# The Stacks Project

## Tag 02M0

Lemma 10.51.12. Let $A$ be a local ring with maximal ideal $\mathfrak m$. Let $B$ be a semi-local ring with maximal ideals $\mathfrak m_i$, $i = 1, \ldots, n$. Suppose that $A \to B$ is a homomorphism such that each $\mathfrak m_i$ lies over $\mathfrak m$ and such that $$[\kappa(\mathfrak m_i) : \kappa(\mathfrak m)] < \infty.$$ Let $M$ be a $B$-module of finite length. Then $$\text{length}_A(M) = \sum\nolimits_{i = 1, \ldots, n} [\kappa(\mathfrak m_i) : \kappa(\mathfrak m)] \text{length}_{B_{\mathfrak m_i}}(M_{\mathfrak m_i}),$$ in particular $\text{length}_A(M) < \infty$.

Proof. Choose a maximal chain $$0 = M_0 \subset M_1 \subset M_2 \subset \ldots \subset M_n = M$$ by $B$-submodules as in Lemma 10.51.11. Then each quotient $M_i/M_{i - 1}$ is isomorphic to $\kappa(\mathfrak m_{j(i)})$ for some $j(i) \in \{1, \ldots, n\}$. Moreover $\text{length}_A(\kappa(\mathfrak m_i)) = [\kappa(\mathfrak m_i) : \kappa(\mathfrak m)]$ by Lemma 10.51.6. The lemma follows by additivity of lengths (Lemma 10.51.3). $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 11970–11988 (see updates for more information).

\begin{lemma}
\label{lemma-pushdown-module}
Let $A$ be a local ring with maximal ideal $\mathfrak m$.
Let $B$ be a semi-local ring with maximal ideals $\mathfrak m_i$,
$i = 1, \ldots, n$.
Suppose that $A \to B$ is a homomorphism such that each $\mathfrak m_i$
lies over $\mathfrak m$ and such that
$$[\kappa(\mathfrak m_i) : \kappa(\mathfrak m)] < \infty.$$
Let $M$ be a $B$-module of finite length.
Then
$$\text{length}_A(M) = \sum\nolimits_{i = 1, \ldots, n} [\kappa(\mathfrak m_i) : \kappa(\mathfrak m)] \text{length}_{B_{\mathfrak m_i}}(M_{\mathfrak m_i}),$$
in particular $\text{length}_A(M) < \infty$.
\end{lemma}

\begin{proof}
Choose a maximal chain
$$0 = M_0 \subset M_1 \subset M_2 \subset \ldots \subset M_n = M$$
by $B$-submodules as in Lemma \ref{lemma-simple-pieces}.
Then each quotient $M_i/M_{i - 1}$ is isomorphic to
$\kappa(\mathfrak m_{j(i)})$ for some $j(i) \in \{1, \ldots, n\}$.
Moreover
$\text{length}_A(\kappa(\mathfrak m_i)) = [\kappa(\mathfrak m_i) : \kappa(\mathfrak m)]$ by
Lemma \ref{lemma-dimension-is-length}. The lemma follows
\end{proof}

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