s21-algebra-1/inputs/projective_spaces.tex

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Let $\mathfrak{l}$ be any field.
\begin{definition}
For a $\mathfrak{l}$-vector space $V$,
let $\mathbb{P}(V)$ be the set of
one-dimensional subspaces of $V$.
Let $\mathbb{P}^n(\mathfrak{l}) \coloneqq \mathbb{P}(\mathfrak{l}^{n+1})$,
the \vocab[Projective space]%
{$n$-dimensional projective space over $\mathfrak{l}$}.
If $\mathfrak{l}$ is kept fixed,
we will often write $\mathbb{P}^n$ for $\mathbb{P}^n(\mathfrak{l})$.
When dealing with $\mathbb{P}^n$,
the usual convention is to use $0$ as the index of the first coordinate.
We denote the one-dimensional subspace generated by
$(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\}$ by
$[x_0,\ldots,x_n] \in \mathbb{P}^n$.
If $x = [x_0,
ldots,x_n] \in \mathbb{P}^n$, the $(x_{i})_{i=0}^n$ are called
\vocab{homogeneous coordinates} of $x$.
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At least one of the $x_{i}$ must be $\neq 0$.
\end{definition}
\begin{remark}
There are points $[1,0], [0,1] \in \mathbb{P}^1$ but there
is no point $[0,0] \in \mathbb{P}^1$.
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\end{remark}
\begin{definition}[Infinite hyperplane]
For $0 \le i \le n$ let $U_i \subseteq \mathbb{P}^n$
denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$.
This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$
and $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the
same point $x \in \mathbb{P}^n$ differ by scaling with a
$\lambda \in \mathfrak{l}^{\times}$,
$x_i = \lambda \xi_i$.
Since not all $x_i$ may be $0$, $\mathbb{P}^n = \bigcup_{i=0}^n U_i$.
We identify $\mathbb{A}^n = \mathbb{A}^n(\mathfrak{l}) = \mathfrak{l}^n$
with $U_0$ by identifying $(x_1,\ldots,x_n) \in \mathbb{A}^n$ with
$[1,x_1,\ldots,x_n] \in \mathbb{P}^n$.
Then $\mathbb{P}^1 = \mathbb{A}^1 \cup \{\infty\} $
where $\infty=[0,1]$.
More generally, when $n > 0$ $\mathbb{P}^n \setminus \mathbb{A}^n$ can be
identified with $\mathbb{P}^{n-1}$ identifying
$[0,x_1,\ldots,x_n] \in \mathbb{P}^n \setminus \mathbb{A}^n$ with
$[x_1,\ldots,x_n] \in \mathbb{P}^{n-1}$.
Thus $\mathbb{P}^n$ is $\mathbb{A}^n \cong \mathfrak{l}^n$ with a copy of
$\mathbb{P}^{n-1}$ added as an \vocab{infinite hyperplane} .
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\end{definition}
\subsubsection{Graded rings and homogeneous ideals}
\begin{notation}
Let $\mathbb{I} = \N$ or $\mathbb{I} = \Z$.
\end{notation}
\begin{definition}
By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$
we understand a ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of
subgroups of the additive group $(A, +)$
such that $A_a \cdot A_b \subseteq A_{a + b}$ for $a,b \in \mathbb{I}$
and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in
the sense that every $r \in A$ has a unique decomposition
$r = \sum_{d \in \mathbb{I}} r_d$ with $r_d \in A_d$ and but finitely many
$r_d \neq 0$.
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We call the $r_d$ the \vocab{homogeneous components} of $r$.
An ideal $I \subseteq A$ is called \vocab{homogeneous} if
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\[r \in I \implies \forall d \in \mathbb{I} ~ r_d \in I_d\]
where $I_d \coloneqq I \cap A_d$.
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By a \vocab{graded ring} we understand an $\N$-graded ring.
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In this case,
\[A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\}\]
is called the \vocab{augmentation ideal} of $A$.
