The Stacks project

Lemma 115.23.1. Let $b : X' \to X$ be the blowing up of a smooth projective scheme over a field $k$ in a smooth closed subscheme $Z \subset X$. Picture

\[ \xymatrix{ E \ar[r]_ j \ar[d]_\pi & X' \ar[d]^ b \\ Z \ar[r]^ i & X } \]

Assume there exists an element of $K_0(X)$ whose restriction to $Z$ is equal to the class of $\mathcal{C}_{Z/X}$ in $K_0(Z)$. Then $[Lb^*\mathcal{O}_ Z] = [\mathcal{O}_ E] \cdot \alpha ''$ in $K_0(X')$ for some $\alpha '' \in K_0(X')$.

Proof. The schemes $X$, $X'$, $E$, $Z$ are smooth and projective over $k$ and hence we have $K'_0(X) = K_0(X) = K_0(\textit{Vect}(X)) = K_0(D^ b_{\textit{Coh}}(X)))$ and similarly for the other $3$. See Derived Categories of Schemes, Lemmas 36.38.1, 36.38.4, and 36.38.5. We will switch between these versions at will in this proof. Consider the short exact sequence

\[ 0 \to \mathcal{F} \to \pi ^*\mathcal{C}_{Z/X} \to \mathcal{C}_{E/X'} \to 0 \]

of finite locally free $\mathcal{O}_ E$-modules defining $\mathcal{F}$. Observe that $\mathcal{C}_{E/X'} = \mathcal{O}_{X'}(-E)|_ E$ is the restriction of the invertible $\mathcal{O}_ X$-module $\mathcal{O}_{X'}(-E)$. Let $\alpha \in K_0(X)$ be an element such that $i^*\alpha = [\mathcal{C}_{Z/X}]$ in $K_0(Z)$. Let $\alpha ' = b^*\alpha - [\mathcal{O}_{X'}(-E)]$. Then $j^*\alpha ' = [\mathcal{F}]$. We deduce that $j^*\lambda ^ i(\alpha ') = [\wedge ^ i(\mathcal{F})]$ by Weil Cohomology Theories, Lemma 45.13.1. This means that $[\mathcal{O}_ E] \cdot \alpha ' = [\wedge ^ i\mathcal{F}]$ in $K_0(X)$, see Derived Categories of Schemes, Lemma 36.38.8. Let $r$ be the maximum codimension of an irreducible component of $Z$ in $X$. A computation which we omit shows that $H^{-i}(Lb^*\mathcal{O}_ Z) = \wedge ^ i\mathcal{F}$ for $i \geq 0, 1, \ldots , r - 1$ and zero in other degrees. It follows that in $K_0(X)$ we have

\begin{align*} [Lb^*\mathcal{O}_ Z] & = \sum \nolimits _{i = 0, \ldots , r - 1} (-1)^ i[\wedge ^ i\mathcal{F}] \\ & = \sum \nolimits _{i = 0, \ldots , r - 1} (-1)^ i[\mathcal{O}_ E] \lambda ^ i(\alpha ') \\ & = [\mathcal{O}_ E] \left(\sum \nolimits _{i = 0, \ldots , r - 1} (-1)^ i \lambda ^ i(\alpha ')\right) \end{align*}

This proves the lemma with $\alpha '' = \sum _{i = 0, \ldots , r - 1} (-1)^ i \lambda ^ i(\alpha ')$. $\square$

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