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\subsection{Finitely generated and Noetherian modules}
\begin{definition}[Generated submodule]
Let $R$ be a ring, $M$ an $R$-module, $S \subseteq M$.
Then the following sets coincide
\begin{enumerate}
\item $\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{finite}, r_s \in R, \right\}$
\item $\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$
\item The $\subseteq$-smallest submodule of $M$ containing $S$
\end{enumerate}
This subset of $N \subseteq M$ is called the \vocab[Module!Submodule]{submodule of $M $ generated by $S$}. If $N= M$ we say that \vocab[Module!generated by subset $S$]{$ M$ is generated by $S$}.
$M$ is finitely generated $:\iff \exists S \subseteq M$ finite such that $M$ is generated by $S$.
\end{definition}
\begin{definition}[Noetherian $R$-module]
$M$ is a \vocab{Noetherian} $R$-module if the following equivalent conditions hold:
\begin{enumerate}
\item Every submodule $N \subseteq M$ is finitely generated.
\item Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates
\item Every set $\mathfrak{M} \neq \emptyset$ of submodules of $M$ has a $\subseteq$-largest element.
\end{enumerate}
\end{definition}
\begin{proposition}[Hilbert's Basissatz]\label{basissatz}
If $R$ is a Noetherian ring, then the polynomial rings $R[X_1,\ldots, X_n]$ in finitely many variables are Noetherian.
\end{proposition}
\subsubsection{Properties of finite generation and Noetherianness}
\begin{fact}[Properties of Noetherian modules]
\begin{enumerate}
\item Every Noetherian module over an arbitrary ring is finitely generated.
\item If $R$ is a Noetherian ring, then an $R$-module is Noetherian iff it is finitely generated.
\item Every submodule of a Noetherian module is Noetherian.
\end{enumerate}
\end{fact}
\begin{proof}
\begin{enumerate}
\item By definition, $M$ is a submodule of itself. Thus it is finitely generated.
\item Since $M$ is finitely generated, there exists a surjective homomorphism $R^n \to M$. As $R$ is Noetherian, $R^n$ is Noethrian as well.
\item trivial
\end{enumerate}
\end{proof}
\begin{fact}
Let $M, M', M''$ be $R$-modules.
\begin{enumerate}
\item Suppose $M \xrightarrow{p} M''$ is surjective. If $M$ is finitely generated (resp. Noetherian), then so is $M''$.
\item Let $M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ be exact. If $M'$ and $M ''$ are finitely generated (reps. Noetherian), so is $M$.
\end{enumerate}
\end{fact}
\begin{proof}
\begin{enumerate}
\item Consider a sequence $M_0'' \subset M_1'' \subset \ldots \subset M''$. Then $p^{-1} M_i''$ yields a strictly ascending sequence.
If $M$ is generated by $S, |S| < \omega$, then $M''$ is generated by $p(S)$.
\item Because of 1. we can replace $M'$ by $f(M')$ and assume $0 \to M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ to be exact. The fact about finite generation follows from EInführung in die Algebra.
If $M', M''$ are Noetherian, $N \subseteq M$ a submodule, then $N' \coloneqq f^{-1}(N)$ and $N''\coloneqq p(N)$ are finitely generated. Since $0 \to N' \to N \to N'' \to 0$ is exact, $N$ is finitely generated.
\end{enumerate}
\end{proof}
\subsection{Ring extensions of finite type}
\begin{definition}[$R$-algebra]
Let $R$ be a ring. An $R$-algebra $(A, \alpha)$ is a ring $A$ with a ring homomorphism $R \xrightarrow{\alpha} A$.
$\alpha$ will usually be omitted. In general $\alpha$ is not assumed to be injective.\\
\\
An $R$-subalgebra is a subring $\alpha(R) \subseteq A' \subseteq A$.\\
A morphism of $R$-algebras $A \xrightarrow{f} \tilde{A}$ is a ring homomorphism with $\tilde{\alpha} = f \alpha$.
\end{definition}
\begin{definition}[Generated (sub)algebra, algebra of finite type]
Let $(A, \alpha)$ be an $R$-algebra.
\begin{align}
\alpha: R[X_1,\ldots,X_m] &\longrightarrow A[X_1,\ldots,X_m] \\
P = \sum_{\beta \in \N^m} p_\beta X^{\beta} &\longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta) X^{\beta}
\end{align}
is a ring homomorphism. We will sometimes write $P(a_1,\ldots,a_m)$ instead of $(\alpha(P))(a_1,\ldots,a_m)$.
Fix $a_1,\ldots,a_m \in A^m$. Then we get a ring homomorphism $R[X_1,\ldots,X_m] \to A$. The image of this ring homomorphism is the $R$-subalgebra of $A$ \vocab[Algebra!generated subalgebra]{generated by the $a_i$}.
$A$ is \vocab[Algebra!of finite type]{of finite type} if it can be generated by finitely many $a_i \in I$.
For arbitrary $S \subseteq A$ the subalgebra generated by $S$ is the intersection of all subalgebras containing $S$ \\
$=$ the union of subalgebras generated by finite $S' \subseteq S$\\
$= $ the image of $R[X_s | s \in S]$ under $P \mapsto (\alpha(P))(S)$.
\end{definition}
\subsection{Finite ring extensions} % LECTURE 2
\begin{definition}[Finite ring extension]
Let $R$ be a ring and $A$ an $R$-algebra. $A$ is a module over itself and the ringhomomorphism $R \to A$ allows us to derive an $R$-module structure on $A$.
$A$ \vocab[Algebra!finite over]{is finite over} $R$ / the $R$-algebra $A$ is finite / $A / R$ is finite if $A$ is finitely generated as an $R$-module.
