Quotients

\(\newcommand{\mb}{\mathbb}\)Let's talk about dividing sets by other sets.

You've probably heard of modular arithmetic. The usual first explanation is either in terms of clocks or remainders, or sometimes both. You say "if I multiply 5 by 9 modulo 13, I get 6" because 5 * 9 = 45, and then we subtract 13s and end up with 6.
Here's another way to look at it: consider a map\(f\) from \(\mb Z\) to a set \(T\) that has 13 elements in it, where \(f\) has the property that if \(a\) and \(b\) differ by an integer multiple of \(13\), then \(f(a) = f(b)\), and if they don't differ by an integer multiple of \(13\) exactly, then \(f\) sends them to different places. So \(f(1) = f(14)\) and \(f(45) = f(6)\), but \(f(1)\) and \(f(45)\) are different. Note that each element of \(T\) has to get hit by something, i.e. \(f\) is a surjection.
Now we give \(T\) some structure. We'll define an addition on \(T\) via addition on \(\mb Z\). In other words, we'll say that \(f(a) + f(b) = f(a + b)\) so that \(f\) is a homomorphism of groups with addition. Since \(f\) hits everything in \(T\), this defines addition for everything in \(T\). The question is whether this is well-defined; if we have \(A \in T\), there are several possible values \(a\) in \(\mb Z\) such that \(f(a) = A\), and similarly for \(B\) in \(T\) there are several values \(b\) such that \(f(b) = B\), and we want to make sure that \(A + B\) gives the same value no matter which \(a\) and \(b\) we pick. The conditions on \(f\) ensure this works in this case.
We can now consider general surjections \(f\) from \(\mb Z\) to other sets \(R\). If we want \(f\) to be a homomorphism, we need to have an addition structure on \(R\), with an identity and additive inverses, but we also need conditions on \(f\). In particular, let's look at \(f(0)\); this has to be the additive identity of \(R\). We call the set of things that get mapped to \(f(0)\) the kernel of \(f\), \(\ker(f)\); for the case above, this kernel is \(13\mb Z = \{13 k | k \in \mb Z\}\). In general, if \(f\) is going to be a homomorphism from \(\mb Z\) to another group , we need that \(\ker(f)\) is a subgroup of \(\mb Z\).

Let's suppose that we're looking at a general group \(G\) and a surjective homomorphism \(f\) from \(G\) to \(H\), and look at \(\ker(f)\), everything that gets sent to the identity \(e\) in \(H\). \(\ker(f)\) needs to be a group, because the identity composed with the identity is the identity.
But it needs to be more than just a subgroup of \(G\).
Consider \(g\) and \(h\) in \(G\) where \(h \in \ker(f)\). \(f(gh) = f(g)f(h) = f(g)\), since \(f\) is a homomorphism and \(f(h)\) is the identity. So everything of the form \(gh\) for \(h \in \ker(f)\) gets sent to the same place. Moreover, suppose that \(f(g') = f(g)\) for some \(g'\). Then we get that \(f(g^{-1}g') = f(g^{-1}f(g') = f(g^{-1})f(g') = f(g)^{-1}f(g') = f(g)^{-1}f(g) = e\). So \(g^{-1}g'\) is in \(\ker(f)\). Multiplying it by \(g\) gives that \(g'\) is of the form \(gh\) for some \(h \in \ker(f)\). We call the set \(gker(f) = \{gh | h \in \ker(f)\}\) a left-coset of \(\ker(f)\) in \(G\).
A similar argument tells us that \(g'\) is of the form \(h'g\) for some \(h' \in \ker(f)\), i.e. \(g'\) must be in the right-coset \(\ker(f)g\). So in order for \(f\) to be a homomorphism, we need that the left and right cosets \(g\ker(f)\) and \(ker(f)g\}\) are the same for all \(g\) in \(G\). We thus say that \(\ker(f)\) is normal in \(G\).
Conversely, if we have a normal subgroup \(N\), i.e. if for each \(g \in G\) the sets \(gN = \{gn | n \in N\}\) and \(Ng = \{ng | n \in N\}\) contain the same elements, then we can make a surjective homomorphism \(f\) from \(G\) to a group denoted \(G/N\) whose elements are cosets \(gN\), with multiplication in \(G/N\) given by \(gNhN = ghN\). We call \(G/N\) the quotient group of \(G\) by \(N\).
In the case of modular arithmetic above, if we treat \(\mb Z\) as a group whose composition operation is addition, then we get that all subgroups of \(\mb Z\) are normal. Indeed, for any group whose composition operation is commutative, all subgroups are normal. So for instance, given a vector space, \(V\) and a subspace \(W\), we can form \(V/W\) whose elements are cosets of the form \(v + W = \{v + w | w \in W\}\). Note that in this case we can also scalar multiply, since if \(w \in W\) then \(cw \in W\) for any scalar \(c\).
For groups that aren't commutative, the situation is more complicated. For example, consider the group \(S_3\) of permutations of 3 elements \(\{a,b,c\}\). The subgroup that contains just the identity and the permutation that swaps \(a\) and \(b\) is not normal. The subgroup \(C\) that contains the identity and the two ways to cycle all 3 elements is normal, and the quotient \(S/C\) is a group with two elements in it.

