... ... @@ -155,23 +155,25 @@ Evidently, this homomorphism is surjective and we have M = Rx \cong R/\ker \imath \colon (x_1, \dotsc, x_n) \to \sum_{i = 1}^{n} x_i \end{align*} % is an isomorphism of\oplus M_i$with$M$. Conversely, in$\oplus M_i$let is an isomorphism of$\bigoplus _{i=1} ^n M_i$with$M$. Conversely, in$\oplus M_ilet % \begin{align*} M_i' = \{(0, \dotsc, 0, x_i, 0, \dotsc, 0) \mid x_i \in M_i\}. \end{align*} % ThenM_i' $is a submodule of$\oplus M_i$isomorphic to$M_i$, and the conditions (i) and (ii) hold for these submodules of$\oplus M_i$. Then$M_i' $is a submodule of$\bigoplus _{i=1} ^n M_i$isomorphic to$M_i$, and the conditions (i) and (ii) hold for these submodules of$\bigoplus _{i=1} ^n M_i$. \end{theorem} \section{Finitely Generated Modules over a PID - Preliminary Results } Consider first the following construction. Let$M$be a module over a pid$D$Consider first the following construction. Let$M$be a module over a PID$D$which is generated by a finite set of elements$x_1, \dotsc, x_n$, so$M = \sum_{i = 1}^{n} D x_i$. Now we introduce the free module$D^{(n)}$with base$(e_1, \dotsc, e_n)$and the surjective homomorphism$\eta: \sum_{i = 1}^n a_i e_i \in D^{(n)} \mapsto \sum_{i = 1}^{n} a_i x_i \in M$. Denote$\ker \eta$by$(e_1, \dotsc, e_n)$and the surjective homomorphism$\eta: D^{(n)} \to M; \sum_{i = 1}^n a_i e_i \mapsto \sum_{i = 1}^{n} a_i x_i$. Denote$\ker \eta$by$K$. Then we have$M \cong D^{(n)}/K$. The core idea is now to examine$K\$. \begin{theorem} ... ...