Complement of a prime ideal is a subset closed under multiplication, containing 1

Question

Let $R$ be a commutative ring and let $I$ be a prime ideal of $R$. Prove that $S = R \setminus I$ is a multiplicative closed subset containing $1$.

Please can anyone help me out here?

Answer 1

Let $I$ be a prime ideal of $R$ and denote $S = R \setminus I$.

Because $I$ is a prime ideal, it is a proper ideal, i.e. $I \neq R$. Therefore, we have that $1 \not\in I$ (do you see why?) and hence $1 \in R \setminus I = S$.

Let $a, b \in S$. Suppose $S$ is not a set closed under multiplication, then we would have that $ab \not\in S = R \setminus I$. Hence we would have that $ab \in I$ but because $I$ is a prime ideal, we would have that $a \in I$ or $b \in I$, a contradiction. Hence the assumption that $ab \not\in S$ is incorrect, so $ab \in S$ and $S$ is closed under multiplication.