Definition of quotient ring I have a question about the definition of quotient ring (I mean a commutative ring  case) . 
Let $(a+I)$ be a coset of $R$ and $(b+I)$ be another left coset of $R$.
Then the multiplication in the quotient ring is defined as 
$(a+I)(b+I)=(ab+I)$
However, if we consider the setwise product of $(a+I)$ and $(b+I)$ we know that their product is just a subset of a fixed coset of $I$. Then why should I believe the multiplication in the quotient ring is well-defined?
 A: The importance of ideals comes from the fact that they're the kernels of a ring homomorphism as consequence of the first isomorphism theorem. From perspective if I have a ring homomorphism $\varphi : R \rightarrow Q$ then we can interpret the cosets as the preimages of singletons, ie $\varphi^{-1}(q)$ for $q \in Q$, which is $q + I$ in the usual quotient construction.
This gives us the motivation to define the multiplication of ideals in a way that respects this construction. So by $(q + I)(p + I)$ we mean there exist $r,s \in R$
with $\varphi(r)=q$ and $\varphi(s)=q$ and we are interested in the preimage $\varphi^{-1}(\varphi(r)\varphi(s))=\varphi^{-1}(\varphi(rs))$ which is $(qp+I)$.
This is why multiplication of cosets is not defined setwise. Even simple ideas like $(2) \subset \mathbb{Z}$ cause trouble immediately since $(2)(2)$ which should simply be $(2)$ since it's $0\cdot 0$ but setwise it would be $(4)$, which as you noted isn't even a coset.
A: An Excellent Question! I am also troubled by this question for a long time. The key point here is to remember why we define the product on the quotient ring this way. You may think this is a set equation, however what really matters is that we can verify the product is well-defined. In another word, it is independent of the choice of the representation element. You can check the definition of quotient group, you will find out the definition there is actually a set equation (What a bad coincidence, which may make you think the set equation is the crucial thing). But if you check a textbook instead of just a definition, you will find out the reason why we define a normal subgroup is to keep the product well-defined!
A: I suppose you mean that you are considering cosets of an ideal $I$ in a ring $R$, then $$(a+I)(b+I) = ab +aI+bI +I*I = ab + I + I + I= ab + I$$. Because $aI = I$, $I*I = I$ and $I + I=I$ as consequences of the definition of an ideal.
