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Say $R$ is an integral domain and $k$ is a field.

Is it true that if $\bar{k}\supseteq R\supseteq k$, then $R$ is a field?

I'm not sure how to show this immediately, and it seems to be implied in the textbook, or else they are using some other facts not listed.

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  • $\begingroup$ How is the textbook implying that? It isn't true, so... $\endgroup$ – Chessanator Oct 1 '18 at 21:21
  • $\begingroup$ @Chessanator It is true. $\endgroup$ – Saucy O'Path Oct 1 '18 at 21:23
  • $\begingroup$ Whoops, yeah. Just ignore me. $\endgroup$ – Chessanator Oct 1 '18 at 21:24
  • $\begingroup$ (by the way, the requirement that $R$ is a domain is superfluous) $\endgroup$ – Saucy O'Path Oct 1 '18 at 21:25
  • $\begingroup$ See also this question, or this one for finite extensions. $\endgroup$ – Dietrich Burde Oct 1 '18 at 21:28
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Yes, it is true. Let $\alpha$ be any element of $R$. Consider the homomorphism $\nu_\alpha:k[x]\to R$, $\nu_\alpha(p)=p(\alpha)$. Since $\alpha$ is algebraic over $k$, $\ker\nu_\alpha$ contains a non-zero polynomial. $\ker\nu_\alpha$ must therefore be a non-zero prime ideal of $k[x]$. Since $k[x]$ is PID, non-zero pime ideals are maximal. So $\operatorname{im}\nu_\alpha$ is a field: in other words, there is some $\beta\in\operatorname{im}\nu_\alpha$ such that $\beta\alpha=1$.

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  • $\begingroup$ Very nice, thank you. $\endgroup$ – Fredrieck Oct 1 '18 at 21:46

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