I want to prove that every nonempty set of prime ideals contain a minimal element.
My attempt is to prove it by using Zorn's lemma and i would like to know if my proof is valid.
Let $\Sigma$ be a nonempty set of prime ideals. It's a partially ordered set with the partial order $a \leq b$ which means $b \subseteq a$ (note the inclusion). I will now prove that any totally ordered subset $S$ of $\Sigma$ has an upper bound.
So let $S$ be such subset and consider $J^{\star} = \underset{J \in S}{\cap}J$, since $J^{\star}$ consist of one single element from $S$ (is this true? im not sure), we have $J^{\star} \in \Sigma$ and for every $s \in S$ we have $s \leq J^{\star}$ since $J^{\star} \subset s$ (so $J^{\star}$ is our upper bound). Thus $\Sigma$ contain a maximal element by zorns lemma, which according to our partial order is of course a minimal element.
Is this proof valid?
Best Answer
Your proof is not quite correct: $J^*$ may not be an element of the set $\Sigma$ under consideration (although $J^*$ is a prime ideal, since $S$ is totally ordered).
Indeed, the claim is false: there exist rings with prime ideals of infinite height. If $p$ is such a prime ideal, then there exist prime ideals $p \supsetneq p_1 \supsetneq p_2 \supsetneq \ldots$, so $\{p, p_1, p_2, \ldots\}$ has no minimal element.