Lemma 15.124.3. Let R be a ring satisfying the equivalent conditions of Lemma 15.124.2. Then every finitely presented R-module is isomorphic to a finite direct sum of modules of the form R/fR.
[Theorem 1, Warfield-Decomposition]
Proof. Let M be a finitely presented R-module. We will use all the equivalent properties of R from Lemma 15.124.2 without further mention. Denote \mathfrak m \subset R the maximal ideal and \kappa = R/\mathfrak m the residue field. Let I \subset R be the annihilator of M. Choose a basis y_1, \ldots , y_ n of the finite dimensional \kappa -vector space M/\mathfrak m M. We will argue by induction on n.
By Nakayama's lemma any collection of elements x_1, \ldots , x_ n \in M lifting the elements y_1, \ldots , y_ n in M/\mathfrak m M generate M, see Algebra, Lemma 10.20.1. This immediately proves the base case n = 0 of the induction.
We claim there exists an index i such that for any choice of x_ i \in M mapping to y_ i the annihilator of x_ i is I. Namely, if not, then we can choose x_1, \ldots , x_ n such that I_ i = \text{Ann}(x_ i) \not= I for all i. But as I \subset I_ i for all i, ideals being totally ordered implies I_ i is strictly bigger than I for i = 1, \ldots , n, and by total ordering once more we would see that \text{Ann}(M) = I_1 \cap \ldots \cap I_ n is bigger than I which is a contradiction. After renumbering we may assume that y_1 has the property: for any x_1 \in M lifting y_1 the annihilator of x_1 is I.
We set A = Rx_1 \subset M. Consider the exact sequence 0 \to A \to M \to M/A \to 0. Since A is finite, we see that M/A is a finitely presented R-module (Algebra, Lemma 10.5.3) with fewer generators. Hence M/A \cong \bigoplus _{j = 1, \ldots , m} R/f_ jR by induction. On the other hand, we claim that A \to M satisfies the property: if f \in R, then fA = A \cap fM. The inclusion fA \subset A \cap fM is trivial. Conversely, if x \in A \cap fM, then x = gx_1 = f y for some g \in R and y \in M. If f divides g, then x \in fA as desired. If not, then we can write f = hg for some h \in \mathfrak m. The element x'_1 = x_1 - hy has annihilator I by the previous paragraph. Thus g \in I and we see that x = 0 as desired. The claim and Lemma 15.124.1 imply the sequence 0 \to A \to M \to M/A \to 0 is split and we find M \cong A \oplus \bigoplus _{j = 1, \ldots , m} R/f_ jR. Then A = R/I is finitely presented (as a summand of M) and hence I is finitely generated, hence principal. This finishes the proof. \square
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