Lemma 42.59.11 (Blow up formula). Let $(S, \delta )$ be as in Situation 42.7.1. Let $i : Z \to X$ be a regular closed immersion of schemes locally of finite type over $S$. Let $b : X' \to X$ be the blowing up with center $Z$. Picture
\[ \xymatrix{ E \ar[r]_ j \ar[d]_\pi & X' \ar[d]^ b \\ Z \ar[r]^ i & X } \]
Assume that the gysin map exists for $b$. Then we have
\[ res(b^!) = c_{top}(\mathcal{F}^\vee ) \circ \pi ^* \]
in $A^*(E \to Z)$ where $\mathcal{F}$ is the kernel of the canonical map $\pi ^*\mathcal{C}_{Z/X} \to \mathcal{C}_{E/X'}$.
Proof.
Observe that the morphism $b$ is a local complete intersection morphism by More on Algebra, Lemma 15.31.2 and hence the statement makes sense. Since $Z \to X$ is a regular immersion (and hence a fortiori quasi-regular) we see that $\mathcal{C}_{Z/X}$ is finite locally free and the map $\text{Sym}^*(\mathcal{C}_{Z/X}) \to \mathcal{C}_{Z/X, *}$ is an isomorphism, see Divisors, Lemma 31.21.5. Since $E = \text{Proj}(\mathcal{C}_{Z/X, *})$ we conclude that $E = \mathbf{P}(\mathcal{C}_{Z/X})$ is a projective space bundle over $Z$. Thus $E \to Z$ is smooth and certainly a local complete intersection morphism. Thus Lemma 42.59.10 applies and we see that
\[ res(b^!) = c_{top}(\mathcal{C}^\vee ) \circ \pi ^! \]
with $\mathcal{C}$ as in the statement there. Of course $\pi ^* = \pi ^!$ by Lemma 42.59.5. It remains to show that $\mathcal{F}$ is equal to the kernel $\mathcal{C}$ of the map $H^{-1}(j^*\mathop{N\! L}\nolimits _{X'/X}) \to H^{-1}(\mathop{N\! L}\nolimits _{E/Z})$.
Since $E \to Z$ is smooth we have $H^{-1}(\mathop{N\! L}\nolimits _{E/Z}) = 0$, see More on Morphisms, Lemma 37.13.7. Hence it suffices to show that $\mathcal{F}$ can be identified with $H^{-1}(j^*\mathop{N\! L}\nolimits _{X'/X})$. By More on Morphisms, Lemmas 37.13.11 and 37.13.9 we have an exact sequence
\[ 0 \to H^{-1}(j^*\mathop{N\! L}\nolimits _{X'/X}) \to H^{-1}(\mathop{N\! L}\nolimits _{E/X}) \to \mathcal{C}_{E/X'} \to \ldots \]
By the same lemmas applied to $E \to Z \to X$ we obtain an isomorphism $\pi ^*\mathcal{C}_{Z/X} = H^{-1}(\pi ^*\mathop{N\! L}\nolimits _{Z/X}) \to H^{-1}(\mathop{N\! L}\nolimits _{E/X})$. Thus we conclude.
$\square$
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