monomorphism on an algebraic field extension

Question

let $E|K$ be an algebraic extension and $\phi:E\rightarrow E$ a $K $-algebra monomorphism,prove that $\phi$ is onto.

i assume $\alpha\in E-\phi(E)$ to make a contradiction and i assume $f(x)$ to be the minimal polynomial of $\alpha$ on $K$,the degree of $f$ is more than $1$,how can i continue?

any hints are welcomed!

Answer 1

Field homomorphisms are always injective; there is no need to specify that $\phi$ is a monomorphism.

If $\alpha \in E$ and $\alpha=\alpha_1,\dotsc,\alpha_n$ are the conjugates of $\alpha$ (i.e. the roots of the minimal polynomial of $\alpha$), then $\phi$ induces a map $\{\alpha_1,\dotsc,\alpha_n\} \to \{\alpha_1,\dotsc,\alpha_n\}$ (why?) which is in fact injective (since $\phi$ is injective). It follows (why?) that the map is surjective and in particular that $\alpha$ has a preimage.

Answer 2

This homework is given in order that you may work on it and learn. Avoid Internet postings .. it deprives you of a self-learning chance.