Till now I visualized Quotient groups as a technique to generate equivalence classes whenever needed, as in $\mathbb{Z}/n\mathbb{Z}$. But now I have a feeling of doubt since I haven't seen a book that introduces the concept of Quotient groups or rings as a technique to generate equivalence classes. Is my visualization correct? What are the other purposes of finding Quotient of an algebraic structure?
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2Yes, "technicaly", as you say, it "generates equivalence classes". But, usually, doing it, it's much more that that. The quotient set can be endowed with a richer structure that the original set, for example, when $p$ is prime, $\mathbb{Z/pZ}$ is a field whereas $\mathbb{Z}$ is only a ring..., or with a simpler structure, etc. – Jean Marie Jun 07 '16 at 10:04
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@JeanMarie so, though equivalence classes are themselves beautiful structures (like ideal classes), but we quotient an algebraic structure with an expectation of creating an algebraically richer structure? – rationalbeing Jun 07 '16 at 10:08
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1You first of all define the relation of equivalence, and then you can define an operation on the set of equivalence classes. If you take $\mathbb Z/p\mathbb Z$, this is only a set, but you cane define what mean $[x]_p+[y]_p$. This is only a trivial example. The question its very hard. – Marco Lecci Jun 07 '16 at 10:16
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This is a standard way for example to build fields by quotienting a ring by a maximal ideal (the case $\mathbb{Z}\rightarrow\mathbb{Z/pZ}$ being a particular case.) – Jean Marie Jun 07 '16 at 10:19
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@rationalbeing : I don't know if this could help you… – Watson Jun 07 '16 at 10:20
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@Watson thanks, but I am very much comfortable with the examples:) – rationalbeing Jun 07 '16 at 10:23
2 Answers
Sure, thinking that quotient structures "generate equivalence classes" is an "okay" perspective. Frankly, it is not the only and not necessarily the most important one. Here is a different one:
Quotient structures (quotient groups, quotient rings, etc.) are examples of coequalizers. Though this fact is usually (I think?) referred to as the "Fundamental theorem on homomorphisms" (due to historical reasons I dare say).
You probably know this theorem (for different kinds of algebraic structures) already. But perhaps you have not realized yet: This theorem basically states exactly what a quotient structure is, up to a canonical isomorphism. This is basically as good as saying "uniquely".
You can say that a quotient group for a congruence relation $\sim$ on (let's say) a group $G$ is a group $\tilde{G}$, such that there is an (epi-)morphism $\varphi : G \to \tilde{G}$, such that $x\sim y \Rightarrow \varphi(x) = \varphi(y)$, such that for all groups $H$ and morphisms $f : G\to H$ with $x\sim y \Rightarrow f(x)=f(y)$, there is a unique morphism $h : \tilde{G} \to H$ such that $h\circ \varphi = f$.
Note that if $N$ is a normal subgroup of $G$ we would get $\sim$ explicitely by $x\sim y \Leftrightarrow xN = yN$. Conversily said $N$ is actually $\ker \varphi$.
For every other group $\tilde{G}_2$ with a morphism $\varphi_2 : G\to \tilde{G}_2$ with the same property as $\varphi$ there is a unique isomorphism $i : \tilde{G} \to \tilde{G}_2$ such that $i \circ \varphi = \varphi_2$.
The only reason why hardly anyone takes this as the definition of a quotient group is because a more explicit description via equivalence classes is already well-known and established.
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We use quotients to introduce relations that might not have been present in the original structure. This has the effect of removing some elements. Here are some examples:
$G/[G,G]$ introduces commutativity in a group.
$K[X]/(f(x))$ introduces a zero of $f$.
$R/N$ removes the nilpotents elements in a ring ($N$ is the nilradical).
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