Quotients of rings and modules

Question

In group theory, for a normal subgroup $H$ of a group $G$, we define coset multiplication by $$ (g_1 H)(g_2 H) := (g_1 g_2) H $$ This definition seems nice to me, since $$ \{ g_1 h g_2 h' | h,h' \in H \}$$ is precisely the right hand side of the above definition, $(g_1 g_2) H$. In the context of rings and modules, we have the definitions $$ (r + I)(s + I) := rs + I \qquad \mathrm{and} \qquad r(u + N) := ru + N$$ where $r,s$ are ring elements, $u$ is a module element, $I$ is a ring ideal, and $N$ is a submodule. However, here it seems that we don't have the same nice interpretation. That is: $$ \{r(u + n)| n \in N \} \neq r(u + N)$$ $$ \{(r + x)(s + y) | x,y \in I \} \neq (r + I)(s + I)$$ My question is simply: am I right when I say this? And if so: are there any problems that arise for rings and modules, which don't arise for groups, on account of not being able to treat coset multiplication as "pairwise" multiplication of elements of each coset?

Answer 1

You seem to confuse things, primarily you seem to think we are working with sets which we are not. Let me explain, when we do a quotient structure we start lumping elements together and proclaim "These are now the same". For various structures the method is slightly different due to how they are. These lumpings are called equivalence classes. To be a bit technical, when an equivalence class respects an algebraic structure, such as groups, modules, rings etc, it's called a congruence relation.

When we write $aN$ what we mean is not necciserily the set $\{an:n\in N\}$ but we mean the equivalence class that $a$ belongs to. Which for groups are all elements $b$ such that $ab^{-1}\in N$, which can be shown implies that the sets $aN=bN$, yes now I mean sets here. That is because we already used set notation for the equivalence definition. We can see that if $an_1=bn_2$ then $b^{-1}a=n_2n_1^{-1}$ which implies the previous.

So when we write $aNbN=abN$ we don't mean that the sets are equal, but that the multiplication can be shown to respect the group structure. Namely that $an_1bn_2$ must lie in the equivalence class of $ab$, that is $$an_1bn_2=abn_3$$ which is trivial to show when $N$ is normal. So to go to ideal and modules, let $I$ be our ideal, of course by definition of ideals do we have that $rI\subseteq I$, again dealing with sets. But if you ask the questio

which equivalence class does any element in $rI$ belong to?

The answer is of course they belong to $I$'s equivalence class, and as such we write $rI=I$, not because their sets are equal but because the equivalence classes all elements in both sets are the same. That is why we write $$(r+I)(s+I)=rs+rI+sI+I^2=rs+I$$ Again this means that the equivalence classes of all elements on all sides are identical, not that the sets are necciserily equal as here, we are not that interested in sets but in the equivalence classes.

The same goes for submodules. Summa summarum

The question is not about sets but about equivalence classes that makes a congruence relation, that is respects our algebraic structure and its operations. As for these they are closely relatede to specific sets that are easily formed from our notation we tend to use that notation.

Answer 2

The two things on the left-hand sides that you have described with set-builder notation are not useful (in any context I've seen) and never come up during a routine algebra course.

I have never seen any problems arise around not having the set-wise version. So I wouldn't lose any sleep over it if I were you. It seems to me that the set version is more complicated than the other one, so it's hard to see why the set version is 'nicer'.