Take the 2-minute tour ×  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.  share|improve this question 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 parti