This is  [ 4 , Proposition V.3.1 ] .

If we are willing to work analytically, we could replace \(X\) with \(\mathbb {A}^2\) and \(P\) with the origin so that we would reduce to the case of Example 4.1. Then it’s clear that \(E\cong \mathbb {P}^1\), and \(\widetilde{X}\) is smooth because is is covered by two affine charts \(U\) and \(V\) both isomorphic to \(\mathbb {A}^2\).

The fact that \(E^2 = -1\) follows from the fact that the ideal sheaf of \(E\) is \(\mathcal{O}_{\widetilde{X}(1)}\). So \(E\cdot E = \deg \mathcal{O}_{\widetilde{X}}(-1)|_E = \deg \mathcal{O}_{\mathbb {P}}(-1) = -1\).

This is  [ 4 , Proposition V.3.2 ] .

Using localization and the fact that \(\operatorname{\mathrm{Pic}}(\widetilde{X}\setminus E) = \operatorname{\mathrm{Pic}}(X\setminus P) = \operatorname{\mathrm{Pic}}X\), we get an exact sequence

where the first map is given by \(1\mapsto [E]\). And \([E]\) is nontorsion since \(E^2 = -1\). So the first map is injective. It splits via the map \(\pi ^*\).

This is  [ 4 , Proposition V.3.3 ] .

We know that \(K_{\widetilde{X}} = \pi ^* K_{X} + n E\) for some \(n\) by the localization sequences. But then the genus formula says that \(\displaystyle 0 = g(C) = 1 + \frac{1}{2} E\cdot (E+K)\). So we get

Therefore, \(n=1\).

The last statement is now obvious.

This is  [ 4 , Propostion V.3.4 ] . The proof uses the theorem on formal functions  [ 4 , Theorem III.11.1 ] .

The assertion for \(i=0\) is basically Zariski’s Main Theorem, which says that if \(f:X\to Y\) is projective birational between integral noetherian schemes and \(Y\) is normal, then, for all \(y\in Y\), \(f^{-1}(y)\) is connected. The proof here can be reduced to the affine case, and then \(B = \Gamma (\widetilde{X}, \mathcal{O}_{\widetilde{X}})\) is a finitely generated \(A\)-module, where \(A = \Gamma (X,\mathcal{O}_B)\). Moreover, \(A\) and \(B\) have the same function fields and \(A\) is integrally closed. So, it’s obvious.

For the rest, first, note that, if \(\pi \) is the blowup of \(X\) at \(P\) and \(E\) is the exceptional divisor in \(\widetilde{X}\), then the restriction of \(\pi \) induces and isomorphism \(\widetilde{X}\setminus E\xrightarrow {\sim } X\setminus P\). For \(i\in \mathbb {Z}\), set \(\mathcal{F}^i := R^i\pi _*\mathcal{O}_{\widetilde{X}}\). Then, we immediately get that \(\mathcal{F}^i_{|X\setminus P} = 0\) for \(i {\gt} 0\) is isomorphic to \(\mathcal{O}_{X\setminus P}\) for \(i=0\).

In this context, that theorem says the following. Let \(E\) be the exceptional divisor, and let \(\mathcal{I}\) denote its ideal sheaf. Let \(E_n = V(\mathcal{I}^n\). Then we have

where \(\widehat{\mathcal{F}}^i\) is the completion of \(\mathcal{F}^i\) at \(P\). But, in fact, since \(P\) is a point and \(\mathcal{F}^i\) is coherent, this completion is just \(\mathcal{F}^i\) itself. So we get that \(\mathcal{F}^i = \lim _{\leftarrow }\operatorname{\mathrm{H}}^i (E_n, \mathcal{O}_{E_n})\).

Now, each one of the sheaves \(\mathcal{O}_n\) sits in an exact sequence

Moreover, \(\mathcal{I}/\mathcal{I}^2 = \mathcal{O}_E(1)\) and \(\mathcal{I}^n/\mathcal{I}^{n+1} = \operatorname{\mathrm{Sym}}(\mathcal{I}/\mathcal{I}^2) = \mathcal{O}_E(n)\). It follows that \(\operatorname{\mathrm{H}}^i(E_n, \mathcal{O}_{E_n}) = \operatorname{\mathrm{H}}^i(E, \mathcal{O}_E)\) for all \(i\). So we get that \(\mathcal{F}^i = 0\) for \(i\).