Exercise 3.10

Question

Let $A$ be a ring. 
\begin{enumerate}[label=\roman*)]
\item If $A$ is absolutely flat (i.e. every A-module is flat) and $S$ is any multiplicatively closed subset of $A$, then $S^{-1}A$ is absolutely flat.
\item $A$ is absolutely flat $\iff$ $A_{\mathfrak{m}}$ is a field for each maximal ideal $\mathfrak{m}$.
\end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $i$] We begin by citing a previous result: 
\begin{proposition} (AM Exercise 2.27) $A$ is absolutely flat $\iff$ Every principal ideal of $A$ is idempotent.
\end{proposition}

Assume $A$ is absolutely flat. Then from the proposition above, for every element $a \in A$ $(a)^2 = (a^2) = (a)$. That is, there exists an element $u \in A^{	imes}$ such that $a^2 = ua$. 
  
  Now let $\frac{a}{s} \in S^{-1}A$. Then there exists $\alpha, \beta \in A^{	imes}$ such that $a^2 = \alpha a$, $s^2 = \beta s$. Then,
  $$\Big{(}\frac{a}{s}\Big{)}^2 = \Big{(}\frac{a^2}{s^2}\Big{)} = \Big{(}\frac{\alpha \beta^{-1} a}{s}\Big{)} \subseteq \Big{(}\frac{a}{s}\Big{)}$$
  
  We note that $$\frac{\alpha^{-1}s^2}{s} \cdot \frac{\alpha \beta^{-1} a}{s} = \frac{a}{s}$$ Hence,$\Big{(}\frac{a}{s}\Big{)}^2 \supseteq \Big{(}\frac{a}{s}\Big{)}$. Thus, Every principal ideal in $S^{-1}A$ is an idempotent, and again from the proposition, $S^{-1}A$ is absolutely flat.

\end{proof}

\begin{proof}[Proof of $ii$]
\end{proof}

Exercise 3.10

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Exercise 3.10

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