Lemma 90.3.4. Let $A$ be a local $\Lambda $-algebra with residue field $k$. Let $M$ be an $A$-module. Then $[k : k'] \text{length}_ A(M) = \text{length}_\Lambda (M)$. In the classical case we have $\text{length}_ A(M) = \text{length}_\Lambda (M)$.

**Proof.**
If $M$ is a simple $A$-module then $M \cong k$ as an $A$-module, see Algebra, Lemma 10.52.10. In this case $\text{length}_ A(M) = 1$ and $\text{length}_\Lambda (M) = [k' : k]$, see Algebra, Lemma 10.52.6. If $\text{length}_ A(M)$ is finite, then the result follows on choosing a filtration of $M$ by $A$-submodules with simple quotients using additivity, see Algebra, Lemma 10.52.3. If $\text{length}_ A(M)$ is infinite, the result follows from the obvious inequality $\text{length}_ A(M) \leq \text{length}_\Lambda (M)$.
$\square$

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