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. In this case, \[A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\}\] is called the \vocab{augmentation ideal} of $A$. \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 $\alpha \in \N^{n+1}$ \[|\alpha| = \sum_{i=0}^{n} \alpha_i \text{ and } x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}.\] \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 all 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 $f,g \in R$ 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 \[ (fg)_{d+e} = f_dg_e + \sum_{\delta = 1}^{\infty} (f_{d + \delta} g_{e - \delta} + f_{d - \delta} g_{e + \delta}) \] 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 $R$. \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*} \end{remark} \pagebreak \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{IEEEeqnarray*}{rCl} \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{IEEEeqnarray*} 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 $A \cap B$ has 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!