The Stacks project

Proof. Suppose that $X$ is quasi-affine. Set $A = \Gamma (X, \mathcal{O}_ X)$. Consider the open immersion

from Lemma 28.18.4. Note that $\mathop{\mathrm{Spec}}(A) = \text{Proj}(A[T])$, see Constructions, Example 27.8.14. Hence we can apply Lemma 28.26.12 to deduce that $\mathcal{O}_ X$ is ample.

Suppose that $\mathcal{O}_ X$ is ample. Note that $\Gamma _*(X, \mathcal{O}_ X) \cong A[T]$ as graded rings. Hence the result follows from Lemmas 28.26.11 and 28.18.4 taking into account that $\mathop{\mathrm{Spec}}(A) = \text{Proj}(A[T])$ for any ring $A$ as seen above. $\square$

Proof. This can be proved directly without making use of the material on ample invertible sheaves; we urge the reader to do this on a napkin. Since $X$ is quasi-affine, we have that $\mathcal{O}_ X$ is ample by Lemma 28.27.1. Then $\mathcal{O}_{X'}$ is ample by Lemma 28.26.14. Then $X'$ is quasi-affine by Lemma 28.27.1. $\square$

Proof. Assume we have $f_1, \ldots , f_ n$ as in the lemma. We may write $1 = \sum g_ i f_ i$ for some $g_ j \in \Gamma (X, \mathcal{O}_ X)$ and hence it is clear that $X = \bigcup X_{f_ i}$. (The $f_ i$'s cannot all vanish at a point.) Since each $X_{f_ i}$ is quasi-compact (being affine) it follows that $X$ is quasi-compact. Hence we see that $X$ is quasi-affine by Lemma 28.27.1 above. Consider the open immersion

see Lemma 28.18.4. The inverse image of the standard open $D(f_ i)$ on the right hand side is equal to $X_{f_ i}$ on the left hand side and the morphism $j$ induces an isomorphism $X_{f_ i} \cong D(f_ i)$, see Lemma 28.18.3. Since the $f_ i$ generate the unit ideal we see that $\mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X)) = \bigcup _{i = 1, \ldots , n} D(f_ i)$. Thus $j$ is an isomorphism. $\square$