Simple question from Kunz's book Introduction to Plane Algebraic Curves I'd like to ask a question that arose when I was reading the book Introduction to Plane Algebraic Curves by Kunz. In the following passage, how can we conclude that ker$\alpha = (q)$? I can see that this would follow if, for example, ker$\alpha$ is prime and $K[X,Y]$ were a PID, but this is not the case.
Thanks!


First $P = \ker \alpha$ is a prime ideal, as $K(T)$ is an integral domain. Second $P$ is a prime ideal of height $1$. Namely, if $P$ would be of height $2$, therefore maximal, then $K[X,Y]/P$ is an algebraic extension of $K$. So $\mathrm{im} \alpha$ would be an algebraic extension of $K$, which is obviously not the case. Now $K[X,Y]$ is a UFD and therefore every prime of height $1$ is principal. As $(q) \subseteq P$ is a prime of height $1$ in $P$ it must be $P = (q)$.