\end{definition}
\begin{fact}[Basic properties of finiteness]
\begin{enumerate}[A]
\item Every ring is finite over itself.
\item A field extension is finite as a ring extension iff it is finite as a field extension.
\item $A$ finite $\implies$ $A$ of finite type.
\item $A / R$ and $B / A$ finite $\implies$ $B / R$ finite.
\end{enumerate}
\end{fact}
\begin{proof}
\begin{enumerate}[A]
\item $1$ generates $R$ as a module
\item trivial
\item Let $A $ be generated by $a_1,\ldots,a_n$ as an $R$-module. Then $A$ is generated by $a_1,\ldots,a_n$ as an $R$-algebra.
\item Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module and $B$ by $b_1,\ldots,b_n$ as an $A$-module.
For every $b$ there exist $\alpha_j \in A$ such that $b = \sum_{j=1}^{n} \alpha_j b_j$. We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$ for some $\rho_{ij} \in R$ thus
$b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij} a_i b_j$ and the $a_ib_j$ generate $B$ as an $R$-module.
\end{enumerate}
\end{proof}
\subsection{Determinants and Caley-Hamilton} %LECTURE 2 TODO: move to int. elements?
This generalizes some facts about matrices to matrices with elements from commutative rings with $1$.
\footnote{Most of this even works in commutative rings without $ 1$, since $1$ simply can be adjoined.}
\begin{definition}[Determinant]
Let $A = (a_{ij}) \Mat(n,n,R)$. We define the determinant by the Leibniz formula \[
\det(A) \coloneqq \sum_{\pi \in S_n} \sgn(\pi) \prod_{i=1}^{n} a_{i, \pi(i)}
\]
Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij} \coloneqq (-1)^{i+j} \cdot M_{ij}$, where $M_{ij}$ is the determinant of the matrix resulting from $A$ after deleting the $i^{\text{th}}$ row and the $j^{\text{th}}$ column.
\end{definition}
\begin{fact}
\begin{enumerate}
\item $\det(AB) = \det(A)\det(B)$
\item Development along a row or column works.
\item Cramer's rule: $A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot A = \det(A) \cdot \mathbf{1}_n$. $A$ is invertible iff $\det(A)$ is a unit.
\item Caley-Hamilton: If $P_A = \det(T \cdot \mathbf{1}_n - A)$ \footnote{$T \cdot \mathbf{1}_n -A \in \Mat(n,n,A[T])$}, then $P_A(A) = 0$.
\end{enumerate}
\end{fact}
\begin{proof}
All rules hold for the image of a matrix under a ring homomorphism if they hold for the original matrix. The converse holds in the case of injective ring homomorphisms.
Caley-Hamilton was shown for algebraically closed fields in LA2 using the Jordan normal form.
Fields can be embedded into their algebraic closure, thus Caley-Hamilton holds for fields. Every domain can be embedded in its field of quotients $\implies$ Caley-Hamilton holds for domains.
In general, $A$ is the image of $(X_{i,j})_{i,j = 1}^{n} \in \Mat(n,n,S)$ where $S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain) under the morphism $S \to A$ of evaluation defined by $X_{i,j} \mapsto a_{i,j}$. Thus Caley-Hamilton holds in general.
\end{proof} %TODO: lernen
\subsection{Integral elements and integral ring extensions} %LECTURE 2
\begin{proposition}[on integral elements]\label{propinte}
Let $A$ be an $R$-algebra, $a \in A$. Then the following are equivalent:
\begin{enumerate}[A]
\item $\exists n \in \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = \sum_{i=0}^{n-1} r_i a^i$
\item There exists a subalgebra $B \subseteq A$ finite over $R$ and containing $a$.
\end{enumerate}
If $a_1, \ldots, a_k \in A$ satisfy these conditions, there is a subalgebra of $A$ finite over $R$ and containing all $a_i$.
\end{proposition}
\begin{definition}\label{intclosure}
Elements that satisfy the conditions from \ref{propinte} are called \vocab{integral over} $R$.
$A / R$ is \vocab[Algebra!integral]{integral}, if all $a \in A$ are integral over $R$.
The set of elements of $A$ integral over $R$ is called the \vocab{integral closure} of $R$ in $A$.
\end{definition}
\begin{proof}
\hskip 10pt
\begin{enumerate}
{\color{gray} \item[B $\implies$ A] Let $a \in A$ such that there is a subalgebra $B \subseteq A$ containing $a$ and finite over $R$.
Let $(b_i)_{i=1}^{n}$ generate $B$ as an $R$-module.
\begin{align}
q: R^n &\longrightarrow B \\
(r_1,\ldots,r_n) &\longmapsto \sum_{i=1}^{n} r_i b_i
\end{align}
is surjective. Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in R^n$ such that $a b_i = q(\rho_i)$. Let $\mathfrak{A}$ be the matrix with the $\rho_i$ as columns.
Then for all $v \in R^n: q(\mathfrak{A} \cdot v) = a \cdot q(v)$. By induction it follows that $q(P(\mathfrak{A}) \cdot v) = P(a)q(v)$ for all $P \in R[T]$. Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \mathfrak{A})$ and using Caley-Hamilton, we obtain $P(a) \cdot q(v) = 0$. $P$ is monic. Since $q$ is surjective, we find $v \in R^{n} : q(v) = 1$. Thus $P(a) = 0$ and $a$ satisfies A.
