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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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For the exact sequence, notice that the kernel of $J/J^2\to I/I^2$ is $(J^2+K)/J^2$. –  user18119 Nov 18 '12 at 23:24
    
@QiL: Thanks, yes, i see that. What exactly is the map $(K/K^2) \otimes_B A \rightarrow J/J^2$? –  Manos Nov 19 '12 at 16:30
    
@QiL: The image of the map $K/K^2 \rightarrow J/J^2$ is precisely $(J^2+K)/J^2$. Why do we need the tensor product with $A$? –  Manos Nov 19 '12 at 16:39
    
@Manos: as $K/K^2$ is canonically a $B$-module, tensoring with $A$ makes it an $A$-module, and the exact sequence becomes an exact sequence of $A$-modules. The map from tensor product is given by $\bar{k}\otimes a\mapsto a\times $ (image of $k$ in $J/J^2$). –  user18119 Nov 19 '12 at 17:27
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@Manos Yes, $J/J^2$ is an $A$-module. By definition, $A=C/J$, so an $A$-module is the same thing as a $C$-module killed by $J$, such as $J/J^2$. The natural map $i:K/K^2\rightarrow J/J^2$ is $B$-linear with $J/J^2$ an $A$-module, so the universal property of base change gives you an $A$-linear map $K/K^2\otimes_BA\rightarrow J/J^2$ with $x\otimes a\mapsto ai(x)$. –  Keenan Kidwell Nov 19 '12 at 20:46
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