The intersection of two minimal prime ideals. 
Let $A$ be a reduced commutative ring (that is, $A$ has no nontrivial nilpotents) and $P_1$, $P_2$ two minimal prime ideals of $A$. Is it true that the intersection of $P_1$ and $P_2$ is zero?

It seems that this has confused some people so let me be clear: a minimal prime ideal is a prime ideal $P$ such that there is no other prime ideal $Q$ strictly contained in $P$. Therefore, in an integral domain the only minimal prime ideal is $(0)$.
I ask this question because of one of the problems in Reid's Book 'Undergraduate Commutative Algebra'. 

 A: Consider $R = \mathbb{C}[x] / ((x-1)(x-2)(x-3))$. The prime ideals in this ring are (the images under the projection $\mathbb{C}[x] \rightarrow R$ of) $(x-1)$, $(x-2)$ and $(x-3)$. But $(x-1)(x-2) \in (x-1) \cap (x-2)$ and $(x-1)(x-2) \neq 0$.
I'm think, though can't remember proof or reference off the top of my head, that the intersection of all the minimal prime ideals must be $0$, but the pairwise intersection of any two doesn't.
A: 
If $P_1, \ldots, P_n$ are finitely many minimal primes such that $\bigcap_{i=0}^n P_i = 0$, then the minimal primes of $R$ are exactly the $P_i$.  

To see this, let $Q$ be a minimal prime of $R$ and note that $P_1 \cdots P_n \subseteq \bigcap_{i=0}^n P_i = 0 \subseteq Q$. 
So by the definition of a prime ideal, $P_k \subseteq Q$ for some $k$ and by minimality, $Q = P_k$.
$\square$
Thus any ring with an infinite number of minimal prime ideals is such that no finite collection of prime ideals intersects to $0$.  
The simplest way to construct such rings is to take an infinite product of fields.
