Non-algebraically closed fixed fields

Posted: May 8, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

Problem.  Suppose that F is a field, \sigma \in \text{Aut}(F) and F^{\sigma} is not algebraically closed in F, prove that there are no field extension K of F and \tau \in \text{Aut}(K)  with \tau|_F = \sigma such that K^{\tau} is algebraically closed in K.

Solution. By contradiction: First note that F \cap K^{\tau}=F^{\sigma}. Now choose x \in F \setminus F^{\sigma} such that x is algebraic over F^{\sigma}. Then clearly x \in K and x would also be algebraic over K^{\tau}. Therefore x \in K^{\tau}, because K^{\tau} is assumed to be algebraically closed in K.  That’s the contradiction we need because we just proved that x \in F \cap K^{\tau} = F^{\sigma}.

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