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In theorem ${3.2}$ of Harsthorne, part $d$ is implicitly using the fact that if ${A}$ is a finitely generated $k$ domain then ${Frac(A)}$ is a finitely generated field extension of $k$. Why is this the case?

CoffeeBean
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    If $a_1,\dots,a_n$ generate $A$ as a $k$-algebra then they generate its fraction field as a field extension. – Eric Wofsey Dec 11 '22 at 22:16
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    Perhaps make it clear that $\Bbb{C}(x)$ is finitely generated as a field extension of $\Bbb{C}$ but not as a $\Bbb{C}$-algebra. – reuns Dec 11 '22 at 22:25
  • @EricWofsey ah I think I know what I am getting confused about: I think I am confusing what it means to be finitely generated as a $k$ algebra and finitely generated as a field extension. Am I right in thinking that in general ${Frac(A)}$ won't be finitely generated as a $k$ algebra, but will as a field extension? – CoffeeBean Dec 11 '22 at 22:25
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    It will be iff $Frac(A)/k$ is a finite algebraic extension https://en.wikipedia.org/wiki/Zariski%27s_lemma – reuns Dec 11 '22 at 22:31
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    Exactly, when generating as a field extension you're allowed to take inverses. – Eric Wofsey Dec 11 '22 at 22:38
  • I see. Thank you both @EricWofsey and reuns. If either of you want to post this as an answer I will give it the green checkmark – CoffeeBean Dec 12 '22 at 20:45

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