# Proving that there only finitely many minimal prime ideals of any ideal in Noetherian commutative ring

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.

(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.