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\end{definition}
\begin{remark}[Decomposition of $1$]
If $1 = \sum_{d \in \mathbb{I}} \varepsilon_d$
is the decomposition into homogeneous components,
then $\varepsilon_a = 1 \cdot \varepsilon_a
= \sum_{b \in \mathbb{I}} \varepsilon_a\varepsilon_b$
with $\varepsilon_a\varepsilon_b \in A_{a+b}$.
By the uniqueness of the decomposition into homogeneous components,
$\varepsilon_a \varepsilon_0 = \varepsilon_a$
and $b \neq 0 \implies \varepsilon_a \varepsilon_b = 0$.
Applying the last equation with $a = 0$ gives
$b\neq 0 \implies \varepsilon_b = \varepsilon_0 \varepsilon _b = 0$.
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Thus $1 = \varepsilon_0 \in A_0$.
\end{remark}
\begin{remark}
The augmentation ideal of a graded ring is a homogeneous ideal.
\end{remark}
% Graded rings and homogeneous ideals (2)
\begin{proposition}
\footnote{This holds for both $\Z$-graded and $\N$-graded rings.}
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\begin{itemize}
\item
A principal ideal generated by a homogeneous element is homogeneous.
\item
The operations $\sum, \bigcap, \sqrt{}$ preserve homogeneity.
\item
An ideal is homogeneous iff it can be generated by a family of homogeneous
elements.
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\end{itemize}
\end{proposition}
\begin{proof}
Most assertions are trivial.
We only show that $J$ homogeneous $\implies \sqrt{J}$ homogeneous.
Let $A$ be $\mathbb{I}$-graded,
$f \in \sqrt{J} $ and
$f = \sum_{d \in \mathbb{I}} f_d$ the decomposition.
To show that all $f_d \in \sqrt{J} $,
we use induction on $N_f \coloneqq \# \{d \in \mathbb{I} | f_d \neq 0\}$.
$N_f = 0$ is trivial.
Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq 0$.
For $l \in \N$, the $le$-th homogeneous component of $f^l$ is $f_e^l$.
Choosing $l$ large enough such that $f^l \in J$ and using the homogeneity of
$J$, we find $f_e \in \sqrt{J}$.
As $\sqrt{J} $ is an ideal,
$\tilde f \coloneqq f - f_e \in \sqrt{J} $.
As $N_{\tilde f} = N_f -1$,
the induction assumption may be applied to $\tilde f$
and shows $f_d \in \sqrt{J} $ for $d \neq e$.
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\end{proof}
\begin{fact}
A homogeneous ideal is finitely generated
iff it can be generated by finitely many of its homogeneous elements.
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In particular, this is always the case when $A$ is a Noetherian ring.
\end{fact}
\subsubsection{The Zariski topology on $\mathbb{P}^n$}
\begin{notation}
Recall that for
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$\alpha \in \N^{n+1}$
\[|\alpha| = \sum_{i=0}^{n} \alpha_i \text{ and }
x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}.\]
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\end{notation}
\begin{definition}[Homogeneous polynomials]
Let $R$ be any ring and
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\[f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n].\]
We say that $f$ is \vocab{homogeneous of degree $d$}
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if
\[|\alpha| \neq d \implies f_\alpha = 0.\]
We denote the subset of homogeneous polynomials of degree $d$ by
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\[R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n].\]
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\end{definition}
\begin{remark}
This definition gives $R$ the structure of a graded ring.
\end{remark}
\begin{definition}[Zariski topology on $\mathbb{P}^n(\mathfrak{k})$]
\label{ztoppn}
Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.%
\footnote{As always, $\mathfrak{k}$ is algebraically closed}
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For $f \in A_d = \mathfrak{k}[X_0,\ldots,X_n]_d$,
the validity of the equation $f(x_0,\ldots,x_{n}) = 0$
does not depend on the choice of homogeneous coordinates, as
\[
f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d f(x_0,\ldots,x_n).
\]
Let $\Vp(f) \coloneqq \{x \in \mathbb{P}^n | f(x) = 0\}$.
