Proof verification: $A$ is absolutely flat if and only if $A_\mathfrak{m}$ is a field

Question

This is a solution to Exercise 3.10 in Atiyah-MacDonald,

(i) If $A$ is absolutely flat and $S$ is any multiplicatively closed subset of $A$, then $S^{-1}A$ is absolutely flat.

(ii) $A$ is absolutely flat$\iff$$A_\mathfrak{m}$ is a field for each maximal ideal $\mathfrak{m}$.

(i) Suppose first that $A$ is absolutely flat. Note first that, for an $S^{-1}A$-module $M$, we have $$M \cong S^{-1}A \otimes_{S^{-1}A} M \cong (S^{-1}A \otimes_A S^{-1}A) \otimes_{S^{-1}A} M \cong S^{-1}A \otimes_A M \cong S^{-1}M.$$ In particular, if $$0 \longrightarrow M' \longrightarrow M \longrightarrow M'' \longrightarrow 0$$ is any short exact sequence of $S^{-1}A$-modules, then the exactness of the tensored sequence $$0 \longrightarrow M' \otimes_{S^{-1}A} N \longrightarrow M \otimes_{S^{-1}A} N \longrightarrow M'' \otimes_{S^{-1}A}N \longrightarrow 0,$$ where $N$ is any $S^{-1}A$-module, follows from the exactness of $$0 \longrightarrow S^{-1}M' \otimes_{S^{-1}A} S^{-1}N \longrightarrow S^{-1}M \otimes_{S^{-1}A} S^{-1}N \longrightarrow S^{-1}M'' \otimes_{S^{-1}A}S^{-1}N \longrightarrow 0,$$ which is equivalent to the exactness of $$0 \longrightarrow S^{-1 }(M' \otimes_A N) \longrightarrow S^{-1}(M \otimes_A N) \longrightarrow S^{-1}(M'' \otimes_A N) \longrightarrow 0. $$ But this is obviously true, since $S^{-1}$ is an exact operation and $A$ is absolutely flat. Thus, $N$ is a flat $S^{-1}A$-module and hence $S^{-1}A$ is absolutely flat.

(ii) Suppose first that $A$ is absolutely flat, and let $\mathfrak{m}$ be a maximal ideal of $A$. Let $\mathfrak{p}$ be the maximal ideal of the local ring $A_\mathfrak{m}$, and denote $k = A_\mathfrak{m}/\mathfrak{p}$ as the residue field. Since $A_\mathfrak{m}$ is absolutely flat by (i), $k$ is flat as an $A_\mathfrak{m}$-module. In particular, if we tensor the exact sequence $$0 \longrightarrow (x) \longrightarrow A_\mathfrak{m} \longrightarrow A_\mathfrak{m}/(x) \longrightarrow 0$$ by $k$ (as $A_\mathfrak{m}$-modules), where $x \in \mathfrak{p}$, we obtain the short exact sequence $$0 \longrightarrow (x)/\mathfrak{p}(x) \longrightarrow k \longrightarrow k \longrightarrow 0.$$ Each element in the sequence below as a $k$-vector space; in particular, $k$ is finite-dimensional as a $k$-vector space, so the surjectivity of the final map $k \to k$ also implies injectivity and the final map is thus an isomorphism. This implies $(x)/\mathfrak{p}(x) = 0$, so $(x) = \mathfrak{p}(x)$. By Nakayama's Lemma, $(x) = 0$. Since $x \in \mathfrak{p}$ was arbitrary, $\mathfrak{p} = 0$ and thus $A_\mathfrak{m}$ is a field.

Conversely, suppose $A_\mathfrak{m}$ is a field for each maximal ideal $\mathfrak{m}$, and let $M$ be any $A$-module. Then $M_\mathfrak{m}$ is an $A_\mathfrak{m}$-vector space for every maximal ideal $\mathfrak{m}$, thus a free $A_\mathfrak{m}$-module, and hence a flat $A_\mathfrak{m}$-module. Thus, $M_\mathfrak{m}$ is a flat $A_\mathfrak{m}$-module for all maximal ideals $\mathfrak{m}$, which implies the flatness of $M$. Thus, $A$ is absolutely flat.

Is this proof correct? I ask because many parts of the proof involve computing tensor products and using the alternate algebraic structures of a particular mathematical object, which I'm not confident about and always queasy about exploiting.

Thank you for your help and advice in advance.