Lemma 38.25.4. If in Situation 38.25.1 the ring $A$ is henselian then the lemma holds.

**Proof.**
It suffices to prove this when $B$ is essentially of finite presentation over $A$ and $N$ is of finite presentation over $B$, see Lemma 38.25.3. Let us temporarily make the additional assumption that $N$ is flat over $A$. Then $N$ is a filtered colimit $N = \mathop{\mathrm{colim}}\nolimits _ i F_ i$ of free $A$-modules $F_ i$ such that the transition maps $u_{ii'} : F_ i \to F_{i'}$ are injective modulo $\mathfrak m_ A$, see Lemma 38.19.5. Each of the compositions $u_ i : F_ i \to M$ is $A$-universally injective by Lemma 38.7.5 wherefore $u = \mathop{\mathrm{colim}}\nolimits u_ i$ is $A$-universally injective as desired.

Assume $A$ is a henselian local ring, $B$ is essentially of finite presentation over $A$, $N$ of finite presentation over $B$. By Theorem 38.24.1 there exists a finitely generated ideal $I \subset A$ such that $N/IN$ is flat over $A/I$ and such that $N/I^2N$ is not flat over $A/I^2$ unless $I = 0$. The result of the previous paragraph shows that the lemma holds for $u \bmod I : N/IN \to M/IM$ over $A/I$. Consider the commutative diagram

whose rows are exact by right exactness of $\otimes $ and the fact that $M$ is flat over $A$. Note that the left vertical arrow is the map $N/IN \otimes _{A/I} I/I^2 \to M/IM \otimes _{A/I} I/I^2$, hence is injective. A diagram chase shows that the lower left arrow is injective, i.e., $\text{Tor}^1_{A/I^2}(I/I^2, M/I^2) = 0$ see Algebra, Remark 10.75.9. Hence $N/I^2N$ is flat over $A/I^2$ by Algebra, Lemma 10.99.8 a contradiction unless $I = 0$. $\square$

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