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If we have a field $K$ such that $K\cong K(t)$ (i.e. it is isomorphic to the field you get if you adjoin one transcendental) then is there necessarily a subfield $L\lt K$ such that $L\ncong L(t)$ and $K$ is a purely transcendental extension of $L$?

The basic idea is that if we have something like $K={\mathbb Q}(u_i,t_i^{1/2^n})_{i,n\in\mathbb N}$ then we are looking for the field ${\mathbb Q}(t_i^{1/2^n})_{i,n\in{\mathbb N}}$.

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I don't know if this is a counterexample or not, but what would you do if $K$ is the field of fractions of the monoid ring $R = \mathbb{Z}[\prod_{i=1}^\infty \mathbb{N}]$? ($R$ is an integral domain because of the lexicographic ordering on the monoid, and $R[t] \cong \mathbb{Z}[\mathbb{N}\times \prod_{i=1}^\infty \mathbb{N}] \cong R$. Note $K$ is the same as the field of fractions of the group ring $\mathbb{Z}[\prod_{i=1}^\infty \mathbb{Z}]$. Baer proved that $\prod_{i=1}^\infty \mathbb{Z}$ is not free, but I don't know if this rules out the possibility of a transcendence basis for $K$ over $\mathbb{Q}$, let alone over any other $L$ satisfying $L\not\cong L(t)$.)

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