*There are likely inaccuracies in this post, as I am just beginning to learn the basics of algebraic geometry. Constructive criticism is strongly encouraged.*

#### There once was a line…

Let’s look at the affine line over $\mathbb{C}$. This is just the complex line with no distinguished element (i.e., a plane which forgot it’s origin — it’s 2 real dimensions, or, equivalently, 1 complex dimension).

#### $\mathbb{A}^1 \simeq \text{Spec } \mathbb{C}[x]$

As we know, the affine line over the field $\mathbb{K}$ is isomorphic to the spectrum of a ring of single variable polynomials (with coefficients in $\mathbb{K}$). If you aren’t familiar with this isomorphism, I recommend popping over to Spectrum of a Ring. For simplicity, let’s work with the field $\mathbb{C}$, although, I’m pretty sure the rest of this post still works for any $\mathbb{K}$.

#### Is there a reasonable way to take in two points, and ask for a third?

This is generally a fun question to answer when you’re handed a space. So, how do we add two points of $\mathbb{A}^1$ to get a third point in $\mathbb{A}^1$?

The same way we usually add two complex numbers! But wait, we need an identity! Where does the identity come from? I think it comes from looking at the affine plane as $\text{Spec }\mathbb{C}[x]$, the maximal ideal $x$ is a distinguished point.

It’s worth considering why $\text{Spec }\mathbb{C}[x] \times\text{Spec }\mathbb{C}[y] \simeq \text{Spec }\mathbb{C}[x,y]$. It seems that this is because $\text{Spec }$ is functorial, that is:

$\text{Spec }\mathbb{C}[x, y] = \text{Spec }(\mathbb{C}[x] \otimes {C}[y]) = \text{Spec }\mathbb{C}[x] \times \text{Spec }\mathbb{C}[y]$.

Another way to think about this is that the cartesian product $\mathbb{A}^1 \times \mathbb{A}^1= \mathbb{A}^2 = \text{Spec }\mathbb{C}[x,y]$.

#### $\mathbb{A}^1 – \{0\} \simeq \text{Spec } \mathbb{C}[x, x^{-1}]$

Can we analogously put a group structure on the punctured affine line? Hell yes we can, here’s one way to do so:

Recall that $\text{Spec }\mathbb{C}[x, x^{-1}] \times \text{Spec }\mathbb{C}[y, y^{-1} =\text{Spec }\mathbb{C}[x, y, x^{-1}, y^{-1}]$.

$\text{Spec }$ is a contravariant functor, so we get a corresponding diagram in $\text{Ring}$:

The explicit maps that define this diagram reveal the desired group law:

That is, $z \mapsto (x+1)(y+1) -1 = xy + x + y$, which is a form of the multiplicative formal group law that we know and love. Intuitively, we can think of it as wanting to multiply $a$ and $b$, but first needing to shift to the multiplicative identity $1$, and afterward translating back again.

#### $\mathbb{A}^1 – \{0\} – \{1\} \simeq \text{Spec } \mathbb{C}[x, x^{-1}]$

What I don’t yet see, dear reader, is how to put group structure on $\text{Spec } \mathbb{C}[x, x^{-1}, (1-x)^{-1}]$ (which is isomorphic to the affine line minus two points). What are your thoughts?

*Thanks to Yifei Zhao for kindly helping me derive the multiplicative formal group law from the group law on the punctured affine line.*