}
\item[B $\implies$ A] if $R$ is Noetherian.\footnote{This suffices in the exam.}
Let $a \in A$ satisfy B. Let $B$ be a subalgebra of $A$ containing $b$ and finite over $R$. Let $M_n \subseteq B$ be the $R$-submodule generated by the $a^i$ with $0 \le i < n$. As a finitely generated module over the Noetherian ring $R$, $B$ is a Noetherian $R$-module. Thus the ascending sequence $M_n$ stabilizes at some step $d$ and $a^d \in M_d$. Thus there are $(r_i)_{i=0}^{d-1} \in R^d$ such that $a^d = \sum_{i=0}^{d-1} r_ia^i$.
\item[A $\implies$ B] Let $a = (a_i)_{i=1}^n$ where all $a_i$ satisfy A, i.e. $a_i^{d_i} = \sum_{j=0}^{d_i - 1} r_{i,j}a_i^j$ with $r_{i,j} \in R$. Let $B \subseteq A$ be the sub-$R$-module generated by $a^\alpha = \prod_{i=1}^n a_i^{\alpha_i}$ with $0 \le \alpha_i < d_i$.
$B$ is closed under $a_1 \cdot $ since \[a_1a^{\alpha} = \begin{cases}
a^{(\alpha_1 + 1, \alpha')} &\text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1\\
\sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} &\text{if } \alpha_1 = d_1 - 1
\end{cases}\]
By symmetry, this hold for all $a_i$. By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$, $B$ is invariant under $a^{\alpha}\cdot $. Since these generate $B$ as an $R$-module, $B$ is multiplicatively closed. Thus A holds. Furthermore we have shown the final assertion of the proposition.
\end{enumerate}
\end{proof}
\begin{corollary}\label{cintclosure}
\begin{enumerate}
\item[Q] Every finite $R$-algebra $A$ is integral.
\item[R] The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$
\item[S] If $A$ is an $R$-algebra, $B$ an $A$-algebra and $b \in B$ integral over $R$, then it is integral over $A$.
\item[T] If $A$ is an integral $R$-algebra and $B$ any $A$-algebra, $b \in B$ integral over $A$, then $b$ is integral over $R$.
\end{enumerate}
\end{corollary}
\begin{proof}
\begin{enumerate}
\item[Q] Put $ B = A $ in B.
\item[R] For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$, hence integral over $R$.
From B it follows, that the integral closure is closed under ring operations.
\item[S] trivial
\item[T] Let $b \in B$ such that $b^n = \sum_{i=0}^{n-1} a_ib^{i}$. Then there is a subalgebra $\tilde{A} \subseteq A$ finite over $R$, such that all $a_i \in \tilde{A}$.
$b$ is integral over $\tilde{A} \implies \exists \tilde{B} \subseteq B$ finite over $\tilde{A}$ and $b \in \tilde{B}$. Since $\tilde{B} / \tilde{A} $ and $\tilde{A} / R$ are finite, $\tilde{B} / R$ is finite and $b$ satisfies B.
\end{enumerate}
\end{proof}
\subsection{Finiteness, finite generation and integrality} %some more remarks on finiteness, finite generation and integrality
\begin{fact}[Finite type and integral $\implies$ finite]\label{ftaiimplf}
If $A$ is an integral $R$-algebra of finite type, then it is a finite $R$-algebra.
\end{fact}
\begin{proof}
Let $A $ be generated by $\left( a_i \right) _{i=1}^{n}$ as an $R$- algebra. By the proposition on integral elements (\ref{propinte}), there is a finite $R$-algebra $B \subseteq A$ such that all $a_i \in B$.
We have $B = A$, as $A$ is generated by the $a_i$ as an $R$-algebra.
\end{proof}
\begin{fact}[Finite type in tower]
If $A$ is an $R$-algebra of finite type and $B$ an $A$-algebra of finite type, then $B$ is an $R$-algebra of finite type.
\end{fact}
\begin{proof}
If $A / R$ is generated by $(a_i)_{i=1}^m$ and $B / A$ by $(b_j)_{j=1}^{n}$, then $B /R$ is generated by the $b_j$ and the images of the $a_i$ in $B$.
\end{proof}
{\color{red}
\begin{fact}[About integrality and fields] \label{fintaf}
Let $B$ be a domain integral over its subring $A$. Then $B$ is a field iff $A$ is a field.
\end{fact}
}
\begin{proof}
Let $B$ be a field and $a \in A \setminus \{0\} $. Then $a^{-1} \in B$ is integral over $A$, hence $a^{-d} = \sum_{i=0}^{d-1} \alpha_i a^{-i}$ for some $\alpha_i \in A$. Multiplication by $a^{d-1}$ yields
$a^{-1} = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$.
On the other hand, let $B$ be integral over the field $A$. Let $b \in B \setminus \{0\}$. As $B$ is integral over $A$, there is a sub-$A$-algebra $\tilde{B} \subseteq B, b \in \tilde{B}$ finitely generated as an $A$-module, i.e. a finite-dimensional $A$-vector space. Since $B$ is a domain, $\tilde{B} \xrightarrow{b\cdot } \tilde{B}$ is injective, hence surjective, thus $\exists x \in \tilde{B} : b \cdot x \cdot 1$.
\end{proof}
\subsection{Noether normalization theorem}
\begin{lemma}\label{nntechlemma}
Let $S \subseteq \N^n$ be finite. Then there exists $\vec k \in \N^n$ such that $k_1 =1$ and $w_{\vec k}(\alpha) \neq w_{\vec k}(\beta)$ for $\alpha \neq \beta \in S$,
where $w_{\vec k}(\alpha) = \sum_{i=1}^{n} k_i \alpha_i$.
