The first part of this question refers to Lemma 33.2 from the chapter "Morphisms of Schemes" of the Stacks Project. In particular, if $i: Z \rightarrow X$ is an immersion and $\mathcal{I}$ is the corresponding ideal sheaf, then the conormal sheaf is $C_{Z/X} = i^*(\mathcal{I})$. What i don't see is why $i^*(\mathcal{I}) = i^{-1}(\mathcal{I}/\mathcal{I}^2)$. My efforts: if i apply the definition of the pullback $i^*$ i get $i^*(\mathcal{I}) = i^{-1} \mathcal{I} \otimes_{i^{-1}O_X} O_Z$. Additionally, i also see that if $R$ is a ring and $I$ some ideal then $I/I^2 = I \times_R R/I$. But i am having trouble combining these two facts to obtain $i^{-1}(\mathcal{I}/\mathcal{I}^2)$.
Conormal Sheaf (Morphisms of Schemes, Stacks Project)
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algebraic-geometry
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0That's quite all right, Manos, don't worry. – 2012-11-18
1 Answers
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By definition, we have $ i^*(\mathcal{I}) = i^{-1}(\mathcal{I})\otimes _{i^{-1}\mathcal O_X} \mathcal O_Z$.
On the other hand $\mathcal O_Z=i^{-1}(\mathcal O_X/\mathcal I)$, so that $ i^*(\mathcal{I}) = i^{-1}(\mathcal{I})\otimes _{i^{-1}\mathcal O_X} i^{-1}(\mathcal O_X/\mathcal I)=i^{-1}[\mathcal I\otimes _{\mathcal O_X} \mathcal O_X/\mathcal I]=i^{-1}[\mathcal I/\mathcal I^2]$ just as is stated in the Stacks Project.
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0Awesome, thanks Georges! – 2012-11-18