Atiyah-Macdonald: Exercise 1.8 
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?
 A: 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.
A: The neat way to argue this is to realise that $\mathscr{Spec}(A)$ is inductively ordered by the dual of the inclusion (so that we may apply Zorn's lemma). More generally, consider a subset $\mathscr{M} \subseteq \mathscr{Spec}(A)$ which is upward directed with respect to the dual of inclusion or equivalently downward directed with respect to inclusion itself. Our objective is to prove that $\mathscr{M}$ has an upper bound with respect to the dual of inclusion, which amounts to a lower bound with respect to inclusion itself.
Since $A$ is not a degenerate ring, $\mathscr{Spec}(A) \neq \varnothing$ is nonempty and any prime ideal serves as a lower bound in the particular case $\mathscr{M}=\varnothing$.
When $\mathscr{M} \neq \varnothing$, let us consider $P\colon=\displaystyle\bigcap\mathscr{M}$ and show it is a prime ideal. As it is the nonempty intersection of a collection of proper ideals, it must also be a proper ideal. Assume by contradiction that it were not prime, which would mean the existence of $a, b \in A \setminus P$ such that $ab \in P$. Since neither $a$ are nor $b$ are in the intersection of all members of $\mathscr{M}$, there must exist ideals $Q, R \in \mathscr{M}$ such that $a \notin Q$ and $b \notin R$. Since $\mathscr{M}$ is donward directed with respect to inclusion, there exists $T \in \mathscr{M}$ such that $T \subseteq Q, R$.
Since $ab \in P \subseteq T$ and $T$ is prime, we must have either $a \in T$ or $b \in T$ which lead either to $a \in Q$ or $b \in R$, both of which are contradictions.

Remark: I have not argued at this level of generality above, but the same claim remains valid for arbitrary non-degenerate rings (the assumption of commutativity is not required).