\end{lemma}
\begin{proof}
Intuitive:
For $\alpha \neq \beta$ the equation $w_{(1, \vec \kappa)}(\alpha) = w_{(1, \vec \kappa)}(\beta)$ ($\kappa \in \R^{n-1}$)
defines a codimension $1$ affine hyperplane in $\R^{n-1}$. It is possible to choose $\kappa$ such that all $\kappa_i$ are $> \frac{1}{2}$ and with Euclidean distance $> \frac{\sqrt{n-1} }{2}$ from the union of these hyperplanes. By choosing the closest $\kappa'$ with integral coordinates, each coordinate will be disturbed by at most $\frac{1}{2}$, thus at Euclidean distance $\le \frac{\sqrt{n-1} }{2}$.
More formally:\footnote{The intuitive version suffices in the exam.}
Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le i \le n\} $. We can choose $k$ such that $k_i > (i-1) M k_{i-1}$.
Suppose $\alpha \neq \beta$. Let $i$ be the maximal index such that $\alpha_i \neq \beta_i$. Then the contributions of $\alpha_j$ (resp. $\beta_j$) with $1 \le j < i$ to $w_{\vec k}(\alpha)$ (resp. $w_{\vec k}(\beta)$) cannot undo the difference $k_i(\alpha_i - \beta_i)$.
\end{proof}
\begin{theorem}[Noether normalization] \label{noenort}
Let $K$ be a field and $A$ a $K$-algebra of finite type. Then there are $a = (a_i)_{i=1}^{n} \in A$ which are algebraically independent over $K$, i.e. the ring homomorphism \begin{align}
\ev_a: K[X_1,\ldots,X_n] &\longrightarrow A \\
P &\longmapsto P(a_1,\ldots,a_n)
\end{align}
is injective. $n$ and the $a_i$ can be chosen such that $A$ is finite over the image of $\ev_a$.
\end{theorem}
\begin{proof}
Let $(a_i)_{i=1}^n$ be a minimal number of elements such that $A$ is integral over its $K$-subalgebra generated by $a_1, \ldots, a_n$. (Such $a_i$ exist, since $A$ is of finite type).
Let $\tilde{A}$ be the $K$-subalgebra generated by the $a_i$.
If suffices to show that the $a_i$ are algebraically independent.
Since $A$ is of finite type over $K$ and thus over $\tilde{A}$, by fact \ref{ftaiimplf} (integral and finite type $\implies$ finite) $A$ is finite over $\tilde{A}$.
Thus we only need to show that the $a_i$ are algebraically independent over $K$.
Assume there is $P \in K[X_1,\ldots,X_n] \setminus \{0\} $ such that $P(a_1,\ldots,a_n) = 0$. Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and $S = \{ \alpha \in \N^n | p_\alpha \neq 0\}$. For $\vec{k} = (k_i)_{i=1}^{n} \in \N^n$ and $\alpha \in \N^n$ we define $w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$.
By \ref{nntechlemma} it is possible to choose $\vec{k} \in \N^n$ such that
$k_1 = 1$ and for $\alpha \neq \beta \in S$ we have $w_{\vec{k}}(\alpha) \neq w_{\vec{k}}(\beta)$.
Define $b_i \coloneqq a_{i+1} - a^{k_{i+1}}_1$ for $1 \le i < n$.
\begin{claim}
$A$ is integral over the subalgebra $B$ generated by the $b_i$.
\end{claim}
\begin{subproof}
By the transitivity of integrality, it is sufficient to show that the $a_i$ are integral over $B$.
For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$. Thus it suffices to show this for $a_1$.
Define $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n}) \in B[T]$.
We have $0 = P(a_1,\ldots, a_n) = Q(a_1)$. Hence it suffices to show that the leading coefficient of $Q$ is a unit.
We have
\[
T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} = T^{w_{\vec k}(\alpha)} + \sum_{l = 0}^{w_{\vec k}(\alpha) - 1} \beta_{\alpha, l} T^l
\]
with suitable $\beta_{\alpha, l} \in B$.
By the choice of $\vec k$, we have \[
Q(T) = p_{\alpha} T^{w_{\vec k}(\alpha)} + \sum_{j=0}^{w_{\vec k}(\alpha) - 1} q_j T^j
\]
with $q_j \in B$ and $\alpha$ such that $w_{\vec k }(\alpha)$ is maximal subject to the condition $p_\alpha \neq 0$.
Thus the leading coefficient of $Q$ is a unit.
\end{subproof}
This contradicts the minimality of $n$, as $B$ can be generated by $< n$ elements $b_i$.
\end{proof}

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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 $x$.
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}
% Lecture 12
\subsection{The projective form of the Nullstellensatz and the closed subsets of $\mathbb{P}^n$}
Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
\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 $J \subseteq A$. \Wlog $J = \sqrt{J}$. 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 $X \in U_i = \mathbb{P}^n \setminus \Vp(X_i)$.
\Wlog $i = 0$. 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!

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\subsection{Sheaves}
\begin{definition}[Sheaf]
Let $X$ be any topological space.
A \vocab{presheaf} $\mathcal{G}$ of sets (or rings, (abelian) groups) on $X$ associates a set (or rings, or (abelian) group) $\mathcal{G}(U)$ to every open subset $U$ of $X$, and a map (or ring or group homomorphism) $\mathcal{G}(U) \xrightarrow{r_{U,V}} \mathcal{G}(V)$ to every inclusion $V \subseteq U$ of open subsets of $X$ such that $r_{U,W} = r_{V,W} r_{U,V}$ for inclusions $U \subseteq V \subseteq W$ of open subsets.
Elements of $\mathcal{G}(U)$ are often called \vocab{sections} of $\mathcal{G}$ on $U$ or \vocab{global sections} when $U = X$.
Let $U \subseteq X$ be open and $U = \bigcup_{i \in I} U_i$ an open covering.
