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 $x$.
    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 $[0,0]
    \in  \mathbb{P}^1$.
\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} .
\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  $r_d
    \neq 0$.

    We call the $r_d$ the \vocab{homogeneous components} of $r$.

    An ideal $I \subseteq A$ is called \vocab{homogeneous}	if $r \in I
    \implies
    \forall d \in \mathbb{I} ~ r_d \in	I_d$ where $I_d \coloneqq I
    \cap A_d$.

    By a \vocab{graded ring}  we understand an $\N$-graded ring.
    Tin this case, $A_{+} \coloneqq \bigoplus_{d=1}^{\infty}
    A_d = \{r \in A | r_0
    = 0\} $ is called the \vocab{augmentation ideal}  of $A$.
\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 $b\neq 0 \implies \varepsilon_b =
    \varepsilon_0 \varepsilon _b = 0$.
    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.}
    \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.
    \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 $d \neq e$.
\end{proof}
\begin{fact}
    A homogeneous ideal is finitely generated iff it can be generated by finitely
    many of its homogeneous elements.
    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 $\alpha \in \N^{n+1}$ $|\alpha| =
    \sum_{i=0}^{n}	\alpha_i$ and
    $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$.
\end{notation}
\begin{definition}[Homogeneous polynomials]
    Let $R$ be any ring and $f = \sum_{\alpha \in \N^{n+1}}
    f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$.
    We say that $f$ is \vocab{homogeneous of degree $d$} if $|\alpha| \neq	d
    \implies f_\alpha = 0$ .
    We denote the subset of homogeneous polynomials of degree $d$  by
    $R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n]$.
\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}
    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
    \[
        X = \bigcap_{i=1}^k \Vp(f_i)
    \]
    where the $f_i \in A_{d_i}$
    are homogeneous polynomials.
\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
    \[
        \{(x_1,\ldots,x_n) \in	\mathfrak{k}^n | f_i(1,x_1,\ldots,x_n) = 0, 1
        \le i \le
        k\}  \subseteq \mathfrak{k}^n
    \]
    Conversely, if $Y \subseteq \mathfrak{k}^n$ is
    closed it has the form
    \[
        \{(x_1,\ldots,x_n) \in	\mathfrak{k}^n |
        g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\}
    \]
    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 $U_i \cong \mathfrak{k}^n$.
\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.
\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
              $\Vp(\sqrt{I}) = \Vp(I)$
    \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.
\end{fact}
\begin{proof}
    By definition, a topological space is Noetherian $\iff$  all open subsets are
    quasi-compact.
\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 $U_i = \mathbb{P}^n \setminus
    \Vp(X_i) \cong
    \mathfrak{k}^n$.
    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:
    \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.
    \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}) 

\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
    $\Vp(I) \subseteq \Vp(f) \iff f \in
    \sqrt{I}$.
\end{proposition}
\begin{proof}
    $\impliedby$ is clear.
    Let $\Vp(I) \subseteq \Vp(f)$.
    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}).
\end{proof}

\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 $\fp \not\supseteq R_+$.
\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}\} $.
\end{remark}
\begin{proposition}
    \label{bijproj}
    There is a bijection
    \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
    \end{align}
    Under this bijection,
    the irreducible subsets correspond to the elements of
    $\Proj(A_\bullet)$.
\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 $X =
    \Vp(J) =
    \emptyset = \Vp(A_+)$.
    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.