We call a subset $X \subseteq \mathbb{P}^n$ Zariski-closed if it
can be represented as
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\[
X = \bigcap_{i=1}^k \Vp(f_i)
\]
where the $f_i \in A_{d_i}$ are homogeneous polynomials.
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\end{definition}
\pagebreak
\begin{fact}
If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq \mathbb{P}^n$ is closed,
then $Y = X \cap \mathbb{A}^n$ can be identified with the closed subset
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\[
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n |
f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\}
\subseteq \mathfrak{k}^n.
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\]
Conversely, if $Y \subseteq \mathfrak{k}^n$ is closed it has the form
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\[
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n |
g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\}
\]
and can thus be identified with $X \cap \mathbb{A}^n$
where $X \coloneqq \bigcap_{i=1}^k \Vp(f_i)$ is given by
\[
f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i
\ge \deg(g_i).
\]
Thus, the Zariski topology on $\mathfrak{k}^n$ can be
identified with the topology induced by the Zariski topology on
$\mathbb{A}^n = U_0$,
and the same holds for $U_i$ with $0 \le i \le n$.
In this sense,
the Zariski topology on $\mathbb{P}^n$ can be thought of as
gluing the Zariski topologies on the $U_i \cong \mathfrak{k}^n$.
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\end{fact}
% The Zariski topology on P^n (2)
\begin{definition}
Let $I \subseteq A = \mathfrak{k}[X_0,\ldots,X_n]$ be a homogeneous ideal.
Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \mathbb{P}^n | \forall f \in I ~%
f(x_0,\ldots,x_n) = 0\}$
As $I$ is homogeneous,
it is sufficient to impose this condition for the homogeneous elements
$f \in I$.
Because $A$ is Noetherian,
$I$ can finitely generated by homogeneous elements $(f_i)_{i=1}^k$ and
$\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as in \ref{ztoppn}.
Conversely, if the homogeneous $f_i$ are given,
then $I = \langle f_1,\ldots,f_k \rangle_A$ is homogeneous.
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\end{definition}
\begin{remark}
Note that $V(A) = V(A_+) = \emptyset$.
\end{remark}
\begin{fact}
For homogeneous ideals in $A$ and $m \in \N$, we have:
\begin{itemize}
\item
$\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Vp(I_\lambda)$.
\item
$\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) = \bigcup_{k=1}^m \Vp(I_k)$.
\item
$\Vp(\sqrt{I}) = \Vp(I)$.
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\end{itemize}
\end{fact}
\begin{fact}
If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an open covering
of a topological space then $X$ is Noetherian
iff there is a finite subcovering and all $U_\lambda$ are Noetherian.
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\end{fact}
\begin{proof}
By definition, a topological space is Noetherian
$\iff$ all open subsets are quasi-compact.
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\end{proof}
\begin{corollary}
The Zariski topology on $\mathbb{P}^n$ is indeed a topology.
The induced topology on the open set
$\mathbb{A}^n = \mathbb{P}^n \setminus \Vp(X_0) \cong \mathfrak{k}^n$
is the Zariski topology on $\mathfrak{k}^n$.
The same holds for all
$U_i = \mathbb{P}^n \setminus \Vp(X_i) \cong \mathfrak{k}^n$.
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Moreover, the topological space $\mathbb{P}^n$ is Noetherian.
\end{corollary}
\subsection{Noetherianness of graded rings}
\begin{proposition}
For a graded ring $R_{\bullet}$,
the following conditions are equivalent:
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\begin{enumerate}[A]
\item
$R$ is Noetherian.
\item
Every homogeneous ideal of $R_{\bullet}$ is finitely generated.
\item
Every chain $I_0\subseteq I_1 \subseteq \ldots$ of
homogeneous ideals terminates.
\item
Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals
has a $\subseteq$-maximal element.
\item
$R_0$ is Noetherian and the ideal $R_+$ is finitely generated.
