For an affine $k$-scheme of finite type, is the underlying ring a finitely generated $k$-algebra?

Question

If $X$ is an affine $k$-scheme of finite type, say, $X=\operatorname{Spec} A$. Can we deduce $A$ is a finitely generated $k$-algebra? Could you prove it or give a counterexample? Thanks!

Answer 1

A morphism $X \to \operatorname{Spec} k$ if of finite type if $X$ can be covered by finitely many affine $U_i = \operatorname{Spec} A_i$ so that each $A_i$ is a finitely generated $k$-algebra.

Suppose $X = \operatorname{Spec} A$. By Nike's trick we can cover $X$ by finitely many basic open affines $D(f_i)$ which are basic open in various $U_i$ so that $ A_{f_i}$ are finitely generated $k$-algebras for each $i$, since the localization of a finitely generated $k$ algebra at an element is again finitely generated.

Then we have reduced this to the following commutative algebra problem. If $(f_1, \dots, f_n) = A$ and $A_{f_i}$ is a finitely generated $k$-algebra for all $i$, then $A$ is a finitely generated $k$-algebra. I'll leave proving this fact to you.