The Stacks project

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. $\square$

Proof. This follows from Lemmas 10.40.5 and 10.40.9. $\square$

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 $M_{\mathfrak p} = 0$. $\square$

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 $V(I) \supset \text{Supp}(M)$. $\square$

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 $1$. $\square$

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. $\square$

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). $\square$