If an ideal $I$ in a unital commutative ring equals the intersection of some prime ideals, then it equals the intersection of all prime ideals.

Question

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.

Answer 1

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.