# Spectrum of a Ring

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!

Sidenote:

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).