$

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 $\mathrm{supp}(M)\subseteq\{\mathfrak{m}\}$.

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).