Clarification on the definition for product on a ring quotient I'm going over my notes on ring theory, and have come across the following definition for the product on a ring quotient $R/I$ for $I \triangleleft R$, 
$$
(a+I)\cdot(b+I) := (a+I)(b+I) + I
$$
We then claim that this indeed is equal to $ab + I$ making the canonical projection a ring morphism from $R$ to the quotient. I can see more or less immediately that:
$$
ab + I  \subseteq (a+0)(b+0) + I \subseteq (a +I)(b + I) + I
$$
However, for the other inclusion I had written  that:
$$
(a+I)(b+I) + I = ab + aI + Ib + I^2 + I \subseteq ab + I
$$
which I'm now not entirely convinced of it being correct, since $z \in (a+I)(b+I)$ if and only if: 
$$
z = \sum_{j = 1}^n(a+x_j)(b+y_j) = \sum_{j = 1}^nab + \sum_{j = 1}^nay_j + \sum_{j = 1}^nx_jb + \sum_{j = 1}^nx_jy_j
$$
for some $x_j,y_j \in I$, and this is in $nab + aI + Ib + I^2$ which may as well not be equal to $ab + aI + Ib + I^2$. Am I computing this the wrong way?
 A: The definition of the product of cosets of an ideal is
$$(a+I)\cdot (b+I)=(a+I)(b+I)+I,$$
where $(a+I)(b+I)$ on the right is the set product $AB:=\{ab\mid a\in A,b\in B\}$, not the ideal product $IJ=\{\sum_{i=1}^nx_iy_i\mid x_i\in I,y_i\in J\}$. Given this, we can show that $(a+I)\cdot (b+I)=ab+I$:
If $i,j\in I$, then $(a+i)(b+j)=ab+(ia+jb+ij)$, where the right term is in $I$ since it is an ideal, and conversely $ab+i=(a+0)(b+0)+i\in (a+I)(b+I)+I$.
(I usually see this done in the opposite direction, where we define $(a+I)(b+I)=ab+I$, and then we must show that this definition does not depend on the member of the coset chosen for the definition, which can be done by rewriting it to $(a+I)(b+I)+I$, where the only thing we need of the set product is that it is a function of $a+I$ and $b+I$ rather than $a$ and $b$.)
If we used the ideal product in that definition, then it would contain $nab$ as you noticed, which is not what we want. In particular it would not be a coset at all if $ab\notin I$, because then it would contain both $0$ and $ab$ and these do not differ by a member of $I$.
