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 $Min(I)$ in sequel). Prove that $Min(I)$ is finite.

There is also a hint : Define $\mathcal{F}$ as set of all Ideals $I$ of $R$ such that $ \vert 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 the contradiction.

So I went along this hint: As $I$ can't be a prime as a prime is the only minimal primes for itself. It means that $\exists a,b \not \in I : ab \in I$. As $R$ is Noetherian there is a finite list of its elements $r$ that generates $I = ( r_1, \ldots, r_n)$ . Then it's possible to set $J_1 = (r_1, \ldots,r_n,a)$, $J_2 =(r_1, \ldots, r_n, b )$ with all required properties. As $I$ is maximal in $\mathcal{F}$ sets $Min(J_1)$ and $Min(J_2)$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.