MCS meet all prime ideals 
let A be a commutative ring, is there any multiplicatively closed subset S (not containning 0), s.t. every prime ideal in A intercept S is not empty?

My thinking is that there is 1-1 corresponding between prime ideals in  A that not meet S, and prime Ideals in $S^{-1}(A)$, which imply there is no prime ideals in  $S^{-1}(A)$, and then can we conclude $S^{-1}(A)$ = $0$ so S contains $0$?
 A: Let $R$ be a commutative ring with unity $1 \neq 0$. Then every proper ideal is contained in a maximal ideal. In particular, starting with the zero ideal we get that the ring $R$ has at least one prime ideal (may be the ideal (0) itself).
Now in this case, $A$ has identity $1 \neq 0$ (we can assume it because if $1 = 0$ then the ring is the zero ring and in this particular case we have nothing to prove!). Let $S$ be a multiplicatively closed subset of $A$ not containing $0.$ Suppose $S \cap \mathfrak{p}$ is non-empty for every prime ideal $\mathfrak{p}$ of $A$. Since the prime ideals of $S^{-1}A$ are in one to one correspondence ($\mathfrak{p} \leftrightarrow S^{-1}\mathfrak{p}$) with the prime ideals of $A$ which does not meet $S$, we get that there is no prime ideal in the ring $S^{-1}A.$ But this is not possible because $S^{-1}A$ is commutative ring with identity $1 \neq 0.$ The identity element of $S^{-1}A$ is just the image of the identity element of $A$ under the natural map $A \rightarrow S^{-1}A, a \mapsto \frac{a}{1}.$
(This is nothing but an explanation of what you wrote above.)
A: If I'm reading this right, you're looking for an MCS which every prime ideal intersects the MCS nontrivially.
There is a very famous lemma that shows the answer is negative.

Lemma: If $M$ is a multiplicatively closed (not containing $0$) subset of a commutative ring $R$ (with identity), then there is an ideal of $R$ maximal with respect to being disjoint from $M$, and this ideal is prime.

It is, for example, theorem #1 on page 1 of Kaplansky's Commutative rings. It's a very easy argument anyway. The poset of ideals disjoint from $M$ is nonempty (because $\{0\}$ is disjoint from $M$) and then Zorn's lemma can be applied to find a maximal element of the poset.
So given any MCS, you can find a prime ideal disjoint from it.
