Ring of fractions

Question

Let $R$ be a commutative ring and let $S \subseteq R$ be a multiplicative closed subset containing 1. Prove that the kernel of the natural homomorphism $\lambda: R\rightarrow R_S$, $r\rightarrow r/1$ is $\ker\lambda=\{x\in R:xs=0$ for some $s$ $\in S\}$, where $R_S$ is the ring of fractions.

Answer 1

To do this question, you need to look carefully at the definition of what it means for two elements in $R_S$ to be equal: Given $r_1, r_2\in R$, $s_1, s_2 \in S$, we say that $r_1/s_1 $ and $ r_2 / s_2$ are equal in $R_S$ if and only if there exists an $s \in S$ such that $$ s s_2 r_1 = s s_1 r_2.$$ (It's like "clearing denominators", except that you are allowed to multiply by an extra $s \in S$ on both sides.)

So given a $r \in R$, when is $r/1$ equal to zero in $R_S$? Writing zero as $0/1$, it follows from our definition that $r/1$ is equal to zero if and only if there exists an $s \in S$ such that $$ s\times 1 \times r = s \times 0 \times 1.$$ But this is precisely the condition that $$ sr = 0.$$