TITLE: Is my understanding of quotient rings correct?
QUESTION [8 upvotes]: 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?

REPLY [3 votes]: For any algebraic structure (group, ring, vector space, etc.) when you form a quotient structure by taking cosets, it is tautological that you can find elements from the original structure that serve as labels for the quotient structure (e.g., $\mathbf Z/n\mathbf Z$ can be represented by $\{0, 1, \ldots, n-1\}$ and ${\mathbf R}[x]/(x^2+1)$ can be represented by polynomials $\{a+bx : a, b \in \mathbf R\}$ in $\mathbf R[x]$), but this is entirely missing the point: the structure of the quotient is often not reflected in how the labels (coset representatives) behave algebraically within the original structure.
For example, $\mathbf Z/5\mathbf Z$ is a ring, even a field, but the set $\{0,1,2,3,4\}$ is not a subring (or a subfield) of $\mathbf Z$. No choice of coset representatives could be a subring since $\mathbf Z$ has no finite subrings. Even if you treat $\mathbf Z$ only as an additive group, and $\mathbf Z/5\mathbf Z$ as an additive quotient group, you can't find coset representatives in $\mathbf Z$ that form a subgroup of $\mathbf Z$: there are no elements of finite order in $\mathbf Z$ besides 0.
In the case of $\mathbf R[x]/(x^2+1)$, the coset representatives $a+bx$ are an additive group in $\mathbf R[x]$ but they are not closed under multiplication.  There is no subring of $\mathbf R[x]$ that looks like the quotient ring $\mathbf R[x]/(x^2+1)$, which is a field, e.g., any subfield of $\mathbf R[x]$ is inside $\mathbf R$. In the quotient ring $\mathbf R[x]/(x^2)$ there is a nonzero element whose square is zero (the additive identity), but there is no such element in the ring $\mathbf R[x]$ itself. 
In the setting of finite groups, the quaternion group $Q_8$ has a (normal) subgroup $\{\pm 1\}$ for which the quotient group $Q_8/\{\pm 1\}$ has order $4$, but $Q_8$ has no subgroup of order $4$ that is isomorphic to that quotient group since the quotient group is not cyclic while every subgroup of order $4$ in $Q_8$ is cyclic.
You need to learn to embrace the new structure found in quotient constructions without trying to shoehorn them all the time into the more elementary idea of a substructure of the original algebraic structure. That's an important step in gaining maturity about what quotient constructions are all about. (Historically, it took mathematicians time to take this step too. See the top answer at Who named "Quotient groups"? where it mentions that Camille Jordan, who introduced quotient groups, thought of them as new group laws on a choice of coset representatives.) This also makes you better appreciate the situations when you can identify the quotient structure (not merely as a set, but with all the algebraic operations too!) with a substructure using well-chosen coset representatives.