Here is the question that I want to answer:
Let $R$ be an integral domain with field of fractions $K,$ and let $S \subset R \setminus \{0\}$ be a multiplicatively closed subset.
Show that if $\mathfrak{p} \subset R$ is a prime ideal, then $S = R \setminus \mathfrak{p}$ is multiplicatively closed.
My thoughts
I know that if $R$ is a commutative ring with unity then the proper ideal $\mathfrak{p}$ is prime iff $R/\mathfrak{p}$ is an integral domain. but how can this help me in my proof? the set $S$ is defined as the setminus not the quotient.
Could anyone help me in solving this please?
This is the definition I am using for multiplicatively closed set.
Definition: A set $S \subset R \setminus \{0\}$ is multiplicatively closed if $x,y \in S \implies xy \in S.$
If you reflect on this, you will start to see that this is true by definition of prime ideals: They are exactly those ideals into which you cannot multiply from outside. Whether you state it as \begin{align*} xy ∈ \mathfrak p &\implies& x ∈ \mathfrak p &~~~\text{or}~~~ y ∈ \mathfrak p, &\quad \text{or as} \\ xy ∈ R \setminus \mathfrak p &\impliedby& x ∈ R \setminus \mathfrak p &~~\text{and}~~ y ∈ R \setminus \mathfrak p \end{align*} is just a matter of giving contrapositions of each other. The latter line says that $S = R \setminus \mathfrak p$ is mutliplicatively closed, which is what you want to prove.
By the way: You could in principle also kind of use the fact that $R' = R / \mathfrak p$ is an integral domain. This means that $S' = R' \setminus \{0\}$ is multiplicatively closed. So as $S = R \setminus \mathfrak p = \mathrm{pr}^{-1} (R' \setminus \{0\}) = \mathrm{pr}^{-1}(S')$ is the preimage of a multiplicatively closed set under a ring homomorphism (so in particular under a multiplicative homomorphism), $S$ itself is multiplicatively closed too. This exposes the core insight behind this to realizing that for a commutative ring $R'$ $$R' ~\text{is integral domain} \iff R'\setminus \{0\} ~\text{is multiplicatively closed},$$ but this line of reasoning is disproportionate to the puzzle at hand.