Confused about quotient ring For $I$ an ideal of a commutative ring $R$, I'm confused about the object $R/I$. My understanding is that this is defined by $R/I := \{rI: r \in R\} = \{\{ri: i \in I\}: r \in R\}$.
However, by the definition of an ideal, if $i \in I$ and $r \in R$ then $ri \in I$. So surely for a given $r$ the set $\{ri: i \in I\}$ will just equal $I$? And hence $R/I=\{I\}$?
Clearly this is wrong somewhere but I can't see where.
 A: As mentioned in the comments already, the elements of the quotient are formed by the group operation, not the monoid operation.
Explicitly, the elements of the quotient are congruence classes of an equivalence relation defined by $r\sim s$ if $r-s\in I$.  In the end, we write the class of $r$ symbolically as "$r+I$".
The group axioms provide everything you need to make this $\sim$ an equivalence relation. Without inverses though, this definition doesn't work, so monoids are different.
That doesn't mean quotients of monoids don't exist, it's just that one can't get a natural one from the binary operation like we did above, because we can't use inverses.
It still makes perfect sense to ask what partitions of a monoid are compatible with the binary operation: these are called congruences of the monoid.  I have not seen this applied in ring theory.
If it helps, then, you can just think of the underlying additive group of $R/I$ as exactly the same group abelian group you would get if you just thought of $R$ and $I$ as abelian groups. The kicker is just that there is also a well-defined monoid operation that makes $R/I$ a ring, if $I$ is a two-sided ideal.
