Fields generated by a collection of elements Question: Is it true that every field generated by a collection of elements is isomorphic to the field of rational polynomials in those elements?  
Here's the context. I've been trying to think about how to show that the following statement is true:
If a field K is generated over a field F by a collection of elements, then any automorphism of K is determined by its action on the generators.
What is bothering me is the case where K is not an algebraic extension since there isn't a neat description of K as F-linear combinations of the powers of the generators.  I don't see how I can show this fact is true without knowing what the field K looks like.  If the question I am asking is true, then this will clearly solve this problem for me.   
 A: The statement you made is not very clear, but here is what is true and what you probably intended.  Let us suppose $K$ is a field with a subfield $F$ and a subset $S$ such that $K$ is generated by $S$ as a field extension of by $F$.  Let $L$ be the set of elements of $K$ that can be written as rational functions in elements of $S$ with coefficients in $F$.  Then I claim $L=K$.  To prove this, note that $L$ is a subfield of $K$ (a sum of two rational functions can be combined as another rational function, and similarly for products and inverses), and $L$ contains $F$ and $S$.  By definition, since $S$ generates $K$ over $F$, this means that $L$ is all of $K$.
Note that this result is entirely unnecessary in order to prove the fact you are trying to prove, since you can just directly use similar reasoning.  If $f,g:K\to K$ are automorphisms which fix $F$ and such that $f(s)=g(s)$ for each $s\in S$, consider the set $L=\{x\in K:f(x)=g(x)\}$.  It is easy to see that $L$ is a subfield of $K$, and by assumption it contains $F$ and $S$.  Since $S$ generates $K$ over $F$, $L$ is therefore all of $K$, and so $f=g$.
