[Theorem 3, Warfield-Decomposition]

Lemma 15.124.4. Let $R$ be a ring such that every local ring of $R$ at a maximal ideal satisfies the equivalent conditions of Lemma 15.124.2. Then every finitely presented $R$-module is a summand of a finite direct sum of modules of the form $R/fR$ for $f$ in $R$ varying.

Proof. Let $M$ be a finitely presented $R$-module. We first show that $M$ is a summand of a direct sum of modules of the form $R/fR$ and at the end we argue the direct sum can be taken to be finite. Let

$0 \to A \to B \to C \to 0$

be a short exact sequence of $R$-modules such that $fA = A \cap fB$ for all $f \in R$. By Lemma 15.124.1 we have to show that $\mathop{\mathrm{Hom}}\nolimits _ R(M, B) \to \mathop{\mathrm{Hom}}\nolimits _ R(M, C)$ is surjective. It suffices to prove this after localization at maximal ideals $\mathfrak m$, see Algebra, Lemma 10.23.1. Note that the localized sequences $0 \to A_\mathfrak m \to B_\mathfrak m \to C_\mathfrak m \to 0$ satisfy the condition that $fA_\mathfrak m = A_\mathfrak m \cap fB_\mathfrak m$ for all $f \in R_\mathfrak m$ (because we can write $f = uf'$ with $u \in R_\mathfrak m$ a unit and $f' \in R$ and because localization is exact). Since $M$ is finitely presented, we see that

$\mathop{\mathrm{Hom}}\nolimits _ R(M, B)_\mathfrak m = \mathop{\mathrm{Hom}}\nolimits _{R_\mathfrak m}(M_\mathfrak m, B_\mathfrak m) \quad \text{and}\quad \mathop{\mathrm{Hom}}\nolimits _ R(M, C)_\mathfrak m = \mathop{\mathrm{Hom}}\nolimits _{R_\mathfrak m}(M_\mathfrak m, C_\mathfrak m)$

by Algebra, Lemma 10.10.2. The module $M_\mathfrak m$ is a finitely presented $R_\mathfrak m$-module. By Lemma 15.124.3 we see that $M_\mathfrak m$ is a direct sum of modules of the form $R_\mathfrak m/fR_\mathfrak m$. Thus we conclude by Lemma 15.124.1 that the map on localizations is surjective.

At this point we know that $M$ is a summand of $\bigoplus _{i \in I} R/f_ i R$. Consider the map $M \to \bigoplus _{i \in I} R/f_ i R$. Since $M$ is a finite $R$-module, the image is contained in $\bigoplus _{i \in I'} R/f_ i R$ for some finite subset $I' \subset I$. This finishes the proof. $\square$

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