## Quotienting stuff out

In this article I’ll talk for a bit and eventually reach a point where I’ll try to describe in simple terms the notion of a quotient ring. The notion of a quotient in ring theory is, simply put, a far-reaching generalization of the idea behind the integers “modulo” $n$. That is, the idea that if you have a subset of a ring with certain characteristics, you can “quotient” it out, and consider a new ring where you only care about what the objects look like relative to that subset. This new ring is called the quotient ring.

Alright, so for starters, I’ll be talking only about commutative, unital rings (I will just call them “rings” here, though, without any further qualification). You can think of a ring like a bunch of objects with binary operations of addition and multiplication defined on it, where every element has an additive inverse $-x$ but not necessarily a multiplicative inverse $x^{-1}$. Also, the laws of algebra we take for granted (associativity, distributivity, etc.) all hold. I’m too lazy to give a formal definition of it, so just go read this if you need to see one. The word “commutative” means that we’re requiring the ring’s multiplication operation to be commutative i.e. $x \cdot y = y \cdot x$ for all elements $x,y$ in the ring. (Addition operation of any ring is always commutative, on the other hand). When I say “unital” I simply mean there exists a “multiplicative identity” element inside the ring, that is, an element in the ring which behaves like the familiar number 1. More precisely, this element $1$ has the property that $1 \cdot x = x \cdot 1 = x$ for any element $x$.

I think I mentioned groups and homomorphisms in an earlier post. There’s also the notion of a ring homomorphism (which we will henceforth just call “hom”s in this article). If $R, S$ are rings, a hom is just a map $\phi : R \to S$ such that for all $x,y \in R$, we have

1. $\phi(x+y) = \phi(x)+\phi(y)$
2. $\phi(xy) = \phi(x)\phi(y)$
3. $\phi(1) = 1$

Note here that “+” is used to refer to both addition in R (on the left hand side) and addition in S (similarly for multiplication) and also “1” is used on the left to refer to $R$‘s 1, whereas it is used on the right to refer to $S$‘s 1.

Now that we know what a hom is, we want to look at a special subset of $R$ which is associated with any hom. The kernel of a hom $\phi$ is the set of all elements $x$ in $R$ which are mapped to 0 under $\phi$, in other words, $\phi(x) = 0$.

This kernel is a pretty interesting set. First off, we can note that it always contains 0 (this is easy to prove). Also, due to the nature of a hom, the difference of any two elements in the kernel is also in the kernel. Furthermore, if we take any element $x$ in the kernel and multiply it with something else (anything else) in $R$, we actually get something which is also in the kernel. All these things are simple to show.

So let’s define an ideal as a subset of $R$ which… well… has exactly all the properties that the kernel of any hom would have, as we listed off above! Clearly then, by definition, the kernel of any hom is an ideal. But… is every ideal the kernel of some hom? To answer this question, for any ideal $I \subseteq R$, we would have to find a target ring $S$ and a hom $\phi : R \to S$ which has $I$ as its kernel…

It turns out that this is indeed the case. The target ring $S$ that we want is actually the quotient ring of $R$ modulo the ideal $I$ (something I’m about to talk about).

The first step is to define a relation on $R$ which (as it will turn out) will be an equivalence relation. So let’s do that. Let’s define our relation by saying that $x$ is “congruent to” $y$ modulo $I$ if and only if the element $x-y$ lies within the ideal $I$. You can check that this relation is reflexive, symmetric, and transitive, and is hence an equivalence relation. By elementary math, every equivalence relation partitions the entire underlying set into disjoint sets called equivalence classes. The equivalence classes of this particular relation are often called cosets, or residue classes mod $I$.

Without further ado, we define the quotient ring. We declare $R/I$ to be the set consisting of all the cosets $a+I = \{ a + i : i \in I \}$ where $a \in R$. So far, this $R/I$ is just a set. How do we turn it into a ring? We define operations in the obvious way.

1. $(a+I) + (b+I) = (a+b)+I$ for all $a+I, b+I \in R/I$
2. $(a+I)(b+I) = (ab) + I$ for all $a+I, b+I \in R/I$
3. Take the multiplicative identity to be the element $1+I \in R/I$

Bang. I’ll leave it to you to check that these operations are well-defined (that they don’t depend on which representatives of the cosets we choose). The fact that $R/I$ is indeed a ring follows immediately from the fact that $R$ is a ring. This pretty much justifies the notation $\mathbb{Z}/n\mathbb{Z}$ people often write to denote the integers mod $n$. Indeed, a quotient ring is really just a generalization of the technique used to construct the integers mod $n$. Here, $n\mathbb{Z}$ is referring to the ideal $\{ nk : k \in \mathbb{Z} \} = \{ \ldots, -3n, -2n, -n, 0, n, 2n, 3n, \ldots \}$. In fact, since this ideal is generated by a single element in that fashion, it has a special name, and is called the principal ideal generated by the element $n$. The principal ideal of a ring $R$ generated by the element $x$ is often denoted by $(x)$ or $xR$. (Check that every principal ideal is indeed an ideal).