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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.

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(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.

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