Discrete valuation ring contains the finite ground field

Question

Let $F/k$ be an algebraic function field in one variable, where $k$ is finite. Then why any discrete valuation ring of $F$ contains $k$?

Answer 1

Every nonzero element $\zeta \in k$ is a root of unity of some order $n$. But for any valuation $v$,

$$ 0 = v(1) = v(\zeta^n) = n v(\zeta) $$

so $v(\zeta) = 0$, and therefore is in the corresponding valuation ring.