Better proof of lemma what-does-it mean

Thanks to Kentaro Inoue
https://stacks.math.columbia.edu/tag/051C#comment-8730

algebra.tex

@@ -23477,43 +23477,28 @@ \section{Criteria for flatness}
 $\text{Tor}_1^R(N, M) = 0$ for any $R$-module $N$ annihilated by $I$.
 
 \medskip\noindent
-In particular, if we apply this to the
-module $I/I^2$, then we conclude that the sequence
+Let us prove (2) by induction on $m$. The case $m = 1$ was
+done in the previous paragraph. For $N$ annihilated by $I^m$
+for $m > 1$ we may choose an exact sequence
+$0 \to N' \to N \to N'' \to 0$ with $N'$ and $N''$ annihilated
+by $I^{m - 1}$. For example one can take $N' = IN$ and $N'' = N/IN$.
+Then the exact sequence
 $$
-0 \to I^2 \otimes_R M \to I \otimes_R M  \to I/I^2 \otimes_R M \to 0
+\text{Tor}_1^R(N', M) \to
+\text{Tor}_1^R(N, M) \to
+\text{Tor}_1^R(N'', M)
 $$
-is short exact. This implies that $I^2 \otimes_R M \to M$ is injective
-and it implies that $I/I^2 \otimes_{R/I} M/IM = IM/I^2M$.
+and induction prove the vanishing we want.
 
 \medskip\noindent
-Let us prove that $M/I^2M$ is flat over $R/I^2$. Let $I^2 \subset J$
-be an ideal. We have to show that
-$J/I^2 \otimes_{R/I^2} M/I^2M \to M/I^2M$ is injective, see
-Lemma \ref{lemma-flat}.
-As $M/IM$ is flat over $R/I$ we know that the map
-$(I + J)/I \otimes_{R/I} M/IM \to M/IM$ is injective.
-The sequence
-$$
-(I \cap J)/I^2 \otimes_{R/I^2} M/I^2M \to
-J/I^2 \otimes_{R/I^2} M/I^2M \to
-(I + J)/I \otimes_{R/I} M/IM \to 0
-$$
-is exact, as you get it by tensoring the exact sequence
-$0 \to (I \cap J) \to J \to (I + J)/I \to 0$ by $M/I^2M$.
-Hence suffices to prove the injectivity of the map
-$(I \cap J)/I^2 \otimes_{R/I} M/IM \to IM/I^2M$. However, the map
-$(I \cap J)/I^2 \to I/I^2$ is injective and as $M/IM$
-is flat over $R/I$ the map
-$(I \cap J)/I^2 \otimes_{R/I} M/IM \to I/I^2 \otimes_{R/I} M/IM$
-is injective. Since we have previously seen that
-$I/I^2 \otimes_{R/I} M/IM = IM/I^2M$ we obtain the desired injectivity.
-
-\medskip\noindent
-Hence we have proven that the assumptions imply:
-(a) $\text{Tor}_1^R(N, M) = 0$ for all $N$ annihilated by $I$,
-(b) $I^2 \otimes_R M \to M$ is injective, and (c) $M/I^2M$ is flat
-over $R/I^2$. Thus we can continue by induction to get the
-same results for $I^n$ for all $n \geq 1$.
+Finally, we prove (1). Given $n \geq 1$ we have to show that $M/I^nM$
+is flat over $R/I^n$. In other words, we have to show that the functor
+$N \mapsto N \otimes_{R/I^n} M/I^nM$ is exact on the category of $R$-modules
+$N$ annihilated by $I^n$. However, for such $N$ we have
+$N \otimes_{R/I^n} M/I^nM = N \otimes_R M$. By the vanishing of
+$\text{Tor}_1$ in (2) we see that the functor $N \mapsto N \otimes_R M$
+is exact on the category of $N$ annihilated by some power of $I$
+and we conclude.
 \end{proof}
 
 \begin{lemma}