Do quotient rings have geometric meaning, or geometric intuition?

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?

Solution 1:

A lot of early algebraic geometry was motivated by trying to view rings as a set of functions on some space. In real analysis, one common ring one studies is the ring $\mathcal C[0,1]$ of all continuous functions $f : [0, 1] \to \mathbf R$ (or $\to \mathbf C$). In fact, this is a $\mathbf R$-algebra, meaning if $f, g \in \mathcal C[0,1]$ and $\lambda, \mu \in \mathbf R$

• $\lambda f + \mu g \in \mathcal C[0,1]$ and

• $fg \in \mathcal C[0,1]$

• the constant zero function acts as an additive identity

• the constant one function acts as a multiplicative identity

More abstractly, one can look at $\mathcal C(X)$ of continuous real-valued functions on a compact metric space or a compact Hausdorff space.

In differential geometry, one looks at $\mathcal C^{\infty}(M)$, the algebra of smooth functions on a smooth manifold.

In algebra we want to do the same thing. Except, we don't want to look at all continuous functions, or even all smooth functions. The functions we want to look at are the functions that appear in algebra: polynomial functions.

Since $\mathcal C[0,1]$ was an $\mathbf R$-algebra, it might seem natural to first look at $\mathbf R$-algebras but in fact, we will want to use the complex numbers at first instead. $\mathbf C$ is algebraically closed, a property which we will see is important later.

The first example of an algebraic ring of functions is $\mathbf C[x_1,x_2,\dots,x_n]$, the algebra of polynomials. Given a polynomial $f(x_1,x_2,\dots,x_n)$, we can treat it as a function $\mathbf C^n \to \mathbf C$.

Now let us say that we don't want our domain to be all of $\mathbf C^n$. Since this is algebra, we want to use an algebraic domain. That domain will be a subset of $\mathbf C^n$ that can be defined by polynomials. Specifically, we are interested in polynomial equations:

$$ f(x_1,x_2,\dots,x_n) = g(x_1,x_2,\dots,x_n). $$

If we subtract, we note that the solution set of this equation can be written as $X = \{(a_1,\dots,a_n \in \mathbf C^n : (f - g)(a_1,\dots,a_n) = 0\}$. $X$ is our second example of what is known as a "variety" (the first example was $\mathbf C^n$). A variety is a subset of $\mathbf C^n$ which can be written as the vanishing set of a set of polynomials. That is, a variety is a set $X$ of the form

$$ X = \mathcal V(S) = \{(a_1,\dots,a_n) \in \mathbf C^n : f(a_1,\dots,a_n) = 0, \forall f \in S\}. $$

Exercise: show that if $I$ is the ideal generated by the set $S$ then $\mathcal V(S) = \mathcal V(I)$. This shows that all varieties show up as vanishing sets of ideals.

For a variety $X$ (in fact, for any subset of $\mathbf C^n$) we can define an important algebraic invariant known as the "ideal of $X$" which is by definition

$$ \mathcal I(X) = \{f \in \mathbf C[x_1,\dots,x_n] : f(a_1,\dots,a_n) = 0, \forall (a_1,\dots,a_n) \in X\}. $$

(Exercise: verify that this is indeed an ideal).

An important theorem due to Hilbert, called the Nullstellensatz (Zero Points Theorem in English) says the following.

Theorem If $k$ is an algebraically closed field, $I \subseteq k[x_1,\dots,x_n]$ is an ideal and $X = \mathcal V(I) \subseteq k^n$ is the corresponding variety. Then the ideal of $X$ is the set of functions $f \in k[x_1,\dots,x_n]$ such that $f^r \in I$ for some $r \ge 1$. (You can verify that if $f^r \in I$ then $f^r$ is identically $0$ on $\mathcal V(I)$ which means that $f$ is identically $0$, the hard part is the convers.)

