Equivalence relation in a commutative ring

Question

Let $R$ be a commutative ring and $I \neq R$ ideal of $R$. For $x,y \in R$, define $$ x \sim y \iff \exists\ a,b \in I\ \text{such that } x(1+a) = y(1+b). $$ One can easily see this is an equivalence relation. I have the following questions:

(1) Define a subset $S \subset R$ by $$ S=\{ x \in R \mid (x,I)=(1) \}. $$ Prove that if $x \in S$ and $x \sim y$ then $y \in S$.

(2) Prove that the map $S/\sim\, \rightarrow (R/I)^{\times}$ is bijective.

My attempt:

(1) Note that $(x,I)=(x)+(I)$ so to prove $y \in S$ we only need to show $(x)=(y)$ given that $x \in S$ and $x \sim y$. However, I cannot go further as $x(1+a) = y(1+b)$ for some $a,b \in I$ is not enough to conclude that $(x)=(y)$.

(2) $x \sim y \iff x(1+a) = y(1+b)$ yielding $x\sim y $ then $x-y=yb-xa\in I$. Set $[x]=:\{y \in S \mid x \sim y\}$. I guess a map $(S/\sim) \rightarrow (R/I)^{\times}$ defined by $[x]\mapsto x+I$ may gives the desired the conclusion. However, I cannot go on with that.

Any help would be much appreciated.

Answer 1

1. $x(1+a)=y(1+b)$ with $a,b\in I$ $\Rightarrow$ $x\equiv y\bmod I$. Then from $(x)+I=R$ we get $x$ invertible in $R/I$. But in $R/I$ we have $x=y$, so $y$ is also invertible in $R/I$, and thus $(y)+I=R$.

2. Yes, that one is the right map. It's obviously well defined and surjective. For injectivity let $x\bmod I=y\bmod I$ and want to show that $x\sim y$. Since $x\bmod I$ is invertible there is $u\in R$ such that $ux\equiv 1\bmod I$. We also have $uy\equiv 1\bmod I$, so $ux-1=b$ and $uy-1=a$ with $a,b\in I$. From the first equation we get $uxy-y=by$, and from the second $uxy-x=ax$. Now subtract them and find $x-y=by-ax$, and we are done.