A family $(f_i)_{i \in I} \in \prod_{i \in I} \mathcal{G}(U_i)$ is called \vocab[Sections!compatible]{compatible} if $r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j)$ for all $i,j \in I$.
Consider the map
\begin{align}
\phi_{U, (U_i)_{i \in I}}: \mathcal{G}(U) &\longrightarrow \{(f_i)_{i \in I} \in \prod_{i \in I} \mathcal{G}(U_i) | r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j) \text{ for } i,j \in I \} \\
f &\longmapsto (r_{U, U_i}( f))_{i \in I}
\end{align}
A presheaf is called \vocab[Presheaf!separated]{separated} if $\phi_{U, (U_i)_{i \in I}}$ is injective for all such $U$ and $(U_i)_{i \in I}$.\footnote{This also called ``locality''.}
It satisfies \vocab{gluing} if $\phi_{U, (U_i)_{i \in I}}$ is surjective.
A presheaf is called a \vocab{sheaf} if it is separated and satisfies gluing.
The bijectivity of the $\phi_{U, (U_i)_{i \in I}}$ is called the \vocab{sheaf axiom}.
\end{definition}
\begin{trivial}+
A presheaf is a contravariant functor $\mathcal{G} : \mathcal{O}(X) \to C$ where $\mathcal{O}(X)$ denotes the category of open subsets of $X$ with inclusions as morphisms and $C$ is the category of sets, rings or (abelian) groups.
\end{trivial}
\begin{definition}
A subsheaf $\mathcal{G}'$ is defined by subsets (resp. subrings or subgroups) $\mathcal{G}'(U) \subseteq \mathcal{G}(U)$ for all open $U \subseteq X$ such that the sheaf axioms still hold.
\end{definition}
\begin{remark}
If $\mathcal{G}$ is a sheaf on $X$ and $\Omega \subseteq X$ open, then $\mathcal{G}\defon{\Omega}(U) \coloneqq \mathcal{G}(U)$ for open $U \subseteq \Omega$ and $r_{U,V}^{(\mathcal{G}\defon{\Omega})}(f) \coloneqq r_{U,V}^{(\mathcal{G})}(f)$ is a sheaf of the same kind as $\mathcal{G}$ on $\Omega$.
\end{remark}
\begin{remark}
The notion of restriction of a sheaf to a closed subset, or of preimages under general continuous maps, can be defined but this is a bit harder.
\end{remark}
\begin{notation}
It is often convenient to write $f \defon{V}$ instead of $r_{U,V}(f)$.
\end{notation}
\begin{remark}
Applying the \vocab{sheaf axiom} to the empty covering of $U = \emptyset$, one finds that $\mathcal{G}(\emptyset) = \{0\} $.
\end{remark}
\subsubsection{Examples of sheaves}
\begin{example}
Let $G$ be a set and let $\mathfrak{G}(U)$ be the set of arbitrary maps $U \xrightarrow{f} G$. We put $r_{U,V}(f) = f\defon{V}$.
It is easy to see that this defines a sheaf.
If $\cdot $ is a group operation on $G$, then $(f\cdot g)(x) \coloneqq f(x)\cdot g(x)$ defines the structure of a sheaf of group on $\mathfrak{G}$.
Similarly, a ring structure on $G$ can be used to define the structure of a sheaf of rings on $\mathfrak{G}$.
\end{example}
\begin{example}
If in the previous example $G$ carries a topology and $\mathcal{G}(U) \subseteq \mathfrak{G}(U)$ is the subset (subring, subgroup) of continuous functions $U \xrightarrow{f} G$, then $\mathcal{G}$ is a subsheaf of $\mathfrak{G}$, called the sheaf of continuous $G$-valued functions on (open subsets of) $X$.
\end{example}
\begin{example}
If $X = \R^n$, $\mathbb{K} \in \{\R, \C\}$ and $\mathcal{O}(U)$ is the sheaf of $\mathbb{K}$-valued $C^{\infty}$-functions on $U$, then $\mathcal{O}$ is a subsheaf of the sheaf (of rings) of $\mathbb{K}$-valued continuous functions on $X$.
\end{example}
\begin{example}
If $X = \C^n$ and $\mathcal{O}(U)$ the set of holomorphic functions on $X$, then $\mathcal{O}$ is a subsheaf of the sheaf of $\C$-valued $C^{\infty}$-functions on $X$.
\end{example}
\subsubsection{The structure sheaf on a closed subset of $\mathfrak{k}^n$}
Let $X \subseteq \mathfrak{k}^n$ be open. Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
\begin{definition}\label{structuresheafkn}
For open subsets $U \subseteq X$, let $\mathcal{O}_X(U)$ be the set of functions $U \xrightarrow{\phi} \mathfrak{k}$ such that every $x \in U$ has a neighbourhood $V$ such that there are $f,g \in R$ such that for $y \in V$ we have $g(y) \neq 0$ and $\phi(y) = \frac{f(y)}{g(y)}$.
\end{definition}
\begin{remark}\label{structuresheafcontinuous}
$\mathcal{O}_X$ is a subsheaf (of rings) of the sheaf of $\mathfrak{k}$-valued functions on $X$.
The elements of $\mathcal{O}_X(U)$ are continuous:
Let $M \subseteq \mathfrak{k}$ be closed. We must show the closedness of $N \coloneqq \phi^{-1}(M)$ in $U$. For $M = \mathfrak{k}$ this is trivial. Otherwise $M$ is finite and we may assume $M = \{t\} $ for some $t \in \mathfrak{k}$. For $x \in U$, there are open $V_x \subseteq U$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $V_x$.
Then $N \cap V_x = V(f_x - t\cdot g_x) \cap V_x)$ is closed in $V_x$. As the $V_x$ cover $U$ and $U$ is quasi-compact, $N$ is closed in $U$.