What about multiplication in quotients of \(\mb Z\)? Well, that always works so it's boring, so let's generalize until it sometimes breaks.
Consider a general associative ring \(R\), i.e. a set with an addition, additive identity, additive inverses, and a multiplication. Think matrices, for instance, or algebras.
We have a surjection \(f\) to another ring and we want to make this a homomorphism; what can we say about \(\ker(f)\)?
\(R\) is a group if we only look at addition, so \(\ker(f)\) must be a subgroup; the addition is commutative, so \(\ker(f)\) is automatically normal for addition. Our elements of \(R/\ker(f)\) are cosets of the form \(r + \ker(f)\).
Recalling that everything in \(\ker(f)\) is equivalent to \(0\) according to our map \(f\), we should get that \(\ker(f)\) times itself should be equivalent to \(0*0 = 0\), so we want that \(\ker(f)^2 = \{ab | a, b \in \ker(f)\}\) should be contained in \(\ker(f)\). So \(\ker(f)\) has to be a ring. But just as we needed more than just a subgroup, we need more than just a subring.
In our quotient object, we need that when we multiply something by anything equivalent to 0, the product is equivalent to 0. In other words, we need that \(a\ker(f) = \{ab | b \in \ker(f)\}\) needs to be contained in \(\ker(f)\). Similarly, we need that \(\ker(f)b = \{ab | a \in \ker(f)\}\) also needs to be contained in \(\ker(f)\). We say that \(\ker(f)\) is therefore a two-sided ideal, which I'll just refer to as an ideal.
Being an ideal in a ring is the equivalent of being a normal subgroup for groups: if we have a ring \(R\) and an ideal \(I\), then we can form the quotient ring \(R/I\) whose elements are of the form \(r + I\) and we have a natural surjective homomorphism \(f\) from \(R\) to \(R/I\) whose kernel is \(I\).
This is how we can think about modular arithmetic: for an integer \(k\), \(k\mb Z\) is an ideal in \(\mb Z\), so \(\mb Z/k\mb Z\) is a ring with addition, subtraction, and multiplication modulo \(k\).

We can think of this in terms of actions: A group \(G\) acts on itself via what is called the adjoint action or conjugation action: \(\rho(g)\) is the map that sends \(h\) to \(ghg^{-1}\). \(H\) being normal is then equivalent to saying that the set \(H\) is fixed as a whole by the adjoint action, although the individual elements of \(H\) might get shuffled around. Note that for commutative groups, the adjoint action is trivial, so all subgroups are fixed under it.
Similarly, a ring \(R\) acts on itself via either the left regular action or right regular action, also known as multiplication: \(lm(a)\) is the map that sends \(b\) to \(ab\), \(rm(b)\) is the map that sends \(a\) to \(ab\). \(I\) being an ideal is then equivalent to saying that \(I\) is invariant under both the left and right regular actions.
This invariance then allows for making quotients.

Given a bunch of elements \(\{x_1,\ldots,x_k\}\) in \(R\), we often write \(\left\langle x_1,\ldots,x_k\right\rangle\) for the smallest ideal in \(R\) that contains all of those elements. For groups, the same notation means the smallest group that contains all of those elements, although this makes no claims to normality. We say that \(\left\langle x_1,\ldots,x_k\right\rangle\) is generated by the elements \(x_1,\ldots,x_k\).

This gives a good way to talk about constructions of groups or algebras: we start with a thing \(F\) with no structures or relations other than the ones guaranteed by the axioms, called a "free" thing. So for instance a set that obeys the group axioms but has no other relations is called a free group. \(\mb Z\) with addition is a free group; everything in \(\mb Z\) has to be there because the group axiom demands that we can form any power of an element, but we don't get "a = b" except in the cases demanded by the associativity and inverse laws.
Then if we want to impose a relation \(A = B\) where \(A\) and \(B\) are expressions, we form the invariant subthing \(I\) generated by the difference between \(A\) and \(B\), and the quotient \(F/I\) thus can't tell the difference between \(A\) and \(B\).

For instance, suppose we start with a vector space \(V\) and form the tensor algebra \(\bigotimes^* V\). This is an associative, unital algebra, but has no other relations between elements; there are no equations that aren't just due to linearity in \(V\), so \(\bigotimes^* V\) is the free associative algebra on \(V\).
Now suppose we want to make a commutative algebra, i.e. \(a\otimes b = b\otimes a\). We start with \(\bigotimes^* V\) and consider the ideal \(I = \left\langle a\otimes b - b\otimes a | a, b \in V\right\rangle\). Then \(S(V) = \bigotimes^* V/I\) is commutative, because we've quotiented out the difference between \(a\otimes b\) and \(b \otimes a\); \(S(V)\) is often called the symmetric algebra on \(V\).

Finally, because of course we should talk about it, we can talk about quotients of coalgebras. Here we define a coideal \(I\) in a coalgebra \(C\) to be a subspace of \(C\) as a vector space with the property that
$$\Delta(I) \subset I \otimes C + C\otimes I$$ which is the dualized version of being an ideal. This is more coaction-y than action-y, using the left and right coregular coactions. Anyway, in this case we have that \(C/I\) is a coalgebra. We'll see an example of a quotient coalgebra in a bit.

We often say that a thing with no proper quotients is simple or irreducible. Such objects are like the atoms of whatever we're studying, because we can't break them into smaller pieces. And just as studying atoms is important for chemistry, studying the simple things is important for the study of things. Also just as there are many ways to put a given set of atoms together to make a molecule, there are often many ways to put a given set of simple things together to make a larger thing.