Prove that in a no trivial ring, the set of all prime ideal has minimal elements (ordered by subset inclusion).
I have thought that we can take as a lower bound of the set the ideal (0)={0} because it's the lower bound of any chain and applying Zorn's lemma we know there's a mininmal element.
I don't know if Zorn's lemma can be used in this way. Thanks.
Lemma (Zorn): Let $\left({X, \preceq}\right), X \ne \varnothing$ be a non-empty ordered set such that every non-empty chain in X has an upper bound (or lower bound) in $X$. Then $X$ has at least one maximal (minimal) element.
We first need to check if every chain has an lower bound. Let $\mathscr{P}$ the set of all prime ideals of our ring $R$. Let $\mathfrak{p}_i\in\mathscr{P}$, $i\in I$ such that $\mathfrak{p}_i \subset \mathfrak{p}_j$ if $i > j$. Then let us define $\mathfrak{p} = \cap_{i} \mathfrak{p}_i$.
Clearly, $\mathfrak{p}$ is a prime ideal (why ?), and constitutes a lower bound of the chain as $\mathfrak{p}$ is contained in every $\mathfrak{p}_i$. Therefore one can apply Zorn's Lemma, yielding us the existence of a minimal element in $\mathscr{P}$.
Edit : As mentioned in the comments below, Zorn's lemma holds if $X$ is a non-empty ordered set. It's actually a non-trivial result that $\mathscr{P}$ is indeed non-empty. To prove this property, one can prove (or admit) for instance that every ring $R\neq 0$ has at least one maximal ideal using Zorn's lemma.