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Prove that there is no intermediate field $K$ with $\mathbb{Q}\subset K\subsetneq \mathbb{C}$ with $\mathbb{C}/K$ purely transcendental.

I guess that $K/\mathbb{Q}$ is algebraic. since $\mathbb{C}/K$ transcendental, then $\mathbb{C}=K(B)$ for some set $B$. Clearly, $\mathbb{C}/\mathbb{Q}$ is not algebraic. Is it true that $K/\mathbb{Q}$ is not algebraic? I really don't know how to do this.

Cille
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    https://math.stackexchange.com/questions/2127437/is-bbb-c-a-purely-transcendental-extension-of-a-proper-subfield –  Mar 19 '20 at 13:11
  • And possibly this: https://math.stackexchange.com/questions/1993393/kx-is-not-algebraically-closed –  Mar 19 '20 at 13:12
  • oh,yes.you are correct,we can show that $K(B)$ is not algebraic closed. – Cille Mar 19 '20 at 13:21
  • What’s your reason for saying $K$ should be algebraic over $\Bbb Q$ ? – Lubin Mar 19 '20 at 17:31
  • because this gives a example that an extension field E/K may not have an intermediate field K with K/k algebraic and E/K pure transcendental. – Cille Mar 22 '20 at 04:01

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