Amidst all the rigorous constructions of quotient rings involving equivalence relations and ideals, I feel that I have finally grasped what a quotient ring is. I have applied this intuition to a few examples and they have served me well, but I would just like to verify that this actually the case.
If $R$ is a ring and $I$ an ideal, then $R/I$ is just what you have leftover if you map all of the elements of $I$ to zero.
So going by that $$\mathbb{Z}/n\mathbb{Z} = \{0, 1,2,3,…,n-1\}=\mathbb{Z}_n$$
Also
$$\mathbb{R}[x]/x = \text{Constant polynomials}=\mathbb{R}$$
It seems to me that the quotient ring is always 'hiding' inside the original ring. So for example $\mathbb{Z}_n \subset \mathbb{Z}$ or $\mathbb{R} \subset \mathbb{R}[x]$. But if we consider this isomorphism
$$\mathbb{R}[x]/(x^2+1) \cong \mathbb{C}$$
Then it doesn't seem to work anymore. I cannot see how $\mathbb{C}$ is 'hiding' inside $\mathbb{R}[x]$. Was my initial observation just a coincidence and hence cannot be applied here? Or am I totally misunderstanding everything?
Best Answer
There is a natural map $\eta$ of sets from $\mathbb{C}$ to $\mathbb{R}[x]$: just send $a+bi$ to $bx+a$. Now, this map is a homomorphism on the underlying additive groups, but multiplicatively it has some problems. Nevertheless, this is a sense in which $\mathbb{C}$ sits inside $\mathbb{R}[x]$, and in fact it turns out to be the right one: the "multiplicative badness" of $\eta$ is killed by quotienting out by $\langle x^2+1\rangle$.
Specifically, the "multiplicative badness" I refer to is the fact that $\eta(z_0\cdot z_1)\color{red}{\not=}\eta(z_0)\cdot \eta(z_1)$ in general. For example, $$\eta(i\cdot i)=\eta(-1)=-1\color{red}{\not=}x^2=x\cdot x=\eta(i)\cdot \eta(i).$$ However, we have for all $z_0, z_1\in\mathbb{C}$ that $$x^2+1\mbox{ divides }\eta(z_0\cdot z_1)-(\eta(z_0)\cdot\eta(z_1)),$$ so - once we kill off $x^2+1$ - the natural way of fitting $\mathbb{C}$ inside $\mathbb{R}[x]$ actually respects multiplication!
OK, now let me try to un-cringe all the algebraists in the room. What's really going on here is that $\eta$ is a left inverse of the composition of the quotient map $j: \mathbb{R}[x]\rightarrow\mathbb{R}[x]/\langle x^2+1\rangle$ with the isomorphism $i: \mathbb{R}[x]/\langle x^2+1\rangle\rightarrow\mathbb{C}$. That is, there are many elements of $\mathbb{R}[x]$ which $i\circ j$ maps to a given complex number $z$; the map $\eta$ picks one "canonical representative."
In the example of $\mathbb{Z}/n\mathbb{Z}$, there's a really natural way to pick such a canonical representative, so we usually don't notice that we did it. For instance, we could have chosen to use $\{n, n+1, n+2, . . . , 2n-1\}$, except that that's weird.
EDIT: By the way, this gets at your statement "$\mathbb{Z}/n\mathbb{Z}\subset\mathbb{Z}$" - while that's false, what is true is that the set of canonical representatives of elements of $\mathbb{Z}/n\mathbb{Z}$ is a subset of $\mathbb{Z}$. And this is a truism about quotients: if $S=R/I$, then by definition each element of $S$ "comes from" many elements of $R$, and the act of picking canonical representatives amounts to defining an appropriate injection of sets of $S$ into $R$. (Note that it's a very bad habit to pretend that this means $S\subset R$ - in particular, the image of $S$ in $R$ won't generally be closed under the operations of $R$!)
Similarly for $\mathbb{R}[x]/\langle x\rangle$. In general, though, a little bit of thought needs to go into picking canonical representatives, and also into how they fit together and relate to the remaining elements of the original ring. For example, here's a good exercise:
$\mathbb{R}[x]/\langle x^2+1\rangle$ is isomorphic to $\mathbb{R}[x]/\langle x^2+2\rangle$.
In fact, they each have the same canonical representatives - the linear polynomials!
Yet clearly something's different about them - specifically, the natural isomorphisms to $\mathbb{C}$ send $x$ to different complex numbers.
Teasing out the details of this example, I think, will help make this business with canonical representatives a lot clearer.