Proposition 1

— Let E be a module over a commutative ring A; suppose E is a direct sum of n cyclic modules $\Lambda/\mathfrak{a}_k$ ($1 \leq k \leq n$), where the $\mathfrak{a}_k$ are ideals of A;

then, for each integer $p > 0$, the $A$-module $\bigwedge^p E$ is isomorphic to the direct sum of the modules $A / a_H$, where for each $p$-element subset $H = \{ k_1, \ldots, k_p \}$ of $[1, n]$, the ideal $a_H$ is $\sum_{j=1}^p a_{k_j}$.

Let $x_k$ be the canonical generator of $A / a_k$, that is the image of the unit element of $A$, so that $E$ is the direct sum of the $Ax_i$ ($1 \leq i \leq n$). Then we know (III, p. 515, Prop. 10) that the exterior algebra $\bigwedge^n E$ is isomorphic as an $A$-module to the tensor product $\bigotimes_{i=1}^n (\bigwedge (Ax_i))$. Now $\bigwedge (Ax_i)$ is just the direct sum $A \oplus Ax_i$, since the exterior product of any two elements of $Ax_i$ is zero, and so $\bigwedge^p E$ is the direct sum of the modules $M_H = (Ax_{k_1}) \otimes \ldots \otimes (Ax_{k_p})$ as $H = \{ k_1, \ldots, k_p \}$ runs over the set of $p$-element subsets of $[1, n]$ (with $k_1 < \ldots < k_p$) ; now $M_H$ is known to be isomorphic to $A / a_{11}$ (II, p. 257, Cor. 4), which completes the proof.

We now see that, in the notation of Prop. 1, if the ideals $a_k$ form an increasing sequence, they are completely determined by the module $E$. More precisely :