Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
show 1 more comment

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
    
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
    
That last canonical morphism is the identity: Stacks Project Lemma 6.32.1 –  Georges Elencwajg Nov 18 '12 at 20:29
    
Awesome, thanks Georges! –  Manos Nov 18 '12 at 20:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.