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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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    What if $K$ is algebraically closed?2013-07-25

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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 $\mathbb{P}^1$).

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    Your example is the same as the one given [here](http://math.stackexchange.com/questions/5278/why-is-mathbbqt-sqrtt3-t-not-a-purely-transcendental-extension-of-m) and pointed out by wxu's comment above.2013-07-25