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 $\neq0$.
\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 $x_{i}\neq0$.
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 \neq0$.
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 \neq0\implies\varepsilon_a \varepsilon_b =0$.
Applying the last equation with $a =0$ gives $b\neq0\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 \neq0\}$.
$N_f =0$ is trivial. Suppose $N_f > 0$ and $e \in\mathbb{I}$ is maximal with $f_e \neq0$.
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
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:
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 $f \in\tilde R$.
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
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 $X$}
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
is a chain of irreducible subsets of $\mathfrak{k}^{n+1}$. Hence $\dim(C(X))\ge1+ d$ 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 $P_d \neq0$ be the decomposition into homogeneous components.
If $P_e $ with $e < d$ was $\neq0$, 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!
