Let $K \subset L$ be an extension of fields of transcendence degree at least $1$. Does there always exist a subfield $K \subset M \subset L$ such that $M/K$ is algebraic and $L/M$ is purely transcendental? If this is not true in general, can some extra conditions on $K$ and $L$ make this work - for example the condition that $L/K$ is finitely generated?
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THere has to be soemthing more here, right? I mean, if $L/K$ is finite then this can never happen for obvious reasons? – Alex Youcis Apr 07 '12 at 22:11
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1See here: http://math.stackexchange.com/a/5285/4396 – wxu Apr 08 '12 at 00:58
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Suppose that $K$ is the algebraic numbers in $\mathbb{C}$, and let $L$ be the algebraic closure of $K(\pi)$. Then $L/K$ has transcendence degree $1$, but surely no such field $M$ exists. – Jim Belk Apr 08 '12 at 04:45
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1What if $K$ is algebraically closed? – Gaston Burrull Jul 25 '13 at 20:17
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Gastón Burrull's comment contains the important idea: if $K$ is algebraically closed, you are asking whether every extension of $K$ is purely transcendental, which is clearly false. For example, when $K = \mathbb{C}$ the fraction field of $\mathbb{C}[x, y]/(y^2 - x^3 - x)$ cannot be purely transcendental because it is the function field of an elliptic curve (which is not birationally equivalent to $\mathbb{P}^1$).
Qiaochu Yuan
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2Your example is the same as the one given here and pointed out by wxu's comment above. – Jul 25 '13 at 21:02