Proposition 7

— Let $A$ be a commutative ring, let $\mathfrak{p}$ be a prime ideal of $A$ (I, p. 117, Def. 3), and let $q$ be an ideal of $A$ contained in $\mathfrak{p}$. Suppose that for every $x \in \mathfrak{p}$ there exists an integer $n > 0$ such that $x^n \in q$. Then the $A$-module $A/q$ is indecomposable.

Let $b$ and $c$ be two ideals of $A$, such that $A = b + c$ and $b \cap c = q$. Then $bc \subset b \cap c = q \subset \mathfrak{p}$; if $x \notin \mathfrak{p}$ and $x \in c$, then $xb \subset p$, so $b \subset \mathfrak{p}$ (I, p. 316, Prop. 4); hence either $b \subset \mathfrak{p}$ or $c \subset p$. Suppose for example that $c \subset \mathfrak{p}$, so $b + p = A$; then there exist $x \in b$ and $y \in \mathfrak{p}$ such that $1 = x + y$; let $n \in \mathbf{N}$ be such that $y^n \in q$; then $1 = (x + y)^n$. so $1 \in xA + y^nA \subset b + q \subset b$, so $b = A$. Lemma 2 now shows that $A/q$ is indecomposable.

Now suppose that $A$ is a principal ideal domain; by VII, p. 2, Prop. 2, the prime ideals of $A$ are the ideals $(p)$, where $p$ is an irreducible element of $A$, and the ideal $0$; by the previous proposition, the modules $A$ and $A/(p^n)$, for $p$ irreducible and $n > 0$, are indecomposable. Since every cyclic module is a direct sum of modules of this type (VII, p. 3, Prop. 4) and since every finitely generated $A$-module is a direct sum of cyclic modules (VII, p. 19, Th. 2), we deduce: