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If $X$ is a prevariety and the open sets $X_f:=\{x:f(x)\neq 0\}$ form a basis for the topology of $X$ as $f$ varies over $\Gamma(X,\mathcal{O}_X)$, then why is $X$ quasi-affine?

The definitions are from Mumford's red book, so prevariety is a scheme with a finite covering by affine varieties (all over the same algebraically closed field), and a quasi-affine variety is an open subset of an affine variety. I found this problem as problem 6 of http://ocw.mit.edu/courses/mathematics/18-725-algebraic-geometry-fall-2003/assignments/ps6.pdf, and it wasn't clear to me how to even start (where would I get the ambient affine variety?).

This isn't homework.

DCT
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1 Answers1

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Given any scheme $X$, there is a canonical morphism $j: X\to \operatorname {Spec}(\Gamma(X,\mathcal O_X))$.
So the ambient affine variety you are asking about in order to start is the very natural $ \operatorname {Spec}(\Gamma(X,\mathcal O_X))$.
Then, under the conditions of the exercise, the morphism $j$ is an open embedding: this is proved in Grothendieck-Dieudonné's EGA, volume II, Proposition (5.1.2), page 94.

Georges Elencwajg
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