The Stacks project

Proof. Omitted. $\square$

Proof. Given filtrations of $M'$ and $M''$ of lengths $n', n''$ it is easy to make a corresponding filtration of $M$ of length $n' + n''$. Thus we see that $\text{length}_ R M \geq \text{length}_ R M' + \text{length}_ R M''$. Conversely, given a filtration $M_0 \subset M_1 \subset \ldots \subset M_ n$ of $M$ consider the induced filtrations $M_ i' = M_ i \cap M'$ and $M_ i'' = \mathop{\mathrm{Im}}(M_ i \to M'')$. Let $n'$ (resp. $n''$) be the number of steps in the filtration $\{ M'_ i\} $ (resp. $\{ M''_ i\} $). If $M_ i' = M_{i + 1}'$ and $M_ i'' = M_{i + 1}''$ then $M_ i = M_{i + 1}$. Hence we conclude that $n' + n'' \geq n$. Combined with the earlier result we win. $\square$

Proof. Assume $\mathfrak m^ n M \not= 0$ for all $n\geq 0$. Choose $x \in M$ and $f_1, \ldots , f_ n \in \mathfrak m$ such that $f_1f_2 \ldots f_ n x \not= 0$. The first $n$ steps in the filtration

are distinct. For example, if $R f_1 x = R f_1 f_2 x$ , then $f_1 x = g f_1 f_2 x$ for some $g$, hence $(1 - gf_2) f_1 x = 0$ hence $f_1 x = 0$ as $1 - gf_2$ is a unit which is a contradiction with the choice of $x$ and $f_1, \ldots , f_ n$. Hence the length is infinite. $\square$

Proof. A filtration of $M$ by $S$-submodules gives rise a filtration of $M$ by $R$-submodules. This proves the inequality. And if $R \to S$ is surjective, then any $R$-submodule of $M$ is automatically an $S$-submodule. Hence equality in this case. $\square$

Proof. The first part is a special case of Lemma 10.52.5. Thus the length is finite if and only if $M$ has a finite basis as a $R/\mathfrak m$-vector space if and only if $M$ has a finite set of generators as an $R$-module. $\square$

Proof. Any submodule $N' \subset S^{-1}M$ is of the form $S^{-1}N$ for some $R$-submodule $N \subset M$, by Lemma 10.9.15. The lemma follows. $\square$

Proof. Consider the filtration $0 = \mathfrak m^ n M \subset \mathfrak m^{n-1} M \subset \ldots \subset \mathfrak m M \subset M$. All of the subquotients are finitely generated $R$-modules to which Lemma 10.52.6 applies. We conclude by additivity, see Lemma 10.52.3. $\square$

Proof. Let $\mathfrak m$ be a maximal ideal of $R$. By Lemma 10.52.6 the module $R/\mathfrak m$ has length $1$. The equivalence of the first two assertions is tautological. Suppose that $M$ is simple. Choose $x \in M$, $x \not= 0$. As $M$ is simple we have $M = R \cdot x$. Let $I \subset R$ be the annihilator of $x$, i.e., $I = \{ f \in R \mid fx = 0\} $. The map $R/I \to M$, $f \bmod I \mapsto fx$ is an isomorphism, hence $R/I$ is a simple $R$-module. Since $R/I \not= 0$ we see $I \not= R$. Let $I \subset \mathfrak m$ be a maximal ideal containing $I$. If $I \not= \mathfrak m$, then $\mathfrak m /I \subset R/I$ is a nontrivial submodule contradicting the simplicity of $R/I$. Hence we see $I = \mathfrak m$ as desired. $\square$

Proof. If $M_ i/M_{i-1}$ is not simple then we can refine the filtration and the filtration is not maximal. Thus we see that $M_ i/M_{i-1}$ is simple. By Lemma 10.52.10 the modules $M_ i/M_{i-1}$ have length $1$ and are of the form $R/\mathfrak m_ i$ for some maximal ideals $\mathfrak m_ i$. By additivity of length, Lemma 10.52.3, we see $n = \text{length}_ R(M)$. Since localization is exact, we see that

is a filtration of $M_{\mathfrak m}$ with successive quotients $(M_ i/M_{i-1})_{\mathfrak m}$. Thus the last statement follows directly from the fact that given maximal ideals $\mathfrak m$, $\mathfrak m'$ of $R$ we have

This we leave to the reader. $\square$

Proof. Choose a maximal chain

by $B$-submodules as in Lemma 10.52.11. Then each quotient $M_ j/M_{j - 1}$ is isomorphic to $\kappa (\mathfrak m_{i(j)})$ for some $i(j) \in \{ 1, \ldots , n\} $. Moreover $\text{length}_ A(\kappa (\mathfrak m_ i)) = [\kappa (\mathfrak m_ i) : \kappa (\mathfrak m)]$ by Lemma 10.52.6. The lemma follows by additivity of lengths (Lemma 10.52.3). $\square$

Proof. The ring map $A \to B$ is faithfully flat by Lemma 10.39.17. Hence if $0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M$ is a chain of length $n$ in $M$, then the corresponding chain $0 = M_0 \otimes _ A B \subset M_1 \otimes _ A B \subset \ldots \subset M_ n \otimes _ A B = M \otimes _ A B$ has length $n$ also. This proves $\text{length}_ A(M) = \infty \Rightarrow \text{length}_ B(M \otimes _ A B) = \infty $. Next, assume $\text{length}_ A(M) < \infty $. In this case we see that $M$ has a filtration of length $\ell = \text{length}_ A(M)$ whose quotients are $A/\mathfrak m_ A$. Arguing as above we see that $M \otimes _ A B$ has a filtration of length $\ell $ whose quotients are isomorphic to $B \otimes _ A A/\mathfrak m_ A = B/\mathfrak m_ AB$. Thus the lemma follows. $\square$

Proof. Follows from Lemma 10.52.13 applied to the ring map $B \to C$ and the $B$-module $M = B/\mathfrak m_ A B$ $\square$