Proposition

Let $I_1, \cdots, I_n$ be ideals and $P$ a prime ideal containing $\cap_{i=1}^{n} I_i$. Then prove that $P \supset I_i$ for some $i$. Further, if $P = \cap_{i=1}^{n} I_i$, show that $P = I_i$ for some $i$.

Solution

Suppose $n = 2$. Then we want to show that $I_1 \cap I_2 \subset P$ implies $I_1 \subset P$ or $I_2 \subset P$. Suppose otherwise. Let $x \in I_1 \setminus P$ and $y \in I_2 \setminus P$. Then $xy \in I_1 \cap I_2$ because $I_1, I_2$ are ideals. This implies that $xy \in P$ but $x \notin P$ and $y \notin P$. This is a contradiction.

Suppose that we have proved the proposition for $n \geq 2$. We will prove the case with $n + 1$ ideals.

Then $\cap_{i=1}^{n + 1} = (\cap_{i=1}^{n} I_i) \cap I_{n + 1}$. Since the intersection of $n$ ideals is also an ideal, $\cap_{i=1}^{n} I_i \subset P$ or $I_{n + 1} \subset P$ by the argument for $n = 2$. If $I_{n + 1} \subset P$, we are done. If $\cap_{i=1}^{n} I_i \subset P$, then by the inductive hypothesis, $I_i \subset P$ for some $1 \leq i \leq n$.

By mathematical induction, the proposition is true for all $n \geq 2$.

Suppose $P = \cap_{i=1}^{n} I_i$. Then by the argument above, $I_{i_0} \subset P$ for some ${i_0}$. Since $P \subset \cap_{i=1}^{n} I_i$, $P \subset I_{i_0}$. Therefore, $P = I_{i_0}$.