\item
$R_0$ is Noetherian and $R / R_0$ is of finite type.
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\end{enumerate}
\end{proposition}
\begin{proof}
\noindent\textbf{A $\implies$ B,C,D} trivial.
\noindent\textbf{B $\iff$ C $\iff$ D} similar to the proof about Noetherianness.
\noindent\textbf{B $\land$ C $\implies $E}
B implies that $R_+$ is finitely generated.
Since $I \oplus R_+$ is homogeneous for any homogeneous ideal $I \subseteq
R_0$, C implies the Noetherianness of $R_0$.
\noindent\textbf{E $\implies$ F}
Let $R_+$ be generated by $f_i \in R_{d_i}, d_i > 0$ as
an ideal.
\begin{claim}
The $R_0$-subalgebra $\tilde R$ of $R$ generated by the $f_i$ equals $R$.
\end{claim}
\begin{subproof}
It is sufficient to show that every homogeneous $f \in R_d$
belongs to $\tilde R$.
We use induction on $d$.
The case of $d = 0$ is trivial.
Let $d > 0$ and $R_e \subseteq \tilde R$ for all $e < d$.
As $f \in R_+$, $f = \sum_{i=1}^{k} g_if_i$.
Let $f_a = \sum_{i=1}^{k} g_{i,
a-d_i} f_i$, where $g_i = \sum_{b=0}^{\infty} g_{i,b}$
is the decomposition into homogeneous components.
Then $f = \sum_{a=0}^{\infty} f_a$ is the decomposition of $f$ into
homogeneous components,
hence $a \neq d \implies f_a = 0 $.
Thus we may assume $g_i \in R_{d-d_i}$.
As $d_i > 0$,
the induction assumption may now be applied to $g_i$,
hence $g_i \in \tilde R$,
hence $f \in \tilde R$.
\end{subproof}
\noindent\textbf{F $\implies$ A}
Hilbert's Basissatz (\ref{basissatz})
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\end{proof}
\subsection{The projective form of the Nullstellensatz and the closed subsets
of $\mathbb{P}^n$} Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
% Lecture 12
\begin{proposition}[Projective form of the Nullstellensatz]
\label{hnsp}
If $I \subseteq A$ is a homogeneous ideal and $f \in A_d$ with $d>0$,
then $\Vp(I) \subseteq \Vp(f) \iff f \in \sqrt{I}$.
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\end{proposition}
\begin{proof}
$\impliedby$ is clear.
Let $\Vp(I) \subseteq \Vp(f)$.
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If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in
which case $f(x) = 0$ since $d > 0$
or the point $[x_0,\ldots,x_n] \in \mathbb{P}^n$ is
well-defined and belongs to $\Vp(I) \subseteq \Vp(f)$,
hence $f(x) = 0$.
Thus $\Va(I) \subseteq \Va(f)$ and $f \in \sqrt{I}$ be the Nullstellensatz
(\ref{hns3}).
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\end{proof}
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\begin{definition}%
\footnote{This definition is not too important, the characterization in the following remark suffices.}
For a graded ring $R_\bullet$,
let $\Proj(R_\bullet)$ be the set of $\fp \in \Spec R$
such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$.
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\end{definition}
\begin{remark}
\label{proja}
As the elements of $A_0 \setminus \{0\}$ are units in $A$
it follows that for every homogeneous ideal $I$
we have $I \subseteq A_+$ or $I = A$.
In particular,
$\Proj(A_\bullet) = \{\fp \in \Spec A \setminus A_+ | \fp \text{ is homogeneous}\} $.
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\end{remark}
\begin{proposition}
\label{bijproj}
There is a bijection
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\begin{align*}
f: \{I \subseteq A_+ | I \text{ homogeneous ideal}, I = \sqrt{I}\} &
\longrightarrow \{X \subseteq \mathbb{P}^n | X \text{ closed}\}\\
I &
\longmapsto \Vp(I)\\
\langle \{f \in A_d | d > 0, X \subseteq \Vp(f)\} \rangle &
\longmapsfrom X
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\end{align*}
Under this bijection,
the irreducible subsets correspond to the elements of
$\Proj(A_\bullet)$.
