# 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).