The Stacks project

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. $\square$