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Let $k$ be a field, and let $A$ be a commutative $k$-algebra which is noetherian.

Suppose that for each prime ideal $p$ of $A$, it holds that the field $k(p)$, the field of fractions of $A/p$ has finite transcendence degree over $k$.

Does this imply that $A$ is a localization of a finite type $k$-algebra?

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    $\begingroup$ No. Take $k=\mathbb{Q}$, $A=\bar{\mathbb{Q}}$. $\endgroup$ – abx Apr 10 '14 at 10:14
  • $\begingroup$ Another problem occurs when $A$ contains infinitely many nilpotents. $\endgroup$ – Will Sawin Apr 10 '14 at 21:33
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The answer is no as was pointed out in the comments by abx (algebraic extensions) and by Will Sawin (non-reducedness).

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