Currently, I'm trying to solve a problem from a textbook:
Let $R$ be a commutative Noetherian ring with identity, and let $I \subset R$ be a proper ideal of $R$. Then we know that set of prime ideals of $R$ containing $I$ has minimal elements by inclusion (I decided to call this set $\mathrm{Min}(I)$ in sequel). Prove that $\mathrm{Min}(I)$ is finite.
There is also a hint: Define $\mathcal{F}$ as set of all ideals $I$ of $R$ such that $ \vert \mathrm{Min}(I) \vert = \infty$. Assume that $\mathcal{F} \neq \emptyset$. Then it must have a maximal element $I$. Find ideals $J_1,J_2$ such that they all strictly include $I$, such that $J_1J_2 \subset I$ and deduce a contradiction.
So I went along this hint: $I$ can't be a prime, as a prime is the only minimal prime over itself. It means that $\exists a,b \not \in I: ab \in I$. As $R$ is Noetherian there is a finite list of elements that generates $I = (r_1, \dots, r_n)$. Then it's possible to set $J_1 = (r_1, \dots,r_n,a)$, $J_2 =(r_1, \dots, r_n, b )$ with all required properties. As $I$ is maximal in $\mathcal{F}$ the sets $\mathrm{Min}(J_1)$ and $\mathrm{Min}(J_2)$ must be finite.
I am failing to find a desired contradiction and will be grateful for any help.
Best Answer
(To continue your argument) But let $P\in Min(I)$, since $ab\in I\subset P$ and $P$ is prime, $a\in P$ or $b\in P$ this implies that $J_1\subset P$ or $J_2\subset P$. Remark that an element $P$ of $Min(I)$ which contains $J_l,l=1,2$ is in $Min(J_l)$ thus $Min(I)\subset Min(J_1)\cup Min(J_2)$. This implies that $Min(J_1)$ or $Min(J_2)$ is infinite. This is in the contradiction with the fact that $I$ is maximal among the ideals such that $Min(I)$ is infinite.