Let $X$ be an integral scheme over a field $k$, and let $r\in K(X)^*$. This gives me the principal divisor $\mathrm{div}(r)$. What can I say about this divisor if $r$ is algebraic over $k$?
I suspect that in this case, $\mathrm{div}(r)$ is trivial, but I don't find a way to show this.
I ask this because in many proofs, one uses that either $\mathrm{div}(r)$ is trivial, or $r$ defines a non-constant function to $\mathbb A^1_k$ (or both), but I don't know how to show this in the case that $k$ is not algebraically closed.