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The definition of affine $k$- variety $X$, I have is that $X$ is an affine scheme that is reduced and of finite type over $k$ ($k$ is a field here). The definition of finite type I have is that $X$ can be covered by a finite number of affine open subschemes of the form $Spec \ B_i$ where $B_i$ is a finitely generated $k$-algebra.

I was wondering does it then follow that every affine $k$-variety must be of the form $Spec \ A$, where $A$ is a finitely generated $k$-algebra?

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Yes, this is true. A $k$-scheme $X$ might be defined to be finite type if it can be covered by finitely many affine opens of the form $\mathrm{Spec}(A)$ with $A$ a $k$-algebra of finite type, but it is in fact true that this implies the stronger condition that every affine open of $X$ is the spectrum of a finite type $k$-algebra (see http://stacks.math.columbia.edu/tag/01T2 for the particular case and http://stacks.math.columbia.edu/tag/01SQ for the general principle behind results like these). In particular if $X$ itself is affine of finite type, it must be the spectrum of a finite type $k$-algebra.

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  • $\begingroup$ Unfortunately, the proof of the relevant algebraic lemma is omitted, but this is a good view of the geometric context. $\endgroup$ – Slade Jun 26 '15 at 19:25
  • $\begingroup$ There is also a good discussion of Vakil's notes. There are tons of "affine-local" properties like this and most of them are probably exercises there, but I think he mostly spells this one out since the algebra is a little harder. $\endgroup$ – Hoot Jun 26 '15 at 19:38
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By (quasi-)compactness, we can rephrase this question as follows:

If $A$ is a $k$-algebra, $f_1, \ldots , f_n \in A$ generate the unit ideal, and each of the localizations $A_{f_i}$ is finitely generated over $k$, then must $A$ be finitely generated over $k$?

The answer is yes—there is a finite set $S\subset A$ that generates each of the localizations $A_{f_i}$, and we claim that $S\cup\{f_1,\ldots , f_n\}$ generates $A$.

Indeed, for any $x\in A$, there is some $N$ such that, for all $i$, $f_i^N x$ is generated by $S\cup \{f_i\}$. Since $f_1^N,\ldots , f_n^N$ also generate the unit ideal, $x$ itself is generated by $S\cup\{f_1,\ldots , f_n\}$.

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  • $\begingroup$ Thank you very much, this one was very helpful too! $\endgroup$ – user249180 Jun 26 '15 at 21:26

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