Q: Let $A$ be a nonzero ring and let $\Sigma$ be the set of multiplicatively closed subsets of $A$ such that $0 \notin S.$ Show that $\Sigma$ has a maximal element, and that $S \in \Sigma$ is maximal iff $A \setminus S$ is a minimal prime ideal of $A$.
It can be shown using Zorn's Lemma $\Sigma$ has a maximal element, be showing every chain has an upper-bound. (for this $0 \notin S$ condition is not required.)
$\underline{Claim}$: $S$ maximal in $\Sigma$ iff $A \setminus S$ is a minimal prime ideal. I proceeded as follows:
$"\impliedby"$
As $A\setminus S$ is a prime ideal, $S \in \Sigma$.
Now how do I show that $S$ is maximal?
$"\implies"$
$S$: a maximal element in $\Sigma$.
-
$0 \in A\setminus S$, and if $xy \in A\setminus S$ then either $(x \in A
\setminus S)$ or $(y \in A \setminus S)$ as $S$ is multiplicatively closed. -
Also, $A \setminus S$ does not contain any prime ideal properly.
If it does, let $\mathfrak{p}$ be one such prime ideal. Then $A \setminus
\mathfrak{p} \in \Sigma$
containing S properly, a contradiction.
It remains to show is $A \setminus S$ is a subgroup of the additive group $A.$ How can I show this?
Edit
Using @SteveD's comment I have completed the above proof, which I have posted as an answer. But the proof is a bit set-theoretic. I would like to see a proof using tools from Commutative Algebra.
Is there any nice application of this result?
Best Answer
Here is a slightly different solution which is also less set-theoretic.
Arbitrary unions of elements of $\Sigma$ are again multiplicatively closed subsets which do not contain 0 and so Zorn's lemma tells us that $\Sigma$ has maximal elements.
Suppose that $S$ is a maximal element of $\Sigma,$ and consider the fact that the ring of fractions $S^{-1}A$ is in particular a ring. Therefore it has at least one maximal ideal, which corresponds to a prime ideal $\mathfrak{p} \subset A\backslash S$. But then $A \backslash \mathfrak{p}$ is a multiplicative system in $A$ and it contains $S$ which was assumed to be maximal so that $A \backslash \mathfrak{p} = S \implies \mathfrak{p} = A \backslash S$. Furthermore, $\mathfrak{p}$ cannot contain any smaller prime ideals for a similar reason and so $\mathfrak{p}$ is minimal.
Conversely, the complement of a minimal prime ideal $\mathfrak{p}$ is a multiplicative system, which must then lie inside a maximal multiplicative system. Then the complement of the maximal multiplicative system would be a prime ideal contained in $\mathfrak{p}$ which was assumed to be minimal and so $A \backslash \mathfrak{p}$ is a maximal multiplicative system.