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 10.51.13. Let $A \to B$ be a flat local homomorphism of local rings. Then for any $A$-module $M$ we have

\[ \text{length}_ A(M) \text{length}_ B(B/\mathfrak m_ AB) = \text{length}_ B(M \otimes _ A B). \]

In particular, if $\text{length}_ B(B/\mathfrak m_ AB) < \infty $ then $M$ has finite length if and only if $M \otimes _ A B$ has finite length.

Proof. The ring map $A \to B$ is faithfully flat by Lemma 10.38.17. 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. $\square$


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