# Meaning of $S^{-1}R$ notation

Here are objects defined in an exercise:

Let $R$ be a commutative ring. Let $A$ be an ideal of $R$ and $S=\{1+a\mid a\in A\}$.

The exercise then makes reference to the prime ideals of $S^{-1}R$. What could be the definition of $S^{-1}R$ ?

I thought it was the set composed of the results of the products of inverses of members in $S$ with arbitrary objects in $R$, but since $R$ is not required to contain multiplicative inverses that does not make sense.

Could someone help me with this notation ?

Let $R$ be a ring. A multiplicatively closed subset of $R$ is a subset $S$ of $R$ such that $1 \in S$ and $S$ is closed under multiplication. Define a relation $\equiv$ on $R \times S$ as follows $$(a, s) \equiv (b, t) \iff (at - bs)u = 0 \text{ for some } u \in S.$$ Let $a/s$ denote the equivalence class $(a, s)$ and let $S^{-1}R$ denote the set of equivalence classes. We put a ring structure on $S^{-1}R$ by defining addition and multiplication of these "fractions" in the same way as we did in elementary algebra: i.e., $$(a/s) + (b/t) = (at + bs)/st \text{ and } (a/s)(b/t) = ab/st.$$
Now, for the particular $S$ you defined as $S := \{1 + a : a \in R\}$, we need to show it's a multiplicatively closed subset for it to agree with the definition above.
1. It is immediate that $1 \in R$ since $1 + 0 \in S$ (choosing $a = 0$).
2. Let $1 + a, 1 + b \in S$, then $(1 + a)(1 + b) = 1 + \underbrace{a + b + ab}_{\in R} \in S$.