The Stacks project

Proof. Denote $j : U \to \mathop{\mathrm{Spec}}(A)$ the inclusion morphism. Then $H^0(U, \mathcal{F}) = H^0(\mathop{\mathrm{Spec}}(A), j_*\mathcal{F})$ and $j_*\mathcal{F}$ is quasi-coherent by Schemes, Lemma 26.24.1. Hence $j_*\mathcal{F} = \widetilde{H^0(U, \mathcal{F})}$ by Schemes, Lemma 26.7.5. Restricting back to $U$ we get the lemma. $\square$

Proof. This is clear as $j$ induces an isomorphism of rings $\Gamma (X, \mathcal{O}_ X)_ f \to \mathcal{O}_ X(X_ f)$ by Lemma 28.17.1 above. $\square$

Proof. If the displayed morphism is a quasi-compact open immersion then $X$ is isomorphic to a quasi-compact open subscheme of $\mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X))$ and clearly $X$ is quasi-affine.

Assume $X$ is quasi-affine, say $X \subset \mathop{\mathrm{Spec}}(R)$ is quasi-compact open. This in particular implies that $X$ is separated, see Schemes, Lemma 26.23.9. Let $A = \Gamma (X, \mathcal{O}_ X)$. Consider the ring map $R \to A$ coming from $R = \Gamma (\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ and the restriction mapping of the sheaf $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$. By Schemes, Lemma 26.6.4 we obtain a factorization:

of the inclusion morphism. Let $x \in X$. Choose $r \in R$ such that $x \in D(r)$ and $D(r) \subset X$. Denote $f \in A$ the image of $r$ in $A$. The open $X_ f$ of Lemma 28.17.1 above is equal to $D(r) \subset X$ and hence $A_ f \cong R_ r$ by the conclusion of that lemma. Hence $D(r) \to \mathop{\mathrm{Spec}}(A)$ is an isomorphism onto the standard affine open $D(f)$ of $\mathop{\mathrm{Spec}}(A)$. Since $X$ can be covered by such affine opens $D(f)$ we win. $\square$

Proof. The diagram is commutative by Schemes, Lemma 26.6.4. Write $A = \Gamma (U, \mathcal{O}_ U)$ and $B = \Gamma (V, \mathcal{O}_ V)$. Let $g \in B$ be such that $V_ g$ is affine and contained in $U$. This means that if $f$ is the image of $g$ in $A$, then $U_ f = V_ g$. By Lemma 28.18.3 we see that $j'$ induces an isomorphism of $V_ g$ with the standard open $D(g)$ of $\mathop{\mathrm{Spec}}(B)$. Thus $V_ g \times _{\mathop{\mathrm{Spec}}(B)} \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A)$ is an isomorphism onto $D(f) \subset \mathop{\mathrm{Spec}}(A)$. By Lemma 28.18.3 again $j$ maps $U_ f$ isomorphically to $D(f)$. Thus we see that $U_ f = U_ f \times _{\mathop{\mathrm{Spec}}(B)} \mathop{\mathrm{Spec}}(A)$. Since by Lemma 28.18.4 we can cover $U$ by $V_ g = U_ f$ as above, we see that $U \to U \times _{\mathop{\mathrm{Spec}}(B)} \mathop{\mathrm{Spec}}(A)$ is an isomorphism. $\square$

Proof. By definition, there exists a ring $A$ such that $X$ is isomorphic to a quasi-compact open subscheme $U \subset \mathop{\mathrm{Spec}}(A)$. Recall that the standard opens $D(f) \subset \mathop{\mathrm{Spec}}(A)$ form a basis for the topology, see Algebra, Section 10.17. Since $U$ is quasi-compact we can choose $f_1, \ldots , f_ n \in A$ such that $U = D(f_1) \cup \ldots \cup D(f_ n)$. Thus we may assume $X = \mathop{\mathrm{Spec}}(A) \setminus V(I)$ where $I = (f_1, \ldots , f_ n)$. We set

The structure morphism $T \to \mathop{\mathrm{Spec}}(A)$ factors through the open $X$ to give the morphism $T \to X$. If $X' = \mathop{\mathrm{Spec}}(A')$ and the morphism $X' \to X$ corresponds to the ring map $A \to A'$, then the images $f'_1, \ldots , f'_ n \in A'$ of $f_1, \ldots , f_ n$ generate the unit ideal in $A'$. Say $1 = f'_1 a'_1 + \ldots + f'_ n a'_ n$. The base change $X' \times _ X T$ is the spectrum of $A'[t, x_1, \ldots , x_ n]/(f'_1 x_1 + \ldots + f'_ n x_ n - 1)$. We claim the $A'$-algebra homomorphism

sending $y_ i$ to $a'_ i t + x_ i$ is an isomorphism. The claim finishes the proof of the lemma. The inverse of $\varphi $ is given by the $A'$-algebra homomorphism

sending $t$ to $-1 + f'_1 y_1 + \ldots + f'_ n y_ n$ and $x_ i$ to $y_ i + a'_ i - a'_ i(f'_1 y_1 + \ldots + f'_ n y_ n)$ for $i = 1, \ldots , n$. This makes sense because $\sum f'_ ix_ i$ is mapped to

To see the maps are mutually inverse one computes as follows:

This finishes the proof. $\square$