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\end{proposition}
\begin{proof}
From the projective form of the Nullstellensatz it follows
that $f$ is injective
and that $f^{-1}(\Vp\left( I \right)) = \sqrt{I} = I$.
If $X \subseteq \mathbb{P}^n$ is closed,
then $X = \Vp(J)$ for some homogeneous ideal $J \subseteq A$.
Without loss of generality loss of generality $J = \sqrt{J}$.
If $J \not\subseteq A_+$, then $J = A$ (\ref{proja}),
hence $X = \Vp(J) = \emptyset = \Vp(A_+)$.
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Thus we may assume $J \subseteq A_+$, and $f$ is surjective.
Suppose $\fp \in \Proj(A_\bullet)$.
Then $\fp \neq A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the
proven part of the proposition.
Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets,
where $X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k = \sqrt{I_k}$.
Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \setminus \fp$.
We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq \Vp(I_k)$
hence $\Vp(f_1f_2) \supseteq \Vp(I_1) \cup \Vp(I_2) = X = \Vp(\fp)$
and it follows that $f_1f_2\in \sqrt{\fp} = \fp \lightning$.
Assume $X = \Vp(\fp)$ is irreducible,
where $\fp = \sqrt{\fp} \in A_+$ is homogeneous.
The $\fp \neq A_+$ as $X = \emptyset$ otherwise.
Assume that $f_1f_2 \in \fp$ but $f_i \not\in A_{d_i} \setminus \fp$.
Then $X \not \subseteq \Vp(f_i)$ by the projective
Nullstellensatz when $d_i > 0$
and because $\Vp(1) = \emptyset$ when $d_i = 0$.
Thus $X = (X \cap \Vp\left( f_1 \right)) \cup (X \cap \Vp(f_2))$ is a
proper decomposition $\lightning$.
By lemma \ref{homprime}, $\fp$ is a prime ideal.
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\end{proof}
\begin{remark}
It is important that $I \subseteq A_{\color{red} +}$,
since $\Vp(A) = \Vp(A_+) = \emptyset$ would be a counterexample.
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\end{remark}
\begin{corollary}
$\mathbb{P}^n$ is irreducible.
\end{corollary}
\begin{proof}
Apply \ref{bijproj} to $\{0\} \in \Proj(A_\bullet)$.
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\end{proof}
\subsection{Some remarks on homogeneous prime ideals}
\begin{lemma}
\label{homprime}
Let $R_\bullet$ be an $\mathbb{I}$ graded ring
($\mathbb{I} = \N$ or $\mathbb{I} = \Z$).
A homogeneous ideal $I \subseteq R$ is a prime ideal
iff $1 \not\in I$ and for all homogeneous elements $f, g \in R$
\[fg \in I \implies f \in I \lor g \in I.\]
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\end{lemma}
\begin{proof}
$\implies$ is trivial.
It suffices to show that for arbitrary $f,g \in R$
we have that $fg \in I \implies f \in I \lor g \in I$.
Let $f = \sum_{d \in \mathbb{I}} f_d, g = \sum_{d \in \mathbb{I}} g_d $ be the
decompositions into homogeneous components.
If $f \not\in I$ and $g \not\in I$ there are $d,e \in I$ with $f_d \in I$,
$g_e \in I$, and they may assumed to be maximal with this property.
As $I$ is homogeneous and $fg \in I$, we have $(fg)_{d+e} \in I$ but
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\[
(fg)_{d+e} = f_dg_e + \sum_{\delta = 1}^{\infty} (f_{d + \delta} g_{e - \delta}
+ f_{d - \delta} g_{e + \delta})
\]
where $f_dg_e \not\in I$ by our assumption
on $I$ and all other summands on the right hand side are $\in I$
(as $f_{d+ \delta} \in I$ and $g_{e + \delta} \in I$ by the maximality
of $d$ and $e$),
a contradiction.
