# If $\bar{k}\supseteq R\supseteq k$, then $R$ is a field? [duplicate]

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.

• How is the textbook implying that? It isn't true, so... – Chessanator Oct 1 '18 at 21:21
• @Chessanator It is true. – Saucy O'Path Oct 1 '18 at 21:23
• Whoops, yeah. Just ignore me. – Chessanator Oct 1 '18 at 21:24
• (by the way, the requirement that $R$ is a domain is superfluous) – Saucy O'Path Oct 1 '18 at 21:25
• See also this question, or this one for finite extensions. – Dietrich Burde Oct 1 '18 at 21:28

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 $$\beta\alpha=1$$.