\end{remark}
\begin{proposition}\label{structuresheafri}
Let $X = V(I)$ where $I = \sqrt{I} \subseteq R$ is an ideal. Let $A = R / I$. Then
\begin{align}
\phi: A &\longrightarrow \mathcal{O}_X(X) \\
f \mod I &\longmapsto f\defon{X}
\end{align}
is an isomorphism.
\end{proposition}
\begin{proof}
It is easy to see that the map $A \to \mathcal{O}_X(X)$ is well-defined and a ring homomorphism.
Its injectivity follows from the Nullstellensatz and $I = \sqrt{I}$ (\ref{hns3}).
Let $\phi \in \mathcal{O}_X(X)$. for $x \in X$, there are an open subset $U_x \subseteq X$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $U_x$.
\begin{claim}
\Wlog we can assume $U_x = X \setminus V(g_x)$.
\end{claim}
\begin{subproof}
The closed subsets $(X \setminus U_x) \subseteq \mathfrak{k}^n$ has the form $X\setminus U_x = V(J_x)$ for some ideal $J_x \subseteq R$.
As $x \not\in X \setminus V_x$ there is $h_x \in J_x$ with $h_x(x) \neq 0$.
Replacing $U_x$ by $X \setminus V(h_x)$, $f_x$ by $f_xh_x$ and $g_x$ by $g_xh_x$, we may assume $U_x = X \setminus V(g_x)$.
\end{subproof}
\begin{claim}
\Wlog we can assume $V(g_x) \subseteq V(f_x)$.
\end{claim}
\begin{subproof}
Replace $f_x$ by $f_xg_x$ and $g_x$ by $g_x^2$.
\end{subproof}
As $X$ is quasi-compact, there are finitely many points $(x_i)_{i=1}^m$ such that the $U_{x_i}$ cover $X$.
Let $U_i \coloneqq U_{x_i}, f_i \coloneqq f_{x_i}, g_i \coloneqq g_{x_i}$.
As the $U_i = X \setminus V(g_i)$ cover $X$, $V(I) \cap \bigcap_{i=1}^m V(g_i) = X \cap \bigcap_{i=1}^m V(g_i) = \emptyset$.
By the Nullstellensatz (\ref{hns1}) the ideal of $R$ generated by $I$ and the $a_i$ equals $R$.
There are thus $n \ge m \in \N$ and elements $(g_i)_{i = m+1}^n$ of $I$ and $(a_i)_{i=1}^n \in R^n$ such that $1 = \sum_{i=1}^{n} a_ig_i$.
Let for $i > m$ $f_i \coloneqq 0$, $F = \sum_{i=1}^{n} a_if_i = \sum_{i=1}^{m} a_if_i \in R$.
\begin{claim}
For all $x \in X $ ~ $f_i(x) = \phi(x) g_i(x)$.
\end{claim}
\begin{subproof}
If $x \in V_i$ this follows by our choice of $f_i$ and $g_i$.
If $x \in X \setminus V_i$ or $i > m$ both sides are zero.
\end{subproof}
It follows that
\[
\phi(x) = \phi(x) \cdot 1 = \phi(x) \cdot \sum_{i=1}^{n} a_i(x) g_i(x) = \sum_{i=1}^{n} a_i(x) f_i(x) = F(x)
\]
Hence $\phi = F\defon{X}$.
\end{proof}
\subsubsection{The structure sheaf on closed subsets of $\mathbb{P}^n$}
Let $X \subseteq \mathbb{P}^n$ be closed and $R_\bullet = \mathfrak{k}[X_0,\ldots,X_n]$ with its usual grading.
\begin{definition}\label{structuresheafpn}
For open $U \subseteq X$, let $\mathcal{O}_X(U)$ be the set of functions $U \xrightarrow{\phi} \mathfrak{k}$ such that for every $x \in U$, there are an open subset $W \subseteq U$, a natural number $d$ and $f,g \in R_d$ such that $W \cap \Vp(g) = \emptyset$ and $\phi(y) = \frac{f(y_0,\ldots,y_n)}{g(y_0,\ldots,y_n)}$ for $y = [y_0,\ldots,y_n] \in W$.
\end{definition}
\begin{remark}
This is a subsheaf of rings of the sheaf of $\mathfrak{k}$-valued functions on $X$.
Under the identification $\mathbb{A}^n =\mathfrak{k}^n$ with $\mathbb{P}^n \setminus \Vp(X_0)$, one has $\mathcal{O}_X \defon{X \setminus \Vp(X_0)} = \mathcal{O}_{X \cap \mathbb{A}^n}$ as subsheaves of the sheaf of $\mathfrak{k}$-valued functions, where the second sheaf is a sheaf on a closed subset of $\mathfrak{k}^n$:
Indeed, if $W$ is as in the definition then $\phi([1,y_1,\ldots,y_n]) = \frac{f(1,y_1,\ldots,y_n)}{g(1,y_1,\ldots,y_n)}$ for $[1,y_1,\ldots,y_n] \in W$.
Conversely if $\phi([1,y_1,\ldots,y_n]) = \frac{f(y_1,\ldots,y_n)}{g(y_1,\ldots,y_n)}$ on an open subset $W $ of $X \cap \mathbb{A}^n$ then
$\phi([y_0,\ldots,y_n]) = \frac{F(y_0,\ldots,y_n)}{G(y_0,\ldots,y_n)}$ on $W$ where $F(X_0,\ldots,X_n) \coloneqq X_0^d f(\frac{X_1}{X_0}, \ldots, \frac{X_n}{X_0})$ and $G(X_0,\ldots,X_n) = X_0^d g(\frac{X_1}{X_0},\ldots, \frac{X_n}{X_0})$ with a sufficiently large $d \in \N$.