\end{proof}
\begin{remark}
    It is important that $I \subseteq A_{\color{red} +}$, since
    $\Vp(A) = \Vp(A_+)
    =  \emptyset$ would be a counterexample.
\end{remark}
\begin{corollary}
    $\mathbb{P}^n$ is irreducible.
\end{corollary}
\begin{proof}
    Apply
    \ref{bijproj} to $\{0\}  \in \Proj(A_\bullet)$.
\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
    homogeneous elements $f, g \in R , fg \in I \implies f \in I \lor g \in I$.
\end{lemma}
\begin{proof}
    $\implies$ is trivial.
    It suffices to show that for arbitrary $f,g \in R fg \in I \implies f \in I
    \lor g \in I$.
    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
    \[
        (fg)_{d+e} = f_dg_e + \sum_{\delta = 1}^{\infty} (f_{d + \delta} g_{e - \delta}
        + f_{d - \delta} g_{e + \delta})
    \]
    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.
\end{proof}

\begin{remark}
    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 $R$.
    \[
        \{\fp  \in \Spec R | \fp \text{ is a homogeneous ideal of }
        R_\bullet\} = \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec
        R_0\}
    \]
\end{remark}

\subsection{Dimension of $\mathbb{P}^n$}
\begin{proposition}
    \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
              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,
              then $\codim(X,Y) = \dim(Y) -
              \dim(X)$.
    \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 $\codim(Y,Z) =
    \codim(Y
    \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
    Thus
    \begin{align}
        \codim(X,Y) + \codim(Y,Z) & = \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{align}
    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
        $X$}
    \[
        C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in  \mathfrak{k}^{n+1} \setminus
        \{0\}  | [x_0,\ldots,x_n] \in X\}
    \]
    If $X = \Vp(I)$ where $I \subseteq A_+ =
    \mathfrak{k}[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) =
    \Va(I)$.
\end{definition}
\begin{proposition}
    \label{conedim}
    \begin{itemize}
        \item
              $C(X)$ is irreducible iff  $X$ is irreducible or $X = \emptyset$.
        \item
              If $X$ is irreducible, then

              $\dim(C(X)) = \dim(X) + 1$ and

              $\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \mathbb{P}^n)$
    \end{itemize}
\end{proposition}
\begin{proof}
    The first assertion follows from
    \ref{bijproj} and
    \ref{bijiredprim} (bijection
    of irreducible subsets and prime ideals in the projective and affine case).

    Let $d = \dim(X)$ and
    \[
        X_0 \subsetneq \ldots \subsetneq X_d = X \subsetneq
        X_{d+1} \subsetneq \ldots \subsetneq X_n =
        \mathbb{P}^n
    \]
    be a chain of
    irreducible subsets of $\mathbb{P}^n$.
    Then
    \[
        \{0\}  \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X)
        \subsetneq \ldots \subsetneq C(X_n) = \mathfrak{k}^{n+1}
    \]
    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 $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) =
    \dim(\mathfrak{k}^{n+1})
    = n+1$, the two inequalities must be equalities.
\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
    $\mathbb{A}^n$ we understand an
    irreducible closed subset of codimension $1$.
\end{definition}

\begin{corollary}
    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.
\end{corollary}
\begin{proof}
    If $H = \Vp(P)$ then $C(H) = \Va(P)$ is a
    hypersurface in $\mathfrak{k}^{n+1}$
    by
    \ref{irredcodimone}.
    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
    $\mathfrak{k}^{n+1}$, 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
    $I =
    \sqrt{I}  \subseteq A$ (
    \ref{antimonbij}), $\fp = P \cdot
    A$.
    Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition
    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 $d$.
\end{proof}
\begin{definition}
    A hypersurface $H \subseteq \mathbb{P}^n$ has
    \vocab{degree $d$} if $H =
    \Vp(P)$ where $P \in A_d$ is an irreducible polynomial.
\end{definition}

\subsubsection{Application to intersections in $\mathbb{P}^n$ and Bezout's
    theorem}
\begin{corollary}
    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 $A \cap B \neq \emptyset$ and every irreducible component
    of $A \cap B$ as dimension $\ge a + b - n$.
\end{corollary}

\begin{remark}
    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}).
\end{remark}

\begin{proof}
    The lower bound on the dimension of irreducible components of $A \cap B$ is
    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
    the dimension of irreducible components of $C(A) \cap C(B)$ (again
    \ref{codimintersection}).
\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.
\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!

