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Unverified Commit 47aea8cb authored by rrueger's avatar rrueger
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Algebra I: Modules. Typset direct sum of modules correctly

parent af069d5c
......@@ -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*}
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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
$\oplus M_i$ let
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\begin{align*}
M_i' = \{(0, \dotsc, 0, x_i, 0, \dotsc, 0) \mid x_i \in M_i\}.
\end{align*}
%
Then $M_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}
......
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