# Conormal Sheaf (Morphisms of Schemes, Stacks Project)

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 $i^{-1}(\mathcal{I}/\mathcal{I}^2)$.

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Dear Manos, you shouldn't add a "second part" to your question: it is irritating for someone who is planning to answer the original question. Wait for an answer and then ask another question. –  Georges Elencwajg Nov 18 '12 at 18:35
Also, you seem to write $\oplus_B$ instead of $\otimes_B$ –  Georges Elencwajg Nov 18 '12 at 18:45
@GeorgesElencwajg: My apologies. Should i remove the second part? Or maybe make it a separate question? The reason i put those two together was that i could not think of a more specific title if these were to be separate questions. –  Manos Nov 18 '12 at 18:47
Dear Manos, no, don't remove the second part because someone might be thinking about answering it right now! My comment was meant as a kind of general rule. Imagine you are taking a written exam in class and suddenly while you are working on it the professor hands you a new sheet with a few supplementary questions: wouldn't that be dreadful? :-) –  Georges Elencwajg Nov 18 '12 at 19:36
@GeorgesElencwajg: I am sorry Georges, but i already moved the second part to a separate question. I'll keep this general rule in mind. I agree it would be dreadful :) If anybody is thinking about answering it, there is a new question posted. –  Manos Nov 18 '12 at 19:38

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.
Dear Georges, thank you for you answer. Please also see my comment above. A question regarding your answer: how do we obtain the equality $O_Z=i^{-1}(O_X/I)$? I can see that if $i:Z\rightarrow X$ is a closed immersion, then we have a surjective morphism of sheaves $i^{\#}:O_X \rightarrow O_Z$ and that $O_Z \cong O_X /\mathcal{I}$. –  Manos Nov 18 '12 at 18:57
Dear Manos, what I wrote and what you say are practically the same thing. It's just that the sheaf $\mathcal O_X/\mathcal I$ is officially a sheaf on $X$, even if its stalks are zero outside $Z$. Since we want $\mathcal O_Z$ to be a sheaf on $Z$ (of course!) we write $i^{-1} (\mathcal O_X/\mathcal I)$ in order to be 100% precise. –  Georges Elencwajg Nov 18 '12 at 19:43
Let me see if i can make this rigorous: we have by definition an exact sequence of sheaves of $X$ given by $0\rightarrow \mathcal{I} \rightarrow O_X \rightarrow i_* O_Z \rightarrow 0$. Now apply the exact functor $i^{-1}$ which is left adjoint to $i_*$ to get $0\rightarrow i^{-1}\mathcal{I} \rightarrow i^{-1}O_X \rightarrow i^{-1} i_* O_Z \rightarrow 0$. This looks about right, the problem is that i am having $i^{-1} i_* O_Z$ instead of $O_Z$...I know though that there is a canonical morphism $i^{-1}i_*O_Z \rightarrow O_Z$. –  Manos Nov 18 '12 at 19:56