K is a field and f is an automorphism of K. We're also given that the order of f, as an element of the group Aut(K), is not finite. Is it always possible to find a field L which contains K and an automorphism of L, say g, such that g = f over K and the fixed field of g is algebraically closed in L?
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Making the fixed field algebraically closedK is a field and f is an automorphism of K. Is it always possible to find a field L which contains K and an automorphism of L, say g, such that g = f over K and the fixed field of g is algebraically closed in L?
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