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Is $\Bbb C$ a purely transcendental extension of a proper subfield $K$ (i.e. there is a set $S \subset \Bbb C$, algebraically independent over $K$, such that $\Bbb C = K(S)$)?

I don't think so, but I wasn't sure how to disprove it. I know that it is not true for $\Bbb R$, because (similar argument as here) $\Bbb R$ has a trivial field automorphism group but if $S$ is non-empty then $K(S) = K(X_i \mid i \in S)$ has many many field (and even $K$-algebra) automorphisms.

This argument doesn't work for $\Bbb C$, which has $2^{2^{\aleph_0}}$ field automorphisms! What would be a correct argument, then?

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For $t\in S$, the polynomial $x^2+t$ has no root in $K(S)$ so it cannot be algebraically closed.

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    That is just adjoining one element, he is allowing for an entire set. – 2017-02-03
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    @Watson If $\sqrt{s}$ is a rational function in the elements of $S$, then this $S$ is not algebraically independent anymore, so I still think my proof works. – 2017-02-03
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    OK, I see, I was just thinking too much about the case with $\Bbb R$. It was quite easy, indeed: if $S = \{s_0\} \sqcup S' \subset \Bbb C$ has at least one element (denoted by $s_0$), then $K(S) = K(S')(s_0) \cong L(X)$ where $L=K(S')$. But $L(X)$ is not algebraically closed, so it can't be $\Bbb C$. – 2017-02-03
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    @Watson Why is not L(X) algebraically closed here, by the way? It differs from C because it does not contain $s_0$, but I do not see how do you prove your stronger claim. – 2017-02-03
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    @heptagon : does [this](http://math.stackexchange.com/questions/1993393/) work? – 2017-02-03
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    @Watson Sorry I meant $K(S')$ instead of $L(X)$. Your comment was correct. – 2017-02-03
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    Maybe $K(S')=L$ is algebraically closed (not if $|S'| \geq 1$), but it doesn't matter. – 2017-02-03
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    @Watson Yes, I agree. Thanks! – 2017-02-03