# Projective Space & Homogeneous Polynomials (Thursday, October 29)

## Projective Space

:::{.definition title="Projective Space"}
Let $n\in \NN$, and define **projective $n\dash$space** over $k$ by \
\[  
\PP^n_{/k} = \ts{\text{lines through the origin in } k^{n+1}}
.\]
:::

:::{.remark}
For notation, given $L\in \PP^n/k$, it is spanned by any nonzero points $\tv{x_0, \cdots, x_n} \in L$, and $L$ is uniquely determined by this point up to scaling by elements in $k\units$. 
In this case, we write $L = \tv{x_0: \cdots : x_n} = \tv{\lambda x_0: \cdots : \lambda x_n}$.
We can then alternatively define $\PP^n{_/k} \da \qty{ k^{n+1}\smz} / \sim$ where we mod out by scalar multiplication $x\sim \lambda x$ for $\lambda\in k\units$.
We call $[x_1 : \cdots : x_n]$ the *homogeneous coordinates* on $\PP^n/k$.
:::

:::{.remark}
Consider the map
\[  
\AA^n \to \PP^n \\
\tv{x_1, \cdots, x_n} &\mapsto \tv{1: x_1 : \cdots : x_n}
.\]
This is injective.
Conversely, consider
\[  
"D(x_0)" \subset \PP^n \da \ts{\tv{x_0 : \cdots : x_n} \st x_0\neq 0}
.\]
This is a well-defined subset of $\PP^n$, since it only depends on the equivalence class of a point.
In this case, there is a unique $\lambda(x_0, \cdots, x_n)$, namely $\lambda = 1/x_0$, such that each point in this set is of the form $\tv{1: {x_1\over x_0} : \cdots : {x_n \over x_0}}$, yielding a copy of $\AA^n\subset \PP^n$ given by points $\tv{{x_1\over x_0}, \cdots, {x_n\over x_0}}$.
What is its complement?
It's given by $\ts{\tv{0: x_1: \cdots : x_n}} \subset \PP^n$, which is equal (as a set) to a copy of $\PP^{n-1}$ defined by the set of lines in $k^n$ defined by $x_0 = 0$.
:::

:::{.example title="?"}
Note that $\PP^1$ contains a copy of $\AA^1$ where $x_0 \neq 0$ and a second copy where $x_1 \neq 0$, yielding maps
\[  
f_1: \PP^1 &\to \AA^1 \\
\tv{x_0: x_1} &\mapsto \tv{{x_0 \over x_1}} \\ \\
f_2: \PP^1 &\to \AA^1 \\
\tv{x_0: x_1} &\mapsto \tv{{x_1 \over x_0}}
,\]
since every point in $\PP^1$ corresponds to some line in $\AA^2$, and thus has either $x_0\neq 0$ or $x_1 \neq 0$.
These two copies cover $\PP^1$, and the "transition map" is inversion.
:::

:::{.remark}
More generally, there are $n+1$ projection $\PP^n \surjects \AA^n$ given by dividing by the $j$th coordinate, and the union of their images is the entire space.
The gluing construction gives $\PP^n$ the structure of a prevariety: we can consider $D(x_j) \subset \PP^n$ where each has the structure of a ringed space $(\AA^n, \OO_{\AA^n})$.
We have $D(x_i) \intersect D(x_j) \subset  D(x_i)$, which has coordinate $\ts{ x_k/x_i,\, k\neq i }$, and similarly $D(x_i) \intersect D(x_j) \subseteq D(x_j)$ with coordinates $\ts{ x_k/x_j, \, k\neq j}$.
Their intersection is $D(x_i / x_j) \cong \AA^{n-1}$.
:::

:::{.example title="?"}
Consider $\PP^1$, then 
$D(x_0) \cong \AA^1$ with which contains a copy of $\AA^1$ with coordinate ring $k[x_1 / x_0]$ and a subset $D(x_1 / x_0)$ with coordinate ring $k[y, y\inv]$, and similarly,
$D(x_1) \cong \AA^1$ has coordinate ring $k[x_0\over x_1]$ and contains $D(x_0/  x_1)$ with coordinate ring $k[z, z\inv]$.
Consider their overlap $D(x_0) \intersect D(x_1)$.
\todo[inline]{Might be mistakes here.}

When do $y, z$ denote the same point in $\PP^1$?
When $y = z\inv$.
We can conclude that the $n+1$ copies $D(x_i) \subset \PP^n$ are affine varieties isomorphic as ringed spaces on the overlaps, so the gluing construction makes $\PP^n$ a prevariety.
:::

## Homogeneous Polynomials

:::{.definition title="Homogeneous Degrees of Polynomials"}
A polynomial $f$ is **homogeneous of degree $d$** if every monomial in $f$ has total degree $d$.
:::

:::{.example title="?"}
The polynomial
\[  
f(x_0, x_1, x_2) = x_0^3 + x_1 x_2^2 + x_0 x_1 x_2
.\]
has homogeneous degree 3.
:::

:::{.remark}
If $f$ is homogeneous of degree $d$, then for all $\lambda \in k\units$, 
\[  
f(\lambda x_0, \cdots, \lambda x_n) = \lambda^d f(x_0, \cdots, x_n)
.\]

If $f$ is homogeneous, $V(f) \subset \PP^n$ is a well-defined subset, since 
\[
f(x_0, \cdots, x_n) = 0 \iff \lambda^d f(x_0, \cdots, x_n) = 0 \iff f(\lambda x_0, \cdots, \lambda x_n) = 0
\]
:::

:::{.definition title="Graded Rings"}
A **graded ring** $R$ is a ring $R$ with abelian subgroups $R_d \subset R$ with 

- $R = \bigoplus_{d\geq 0} R_d$, and
- For all $f\in R_d$ and $g\in R_{d'}$, we have $fg \in R_{d+d'}$ and $R_d + R_{d} \subset R_d$.
:::