# Prevarieties (Thursday, October 15)

> This corresponds to the end of Chapter 4.

Recall that we had a proposition: morphisms between affine varieties are in bijection with $k\dash$algebra morphisms between their coordinate rings.
As a result, we'll redefine an *affine variety* to be a ringed space isomorphic to an affine variety (using the previous definition of affine variety).
This provides a way of saying when affine varieties embedded in different ways are the "same".

:::{.example}
$\AA^2$ vs $V(x) \subset  \AA^n$.
In fact, the map
\[  
f: \AA^2 &\to \AA^3
(y,z) &\mapsto (0, y, z)
.\]
This is continuous and the pullback of regular functions are again regular.
:::

:::{.remark}
With the new definition, there is a bijection between affine varieties up to isomorphisms and finitely generated $k\dash$algebras up to algebra isomorphism.
:::

:::{.proposition title="Distinguished opens are ringed spaces"}
Let $D(f) \subset X$ be a distinguished open, then $D(f)$ is a ringed space.
This follows because $(X, \OO_X)$ is a ringed space, and we can restrict the structure sheaf to any open subset of $X$.
:::

:::{.proof}
Set
\[  
Y \da \ts{(x, t) \in X\cross \AA^1 \st tf(x) = 1} \subset X\cross \AA^1
.\]
This is an affine variety, since $Y = V(I + \gens{ft-1})$.
This is isomorphic to $D(f)$ by the map
\[  
Y &\to D(f) \\
\tv{x, t} &\mapsto x
.\]
with inverse 
\[
D(f)  &\to Y \\
x     &\mapsto \tv{x, {1\over f(x)} }
\]

![Image](figures/image_2020-10-15-09-50-03.png){width=350px}

Note that $\pi: X\cross \AA^1 \to X$ is regular, using prop 3.8 in [@AndreasGathmann515], if the coordinates of a map are regular functions, then the entire map is a morphism of ringed spaces.
We can then note that $1\over f(x)$ is regular on $D(f)$, since $f\neq 0$ on this set.
:::

:::{.example}
$\AA^2 \smz$ is not an affine variety.
Note that this is also not a distinguished open.
We showed on a HW problem that the regular functions on $\AA^2\smz$ are $k[x, y]$, which are also the regular functions on $\AA^2$.
So there is a map inducing a pullback
\[  
\iota: \AA^2\smz &\to \AA^2 \\ \\
\iota^*: k[x, y]  &\mapsvia{\sim} k[x, y]
.\]
Note that $\iota^*$ is an isomorphism on the space of regular functions, but $\iota$ itself is not an isomorphism of topological spaces.
Why? $i^{-1}$ is not defined at zero.
:::

## Prevarieties (Chapter 5)

:::{.definition title="Prevariety"}
A **prevariety** is a ringed spaced $X$ with a finite open cover by affine varieties.
This is a topological space $X$ with an open cover $\ts{U_i}_{i=1}^n \covers X$ such that $(U_i, \ro{\OO_X}{U_i} )$ is isomorphic to an affine variety.
We'll call $\OO_X$ the sheaf of *regular functions* and $U_i\subset X$ *affine open sets*.
:::

One way to construct prevarieties from affine varieties is by *gluing*:

:::{.definition title="Glued Spaces"}
Let $X_1, X_2$ be prevarieties which are themselves actual varieties,
and let $U_{12} \subset X_1, U_{21} \subset X_2$ be opens with $f: U_{12} \to U_{21}$ an isomorphism of ringed spaces.

![Image](figures/image_2020-10-15-10-08-59.png){width=350px}

As a set, take $X = X_1 \disjoint X_2/\sim$ where $a\sim f(a)$ for all $a\in U_{12}$.
As a topological space, $U \subset X$ is open iff $U_i \da U\intersect X_i$ are open in $X_i$.
As a ringed space, we take 
\[
\OO_X(U) \da \ts{\phi: U\to k \st \ro{\phi}{U_i} \in \OO_{X_i}}
\]
:::

:::{.example}
The prototypical example is $\PP^1_{/k}$ constructed from two copies of $\AA^1_{/k}$.
Set $X_1 = \AA^1, X_2 = \AA^2$, with $U_{12} \da D(x) \subset X_1$ and $U_{21} \da D(y) \subset X_2$.
Then let
\[  
f: U_{12} &\to U_{21} \\
x & \mapsto {1\over x}
.\]
This defines a regular function on $U_{12}$ so defines a morphism $U_{12} \mapsvia{\sim} \AA^1$.

![Gluing two affine lines to obtain projective space.](figures/image_2020-10-15-10-20-32.png){width=350px}

Over $\CC$, topologically this yields a sphere

![Complex projective space is a sphere.](figures/image_2020-10-15-10-23-24.png){width=350px}

Given a ringed space $X = X_1\union X_2$ with a structure sheaf $\OO_X$, what is $\OO_X(X)$? 
By definition, it's 
\[  
\OO_X(X) \da \ts{\phi: X\to k \st \ro{\phi}{X_1}, \ro{\phi}{X_2} \text{ are regular} }
.\]

Then if $\ro{\phi}{X_1} = f(x)$ and $\ro{\phi}{X_2} = g(y)$, we have $y=1/x$ on the overlap and so 
\[
\ro{f(x)}{D(x)} = \ro{g(1/x)}{D(x)}
\]
Since $f, g$ are rational functions agreeing on an infinite set, $f(x) = g(1/x)$ both being polynomial forces $f = g = c$ for some constant $c \in k$.
Thus $\OO_X(X) = k$.

What about $\OO_X(X_1)$?
This is just $k[x]$, and similarly $\OO_X(X_2) = k[y]$.
We can also consider $\OO_X(X_1\intersect X_2) = D(x) \subset X$, so this yields $k[x, 1/x]$.
We thus have a diagram
\begin{tikzcd}
                         & \OO_X(X_1) = k[x] \ar[rd, "x\mapsto x"]   & \\
\OO_X(X) \ar[ru] \ar[rd] &                                           & \OO_X(X_1\intersect X_2) = k[x, 1/x] \\
                         & \OO_X(X_2) = k[y] \ar[ru, "y\mapsto 1/x"] &
\end{tikzcd}
:::