Characterization of maximal multiplicatively closed subsets

Let $A$ be a non-zero ring, and let $\Sigma $ be the set of the multiplicatively closed subsets properly contained in $A$. Show that $\Sigma$ admits a maximal element respect to inclusion. Then prove that $S\in \Sigma$ is maximal if and only if $A\setminus S$ is a minimal prime ideal.

For the first task I'd use Zorn's lemma: if $S_1\subseteq S_2\subseteq\dots $ is an ascending chain in $\Sigma$, define $S$ as the set of finite products of elements belonging to some $S_n$. ​$S_n\subseteq S$ for all $n$, so any ascending chain has an upper bound. (Is this argument valid even if $S_1,S_2,\dots$ is just a subset of $\Sigma$, not totally ordered?). Since $\Sigma$ contains $\{1\}$ it is non-empty, and admits a maximal element.

Take $S\in\Sigma$. $S^{-1}A$ has a maximal ideal, whose contraction $p\subseteq A\setminus S$ is prime. So $T:=A\setminus p\in \Sigma$, and $S\subseteq T$; the equality holds only if $A\setminus p=S$. Also, if $q\subset p$ is another prime ideal, $S\subset A\setminus q$, proving that if $S$ is maximal, necessarily $A\setminus S$ is a minimal prime.

If $p$ is a minimal prime, set $S:=A\setminus p$. If $S$ was contained in a maximal $T\in \Sigma$, we could conclude that $A\setminus T\subseteq p$ is a prime ideal, and being $p$ minimal $S=T$. However I'm not sure about the sentence in italics: is it true? If not, I also thought of this way: if, again, $p$ is a minimal prime and $S:=A\setminus p$, any $S\subset T \in \Sigma$ meets all the prime ideals of $A$. This means that $T^{-1}A=0$, i.e. $0\in T$. I don't see if we can get a contradiction from here, anyway. Would you help me with this last part? Thanks

Solution 1:

First: the statement is false unless your definition of "multiplicatively closed subset" includes $0\notin S$. For example, in $\mathbb{Z}/(6)$, $\{0,1,2,3,4\}$ is a maximal proper multiplicatively closed subset, but its complement $\{5\}$ is not even an ideal.

So let's agree that we want to show that the set $\Sigma$ of multiplicatively closed sets which do not contain $0$ (the condition "properly contained in $A$" is now redundant) has maximal elements with respect to inclusion, and that $S$ is maximal if and only if $A\setminus S$ is a minimal prime ideal.

Second: In your attempted proof, you forgot to check that the set of finite products $S$ is proper. In fact, this is false in general.

To correct this, note that since the $S_n$ form a chain, we don't need to take finite products from the union: their union $S=\bigcup S_n$ is already multiplicatively closed. And while the union of a chain of proper multiplicatively closed sets may fail to be proper, the union of a chain of multiplicatively closed sets which do not contain $0$ also does not contain $0$.

This is why it's important that the $S_n$ form a chain. For example, in $\mathbb{Z}/(6)$ again, the smallest multiplicatively closed set containing $\{1,2,4\}$ and $\{1,3\}$ also contains $0$.

Third: Your proof about the equivalence with minimal primes is basically fine, including the sentence in italics. A slight strengthening of Zorn's Lemma is that any $S\in \Sigma$ is below a maximal element of $\Sigma$. This is just Zorn's Lemma applied to the poset $\Sigma_S =\{T\in \Sigma\mid S\leq T\}$. You should be familiar with this idea from the proof that every ideal is contained in a maximal ideal.