Exercise 3.1

Question

Let $S$ be a multiplicatively closed subset of a ring $A$, and let $M$ be a finitely generated $A$-module. Prove that $S^{-1}M = 0$ if and only if there exists $s \in S$ such that $sM = 0$.

Amit Awasthi · Accepted Answer

\begin{proof}
  Assume that $S^{-1}M = 0$, and let $\{x_1, \ldots, x_n\}$ be a set of
  generators of $M$, as an $A$-module. Then as $S^{-1}M = 0$, for each $i \in
  \{1, \ldots, n\}$, there exists $s_i \in S$ such that $s_i \cdot x_i=0$. Let $s= \prod_{i} s_i$. Then, for any $\sum_{i}a_ix_i \in M$,
  \begin{equation*}
    s \cdot \sum_{i}a_ix_i = \sum_{i}a_isx_i = 0,
  \end{equation*}
  hence $sM = 0$.
  \par Conversely, if there exists $s \in S$ such that $sM = 0$, then
  $S^{-1}M = 0$ since for each generator $x_i$, $sx_i = 0$.
\end{proof}

Exercise 3.1

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Exercise 3.1

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