## Non-algebraically closed fixed fields

Posted: May 8, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
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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 $x \in F \cap K^{\tau} = F^{\sigma}.$