Form of an element in an algebraic field extension

Question

Let $e_1,...,e_n$ be algebraic elements over a prime field $F$. Since $F$ is a prime field, we know that the automorphism group of $F$ is trivial. Let $K = F(e_1,...,e_n)$ and consider the extension $F \subset K$. Is it true that every $F$-automorphism of $K$ is determined by the images of the $e_i$? I can also reformulate the question in the following sense: Can every element of an $F$-basis of $K$ be expressed in terms of the $e_i$?

I personally believe it is true due to all the examples i tried out, but cannot seem to find a good formal argument.

Answer 1

Using the fact that the $e_i$ are algebraic over $F$, it is easy to show that $K=F[e_1,\dots,e_n]$, i.e. $K$ coincides with the smallest sub-$F$-algebra of $K$ containing the $e_i$. Now this shows that we have $K\cong F[X_1,\dots,X_n]/\mathfrak m$ for some maximal ideal $\mathfrak m$. In particular we see that any automorphism of $K$ is determined by the image of the $e_i$ since any automorphism of the above quotient comes from a map $F[X_1,\dots,X_n]\to K$ subject to the condition that $\mathfrak m$ is the kernel of that map.