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Let $Z \subset Y \subset \mathbb{A^n}$ be a smooth subvarieties of $\mathbb{A^n}$. I'm trying to show that there is an exact sequence of normal bundles.

$0 \rightarrow N_{Z/Y} \rightarrow N_{Z} \rightarrow N_{Y}|_{Z} \rightarrow 0$

It seems obvious, but I can't figure out how things work in algebraic setting.

More precisely, let $I \subset J \subset k[x_1, ...x_n]$ be ideals defining Y and Z. Then,

$N_Z = \mathcal{Hom_Z(J/J^2, O/J)}$

$N_Y|_Z = \mathcal{Hom_Y(I/I^2, O/I) \otimes O/J}$

I need a natural map $N_Z \to N_Y|_Z$. And I think the natural restriction map

$\phi \mapsto \phi|_{I/I^2} \otimes 1$

is a candidates. But it is not well defined, and I stuck.

What's the problem? I appreciate any help.

share|cite|improve this question
Perhaps it is easier to look at the sequence for conormal bundles first? – J.C. Ottem Apr 16 '12 at 14:55
See IV, par. 3, Prop. 3.4, p. 79 in "Riemann-Roch algebra" by Fulton-Lang (Springer). – Damian Rössler Apr 16 '12 at 19:35
EGA IV, 19.1.5 (iii) – Parsa Apr 17 '12 at 5:38
It seems that EGA or Fulton's book prove it for conormal case.(which is easy for me too). In that case, restrict comes first and dualizing follows. In my question, dualization comes first. Do Dualization and restriction commute? It may needs flatness or something..... – Choa Apr 17 '12 at 11:27
@Choa Once you have it for conormal sheaves you just dualize, and use that for locally free sheaves dualization and restriction commute. All three sheaves in question are locally free, since a smooth subvariety of a smooth variety is automatically a local complete intersection. – Parsa Apr 20 '12 at 10:35

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