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Let $C\rightarrow B \rightarrow A$ be sequence (not exact) of surjective ring homomorphisms with $K=ker(C \rightarrow B)$, $I=ker(B \rightarrow A), J=ker(C \rightarrow A)$. I can see that we have an exact sequence $J/J^2 \rightarrow I/I^2 \rightarrow 0$ and also that $I=J/K$. But why do we have an exact sequence $K/K^2 \otimes_B A \rightarrow J/J^2 \rightarrow I/I^2 \rightarrow 0$ and why is $K/K^2 \otimes_B A$ a $B$-module? In particular, $A$ is a $B$ module, but why is $K/K^2$ a $B$-module?

The motivation of this question comes from Lemma 33.5 of the chapter of Morphisms of Schemes of the Stacks Project.

Edited: What exactly is the map $(K/K^2) \otimes_B A \rightarrow J/J^2$?

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The reason $K/K^2$ is a $B$-module is that the surjective map $C\rightarrow B$ with kernel $K$ induces an isomorphism $C/K\cong B$. Thus any $C$-module which is annihilated by $K$ is naturally a $B$-module. In particular, since $K/K^2$ is annihilated by $K$, it is a $B$-module. Explicitly, if $x\in K/K^2$ and $b\in B$, $b\cdot x$ is, by definition, equal to $cx$, and $c\in C$ is any element with $c\mapsto b$ under $C\rightarrow B$.

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    @KeenanKidwell: Thanks Keenan, i think i understand. Let me verify this: we have an exact sequence of $B$-modules $K/K^2 \rightarrow J/J^2 \rightarrow I/I^2 \rightarrow 0$. Since $A$ is a $B$ module, we can tensor the exact sequence to get another exact sequence $K/K^2 \otimes_B A \rightarrow J/J^2 \otimes_B A \rightarrow I/I^2 \otimes A \rightarrow 0$ of $A$-modules. But both $J/J^2, I/I^2$ are also $A$-modules and so $J/J^2 \otimes_B A \cong J/J^2$ and similarly for $I/I^2$. Do i have that right?2012-11-20