Lemma 115.4.8. Let $A \to B$ be a ring map. Let $f \in B$. Assume that

$A \to B$ is flat,

$f$ is a nonzerodivisor, and

$A \to B/fB$ is flat.

Then for every ideal $I \subset A$ the map $f : B/IB \to B/IB$ is injective.

Lemma 115.4.8. Let $A \to B$ be a ring map. Let $f \in B$. Assume that

$A \to B$ is flat,

$f$ is a nonzerodivisor, and

$A \to B/fB$ is flat.

Then for every ideal $I \subset A$ the map $f : B/IB \to B/IB$ is injective.

**Proof.**
Note that $IB = I \otimes _ A B$ and $I(B/fB) = I \otimes _ A B/fB$ by the flatness of $B$ and $B/fB$ over $A$. In particular $IB/fIB \cong I \otimes _ A B/fB$ maps injectively into $B/fB$. Hence the result follows from the snake lemma applied to the diagram

\[ \xymatrix{ 0 \ar[r] & I \otimes _ A B \ar[r] \ar[d]^ f & B \ar[r] \ar[d]^ f & B/IB \ar[r] \ar[d]^ f & 0 \\ 0 \ar[r] & I \otimes _ A B \ar[r] & B \ar[r] & B/IB \ar[r] & 0 } \]

with exact rows. $\square$

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