Let $R$ be a (commutative) ring not equal to $0$. I want to show that the set of prime ideals of $R$ has a minimal element w.r.t. inclusion.
This may be a wholeheartedly wrong attempt, but I thought I’d try this:
Let $P$ be the set of all prime ideals of $R$. We have chains of prime ideals with respect to inclusion, i.e.
$mathfrak{p}_1supset mathfrak{p}_2supsetldotssupset mathfrak{p}_nsupsetldots $
Now consider the chain $R/mathfrak{p}_1subset R/mathfrak{p}_2subsetldotssubset R$.
This has a maximal element, namely $R$, so if we let $U$ be the set ${R/mathfrak{p}_i : mathfrak{p}_i: text{is prime}}$, then every chain has a maximal element, so if we take the order-reversing chains, there must be a minimal element for each chain, hence $P$ has a minimal element w.r.t. inclusion.
I suspect this may be pretty dodgy logic, but I think I don’t have a full understanding of when best to apply the lemma.