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\end{proof}
\begin{remark}
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If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, then
\[\fp \oplus R_+ = \{r \in R | r_0 \in \fp\}\]
is a homogeneous prime ideal of $R$.
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\begin{IEEEeqnarray*}{rCl}
&&\{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } R_\bullet\}\\
&=&\Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec R_0\}.
\end{IEEEeqnarray*}
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\end{remark}
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\pagebreak
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\subsection{Dimension of $\mathbb{P}^n$}
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\begin{proposition}\,
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\begin{itemize}
\item
$\mathbb{P}^n$ is catenary.
\item
$\dim(\mathbb{P}^n) = n$.
Moreover, $\codim(\{x\} ,\mathbb{P}^n) = n$ for every $x \in \mathbb{P}^n$.
\item
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If $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$, then
\[\codim(\{x\}, X) = \dim(X) = n - \codim(X, \mathbb{P}^n).\]
\item
If $X \subseteq Y \subseteq \mathbb{P}^n$ are irreducible subsets,
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then \[\codim(X,Y) = \dim(Y) - \dim(X).\]
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\end{itemize}
\end{proposition}
\begin{proof}
Let $X \subseteq \mathbb{P}^n$ be irreducible.
If $x \in X$, there is an integer $0 \le i \le n$ and
$X \in U_i = \mathbb{P}^n \setminus \Vp(X_i)$.
Without loss of generality loss of generality $i = 0$.
Then $\codim(X, \mathbb{P}^n) = \codim(X \cap \mathbb{A}^n, \mathbb{A}^n)$ by
the locality of Krull codimension (\ref{lockrullcodim}).
Applying this with $X = \{x\}$
and our results about the affine case gives the second assertion.
If $Y$ and $Z$ are also irreducible with $X \subseteq Y \subseteq Z$,
then
$\codim(X,Y) = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)$,
$\codim(X,Z) = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$
and $\codim(Y,Z) = \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
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Thus
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\begin{IEEEeqnarray*}{rCl}
\codim(X,Y) + \codim(Y,Z)
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& = &\codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)\\
& & + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\
& = &\codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)\\
& = &\codim(X, Z)
\end{IEEEeqnarray*}
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because $\mathfrak{k}^n$ is catenary and the first point follows.
The remaining assertions can easily be derived from the first two.
\end{proof}
\subsection{The cone $C(X)$}
\begin{definition}
If $X \subseteq \mathbb{P}^n$ is closed, we define the
\vocab{affine cone over $X$}
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\[
C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\} | [x_0,\ldots,x_n] \in X\}
\]
If $X = \Vp(I)$ where $I \subseteq A_+ = \mathfrak{k}[X_0,\ldots,X_n]_+$
is homogeneous, then $C(X) = \Va(I)$.
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\end{definition}
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\begin{proposition}\,
\label{conedim}
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\begin{itemize}
\item
$C(X)$ is irreducible iff $X$ is irreducible or $X = \emptyset$.
\item
If $X$ is irreducible, then
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\[\dim(C(X)) = \dim(X) + 1\] and
\[\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \mathbb{P}^n).\]
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\end{itemize}
\end{proposition}
\begin{proof}
The first assertion follows from \ref{bijproj} and \ref{bijiredprim}
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(bijection of irreducible subsets and prime ideals in the projective
and affine case).
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Let $d = \dim(X)$ and
\[
X_0 \subsetneq \ldots \subsetneq X_d = X
\subsetneq X_{d+1} \subsetneq \ldots \subsetneq X_n = \mathbb{P}^n
\]
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be a chain of irreducible subsets of $\mathbb{P}^n$.
Then
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\[
\{0\} \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X)
\subsetneq \ldots \subsetneq C(X_n) = \mathfrak{k}^{n+1}
\]
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is a chain of irreducible subsets of $\mathfrak{k}^{n+1}$.
Hence $\dim(C(X)) \ge 1 + d$ and
$\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$.
