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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!

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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 $P = (q)$.

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  • $\begingroup$ Thanks for your reply. How does $P=$ker$\alpha$ follow from $K(T)$ being an integral domain? $\endgroup$ – ougoah Oct 8 '15 at 10:55
  • $\begingroup$ I assume you mean to ask why $P$ is prime. Now if you have a ring map $f:A \to B$ then $A/kerf$ can be considered a subring of $B$. If $B$ is integral, then the subring $A/\ker f$ is integral too. But an ideal $I \subseteq A$ is prime exactly when $A/I$ is integral. $\endgroup$ – Jürgen Böhm Oct 8 '15 at 13:58
  • $\begingroup$ That was indeed my question. Thanks, I understand now. $\endgroup$ – ougoah Oct 8 '15 at 22:51
  • $\begingroup$ Very nice answer, although I can't help but feel that there is a shorter explanation, given the way the author stated the result. $\endgroup$ – ougoah Oct 10 '15 at 5:22

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