Lemma 38.25.2. If in Situation 38.25.1 the ring $A$ is Noetherian then the lemma holds.
Proof. Applying Algebra, Lemma 10.99.1 we see that $u$ is injective and that $N/u(M)$ is flat over $A$. Then $u$ is $A$-universally injective (Algebra, Lemma 10.39.12) and $N$ is $A$-flat (Algebra, Lemma 10.39.13). Since $B$ is Noetherian in this case we see that $N$ is of finite presentation. $\square$
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