Let us look at the equation $x^2 + y^2 = 1$ (a circle). The corresponding variety is

$$ X = \mathcal V(\{x^2 + y^2 - 1\}) = \{(a,b) \in \mathbf C^2 : a^2 + b^2 - 1 = 0\}. $$

By the Nullstellensatz, the ideal of $X$ is $I = \langle x^2 + y^2 - 1 \rangle$ (if you believe me that this ideal has the property that $f^r \in I \implies f \in I$).

Now we ask ourselves: what are the polynomial functions $f : X \to \mathbf C$? Since the domain is $X$, we agree that two functions $f, g$ are equal if $f(a,b) = g(a,b)$ for all $(a,b) \in X$ (i.e. in the domain). Well, if $f(a,b) = g(a,b)$ for all $(a,b) \in X$ then $(f - g)(a,b) = 0$ for all $(a,b) \in X$. I.e. $f - g \in I$ (i.e. in the ideal of $X$). Therefore, the polynomial functions $X \to \mathbf C$ is the ring

$$ \mathbf C[X] = \mathbf C[x,y]/\langle x^2 + y^2 - 1 \rangle. $$

Following this example, we define the coordinate ring of a variety $X$ to be the ring

$$ \mathbf C[X] = \mathbf C[x_1,\dots,x_n]/\mathcal I(X) $$

since two functions $X \to \mathbf C$ agree on $X$ iff their difference is in $\mathcal I(X)$.

Classical algebraic geometry was all about studying these coordinate rings and their corresponding varieties. Eventually, people asked: do coordinate rings have to look like a quotient of $\mathbf C[x_1,\dots,x_n]$, or can we use any ring? This is where the theory of Schemes starts.

If we have a univariate polynomial $f(x)$ over $\mathbf C$, we know that we can write it as product of linear polynomials:

$$f(x) = c(x - r_1)(x - r_2) \dots (x - r_n).$$

As you may have noticed, much of what we were doing with varieties was simply determining whether or not $f(a) = 0$ or not. With this factorization we can say that $f(a) = 0$ iff $a \in \{r_1,\dots,r_n\}$.

But this isn't the only ring where we have factorization. The integers have it as well. If I take an integer, such as $4020$. I can factor it into primes:

$$ 4020 = 2^2 \cdot 3 \cdot 5 \cdot 67. $$

In a moment of inspiration, I might even go so far as to define the vanishing set of $4020$ as

$$ \mathcal V(4020) = \{2, 3, 5, 67\}.$$

You see, our points of $\mathbf C^1$ were really prime ideals all along: the point $a \in \mathbf C^1$ was really the ideal $\langle x - a \rangle$ in disguise. The point $(a_1,\dots,a_n) \in \mathbf C^n$ was really the ideal $\langle x_1 - a_1, \dots, x_n - a_n \rangle$. Our functions, were really functions on the set of prime ideals.

If $f \in \mathbf C[x_1,\dots,x_n]$ then

$$ f(x_1,\dots,x_n) \equiv f(a_1,\dots,a_n) \mod \langle x_1 - a_1, \dots, x_n - a_n \rangle. $$

Saying that $f(a_1,\dots,a_n) = 0$ is the same as saying that

$$ f(x_1,\dots,x_n) \equiv 0 \mod \langle x_1 - a_1, \dots, x_n - a_n \rangle. $$

Given an integer $f$, I can view $f$ as a function on the set of prime numbers where

$$ f(p) = f \bmod p. $$

For example if $f = 4020$ then $f(2) = 4020 \mod 2 = 0 \mod 2$. Likewise $f(3) = 0 \mod 3$ and $f(5) = 0 \mod 5$ and $f(67) = 0 \mod 67$. We can also say that $f(7) \ne 0 \mod 7$.

As you continue to study Schemes, you will see how to take into account different rings. For example, what if we use $\mathbf R$ instead of $\mathbf C$? Or any field for that matter. How to we account for nilpotents?

A nilpotent $f$ in a ring $R$ is a non-zero element such that $f^r = 0$ for some power $r \ge 2$. The problem with nilpotents is they don't behave well with the function analogy. If you have a function $f : X \to \mathbf C$ such that $f^r = 0$, you would generally prefer if $f = 0$ as well.

Solution 2:

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.