**Proof.**
In the proof we will use Lemma 15.6.4 without further mention.

Proof of (1). Set $J = \mathop{\mathrm{Ker}}(B' \to B)$. This is an ideal of $B'$ mapping isomorphically to $I = \mathop{\mathrm{Ker}}(A' \to A)$. Let $\mathfrak b' \subset B'$ be an ideal. We have to show that $\mathfrak b' \otimes _{B'} N' \to N'$ is injective, see Algebra, Lemma 10.38.5. We know that

\[ \mathfrak b'/(\mathfrak b' \cap J) \otimes _{B'} N' = \mathfrak b'/(\mathfrak b' \cap J) \otimes _ B N \to N \]

is injective as $N$ is flat over $B$. As $\mathfrak b' \cap J \to \mathfrak b' \to \mathfrak b'/(\mathfrak b' \cap J) \to 0$ is exact, we conclude that it suffices to show that $(\mathfrak b' \cap J) \otimes _{B'} N' \to N'$ is injective. Thus we may assume that $\mathfrak b' \subset J$. Next, since $J \to I$ is an isomorphism we have

\[ J \otimes _{B'} N' = I \otimes _{A'} A' \otimes _{B'} N' = I \otimes _{A'} M' \]

which maps injectively into $M'$ as $M'$ is a flat $A'$-module. Hence $J \otimes _{B'} N' \to N'$ is injective and we conclude that $\text{Tor}_1^{B'}(B'/J, N') = 0$, see Algebra, Remark 10.74.9. Thus we may apply Algebra, Lemma 10.98.8 to $N'$ over $B'$ and the ideal $J$. Going back to our ideal $\mathfrak b' \subset J$, let $\mathfrak b' \subset \mathfrak b'' \subset J$ be the smallest ideal whose image in $I$ is an $A'$-submodule of $I$. In other words, we have $\mathfrak b'' = A' \mathfrak b'$ if we view $J = I$ as $A'$-module. Then $\mathfrak b''/\mathfrak b'$ is killed by $J$ and we get a short exact sequence

\[ 0 \to \mathfrak b' \otimes _{B'} N' \to \mathfrak b'' \otimes _{B'} N' \to \mathfrak b''/\mathfrak b' \otimes _{B'} N' \to 0 \]

by the vanishing of $\text{Tor}_1^{B'}(\mathfrak b''/\mathfrak b', N')$ we get from the application of the lemma. Thus we may replace $\mathfrak b'$ by $\mathfrak b''$. In particular we may assume $\mathfrak b'$ is an $A'$-module and maps to an ideal of $A'$. Then

\[ \mathfrak b' \otimes _{B'} N' = \mathfrak b' \otimes _{A'} A' \otimes _{B'} N' = \mathfrak b' \otimes _{A'} M' \]

This tensor product maps injectively into $M'$ by our assumption that $M'$ is flat over $A'$. We conclude that $\mathfrak b' \otimes _{B'} N' \to N' \to M'$ is injective and hence the first map is injective as desired.

Proof of (2). This follows by tensoring the short exact sequence $0 \to B' \to B \oplus A' \to A \to 0$ with $L'$ over $B'$.

Proof of (3). Immediate consequence of (1) and (2).
$\square$

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