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Assume that $k\subset K$ are two algebraically closed fields with $k\neq K$. Can $K$ have finite transcendence degree over $k$? Equivalently, for an algebraically closed field $k$ can $k(x_1,...,x_n)$ be algebraically closed for some $n$? I believe that the answer should be no. At least for zero characteristic, there is no solution of the polynomial $t^2-x_1$ in $k(x_1,...,x_n)[t]$ which makes sense considering the theorem of Puiseux characterizing the algebraic closure of $\mathbb{C}(x)$. I guess even with prime characteristic same polynomial should work.

Edit : The question comes to my mind while I was reading Mumford's Complex Projective Varieties book. He shows a proof of Hilbert's Nullstellensatz where the field is $\mathbb{C}$ but the fact he uses is that the transcendence degree of $\mathbb{C}$ over $\mathbb{Q}$ (also hence over $\mathbb{\overline{Q}}$) is infinite. I just wondered to which fields it can be generalized and it seems like to any field except the algebraic closures of prime fields (if this question has a negative answer as I expect).

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    $\begingroup$ That sounds right $\endgroup$ – D_S Mar 7 '18 at 0:14
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Sure: every field has an algebraic closure. So even though $k(x_1,\dots,x_n)$ is not algebraically closed, it has an algebraic closure $K$ which has transcendence degree $n$ over $k$.

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  • $\begingroup$ Oh God. Stupid question is stupid. Thank you very much! $\endgroup$ – Levent Mar 7 '18 at 0:46

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