Show that a field which is algebraic over a finite field has no non-trivial valuations.
Due to lulu's comment I think I should clearly specify that I am working with an additive valuation.
Let $\Bbb F$ be a finite field (of characteristic say $p$) and $L$ be an algebraic extension over $\Bbb F$ .
I know that valuations correspond to valuation rings and hence it suffices to show that any proper subring of $L$ fails to be a valuation ring. Intuitively it seems to attack with a non-normal argument, but can't really see it coming.
Or maybe something like defining a valuation on $L$ and then restricting it onto $\Bbb F$ and
I am aware of the following informations: For any $\alpha \in L$ the minimal polynomial of $\alpha $ over $\Bbb F$ is of the form $X^{p^n}-X$ for some $n \in \Bbb N$ and also that $\Bbb F$ is never Algebraically closed.
Or maybe something like defining a valuation on $L$ and then restricting it onto $\Bbb F$ and then somehow using that $\Bbb F$ is not Algebraically closed?
Can I argue somehow with the above informations (or elsewise) to answer the question? Thanks in advance for help!
Take $\alpha \in L^*$. Then, since $\alpha$ is algebraic over a finite field, it must satisfy an equation of the form $x^{p^k-1}=1$. But then $$0=v(1)=v(\alpha^{p^k-1})=(p^k-1)v(\alpha)\implies v(\alpha)=0$$
Hence the valuation is trivial.