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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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Watson
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2 Answers2

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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.

heptagon
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  • That is just adjoining one element, he is allowing for an entire set. – Adam Hughes Feb 03 '17 at 14:11
  • @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. – heptagon Feb 03 '17 at 14:24
  • 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 $\Bbb C$. – Watson Feb 03 '17 at 14:35
  • @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. – heptagon Feb 03 '17 at 14:38
  • @heptagon : does this work? – Watson Feb 03 '17 at 14:42
  • @Watson Sorry I meant $K(S')$ instead of $L(X)$. Your comment was correct. – heptagon Feb 03 '17 at 14:47
  • Maybe $K(S')=L$ is algebraically closed (not if $|S'| \geq 1$), but it doesn't matter. – Watson Feb 03 '17 at 14:48
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    @Watson Yes, I agree. Thanks! – heptagon Feb 03 '17 at 14:49
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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.

Hetong Xu
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