I know that a finite field extension $E/F$ is integral. I think the converse is not true. Somebody told me it was true under the hypothesis that $E$ is a finitely-generated $F$-module. I've thought about this for about an hour and I am stuck. Maybe somebody can lend a hint?


closed as off-topic by Rob Arthan, user26857, Namaste, Leucippus, hardmath Feb 23 '18 at 4:55

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  • $\begingroup$ What about taking an introductory algebra textbook and read it? $\endgroup$ – user26857 Feb 22 '18 at 22:04
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    $\begingroup$ What are you stuck on? "Finite field extension of $F$" means "finitely generated $F$-module, so I suspect "somebody" meant to say "$F$-algebra". Are you asking for an example of an integral extension that is not finite or for help with the proof of some version of what "somebody" said? $\endgroup$ – Rob Arthan Feb 22 '18 at 22:12
  • $\begingroup$ @user26857 There is no need to be condescending. I apologize for phrasing the question poorly. I meant to write "F-algebra" instead of "F-module." $\endgroup$ – weissss Feb 23 '18 at 19:36
  • $\begingroup$ I can state the question like this: Let $F$ be a field and let $E$ be a finitely-generated $F$-algebra integral over $F$. Is it true that $E$ is finitely-generated as an $F$-module? $\endgroup$ – weissss Feb 23 '18 at 19:45
  • $\begingroup$ @weissss Isn't the way you formulated the question.This is a result you find in every algebra textbook, but needs to open it. $\endgroup$ – user26857 Feb 23 '18 at 22:21

Let $\alpha \in E$. If $E$ is a finitely-generated $F$-module, then $E$ is a finite-dimensional vector space over $F$ and so is $F(\alpha)$. This implies that $\alpha$ is integral over $F$.


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