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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?2012-04-07
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    See here: http://math.stackexchange.com/a/5285/43962012-04-08
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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.2012-04-08
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    What if $K$ is algebraically closed?2013-07-25

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