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**Sampras** Let $\displaystyle F(\alpha_1, \dots, \alpha_n) $ be an extension field of $\displaystyle F $. Show that any automorphism $\displaystyle \sigma $ of $\displaystyle F(\alpha_1, \dots, \alpha_n) $ leaving $\displaystyle F $ fixed is completely determined by the $\displaystyle n $ values of $\displaystyle \sigma(a_i) $.

So suppose we have some isomorphism $\displaystyle \sigma_1: F \to F' $. We can extend this to isomorphism to $\displaystyle \sigma: F(\alpha_1, \dots, \alpha_n) \to L $ where $\displaystyle L $ is some subfield of $\displaystyle \bar{F'} $.

So consider $\displaystyle \text{irr}(\alpha_i, F) $ and show that $\displaystyle \sigma(a_i) $ must be one of the zeros in $\displaystyle \bar{F'} $?