TITLE: Atiyah-Macdonald: Exercise 1.8
QUESTION [5 upvotes]: Let $A$ be a ring $\neq0$. Show that the set of prime ideals of $A$ has minimal elements with respect to inclusion.

I am trying to do this exercise from Atiyah-Macdonald.
Attempt:
We should assume that there is no such minimal prime ideal. Then we have a chain $P_1 \supset P_2 \supset P_3 \supset ...$. Then we should set $P= \bigcap_i P_i$, This would be a minimal element but I can't see why it should be prime?

REPLY [5 votes]: Hint: Prove it by contraposition: if neither $x$ nor $y$ belongs to $\mathfrak p$, then $xy$  is not in  $\mathfrak p$.
You'll have to show first that, with this hypothesis on $x$ and $y$, there exists a prime ideal $\mathfrak p_i$ in the chain which contains neither $x$ nor $y$.
A last remark: to show the existence of minimal prime ideals, you should consider a totally ordered (by inclusion) family of prime ideals, not a mere sequence. You have no reason to suppose this family is countable.