# Spectrum of a Ring

#### Approaching Scheme Theory as a Beginner

Thank you to James Tao for explaining the following, geometrically intuitive, interpretation to me.

A scheme is the result of patching affine schemes together, an affine scheme is any locally ringed space that is equivalent to Spec of a ring. To make an analogy, a manifold is the result of patching together copies of $\mathbb{R}^n$, in a way that preserves the smooth structure.

And the smooth structure of $\mathbb{R}^n$ is essentially the fact that it is a sheaf: over any open set, we have the space of smooth functions on that set, and these functions restrict and glue nicely, etc.

So when you go from $\mathbb{R}^n$ to “smooth manifolds,” you are creating spaces that locally look like $\mathbb{R}^n$, respect the same kind of smooth structure (as captured by the fact that smooth functions are defined on open sets in a way independent of the patching), yet have some global structure that might prevent them from just being the same as $\mathbb{R}^n$.

If you care about algebraic (polynomial / rational) functions defined on algebraic sets (zeroes of systems of polynomials), then you can do the same program! The “basic patch” is “affine scheme,” and you patch affine schemes together to get a scheme.

The category of affine schemes is equivalent to $\text{Ring}^{op}$. In other words, telling you how to map forward is the same as telling you how to pull back functions. However, this is NOT true for the larger category of schemes. $\text{Hom}_{Ring}(A, \Gamma(\mathcal{O}_B)) = \text{Hom}_{Schemes} (B, \text{Spec } A)$ (note that $\Gamma(\mathcal{O}_B)$ := the global sections of the structure sheaf $\mathcal{O}_B$).

In particular, given a scheme, it’s unclear how to get a ring that encodes all the information in that scheme. Consider, for example, projective space, whose only global sections are constants, yet the field $k$ of constant functions certainly doesn’t single out any projective space!

#### What are these affine Spec R things?

We have a functor Spec from Ring to Schemes:

$\text{Ring} \xrightarrow{\text{Spec}} \text{Schemes}$

Schemes look like Fun(Ring, Set). So Spec sends a ring to a functor from Ring to Set.

$\text{Ring} \xrightarrow{\text{Spec}} (\text{Ring} \to \text{Set})$

Honestly, whenever I see “scheme” I replace it mentally with “algebraic curve,” but this is just because I don’t really get what a scheme is yet.

$R \xrightarrow{\text{Spec} (-)} \text{Spec(R)}$

$R \xrightarrow{\text{Spec} (-)} (S \mapsto \text{hom}(R,S))$

$\text{Spec}$ $R$ $S$ = $\text{Hom}_{\text{Ring}}(R, S)$

Let say we have:

1. a ring $\mathbb{Z}[t]$ (this is just the ring of polynomials with $t$ as a variable and coefficients in $\mathbb{Z}$)
2. an arbitrary ring $S$.

What is $\text{hom}(\mathbb{Z}[t], S)$?

Notice that we’re in Ring, so the members of $\text{hom}(\mathbb{Z}[t], S)$ must be Ring homomorphisms. Let’s say we’re looking at a ring homomorphism $\phi: \mathbb{Z}[t] \to S$.

I learned from Cris Moore that demanding examples is a useful practice, so: What is $\phi(7t^2-4t+3)$ in simpler terms?

$\phi$ is a ring homomorphism, so we know that $\phi(1) = 1_s$. This implies that $\phi(n) = n_s$.

$\phi(7t^2-4t+3) = \phi(7t^2) – \phi(4t) + \phi(3)$
$= \phi(t)(\phi(7t) – \phi(4)) + \phi(3)$
$= \phi(t) (\phi(t) 7_s – 4_s) + 3_s$

$\phi(t)$ is not determined! We can pick it to be any element of S!

$\text{hom}(\mathbb{Z}[t], S) \simeq \{\phi(t) \in S\} = S$

In other words, $\text{Spec}(\mathbb{Z}[t])$ is a forgetful functor from $S$ as an object in $\text{Ring}$ to its underlying set.

Exercise: show that $\mathbb{Z}[t]$ is an initial object in an appropriate category, and that the trivial ring is the terminal object in the appropriate category.

#### But what is Spec?

It’s the spectrum of a ring! Intuitively, $\text{Spec} R$ is the geometric object canonically associated to $R$.

For example:

• $\text{Spec}(\mathbb{Z})$ is “The Point”
• $\text{Spec}(\mathbb{Z}[x,y]/ (x^2 + y^2 – 1))$ is “The Circle”
• $\text{Spec}(\mathbb{Z}[x,x^{-1}]$ is “Pairs of Invertible Elements” (e.g. a field with the origin removed)

Let’s look at $\text{Spec}(\mathbb{Z}[x,y]/ (x^2 + y^2 – 1))(S)$ when $S$ is $\mathbb{R}$, it’s obviously the circle BUT IT WORKS FOR (almost) ANY S. The concept of there being a canonical geometry associated to every ring is very exciting!

#### Murky Afterthoughts

When people say “elliptic curve” they might mean “a family of elliptic curves”, this is commonly written $C \to \text{Spec} R$ where $R$ is the underlying coefficient ring.

For example, $y^2=4x^3+ax+b \mapsto a, b \in R$ is the family of elliptic curves of the form $y^2=4x^3+ax+b$ over the coefficient ring $R$.

I’m trying to figure out what happens when $S$ is not something nice like $\mathbb{R}$ or $\mathbb{C}$. What is a circle in the coefficient ring of an elliptic curve?

Thanks to Aaron Mazel-Gee for walking me through the concept of $\text{Spec}$ (all errors in this post are mine and not his).

## One thought on “Spectrum of a Ring”

1. Josh G says:

I think you do yourself a disservice by jumping right from “algebraic curves” to the viewpoint of (affine) schemes as functors. To make the connection more concrete, one should first understand the connection between varieties (or even just algebraic curves) and their coordinate rings, and then understand how Spec of the coordinate ring recovers the Zariski topology on the original variety. *Then* one can take the more abstract viewpoint, whereby one wishes to treat not just integral domains over a field, but *any* ring as the coordinate ring of some “geometric object”, namely, a scheme. And *then* one can generalize from schemes as the categorical duals of rings to the “functor of points” viewpoint. Taking this route will give you a much more intuitive, geometric picture of what is going on (and should enable you to connect your intuitive geometric picture of algebraic curves to the highly abstract picture of schemes as functors.)

From this viewpoint, it is clear why $hom(\mathbb{Z}[t], S)$ should just be $S$. Imagine every ring is the ring of “well-behaved” functions on some geometric object. Then certainly $\mathbb{Z}[t]$ is the ring of polynomial functions on the line (where, here, we haven’t specified *which* line – real, complex, or what have you – but that’s okay because we’re only considering polynomials with integer coefficients, which make sense on the line over any ring). And $S$ is the ring of well-behaved functions on some geometric object which we’ll call $Spec(S)$. Since rings are categorically dual to schemes, $hom(\mathbb{Z}[t], S)$ should be dual to $hom(Spec(S), line)$. But what are maps from $Spec(S)$ to the line? They are exactly the (well-behaved) functions on $Spec(S)$ – i.e., the elements of $S$!