Is the finite intersection of prime ideals radical?

Question

Does there exist a ring $R$ and finitely many prime ideals $P_i$ such that $\cap_{i = 1}^n P_i$ is not radical ideal?

In other words, is the finite intersection of prime ideals a radical ideal?

Answer 1

Consider a commutative unital ring $R$ and some prime ideals $P_1, \dots, P_n.$

Claim. We have that $I = P_1 \cap \cdots \cap P_n$ is radical, i.e., $\sqrt I = I.$

Proof. Considering that radicals distribute over intersections, we have that $$\sqrt I = \sqrt{P_1 \cap \cdots \cap P_n} = \sqrt{P_1} \cap \cdots \cap \sqrt{P_n}.$$ But prime ideals are radical, so $\sqrt{P_i} = P_i$ implies that $\sqrt I = I.$ QED.

Answer 2

Carlo has given a perfect answer. You can also argue directly as below. This, of course, just proves that the radical commutes with intersections, as Carlo said.

Suppose $a^{m} \in \bigcap_{i=1}^{n} P_i$ where $P_i$ are the prime ideals under consideration. Then $a^m \in P_i$ for each $1\leq i\leq n$. Since $P_i$ is a prime ideal, we get $a\in P_i$ for each $1\leq i\leq n$, and so $a\in \bigcap_{i=1}^{n} P_i$. We have shown that $a^{m} \in \bigcap_{i=1}^{n} P_i \ \Rightarrow a\in \bigcap_{i=1}^{n} P_i$. Thus, the intersection $\bigcap_{i=1}^{n} P_i$ is radical.

Note that we didn't use anywhere the fact that intersection has to be finite. Thus, intersection of arbitrary collection of prime ideals (or even radical ideals in general) must be radical.