# Homogenization and Dehomogenization (Tuesday, November 10)


Last time: projective varieties $V(f_i) \subset \PP^n_{/k}$ with $f_i$ homogeneous.
We proved the projective nullstellensatz: for any projective variety $X$, we have $V_p(I_p(X))$ and for any homogeneous ideal $I$ with $\sqrt{I} \neq I_0$ the irrelevant ideal, $I_p(V_p(I)) = \sqrt{I}$.
Recall that $I_0 = \gens{x_0, \cdots, x_n}$.
We had a notion of a projective coordinate ring, $S(X) \da \kx{n} / I_p(X)$, which is a graded ring since $I_p(X)$ is a homogeneous ideal.

:::{.remark}
Note that $S(X)$ is not a ring of functions on $X$: e.g. for $X= \PP^n$, $S(X) = \kx{n}$ but $x_0$ is not a function on $\PP^n$.
This is because $f\qty{\tv{x_0: \cdots : x_n}} = f\qty{\tv{\lambda x_0: \cdots : \lambda x_n}}$ but $x_0\neq \lambda x_0$.
It still makes sense to ask when $f$ is zero though, so $V_p(f)$ is a well-defined object.
:::


## Dehomogenization

:::{.definition title="Dehomogenization of functions and ideals"}
Let $f\in \kx{n}$ be a homogeneous polynomial, then we define its **dehomogenization** as
\[  
f^i \da f(1, x_1, \cdots, x_n) \in k[x_1,\cdots, x_n]
.\]

For a homogeneous ideal, we define
\[  
J^i \da \ts{f^i \st f\in J}
.\]
:::

:::{.example}
The dehomogenization is usually not homogeneous. 
Take
\[  
f &\da x_0^3 + x_0 x_1^2 + x_0 x_1 x_2 + x_0^2 + x_1 \\
\implies f^i &= 1  +x_1^2 + x_1 x_2 + x_1
,\]
which has terms of mixed degrees.
:::

:::{.remark}

\[  
(fg)^i &= f^i g^i \\
(f+g)^i &= f^i + g^i
.\]

In other words, evaluating at $x_0 = 1$ is a ring morphism.
:::

## Homogenization

:::{.definition title="Homogenization of a function"}
Let $f\in \kx{n}$, then the **homogenization** of $f$ is defined by
\[  
f^h \da x_0^d f\qty{ {x_1 \over x_0}, \cdots, {x_n \over x_0} }
\]
where $d\da \deg(f)$.
:::


:::{.example title="?"}
Set 
\[  
f(x_1, x_2) &\da  1 + x_1^2 + x_1 x_2 + x_2^3 \\
\implies f^h(x_0, x_1, x_2) &= x_0^3 + x_0 x_1^2 + x_0 x_1 x_2 + x_2^3
,\]
which is a homogeneous polynomial of degree $3$.
Note that $(f^h)^i = f$.
:::

:::{.example title="?"}
It need not be the case that $(f^i)^h = f$.
Take $f = x_0^3 + x_0 x_1 x_2$, then
$f^i = 1 + x_1 x_2$ and $(f^i)^h = x_0^2 + x_1 x_2$.
Note that the total degree dropped, since everything was divisible by $x_0$.
:::

:::{.remark}
\[  
(f^i)^h = f \iff x_0 \notdivides f
.\]
:::

:::{.definition title="Homogenization of an ideal"}
Given $J\subset \kx{n}$, define its **homogenization** as 
\[  
J^h \da \ts{f^h \st f\in J}
.\]

:::

:::{.example}
This is not a ring morphism, since $(f+g)^h \neq f^h + g^h$ in general.
Taking $f = x_0^2 + x_1$ and $g= -x_0^2 + x_2$, we have 
\[  
f^h + g^h &= x_0 x_1 + x_0 x_2 \\ 
(f+g)^h &= x_1 + x_2
.\]
:::

What is the geometric significance?

:::{.proposition title="Geometric significance of homogenization"}
Set 
\[
U_0 \da \ts{\tv{x_0: \cdots :x_n} \in \PP^n_{/k} \st x_0 \neq 0 } \cong \AA^n_{/k}
\]
with coordinates $\tv{{x_1\over x_0} : \cdots : {x_n \over x_0}}$.
Then $U_0$ with the subspace topology is equal to $\AA^n$ with the Zariski topology.
:::

:::{.proof title="?"}

If we define the Zariski topology on $\PP^n$ as having closed sets $V_p(I)$, we would want to check that $\AA^n\cong U_0 \subset \PP^n$ is closed in the subspace topology.
This amounts to showing that $V_p(I) \intersect U_0$ is closed in $\AA^n \cong U_0$.
We can check that
\[  
V_p\qty{f \st f\in I} = \ts{\vector x \da \tv{x_0:\cdots:x_n} \st f(\vector x) = 0 \,\, \forall f\in I}
.\]
Intersecting with $U_0$ yields 
\[  
V_p\qty{f\st f\in I} \Intersect U_0 = \ts{\tv{x_1:\cdots:x_n} \st f(\vector x) = 0,\, x_0\neq 0}
.\]
Equivalently, we can rewrite this set $S$ as 
\[
S = \ts{\tv{x_1:\cdots:x_n} \st f\qty{\tv{1, {x_1 \over x_0}, \cdots,{x_n \over x_0} }} = 0,\, f \text{ homogeneous}}
\]
Since these are coordinates on $\AA^1$, we have $V_p(I) \intersect U_0 = V_a(I^i)$ which is closed.
Conversely, given a closed set $V(I)$, we can write this as $V(I) = U_0 \intersect V_p(I^h)$.
:::

