0

Let $I \subset R$ be an ideal of a unital commutative ring $R$.

I have in my notes that TFAE:

  • $I$ equals the intersection of some prime ideals containing $I$.
  • $I$ equals the intersection of all prime ideals containing $I$.

Now, I think the direction from the second point to the first is obvious. I don´t think the other direction is as obvious. Any suggestions for how to prove this? The lecture notes said that this equivalence was obvious, and hence skipped the proof.

Any links to sources where a proof is given would make me happy.

Ben123
  • 952
  • 5
  • 10

1 Answers1

0

This is purely set-theoretical, which has little to do with ring theory or prime ideals.

If we have a family of sets $\{P_j\mid j\in J\}$, such that $I\subseteq P_j$ for all $j\in J$, then the following are equivalent:

  1. There is $J'\subseteq J$, such that $I=\cap_{j\in J'} P_j$
  2. $I = \cap_{j\in J} P_j$

Obvoiusly (2)$\Rightarrow$(1) is easy by taking $J'=J$.

As for the other direction, $I\subseteq P_j$ for all $j$ implies $I\subseteq\cap_{j\in J} P_j$, combined with (1), we have $$\cap_{j\in J}P_j\subseteq\cap_{j\in J'}P_j = I \subseteq \cap_{j\in J}P_j$$

So they are all equal.

Just a user
  • 14,899
  • Thanks for the answer. I don´t follow why one can say $\cap_{j \in J}P_j \subseteq \cap_{j \in J'} P_j$ in your last statement? – Ben123 Dec 19 '23 at 04:38
  • Intuitively the intersection of a bigger family is contained in the intersection of a smaller family. If you really have to, $\cap_{j\in J}P_j=(\cap_{j\in J'}P_j)\cap (\cap_{j\in J\setminus J'}P_j)\subseteq \cap_{j\in J'}P_j$. – Just a user Dec 19 '23 at 04:46
  • Oh, my bad. Yes. – Ben123 Dec 19 '23 at 04:47