Intuition for quotients through exercise on rings

Question

Show that $\Bbb Z \times \{0 \}$ is an ideal of $\Bbb Z \times \Bbb Z$ and describe the elements of the quotient ring $(\Bbb Z \times \Bbb Z)/(\Bbb Z \times \{0 \})$.

Pick $a,b \in \Bbb Z \times \{0 \} $, then $a + (-b)=(x,0)+(-(y,0)) = (x-y,0) \in \Bbb Z \times \{0 \} $ for some $x,y \in \Bbb Z$. Also let $c \in \Bbb Z \times \Bbb Z$, then $$ac =(x,0) \cdot (u,v) = (xu,0) \in \Bbb Z \times \{0 \} $$ and similarly $ca \in \Bbb Z \times \{0 \} $. Thus $\Bbb Z \times \{0 \} $ is an ideal of $\Bbb Z \times \Bbb Z$.

I don’t really understand the idea behind quotients in general. They’re used in lots of fields such as topology, but I’ve never really understood the ”motivation” behind them so I’m stuck here.

Is the idea to basically ”squish” everything to a point in the ”dividing” set or how should I think about this?

Answer 1

The idea is not just to squish the "dividing set" into a point. More than that: the idea is to completely kill it and make it the zero element of a new ring. Essentially, a quotient $R/I$ of a ring $R$ behaves exactly as $R$ itself. With one exception: Every element of $I$ is now $0$ instead. You can think of it as imposing additional algebraic relations. For instance, $\mathbb Z$ has a minimum of algebraic relations. The only equalities we can find are those which are forced by the ring axioms. Like $2+3=3+2$ is the law of commutativity. But we don't have $2+3=2$, because that doesn't follow from the axioms. Such a ring is called "free". But we could impose the relation $2+3=2$. Though it would lead to other relations as well. For instance, if $2+3=2$, then $3=0$. But then also $6=3+3=0+0=0$. Or $4=1+3=1+0=1$, or $8=2+2\cdot3=2+2\cdot0=2+0=2$. Essentially, any two numbers would suddenly be equal if they previously differed by a multiple of $3$. They now differ by $0$ instead. Essentially, all multiples of 3, that is the set $3\mathbb Z$, are now $0$. And the ring we obtain this way is $\mathbb Z/3\mathbb Z$. It's $\mathbb Z$, but with the additional condition that all elements of $3\mathbb Z$ are $0$. Apart from that, the quotient ring behaves the same way. This is formalized by the existence of a homomorphism $\pi:\mathbb Z\to\mathbb Z/3\mathbb Z$, the natural projection. It tells us what we turned our elements into. It assigns to each element of $\mathbb Z$ the corresponding element of the new ring with the additional relation. We didn't add any new elements, which means that every element of the codomain has a preimage. That is, $\pi$ is surjective. And we made all elements of $3\mathbb Z$ into $0$ (but no other element!). So the kernel should be $3\mathbb Z$. That's essentially it. The fundamental theorem on homomorphisms will now tell us that up to natural isomorphism, these conditions already fully determine how the image looks.

Which is why I recommend the following way of understanding quotients: a quotient of $R$ modulo $I$ is the codomain of a surjective ring homomorphism with $I$ as its kernel.

Answer 2

The name comes from investigating the integers modulo $n$. So if I take some integer $k$ and divide it by $n$ I will get some remainder $r$. Said another way there exists a quotient $q$ such that $k = qn +r$ with $r < n$. So we're doing a division operation.

So where does the squishing come in? Well, many different numbers have have the same remainder modulo $n$ and so if we group all the numbers with the same remainder together there will be one for each possible remainder between $0$ and $n-1$. You can think of every element being squished down to its remainder, or to one of these sets. The sets will not have any elements in common because a number can't have two remainders when divided by $n$. Also, every integer will be in one of these sets, and so they form a partition of the integers. This partition comes with an equivalence relation by the fundamental theorem of equivalence relations and we call them cosets.

What's of particular interest here is that this partitioning respects the algebraic structure in a meaningful way. If I do addition or multiplication with the remainders, or the whole coset, everything just works. So it's this specific example that motivates the nomenclature.

However the idea of a quotient operation generalizes to a much larger collection of algebraic structures as you're investigating now. For example in a vector space an orthogonal projection will be a quotient operation. Using the example you're working on you can generalize this result to see that $\mathbb{Z}^n$ modulo $\mathbb{Z}$ is isomorphic to $\mathbb{Z}^{n-1}$ in a meaningful way. This will not be the only result that resembles division that also reinforce this intuition such as Noether's isomorphism theorems. So that's where the name comes from.