monomorphism on an algebraic field extension 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!
 A: 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.
A: 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.
A: Assume $a\in E$ is not hit by $\sigma$. Let $p\in K[x]$ be an irreducible polynomial such that $p(a)=0$ and let $a=a_1, a_2, \ldots, a_m$ be all the roots of $p$ in $E$. It is easy to see that $p(\sigma(a_i))=\sigma(p(a_i))=\sigma(0)=0$ for any $i$, and thus $\sigma$ carries $\{a_1,\ldots, a_m\}$ into $\{a_1,\ldots,a_m\}$. Moreover, $\sigma$ is clearly injective and so $a_1=\sigma(a_i)$ for some $i$ (because $m$ is finite), our desired contradiction.
