We have the rational function field $K(x)$ and the algebraically closed $K$. One of the statements in my lessons was:
Then every Divisor $D \in $Div$(K(x)|K)$ is a principal divisor, so there exists a rational function $z \in K(x)$ such as $(z) := \sum_{P \text{ place}}^{} v_p(z) \cdot (P)$ where $v_p(z)$ is the valuation which gives us the order of the zero or pole of the function $z$ and (P) is the prime divisor generated by the place $P$.
My objection is that we know that every algebraic function has at least one zero and one pole, so a prime divisor could never be a principal one. Why should this be possible only because of the fact we consider the rational function field?
We would appreciate every kind of advice! Thanks a lot!