I was trying to prove a theorem in algebraic geometry which basically held if and only if this lemma held. Here’s the lemma:

**Lemma: **Given any ring $A$, a prime ideal $ \mathfrak{p} \subset A$, and a finite collection of ideals $I_j,$ where $j \in \{1, 2, … , n\}$, then if $I$ is the intersection of the ideals, then $I \subset \mathfrak{p}$ implies that $I_j \subset \mathfrak{p}$ for some $j \in \{1, 2, … , n\}$.

A couple of things before we give the proof—the first being that if you have experience in ring theory, half the fun is to try this yourself! It’s a pretty fun thing. Note that the lemma holds trivially true for principal ideal domains by the definition of prime ideals. Moreover, you can use this lemma in scheme theory to show that if you take the finite union of “closed subschemes” in a scheme using a more “scheme/ring theoretic” definition of a closed subscheme is topologically the finite union of the subsets. (Chapter 8 of Ravi Vakil’s excellent, free lecture notes give more information on this topic!)

**Proof of Lemma:** We prove this lemma by induction, noting that the base case is trivial. For the inductive step, by clever use of parenthesis, we may also reduce to the case that there are two ideals, i.e. $n = 2$. Swapping if necessary, assume that $I_1 \not\subset \mathfrak{p}$. Then there exists some $f \in I_1$ such that $f \notin \mathfrak{p}$. Let $g$ denote an arbitrary element of $I_2$. Then $fg \in I_1 \cap I_2 \subset \mathfrak{p}$ by assumption, so either $f \in \mathfrak{p}$ or $g \in \mathfrak{p}$ by the definition of prime ideal. But we assumed that the former didn’t hold, so we conclude $g \in \mathfrak{p}$, that is, any arbitrary element of $I_2$ is in $\mathfrak{p}$. Thus $I_2 \subset \mathfrak{p}$.

Pretty cool right?! For a fun bonus problem, you can check if this holds if $n$ is allowed to take on infinite values. For a really fun bonus problem, you can attempt to prove the Birch and Swinnerton-Dyer conjecture. Best of luck on both!

Also, a proof by contrapositive:

Suppose for each $j$ there exists $a_jin I_jsetminus mathfrak{p}$. Let $a = a_1…a_n$. Then $ain I$, but $anotin mathfrak{p}$.

A useful lemma. This is the second half of Proposition 1.11 in Atiyah-MacDonald.

Simple counterexample to the infinite case:

$A = mathbb{Z}$, $mathfrak{p} = 2mathbb{Z}$, $I_j = p_j mathbb{Z}, j > 1$.

The intersection of the $I_j$ is ${ 0 } subset mathfrak{p}$, but no $I_j$ is a subset of the initial prime ideal.