Is my understanding of quotient rings correct?

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?

Solution 1:

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.

Solution 2:

Another minor clarification: You are not just mapping evertyhing in $I$ to zero, you are also mapping everything NOT in $I$ to 'whatever part is left over after mapping I to zero'.

Example: Consider $2x^2 + 3x + 3$ in $R[x]$. Map $I$ = <$x^2 + 1$> to zero, then this element is $2(x^2 + 1) + x + 3$, and what is left of it after mapping $I$ to zero is of course $x + 3$.

Solution 3:

While quotient rings are constructed from $R$, it is not always true that the quotient ring "hides inside $R$" in the sense that the quotient ring is isomorphic to a subring of the original ring so this is not a good intuition to hold. You can always pick a representative for each equivalence class and identify the quotient ring with a subset of $R$ but the addition/multiplication between the elements of this subset considered as elements in the quotient ring will be different from the addition/multiplication between the elements of this subset considered as elements in the original ring.

For example, the quotient ring $\mathbb{Z}/n\mathbb{Z}$ consists of equivalence classes $[n]$ of integers $\mod n$. If you want, you can arbitrary choose a unique representative - for example $0 \leq k < n$ - from each equivalence class and then identify $\mathbb{Z}/n\mathbb{Z}$ with the set $\{0, 1, \dots, n - 1\} = \mathbb{Z}_n$ which is a subset of the original ring $\mathbb{Z}$ but this set is not a subring (nor a subgroup) of $\mathbb{Z}$ as it is not closed under the original addition/multiplication.

Something similar happens with $\mathbb{R}[x] / (x^2 + 1)$. If you choose for each equivalence class a unique representative $a + bx$ then you get a subset of $\mathbb{R}[x]$ which is a subgroup but not a subring. The addition here plays well but the multiplication doesn't. In $\mathbb{R}[x]$ we have $(a + bx)(c + dx) = ac + (ad + bc)x + (bd) x^2$ and this doesn't lie in the subset so you don't get a subring. The multiplication $(a + bx)(c + dx)$ in the quotient results in $(ac - bd + (ad + bc)x)$.

Sometimes, it does happen that the quotient ring is isomorphic to a subring of the original ring in such a way that you can choose your representatives and get an isomorphic subring. This happens with $\mathbb{R}[x] / (x)$ where if you choose the representatives of each equivalence class to be the constant polynomials, then you would indeed get not only a subset but a subring (you have closure under multiplication) that is isomorphic to the quotient ring. However, if you would have chosen the representatives differently, this wouldn't necessarily be the case.