Commutative rings and ideals, showing a map is well defined

Question

Let $R$ be a commutative ring with an ideal $I$. The additive group $R/I$ is the set of cosets of $I$ with respect to addition in $R$.

Let

$\cdot : R/I \times R/I \to R/I$ be defined by $$(x+I)\cdot (y+I) = xy + I$$

I want to show that $\cdot$ is well defined, i.e. if $x + I = x' + I$ and $y + I = y' + I$ then $xy + I = x'y' + I$

my progress:

$x+I = x' + I \iff x-x' \in I$

$y + I = y' + I \iff y - y' \in I$

write $xy - x'y' = (x-x')y + x'(y-y')$ then using the fact that $y - y' \in I$ and $x-x' \in I$ I could use this to show that$xy - x'y'\in I$ and be done however I have one problem.

How do I know that $(x-x')y \in I $ and $x'(y-y') \in I$? We have that $(x-x') \in I$ but $ y \in R$ so not necessarily that $(x-x')y \in I $. Could somoene explain how I can show $(x-x')y \in I $?

Answer 1

You know this because $I$ is an ideal. By definition $I$ being an ideal of $R$ means that $RI \subseteq I$. So $(x - x') \in I$ and $y\in R$ implies $y(x-x') \in I$.