Why the ideal generated by $\varphi(I)$ is $IS^{-1}R$?

Question

Let $\varphi:R\longrightarrow S^{-1}R$ defined by $\varphi(r)=\frac{r}{1}$ where $R$ is a commutative ring and $S$ a subset of $R$ that is closed under multiplication. We denote $I^e$ the ideal generated by $\varphi(I)$ where $I\lhd R$. Why $$I^e= IS^{-1}R\ \ ?$$ To me, $I^e=S^{-1}R\varphi(I)$ and $\varphi(I)=\{\frac{r}{1}\mid r\in I\}$, but is this $I$ ?

Answer 1

If $\varphi\colon R\to T$ is a ring homomorphism (with $R$ and $T$ commutative) and $I$ is an ideal of $R$, then the ideal of $T$ generated by $\varphi(I)$ is $$ IT=\Bigl\{\sum_{k=1}^n x_kt_k: n\ge 1, x_k\in I, t_k\in T\Bigr\} $$ The case of $T=S^{-1}R$ is no different, except that the description of the ideal is simpler: $$ IS^{-1}R=\Bigl\{\frac{x}{s}:x\in I,s\in S\Bigr\} $$ That the set on the right hand side is contained in $IS^{-1}R$ is obvious; since it is easy to prove it's an ideal, we are done.