:::{.corollary title="Projective space is irreducible"}
$\PP^n_{/k}$ is irreducible of dimension $n$.
:::

:::{.proof title="?"}
This follows from the fact that $\PP^n$ is covered by irreducible topological spaces of dimension $n$ with nonempty intersection, along with a fact from the exercises.
:::

:::{.example title="?"}
Consider $f(x_1, x_2) = x_1^2 - x_2^2 - 1$ and consider $V(f) \subset \AA^2_{/\CC}$:

![The variety $V(x_1^2 - x_2^2 - 1)$](figures/image_2020-11-10-10-10-42.png){width=350px}

Note that for real projective space, we can view this as a sphere with antipodal points identified. 
We can thus visualize this in the following way:

![Projective 2-space as sphere with anitpodal points identified](figures/image_2020-11-10-10-12-20.png){width=350px}

We can normalize the $x_0$ coordinate to one, hence the plane.
We can also project $V(f)$ from the plane onto the sphere, mirroring to antipodal points:

![Projecting a variety onto a sphere](figures/image_2020-11-10-10-14-09.png){width=350px}

This misses some points on the equator, since we aren't including points where $x_0 = 0$.
Consider the homogenization $V(f^h) \subset \PP^2_{/\CC}$:
\[  
V(f^h) = V(x_1^2 - x_2^2 - x_0^2)
.\]
Then
\[  
V(f^h) \intersect V(x_0) = \ts{\tv{0:x_1:x_2} \st f^h(0, x_1, x_2) = 0 } = \ts{\tv{0:1:1}, \tv{0:1:-1}}
,\]
which can be seen in the picture as the points at infinity:

![Homogenization and points at infinity](figures/image_2020-11-10-10-19-19.png){width=550px}

Note that the equator is $V(x_0) = \PP^2_{/\CC}\sm U_0 \cong \PP^2\sm \AA^2$.
So we get a circle of points at infinity, i.e. $V(x_0) = \PP^1 = \ts{\tv{0:v_1:v_2}}$.
:::

:::{.example title="?"}
Consider $V(f)$ where $f$ is a line in $\AA^2_{/\CC}$, say $f(x_1, x_2) = ax_1 + bx_2 + c$.
This yields $f^h = ax_1 + bx_2 + cx_0$ and we can consider $V(f^h) \cong \PP^2_{\CC}$.
We know $\PP^1_{\CC}$ is topologically a sphere and $\AA^1_{/\CC}$ is a point:

![$\PP^1_{\CC}$](figures/image_2020-11-10-10-26-40.png){width=250px}

The points at infinity correspond to 
\[  
V(f^h) = V(f^h) \intersect V(x_0) = \ts{\tv{0:-b:a}}
,\]
which is a single point not depending on $c$.
:::

:::{.remark}
$\PP^2_{/k}$ for any field $k$ is a **projective plane**, which satisfies certain axioms:

1. There exists a unique line through any two distinct points,

2. Any two distinct lines intersect at a single point.

A famous example is the *Fano plane*:

![Fano Plane](figures/image_2020-11-10-10-29-34.png)

Why is this true?
$\PP^2_{/k}$ is the set of lines in $k^3$, and the lines in $\PP^2_{/k}$ are the vanishing loci of homogeneous polynomials and also planes in $k^3$, since any two lines determine a unique plane and any two planes intersect at the origin.
:::


:::{.proposition title="?"}
Let $J\subset \kx{n}$ be an ideal.
Let $X \da V_a(J) \subset \AA^n$ where we identify $\AA^n = U_0 \subset \PP^n$.
Then the closure $\bar{X} \subset \PP^n$ is given by $\bar{X} = V_p(J^h)$.
In particular, 
\[
V_a(J) = V_p(J^h)
\]
:::

:::{.proof title="?"}
$\supseteq$:
It's clear that $V_p(J^h)$ is closed and contains $V_a(J)$.

$\subseteq$
Let $Y\containing X$ be closed; we want to show that $Y\contains V_p(J^h)$.
Since $Y$ is closed, $Y = V_p(J')$ where $J'$ is some homogeneous ideal.
Any element $f'\in J'$ satisfies $f' = x^d f$ for some maximal $d$ where $x_0^d f$ vanishes on $X$.
We also have $f=0$ on $X$ since $X\subset U_0$.
We can compute
\[  
f\in I_a(X) = I_a(V_a(J)) = \sqrt J
,\]
so $f^m\in J$.
Then $(f^h)^m \in J^h$ for some $m$, and this $f^h \in \sqrt{J^h}$.
So $J'\subset \sqrt J$, which shows that $V_p(J') \contains V_p(J^h)$ as desired.
:::

:::{.definition title="Projective Closure"}
The **projective closure** of $X = V_a(J)$ is the smallest closed subset containing $X$ and is given by
\[  
\bar{X} = V_p(J^h)
.\]
:::