I read a proof of the following theorem in "Basic abstract algebra" by Bhattacharya, Jain and Nagpaul and I thought that the proof looked overcomplicated. I have written my own proof of the theorem and would like to know if there are any mistakes, or if the proof seems legit.
Theorem:
Let $E$ be an algebraic extension of a field F, and let $\sigma : E \rightarrow E$ be a 1-1 homomorphism of $E$ into itself over $F$. Then $\sigma$ is onto and, hence, an automorphism of $E$.
Proof:
Let $a \in E$ and let $p(x)$ be its minimal polynomial over $F$. Now, $\sigma$ must map a root of $p(x)$ onto a root of $p(x)$. Let $\{a_i\}$ be the set of distinct roots of $p(x)$ that lie in $E$. Then since $\sigma$ is 1-1 the set $\{\sigma(a_i)\}$ of roots of $p(x)$ that lie in E must have the same size as $\{a_i\}$ and therefore $\{\sigma(a_i)\} = \{a_i\}$. This means that since $a \in \{a_i\}$ then $a \in \{\sigma(a_i)\} $. And therefore $ a \in \sigma(E)$ for all $a \in E$,and so $\sigma$ is onto.
Is this good enough? Are there any mistakes?