Proposition 8

*Let $\mathbf{A}$ be a ring and $m$ an ideal of $\mathbf{A}$ such that the ordered pair $(\mathbf{A}, m)$ satisfies Hensel’s conditions. Let $\mathbf{B}$ be the quotient ring $\mathbf{A}/m$ and $\pi : \mathbf{A} \to \mathbf{B}$ the canonical homomorphism. If B is the direct composition of a finite family $(\mathfrak{b}_i)_{i \in I}$ of ideals, there exists a unique family $(\mathfrak{a}_i)_{i \in I}$ of ideals of A such that $\pi(\mathfrak{a}_i) = \mathfrak{b}_i$ for all $i \in I$ and A is the direct composition of the family $(\mathfrak{a}_i)$.

Let $I = \sum_i c_i$ where $c_i \in \mathfrak{b}_i$ for all $i$; the $c_i$ are idempotents of $B$ such that $c_i c_j = 0$ for $i \neq j$. By Lemma 2 there therefore exist idempotents $e_i$ of $A$ ($i \in I$) such that $\pi(e_i) = c_i$ for all $i$; as $e_i e_j$ is an idempotent such that

for $i \neq j$, $e_i e_j = 0$ for $i \neq j$ (Lemma 2); as $1 - \sum_i e_i$ is an idempotent such that similarly $1 = \sum_i e_i$. It follows that $A$ is the direct composition of the ideals $\mathfrak{a}_i = e_i A$ and that $\pi(\mathfrak{a}_i) = \pi(e_i) B = \mathfrak{b}_i$.

It remains to show the uniqueness of such a decomposition. Now, suppose that $A$ is the direct composition of another family $(\mathfrak{a}'_i)_{i \in I}$ of ideals such that $\pi(\mathfrak{a}'_i) = \mathfrak{b}_i$ for all $i$; then $1 = \sum_i e'_i$ where $e'_i \in \mathfrak{a}'_i$, whence, in $B$, $1 = \sum_i \pi(e'_i)$ where $\pi(e'_i) \in \mathfrak{b}_i$, which implies $\pi(e'_i) = c_i$; as $e'_i$ and $e_i$ are idempotents, necessarily $e'_i = e_i$ (Lemma 2), which completes the proof.