Maximal multiplicative set implies complement is a prime ideal

Let $R$ be a commutative ring with identity. I've been trying to prove the following:

If $M \subset R$, i.e. $M \neq R$, is a maximal multiplicative set, then $R \setminus M$ is a prime ideal of $R$.

I have spent some time in it, but nothing is coming to my mind. Any idea/solution will be appreciated.

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I think you should be clear here that you want all your multiplicative subsets to be disjoint from $\{0\}$ (otherwise in my argument $S^{-1} R$ is the zero ring, and in general things don't behave very well). – Pete L. Clark Mar 8 '11 at 8:30
I understand that a set is multiplicative if it is subset of $R$ that is closed under multiplication and contains the identity. Does this implies that it does not contains 0? (I don't see it) Or is it something that you have to add to the definition for things to work? – 4math Mar 8 '11 at 9:15
no, a multiplicative subset can contain zero, and as far as the general definition goes you do want to allow this: in particular for any element $f \in R \setminus \{0\}$ you want to be able to localize at the multiplicative subset generated by $f$, and if $f$ is nilpotent, this contains zero.  But in practice the case of $0 \in S$ is often excluded. To see that your assertion is not true unless zero is excluded, consider $\mathbb{Z} \setminus \{-1\}$: this is a maximal proper multiplicative subset of $\mathbb{Z}$ and it is not of the form you want. – Pete L. Clark Mar 8 '11 at 13:55

3 Answers

I assume that all multiplicative subsets discussed here are disjoint from $\{0\}$.

Let me write $S$ for the maximal multiplicative subset of $R$.

Step 1: I claim that $S^{-1} R$ is a local ring.

Indeed, if not, there would be at least two maximal ideals in the localization, i.e., two prime ideals $\mathfrak{p}_1 \neq \mathfrak{p}_2$, each maximal with respect to being disjoint from $S$. Thus $S$ is contained in both $R \setminus \mathfrak{p}_1$ and $R \setminus \mathfrak{p}_2$, but these are two different sets so at least one of the containments must be proper, contradicting the maximality of $S$.

Step 2: Therefore there is a unique maximal ideal in the localization, which corresponds to a prime ideal $\mathfrak{p}$ of $R$ which is disjoint from $S$. Arguing as above, maximality of $S$ implies $S = R \setminus \mathfrak{p}$.

Remark 1: When $R$ is a domain, the unique maximal multiplicative subset is obviously $R \setminus \{0\}$. In this case it is tempting to construe "maximal multiplicative subset" to mean "multiplicative subset maximal with respect to being properly contained in $R \setminus \{0\}$."

Remark 2: Conversely, the primes $\mathfrak{p}$ such that $R \setminus \mathfrak{p}$ is maximal are precisely the primes which are minimal with respect to containing $\{0\}$. (When $R$ is a domain and the question is reconstrued as above, we want $\mathfrak{p}$ to be minimal with respect to properly containing $\{0\}$.)

Mariano's answer makes this clear as well.

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Very instructive answer. I had come up to something similar, but I hadn't realized your step 1. Thank you! – 4math Mar 8 '11 at 8:19
@PeteL.Clark +1 Very nice answer. I was stuck on a similar problem in Atiyah - Macdonald. – BenjaLim Apr 25 '12 at 23:30

Another way to proceed is to show

• First, that every multiplicative subset $S$ of a ring is contained in a saturated multiplicative subset $S'$, that is, one such that $$ab\in S'\implies a\in S'\wedge b\in S'.$$ In fact, there is a smallest saturated multiplicative subset containing $S$, called the saturation.

• Second, that a saturated multiplicative subset $S$ of a ring (which does not contain $0$) is the intersection of the complements of the prime ideals which are disjoint from it. (The hard part here is to show that such primes actually exist: you can construct them as the maximal elements in the family of ideals disjoint from $S$)

Once you have these two facts, your claim follows easily.

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+1: I like this argument better than mine. – Pete L. Clark Mar 8 '11 at 8:20
+1 @MarianoSuárez-Alvarez Very nice answer. – BenjaLim Apr 26 '12 at 0:00

We must assume $\rm 0\not\in M$ since $\rm\: 0\in M\: \Rightarrow\: 0 \not\in M' := R\setminus M\:,\:$ so $\rm\:M'\:$ is not an ideal. A simple Zorn lemma argument (see below) shows that, since $\rm\:M\:$ is multiplicatively closed, the ideal $\rm\{0\}\subset M'\:$ can be enlarged to an ideal $\rm\:P\:$ maximal w.r.t. to exclusion of $\rm\:M\:,\:$ and such an ideal must be prime. Hence $\rm\: P = M'\:,\:$ else we could enlarge $\rm\:M\:$ to the monoid $\rm\:P'\:,\:$ contra maximality of $\rm\:M\:.\quad$ QED

Note $\:$ The prime $\rm\:P\:$ above may be alternatively constructed as the contraction of a maximal ideal $\rm\:Q\:$ of the localization $\rm\: R_M = M^{-1} R\:.\:$ $\rm\:Q\:$ exists since $\rm\ 0\not\in M\ \Rightarrow\ R_M \ne \{0\}\:.\:$ Generally, there is a bijective order-preserving correspondence between all prime-ideals in $\rm\:R_M\:$ and all prime ideals in $\rm R$ disjoint from $\rm\:M\:,\:$ see Theorem 34 in Kaplansky's Commutative Rings. His Theorem 1, pp. 1-2, is the above-invoked form of this result employing no localization theory. I've appended it below.

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