Lemma 57.29.2. In the situation of Lemma 57.29.1 assume that $H^ i(V, \mathcal{O}_ V) = 0$ for $i \geq \dim (A) - 1$. Then $V$ is affine.

Proof. Let $k = A/\mathfrak m$. Since $V \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(k) = \emptyset$, by cohomology and base change we have

$R\Gamma (V, \mathcal{O}_ V) \otimes _ A^\mathbf {L} k = 0$

See Derived Categories of Schemes, Lemma 36.22.5. Thus there is a spectral sequence (More on Algebra, Example 15.61.4)

$E_2^{p, q} = \text{Tor}_{-p}(k, H^ q(V, \mathcal{O}_ V)),\quad d_2^{p, q} : E_2^{p, q} \to E_2^{p + 2, q - 1}$

and $d_ r^{p, q} : E_ r^{p, q} \to E_ r^{p + r, q - r + 1}$ converging to zero. By Lemma 57.29.1, Dualizing Complexes, Lemma 47.21.9, and our assumption $H^ i(V, \mathcal{O}_ V) = 0$ for $i \geq \dim (A) - 1$ we conclude that there is no nonzero differential entering or leaving the $(p, q) = (0, 0)$ spot. Thus $H^0(V, \mathcal{O}_ V) \otimes _ A k = 0$. This means that if $\mathfrak m = (x_1, \ldots , x_ d)$ then we have an open covering $V = \bigcup V \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A_{x_ i})$ by affine open subschemes $V \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A_{x_ i})$ (because $V$ is affine over the punctured spectrum of $A$) such that $x_1, \ldots , x_ d$ generate the unit ideal in $\Gamma (V, \mathcal{O}_ V)$. This implies $V$ is affine by Properties, Lemma 28.27.3. $\square$

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