\end{remark}
\begin{remark}
It follows from the previous remark and the similar result in the affine case that the elements of $\mathcal{O}_X(U)$ are continuous on $U \setminus V(X_0)$.
Since the situation is symmetric in the homogeneous coordinates, they are continuous on all of $U$.
\end{remark}
The following is somewhat harder than in the affine case:
\begin{proposition}
If $X$ is connected (e.g. irreducible), then the elements of $\mathcal{O}_X\left( X \right) $ are constant functions on $X$.
\end{proposition}
% Lecture 14
\subsection{The notion of a category}
\begin{definition}
A \vocab{category} $\mathcal{A}$ consists of:
\begin{itemize}
\item A class $\Ob \mathcal{A}$ of \vocab[Objects]{objects of $\mathcal{A}$}.
\item For two arbitrary objects $A, B \in \Ob \mathcal{A}$, a \textbf{set} $\Hom_\mathcal{A}(A,B)$ of \vocab[Morphism]{morphisms for $A$ to $B$ in $\mathcal{A}$}.
\item A map $\Hom_\mathcal{A}(B,C) \times \Hom_\mathcal{A}(A,B) \xrightarrow{\circ} \Hom_\mathcal{A}(A,C)$, the composition of morphisms, for arbitrary triples $(A,B,C)$ of objects of $\mathcal{A}$.
\end{itemize}
The following conditions must be satisfied:
\begin{enumerate}[A]
\item For morphisms $A \xrightarrow{f} B\xrightarrow{g} C \xrightarrow{h} D$, we have $h \circ (g \circ f) = (h \circ g) \circ f$.
\item For every $A \in \Ob(\mathcal{A})$, there is an $\Id_A \in \Hom_{\mathcal{A}}(A,A)$ such that $\Id_A \circ f = f$ (reps. $g \circ \Id_A = g$) for arbitrary morphisms $B \xrightarrow{f} A$ (reps. $A \xrightarrow{g} C).$
\end{enumerate}
A morphism $X \xrightarrow{f} Y$ is called an \vocab[Isomorphism]{isomorphism (in $\mathcal{A} $)} if there is a morphism $Y \xrightarrow{g} X$ (called the \vocab[Inverse morphism]{inverse $f^{-1}$ of $f$)} such that $g \circ f = \Id_X$ and $f \circ g = \Id_Y$.
\end{definition}
\begin{remark}
\begin{itemize}
\item The distinction between classes and sets is important here.
\item We will usually omit the composition sign $\circ$.
\item It is easy to see that $\Id_A$ is uniquely determined by the above condition $B$, and that the inverse $f^{-1}$ of an isomorphism $f$ is uniquely determined.
\end{itemize}
\end{remark}
\subsubsection{Examples of categories}
\begin{example}
\begin{itemize}
\item The category of sets.
\item The category of groups.
\item The category of rings.
\item If $R$ is a ring, the category of $R$-modules and the category $\Alg_R$ of $R$-algebras
\item The category of topological spaces
\item The category $\Var_\mathfrak{k}$ of varieties over $\mathfrak{k}$ (see \ref{defvariety})
\item If $\mathcal{A}$ is a category, then the \vocab{opposite category} or \vocab{dual category} is defined by $\Ob(\mathcal{A}\op) = \Ob(\mathcal{A})$ and $\Hom_{\mathcal{A}\op}(X,Y) = \Hom_\mathcal{A}(Y,X)$.
\end{itemize}
In most of these cases, isomorphisms in the category were just called `isomorphism'. The isomorphisms in the category of topological spaces are the homeomophisms.
\end{example}
\subsubsection{Subcategories}
\begin{definition}[Subcategories]
A \vocab{subcategory} of $\mathcal{A}$ is a category $\mathcal{B}$ such that $\Ob(\mathcal{B}) \subseteq \Ob(\mathcal{A})$, such that $\Hom_\mathcal{B}(X,Y) \subseteq \Hom_\mathcal{A}(X,Y)$ for objects $X$ and $Y$ of $\mathcal{B}$, such that for every object $X \in \Ob(\mathcal{B})$, the identity $\Id_X$ of $X$ is the same in $\mathcal{B}$ as in $\mathcal{A}$, and such that for composable morphisms in $\mathcal{B}$, their compositions in $\mathcal{A}$ and $\mathcal{B}$ coincide.
We call $\mathcal{B}$ a \vocab{full subcategory} of $\mathcal{A}$ if in addition $\Hom_\mathcal{B}(X,Y) = \Hom_\mathcal{A}(X,Y)$ for arbitrary $X,Y \in \Ob(\mathcal{B})$.
\end{definition}
\begin{example}
\begin{itemize}
\item The category of abelian groups is a full subcategory of the category of groups.
It can be identified with the category of $\Z$-modules.
\item The category of finitely generated $R$-modules as a full subcategory of the category of $R$-modules.
\item The category of $R$-algebras of finite type as a full subcategory of $\Alg_R$.
\item The category of affine varieties over $\mathfrak{k}$ as a full subcategory of the category of varieties over $\mathfrak{k}$.
