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Let $k$ be an algebraically closed field and let $K/k$ be a field extension of infinite transcendence degree where $K$ is algebraically closed. Is it true that $\mathrm{Aut}_k(K)$ is a simple group?

The above question is motivated by work of Lascar. Lascar gives a proof of the above statement when $k=\overline{\mathbb{Q}}$ and $K=\mathbb{C}$ in his article The group of automorphisms of the field of complex numbers leaving fixed the algebraic numbers is simple.

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I suppose you want $k$ to be also algebraically closed, otherwise the subgroup of elements fixing $\overline k \subseteq K$ would be a proper normal subgroup. I don't have access to Lascar's paper, but from the title I would guess he takes $k = \overline{\mathbb Q}$. –  Angelo May 12 '13 at 11:43
@Angelo, thank you for noticing. I meant the case where $k$ is algebraically closed as well. –  Michiel Kosters May 12 '13 at 11:56
A MathSciNet search gives the article Automorphism groups of fields, and their representations by Rovinskii, according to which Lascar actually proves the result for any extension of algebraically closed fields of uncountable transcendence degree. –  François Brunault May 12 '13 at 18:11
I don't have access to Lascar's article neither, so I cannot check this. –  François Brunault May 12 '13 at 18:18

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