The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.118.9. Let $R$ be a domain with fraction field $K$. Let $M$ be an $R$-submodule of $K^{\oplus r}$. Assume $R$ is local Noetherian of dimension $1$. For any nonzero $x \in R$ we have $\text{length}_ R(R/xR) < \infty $ and

\[ \text{length}_ R(M/xM) \leq r \cdot \text{length}_ R(R/xR). \]

Proof. If $x$ is a unit then the result is true. Hence we may assume $x \in \mathfrak m$ the maximal ideal of $R$. Since $x$ is not zero and $R$ is a domain we have $\dim (R/xR) = 0$, and hence $R/xR$ has finite length. Consider $M \subset K^{\oplus r}$ as in the lemma. We may assume that the elements of $M$ generate $K^{\oplus r}$ as a $K$-vector space after replacing $K^{\oplus r}$ by a smaller subspace if necessary.

Suppose first that $M$ is a finite $R$-module. In that case we can clear denominators and assume $M \subset R^{\oplus r}$. Since $M$ generates $K^{\oplus r}$ as a vectors space we see that $R^{\oplus r}/M$ has finite length. In particular there exists an integer $c \geq 0$ such that $x^ cR^{\oplus r} \subset M$. Note that $M \supset xM \supset x^2M \supset \ldots $ is a sequence of modules with successive quotients each isomorphic to $M/xM$. Hence we see that

\[ n \text{length}_ R(M/xM) = \text{length}_ R(M/x^ nM). \]

The same argument for $M = R^{\oplus r}$ shows that

\[ n \text{length}_ R(R^{\oplus r}/xR^{\oplus r}) = \text{length}_ R(R^{\oplus r}/x^ nR^{\oplus r}). \]

By our choice of $c$ above we see that $x^ nM$ is sandwiched between $x^ n R^{\oplus r}$ and $x^{n + c}R^{\oplus r}$. This easily gives that

\[ r(n + c) \text{length}_ R(R/xR) \geq n \text{length}_ R(M/xM) \geq r (n - c) \text{length}_ R(R/xR) \]

Hence in the finite case we actually get the result of the lemma with equality.

Suppose now that $M$ is not finite. Suppose that the length of $M/xM$ is $\geq k$ for some natural number $k$. Then we can find

\[ 0 \subset N_0 \subset N_1 \subset N_2 \subset \ldots N_ k \subset M/xM \]

with $N_ i \not= N_{i + 1}$ for $i = 0, \ldots k - 1$. Choose an element $m_ i \in M$ whose congruence class mod $xM$ falls into $N_ i$ but not into $N_{i - 1}$ for $i = 1, \ldots , k$. Consider the finite $R$-module $M' = Rm_1 + \ldots + Rm_ k \subset M$. Let $N'_ i \subset M'/xM'$ be the inverse image of $N_ i$. It is clear that $N'_ i \not=N'_{i + 1}$ by our choice of $m_ i$. Hence we see that $\text{length}_ R(M'/xM') \geq k$. By the finite case we conclude $k \leq r\text{length}_ R(R/xR)$ as desired. $\square$


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