Isomorphism of field extension

Question

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?

Answer 1

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 $Fix(\sigma)=F(\alpha_1,\ldots,\alpha_n)$.

Answer 2

An element of that field looks as $\;f(\alpha_1,...,\alpha_n)\;$ , with $\;f(x_1,...,x_n)\in F(x_1,...,x_n)\;$

Thus, we have that

$$f(\alpha_1,...,\alpha_n)=\sum_{i_1,...,i_n}a_{i_1,...,i_n}\alpha_1^{i_1}\cdot\ldots\cdot\alpha_n^{i_n}\implies$$

$$\sigma f(\alpha_1,...,\alpha_n)=\sum_{i_1,...,i_n}\sigma a_{i_1,...,i_n}\;\sigma\alpha_1^{i_1}\cdot\ldots\cdot\sigma\alpha_n^{i_n}=\sum_{i_1,...,i_n}a_{i_1,...,i_n}\alpha_1^{i_1}\cdot\ldots\cdot\alpha_n^{i_n}=f(\alpha_1,...,\alpha_n)$$

Answer 3

HINT:

The elements $r$ of the field $F(\alpha_1, \ldots , \alpha_n)$ can be written as $$r = \frac{P[\alpha_1, \ldots, \alpha_n]}{Q[\alpha_1, \ldots, \alpha_n] }$$ where $P[\alpha_1, \ldots, \alpha_n]$, $Q[\alpha_1, \ldots, \alpha_n]$ are polynomials in $\alpha_1$, $\ldots$ , $\alpha_n$ with coefficients in $F$ and $Q[\alpha_1, \ldots, \alpha_n]\ne 0$. Apply $\sigma$ to $r$.