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Unverified Commit 108dcfc2 authored by rrueger's avatar rrueger
Browse files

Algebra I: Modules. Update grammar and whitespace typo

parent 47d3df82
......@@ -91,7 +91,7 @@ Evidently, this homomorphism is surjective and we have $M = Rx \cong R/\ker
\begin{definition}
Let $M = Rx$ be a cyclic $R$-module. Let $\mu_x \colon R \to Rx$ be defined by
$r \mapsto rx$. Then we define
$r \mapsto rx$. Then we define
\begin{align*}
\ann x = \ker \mu_x = \{d \in R \mid dx = 0\}.
\end{align*}
......@@ -121,9 +121,9 @@ Evidently, this homomorphism is surjective and we have $M = Rx \cong R/\ker
\begin{definition}
Let $M_1, \dotsc, M_n$ be modules over the same ring $R$. Let $M$ be the
product set $M_ 1 \times \dotsm \times M_n$. As in the special case of the
free module we define addition, the zero element and multiplication by
elements in $R$ by
product set $M_ 1 \times \dotsm \times M_n$. In the special case of the free
module we define addition, the zero element and multiplication by elements in
$R$ by
%
\begin{align*}
(x_1, \dotsc, x_n)+(y_1, \dotsc, y_n) & = (x_1 + y_1, \dotsc, x_n + y_n) \\
......
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