Since
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\[\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) = n+1,\]
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the two inequalities must be equalities.
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\end{proof}
\subsubsection{Application to hypersurfaces in $\mathbb{P}^n$}
\begin{definition}[Hypersurface]
Let $n > 0$.
By a \vocab{hypersurface} in $\mathbb{P}^n$ or
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$\mathbb{A}^n$ we understand an irreducible closed subset
of codimension $1$.
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\end{definition}
\begin{corollary}
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If $P \in A_d$ is a prime element,
then $H = \Vp(P)$ is a hypersurface in $\mathbb{P}^n$
and every hypersurface $H$ in $\mathbb{P}^n$ can be obtained in this way.
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\end{corollary}
\begin{proof}
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If $H = \Vp(P)$ then $C(H) = \Va(P)$
is a hypersurface in $\mathfrak{k}^{n+1}$ by \ref{irredcodimone}.
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By \ref{conedim}, $H$ is irreducible and of codimension $1$.
Conversely, let $H$ be a hypersurface in $\mathbb{P}^n$.
By \ref{conedim}, $C(H)$ is a hypersurface in $\mathfrak{k}^{n+1}$,
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hence $C(H) = \Vp(P)$ for some prime element $P \in A$
(again by \ref{irredcodimone}).
We have $H = \Vp(\fp)$ for some $\fp \in \Proj(A)$ and $C(H) = \Va(\fp)$.
By the bijection between closed subsets of $\mathfrak{k}^{n+1}$ and ideals
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$I = \sqrt{I} \subseteq A$ (\ref{antimonbij}),
$\fp = P \cdot A$.
Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition
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into homogeneous components.
If $P_e $ with $e < d$ was $\neq 0$,
it could not be a multiple of $P$ contradicting the homogeneity of
$\fp = P \cdot A$.
Thus, $P$ is homogeneous of degree $d$.
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\end{proof}
\begin{definition}
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A hypersurface $H \subseteq \mathbb{P}^n$ has \vocab{degree $d$}
if $H = \Vp(P)$,
where $P \in A_d$ is an irreducible polynomial.
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\end{definition}
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\subsubsection{Application to intersections in $\mathbb{P}^n$
and Bezout's theorem}
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\begin{corollary}
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Let $A \subseteq \mathbb{P}^n$ and $B \subseteq \mathbb{P}^n$ be
irreducible subsets of dimensions $a$ and $b$.
If $a+ b \ge n$, then $A \cap B \neq \emptyset$
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and every irreducible component of $A \cap B$
has dimension $\ge a + b - n$.
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\end{corollary}
\begin{remark}
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This shows that $\mathbb{P}^n$ indeed fulfilled the goal of allowing for
nicer results of algebraic geometry because ``solutions at infinity''
to systems of algebraic equations are present in $\mathbb{P}^n$
(see \ref{affineproblem}).
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\end{remark}
\begin{proof}
The lower bound on the dimension of irreducible components of $A \cap B$ is
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easily derived from the similar affine result
(corollary of the principal ideal theorem, \ref{codimintersection}).
From the definition of the affine cone it follows that
$C(A \cap B) = C(A) \cap C(B)$.
We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by \ref{conedim}.
If $A \cap B = \emptyset$,
then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an irreducible component,
contradicting the lower bound $a + b + 1 - n > 0$ for
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the dimension of irreducible components of $C(A) \cap C(B)$
(again \ref{codimintersection}).
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\end{proof}
\begin{remark}[Bezout's theorem]
If $A \neq B$ are hypersurfaces of degree $a$ and $b$
in $\mathbb{P}^2$,
then $A \cap B$ has $ab$ points counted by (suitably defined) multiplicity.
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\end{remark}
%TODO Proof of "Dimension of P^n"
% SLIDE APPLICATION TO HYPERSURFACES IN $\P^n$
%ERROR: C(H) = V_A(P)
%If n = 0, P = 0, V_P(P) = \emptyset is a problem!