1. ## Extension

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

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

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

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

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

So consider $\text{irr}(\alpha_i, F)$ and show that $\sigma(a_i)$ must be one of the zeros in $\bar{F'}$?
I don't see why you need to look at "any subfield of F" or " $irrF(\alpha_i,F)$. Just use the definitions.

Any number in the extension field, $F(\alpha_1, \dots, \alpha_n)$ can be written as $a_1\alpha_1+ a_2\alpha_2+ \cdot\cdot\cdot+ a_n\alpha_n$ where $a_1, a_2, \cdot\cdot\cdot, a_n$ are members of F.

Since $\sigma$ is an "automorphism", it is, by definition, an isomorphism. In particular that means $\sigma(ax+ by)= \sigma(a)\sigma(x)+ \sigma(b)\sigma(y)$. Since it "leaves F fixed", $\sigma(a_i)= a_i$. Therefore $\sigma(x)= \sigma(a_1\alpha_1+ a_2\alpha_2+ \cdot\cdot\cdot+ a_n\alpha_n)= a_1\sigma(\alpha_1)+ a_2\sigma(\alpha_2)+ \cdot\cdot\cdot+ a_n\sigma(\alpha_n)$.