Isomorphism of field extension 
Let $F$ be a field. I need to prove that if $\sigma$ is an isomorphism of $F(\alpha_1,...,\alpha_n)$ with itself such that $\sigma|_F = id_F$ and $\sigma(\alpha_i)=\alpha_i$ for $i=1,...n$, then $\sigma$ is the identity.

I've tried many things but nothing worked. The main problem is that I do not have a nice description of the elements in $F(\alpha_1,...,\alpha_n)$... Any help?

Let $Fix(\sigma):=\{x\in F(\alpha_1,\ldots,\alpha_n)|\sigma(x)=x\}$. It follows straight from the definition that $Fix(\sigma)$ is a field. By assumption, $F\subset Fix(\sigma)$ and for all $i$ $\alpha_i\in Fix(\sigma)$, hence $Fix(\sigma)=F(\alpha_1,\ldots,\alpha_n)$.
