Proving operations of a quotient ring are well-defined I'm a little confused about a specific bit of reasoning in proving that the $\times$ operator is well-defined for a quotient ring. I looked up a proof and it omits, without any mention, the part I'm confused about.
We want to show that for a two-sided ideal $I$ of a ring $R$, for any $r, s \in  R$
$$ (r + I) \times (s + I) = (rs + I) $$
is a well-defined operation. Since this is an equivalence of sets, I first start off with showing that any $x$ in the left hand-side must be a member in the right hand side. That's straightforward, since for some $i_1, i_2 \in I$,
$$ x = (r + i_1) \times (s + i_2) = rs + i_1s + ri_2 + i_1i_2$$
$i_1s, ri_2, i_1i_2 \in I$ from our assumption of an ideal, as is their sum. So $x \in (rs + I) $ and  $(r + I) \times (s + I) \subseteq (rs + I)$.
How do we prove the reverse? That is, $(rs + I) \subseteq (r + I) \times (s + I)$. We'd want $x \in (rs + I) \implies x \in (r + I) \times (s + I)$. I'm assuming the ring doesn't necessarily have a unit element over $\times$. We'd have to show something like for $i_0 \in I$, $rs + i_0 = rs + i_1s + ri_2 + i_1i_2 $, finding some combination of $i_1, i_2$ to match any $r, s, i_0$.
I feel like I'm missing something since the proof in the text I have completely ignores this direction.
Thanks for any help!
 A: The expression $(r+I)\times (s+I) = rs + I$ is not an equality of sets, it's a mapping rule for the corresponding equivalence classes. That it's not an equality of sets you can check by taking $R=\mathbb Z$, $I=(5)$, $r=2$ and $s=3$. Then $rs+I=6+(5)$ which as a set contains $1$, so if we were to write $1=(2+k_1\cdot5)(3+k_2\cdot5)$, where $k_1$ and $k_2$ are some integers, and expand, this eventually leads to $1=6+5(3k_1+2k_2+5k_1k_2)$, equivalently
$$3k_1+2k_2+5k_1k_2=-1 \iff k_1=-\frac{1+2k_2}{3+5k_2}$$
However the right-hand-side is never an integer when $k_2$ is an integer.
The notation $r+I$ denotes the coset represented by $r$. For a moment, let's change notation so that $[r]$ denotes the coset represented by $r$. Then $R$ is decomposed into a disjoint union of cosets $\cup_j[r_j]$ where $j$ ranges over some index set.
We now can define a function $\operatorname{prod}\colon R/I \times R/I\to R/I$ by $\operatorname{prod}([r_i],[r_j])=[r_i r_j]$. Now one has to verify that this function is well defined, that is no matter which representatives we choose for cosets $[r_i]$ and $[r_j]$, we get the coset $[r_i r_j]$ back after applying $\operatorname{prod}$.
To go back to the example $R=\mathbb Z$, $I=(5)$, $r=2$ and $s=3$, we have $\operatorname{prod}([2],[3])=[6]$ by definition. But now we also have $[-2]=[2]$, so $$[6]=\operatorname{prod}([2],[3])=\operatorname{prod}([-2],[3])=[-6],$$ by definition. But this isn't contradictory because $-6$ and $6$ represent the same coset, that is $[-6]=[6]$.
