Let $R$ be a nontrivial ring and $I\leq R$ a nontrivial left ideal. Is it possible to have an element $r\in R\setminus \{0\}$ such that $rI\subsetneq I$?
I'm just thinking, if such an $r$ exists for some ring $R$ and ideal $I$, how can we say for sure $(x+I))(y+I)=xy+I$, for all $x,y\in R$? It'd be possible $(x+I)(y+I)\subsetneq xy+I$. Thanks.
On
It is entirely possible that when multiplying (in the sense of $AB = \{ab \mid a\in A, b\in B\}$) the two sets $(x+I)$ and $(y+I)$, this yields something strictly contained in $xy + I$. But the definition of quotient ring multiplication isn't to use that multiplication. The definition of quotient multiplication is that the product of $x+I$ and $y+I$ is $xy+I$.
You still have to show that this quotient ring product is well defined, of course (as in, making sure that choosing different representatives of the two factor cosets still lands you in the same product coset). But you don't have to show that the quotient ring product of two cosets of $I$ is another coset of $I$. This is part of the definition.
In group theory, using normal subgroups to make quotient groups the standard way, you can't get strict containment, and thus the two notions of multiplication coincide. But not so for rings and ideals, as you have seen.
$R=\mathbb{Z}$, $I=3\mathbb{Z}$, $r=2$.
No, if $R$ is a ring and $I\subseteq R$ an ideal then the multiplication $(x+I)(y+I):=xy+I$ for $x,y\in R$ is well defined.
I don't think I completely understand your doubt.
The point is that you define multiplication by $(x+I)(y+I):=xy+I$ and then you have to check that this is well defined. Multiplication to be well defined means that if $x,x',y,y'\in R$ are such that $x-x'\in I$ and $y-y'\in I$, then $xy-x'y'$ is in $I$ and this is quite straightforward from the definitions.