TITLE: Do quotient rings have geometric meaning, or geometric intuition?
QUESTION [16 upvotes]: I am wondering if quotient rings have any geometric meaning. For example, I am trying to identify the ring $\mathbb Z[x]/(x^2-3, 2x+4)$ and I am trying to think about lattice points and polynomial graphs but it is a bit overwhelming. 
But even for something simpler like $\mathbb Z[x]/(x^2)$. Since we are modding out by $x^2$, then we will be left with linear integer polynomials. What do these rings look like in a geometric sense?

REPLY [7 votes]: I'm adding a second answer because my first answer is already quite long and I want to explain a different picture and nobody else has responded yet.
You may of heard that $\mathbf R/2\pi\mathbf Z$ or $\mathbf R/\mathbf Z$ is a circle. Let us see why that is. Let $$\mathbf S^1 = \{(a,b) \in \mathbf R^2 : a^2+b^2=1\}$$ be the unit circle. What does it mean to give a continuous function
$$ f : \mathbf S^1 \to \mathbf R? $$
We can view this a function in two variables $f(x,y)$ but somehow this isn't what we want. When you draw the circle, you see a 1-dimensional figure:
                                                                
So ideally what we would like is a function in one variable. Luckily, the circle has a 1-parameter parameterization $(\cos t, \sin t)$. So our function in two variables, $f(\cos t, \sin t)$, becomes a function in one variable: $f(t)$ where $t$ is the angle. Since the angles $t$ and $t + 2\pi$ and $t + 4\pi$, etc. all represent the same point, we have
$$ f(t + 2\pi n) = f(t), \forall n \in \mathbf Z. $$
That is, a function
$$ f : \mathbf S^1 \to \mathbf R $$
is a function
$$ f : \mathbf R \to \mathbf R $$
that is periodic. To describe a periodic function, one only needs to give the values of the functions in the interval $[0,2\pi)$ since any angle is equivalent to an angle in $[0,2\pi)$. The quotient $\mathbf R/2\pi\mathbf Z$ consists of equivalence classes under the relation $x \sim x + 2\pi n$. That is,
$$\mathbf R/2\pi\mathbf Z = \{ \{x + 2 \pi n : n \in \mathbf Z\} : x \in [0,2\pi)\}. $$
A periodic function $f : \mathbf R \to \mathbf R$ is constant on each equivalence class. I.e. for all $a, b \in \{x + 2 \pi n : n \in \mathbf Z\}$, $f(a) = f(b)$. So it makes sense to talk about $f$ as a function that takes an equivalence class $\{x + 2 \pi n : n \in \mathbf Z\}$ to $f(x)$ since the number of multiples of $2\pi$ we add, do not affect the value of $f$. Thus a function on the circle is a function
$$ f : \mathbf R/2\pi\mathbf Z \to \mathbf R. $$
We can picture this in a different way, by saying that $\mathbf S^1$ is obtained by taking the line segment $[0,2\pi]$ and glueing the two endpoints together:
                                            
If you like, this is a continuous analogue of the discrete quotient group, $\mathbf Z/n\mathbf Z$.
We can extend this construction to higher dimensions. I will use just $2$ dimensions because it is easier to draw. Let us take the plane $\mathbf R^2$ and consider two vectors $\vec{v}$ and $\vec{2}$ with integer coordinates. E.g. $\vec v = (2,1)$ and $\vec{w} = (-1,3)$:
                           
Now we take all integer combinations of these vectors (i.e. $m\vec v + n \vec w, m, n \in \mathbf Z$).:
                           
These integer combinations form a lattice, which we will donote by the Greek letter lambda, $\Lambda$. A function
$$ f : \mathbf R^2/\Lambda \to \mathbf R $$
is a function of two variables, $f(x,y)$, which is periodic in two directions. For the lattice $\Lambda$ generated by $(2,1)$ and $(-1,3)$, that means that
$$ f(x,y) = f(x + 2, y + 1) = f(x - 1, y + 3) = f((x,y) + m\vec{v} + n \vec{w}) $$
for all $(x,y) \in \mathbf R^2$ and $(m,n) \in \mathbf Z^2$.
This corresponds to glueing the opposite edges of one of the parallelograms in our lattice to form a torus. When we glue two opposite sides we first obtain a cylinder, and when we glue the two ends of the cylinder together it forms a torus.
For $\mathbf Z[x]/(x^2)$ we are doing a similar kind of gluing, except that instead of gluing points in $\mathbf R^2$ together, we are glueing points in $\mathbf Z[x]$ (i.e. polynomials) together. The set of linear polynomials, $a + bx$ corresponds to the parallelogram in the lattice. Any polynomial $a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n$ is glued to its linear piece:
$$ a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n \equiv a_0 + a_1 x \bmod x^2. $$
You should also think of $\mathbf Z[x]/(x^2)$ as a ring of first order (i.e. linear) Taylor approximations to functions. For example
\begin{align}
  e^x &\approx 1 + x \\
  \log(1 - x) &\approx  x \\
  (1 + x)^n &\approx 1 + nx.
\end{align}
This is how the (co)tangent space of a smooth manifold arises. In smooth manifold theory, one is interested in Taylor approximations of smooth functions, in algebraic geometry we restrict ourselves to polynomial functions. The idea, however, remains the same.
Nilpotents in a ring correspond to tangent vectors. Or more generally, to jets, which are the higher order Taylor approximations.