Is $\Bbb C$ a purely transcendental extension of a proper subfield? 
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?
 A: For $t\in S$, the polynomial $x^2+t$ has no root in $K(S)$ so it cannot be algebraically closed.
A: I have tried to fill up some details. Hope I got this correct.
Suppose $\mathbb{C}/K$ is purely transcendental, i.e. $\mathbb{C}=K(S)$ for some $K$-algebraically independent set $K \subseteq \mathbb{C}$. Let $t \in S$ be nonzero and consider the polynomial $X^2 - t \in K(S)[X]=\mathbb{C}[X]$. Since $\mathbb{C}$ is algebraically closed, the polynomial has a root $\alpha \in \mathbb{C}$.
Since we supposed $\mathbb{C}=K(S)$, we can write $$\alpha = \dfrac{P(s_1, \ldots, s_n)}{Q(t_1, \ldots , t_m)}$$ for some polynomial $P \in K[X_1, \ldots, X_n]$, $Q \in K[X_1, \ldots, X_m]$ and $s_1, \ldots, s_n, t_1, \ldots, t_m \in S$. Pluging this into $\alpha^2-t=0$, we get an algebraic relation of $s_1, \ldots, s_n, t_1, \ldots, t_m$, contradicting to the $K$-algebraically independence of $S$. Hence we are done.
So by exactly the same argument as above, it seems that we have the following result:

"Result": Let $E/F$ be a field extension, $E$ is algebraically closed. Then $E$ is not purely transcendental over any proper subfield $K$ (i.e. $F \subseteq K \subsetneq E$).

and as a corollary of this, we see

"Corollary" Let $E/F$ be a field extension, $S \subset E$ be an $F$-algebraically independent set. Then the purely transcendental extension $F(S)/F$ is never algebraically closed.

This corollary may lead to the discussion here and here.
Back to the argument above, we may also show that $\mathbb{R}$ is not purely transcendental over any proper subfield $K$. In fact, if $\mathbb{R}=K(S)$ for some $K$-algebraically independent set. Then we may pick $t \in K(S)$ to satisfy $t>0$. Then $X^2-t \in K(S)[X]=\mathbb{R}[X]$ does have a root $\alpha = \sqrt{t}$ in $\mathbb{R}$. The the same arguments as above shows that this contradicts to the algebraically independence of $S$ over $K$.
But for the $\mathbb{R}$-case above, we rely on the order structure on $\mathbb{R}$ and some other things, so it seems hard to generalize in this way.
Results above are quite broad but I haven't seen this on any textbook before. So I'm not quite certain on this "result". Hope I got this right.
