Lemma 15.121.3. Let $R$ be a ring satisfying the equivalent conditions of Lemma 15.121.2. 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 15.121.2 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 15.121.1 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (3)

Comment #2061 by Kestutis Cesnavicius on

Comment #2063 by Pieter Belmans on

Comment #2091 by Johan on