2.6 Left-exactness of conormal bundle
Let \(X\) be a regular variety, and let \(Y\) be a regular subvariety. Let \(\mathcal{I}\) be the ideal sheaf of \(Y\). Then \(\mathscr {I} / \mathscr {I}^{2}\) is a locally free sheaf of rank \(\dim X - \dim Y\).
Let \(U = \operatorname {Spec}R\) be an affine open neighborhood of \(y\) in \(X\).
4. \(U \cap Y\) is also affine and irreducible (\(Y\) is irred) By Lemmas 2.5.2
5. As \(U \cap Y\) is irreducibe, it corresponds to a prime ideal \(\mathfrak {p}\). By Lemma 2.5.5.
6. We work in the local ring of \(X\) at \(\mathfrak {p}\), which is \(R_{\mathfrak {p}}\). Let \(\mathfrak {m}\) be the maximal ideal \(\mathfrak {p} R_{\mathfrak {p}}\). Then \(\mathfrak {m} / \mathfrak {m}^2\) is a free \(R_{\mathfrak {p}} / \mathfrak {m}\) module of rank \(\dim R_{\mathfrak {p}} = \text{ht} P = n-s\), where \(n = \dim X\). By 2.1.2 and 2.1.1
7. Thus, by Hartshorne II.5.7 (2.4.7) we have some \(f\) not in \(\mathfrak {p}\) such that \((\mathfrak {p} / \mathfrak {p}^2)_f\) is free of rank \(r\).
The conormal sequence is exact on the left.
1. \(Y\) is regular, therefore \(\Omega _{Y/k}\) is free of rank \(s = \dim Y\). Likewise, \(X\) is regular and \(\Omega _{X/k}\) is free of rank \(n = \dim X\). By Theorem 2.3.21
2. We have the conormal exact sequence
3. Take the stalk of this sequence at an arbitrary closed point \(y \in Y\). By Theorem ??
3a. The stalk-taking/localization pulls into the \(\mathcal{I} / \mathcal{I}^2\). By Corollary 2.4.4
3b. The stalk-taking/localization pulls into the restriction/tensor product in the middle of the sequence By Lemma 2.4.2 (tensor is a colimit, in fact a pushout).
Steps 4-7: Apply the previous lemma to conclude that \(\mathscr {I} / \mathscr {I}^{2}\) is locally free.
Now, we have the conormal exact sequence localized at \(y\):
8. The module in the middle of the (localized) exact sequence is free as it is the localization of a free module.
8a. First, we see that \(\Omega _{X/k} \otimes _{\mathcal{O}_X} \mathcal{O}_Y\) is a free module, first by using part 1 to see that \(\Omega _X\) is free, then using that tensor products commute with direct sums to reduce to the case of \(\mathcal{O}_X\), and finally using the identity property of the tensor product. (tensor product identity property will need to be proved, but it’s not currently written down)
8b. Then, we use the fact that localization pulls in/out of the construction (3b).
Let \(u\) be the surjective map in the (localized) conormal exact sequence.
9. Then \(\operatorname {Ker}u\) is a projective module over \(R_{\mathfrak {p}}\), as the exact sequence
splits (both of the second components are free). By 2.2.1
10. \(\operatorname {Ker}u\) is a (finitely generated) projective module over a local ring, it is free. By 1.1.3
11. By the additivity of rank on exact sequences, \(\operatorname {Ker}u\) has rank \(n-s\). By 2.2.2
12. The first map of the localized conormal exact sequence is a surjection from \(\mathcal{I}_y / \mathcal{I}^2_y\) and \(\operatorname {Ker}u\). Both of these are free modules of rank \(n-s\). By Nakayama’s lemma (the endomorphism from free to free corollary), this is an isomoprhism. By 1.1.4
13. Thus, on stalks the conormal sequence is exact on the left, so it is exact globally. By Theorem 2.4.3.