Let $A$ be an $1$-dimensional Noether integral domain. Let $I$ be an nonzero ideal of $A$. Then, I want to prove the number of prime ideals of $A$ containing $I$ is finite.
I tried to prove with contradiction. Let $p_1, p_2, \dots$ be infinitely many prime ideals containing $I$. I couldn't proceed. I may use Krull dimension $1$, but in vain. Thank you in advance.
What we want to show is equivalent to showing that $A/I$ only has finitely many prime ideals. Because $(0) \subsetneq I$ and $(0)$ is a prime ideal (as $A$ is an integral domain), we have that $\mathrm{dim}(A/I) \leq 0$. Also, being a quotient of a Noetherian ring, $A/I$ is Noetherian. Thus $A/I$ is (at most) zero-dimensional and Noetherian, thus $A/I$ is Artinian and has only finitely many prime ideals.