The Stacks project

Lemma 10.119.9. 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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00PE. Beware of the difference between the letter 'O' and the digit '0'.