\end{itemize}
\end{example}
\subsubsection{Functors and equivalences of categories}
\begin{definition}
A \vocab[Functor!covariant]{(covariant) functor} (resp. \vocab[Functor!contravariant]{contravariant functor}) between categories $\mathcal{A} \xrightarrow{F} \mathcal{B}$ is a map $\Ob(\mathcal{A}) \xrightarrow{F} \Ob(\mathcal{B})$ with a family of maps $\Hom_\mathcal{A}(X,Y) \xrightarrow{F} \Hom_\mathcal{B}(F(X),F(Y))$ (resp. $\Hom_\mathcal{A}(X,Y) \xrightarrow{F} \Hom_\mathcal{B}(F(Y),F(X))$ in the case of contravariant functors), where $X$ and $Y$ are arbitrary objects of $\mathcal{A}$, such that the following conditions hold:
\begin{itemize}
\item $F(\Id_X) = \Id_{F(X)}$
\item For morphisms $X \xrightarrow{f} Y \xrightarrow{g} Z$ in $\mathcal{A}$, we have $F(gf) = F(g)F(f)$ ( resp. $F(gf) = F(f)F(g)$)
\end{itemize}
A functor is called \vocab[Functor!essentially surjective]{essentially surjective} if every object of $\mathcal{B}$ is isomorphic to an element of the image of $\Ob(\mathcal{A}) \xrightarrow{F} \Ob(\mathcal{B})$.
A functor is called \vocab[Functor!full]{full} (resp. \vocab[Functor!faithful]{faithful}) if it induces surjective (resp. injective) maps between sets of morphisms.
It is called an \vocab{equivalence of categories} if it is full, faithful and essentially surjective.
\end{definition}
\begin{example}
\begin{itemize}
\item There are \vocab[Functor!forgetful]{forgetful functors} from rings to abelian groups or from abelian groups to sets which drop the multiplicative structure of a ring or the group structure of a group.
\item If $\mathfrak{k}$ is any vector space there is a contravariant functor from $\mathfrak{k}$-vector spaces to itself sending $V$ to its dual vector space $V\subseteq$ and $V \xrightarrow{f} W$ to the dual linear map $W^{\ast} \xrightarrow{f^{\ast}} V^{\ast}$.
When restricted to the full subcategory of finite-dimensional vector spaces it becomes a contravariant self-equivalence of that category.
\item The embedding of a subcategory is a faithful functor. In the case of a full subcategory it is also full.
\end{itemize}
\end{example}
\subsection{The category of varieties}
\begin{definition}[Algebraic variety]\label{defvariety}
An \vocab{algebraic variety} or \vocab{prevariety} over $\mathfrak{k}$ is a pair $(X, \mathcal{O}_X)$, where $X$ is a topological space and $\mathcal{O}_X$ a subsheaf of the sheaf of $\mathfrak{k}$-valued functions on $X$ such that for every $x \in X$, there are a neighbourhood $U_x$ of $x$ in $X$, an open subset $V_x$ of a closed subset $Y_x$ of $\mathfrak{k}^{n_x}$\footnote{By the result of \ref{affopensubtopbase} it can be assumed that $V_x = Y_x$ without altering the definition.} and a homeomorphism $V_x \xrightarrow{\iota_x} U_x$ such that for every open subset $V \subseteq U_x$ and every function $V\xrightarrow{f} \mathfrak{k}$, we have $f \in \mathcal{O}_X(V) \iff \iota^{\ast}_x(f) \in \mathcal{O}_{Y_x}(\iota_x^{-1}(V))$,
In this, the \vocab{pull-back} $\iota_x^{\ast}(f)$ of $f$ is defined by $(\iota_x^{\ast}(f))(\xi) \coloneqq f(\iota_x(\xi))$.
A morphism $(X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ of varieties is a continuous map $X \xrightarrow{\phi} Y$ such that for all open $U \subseteq Y$ and $f \in \mathcal{O}_Y(U)$, $\phi^{\ast}(f) \in \mathcal{O}_X(\phi^{-1}(U))$.
An isomorphism is a morphism such that $\phi$ is bijective and $\phi^{-1}$ also is a morphism of varieties.
\end{definition}
\begin{example}
\begin{itemize}
\item If $(X, \mathcal{O}_X)$ is a variety and $U \subseteq X$ open, then $(U, \mathcal{O}_X\defon{U})$ is a variety (called an \vocab{open subvariety} of $X$), and the embedding $U \to X$ is a morphism of varieties.
\item If $X$ is a closed subset of $\mathfrak{k}^n$ or $\mathbb{P}^n$, then $(X, \mathcal{O}_X)$ is a variety, where $\mathcal{O}_X$ is the structure sheaf on $X$ (\ref{structuresheafkn}, reps. \ref{structuresheafpn}).
A variety is called \vocab[Variety!affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of this form, with $X $ closed in $\mathfrak{k}^n$ (resp. $\mathbb{P}^n$).
A variety which is isomorphic to and open subvariety of $X$ is called \vocab[Variety!quasi-affine]{quasi-affine} (resp. \vocab[Variety!quasi-projective]{quasi-projective}).
\item If $X = V(X^2 - Y^3) \subseteq \mathfrak{k}^2$ then $\mathfrak{k} \xrightarrow{t \mapsto (t^3,t^2)} X$ is a morphism which is a homeomorphism of topological spaces but not an isomorphism of varieties.
% TODO
\item The composition of two morphisms $X \to Y \to Z$ of varieties is a morphism of varieties.
\item $X\xrightarrow{\Id_X} X$ is a morphism of varieties.
\end{itemize}
\end{example}
\subsubsection{The category of affine varieties}
\begin{lemma}\label{localinverse}
Let $X$ be any $\mathfrak{k}$-variety and $U \subseteq X$ open.
\begin{enumerate}[i)]
\item All elements of $\mathcal{O}_X(U)$ are continuous.
\item If $U \subseteq X$ is open, $U \xrightarrow{\lambda} \mathfrak{k}$ any function and every $x \in U$ has a neighbourhood $V_x \subseteq U$ such that $\lambda \defon{V_x} \in \mathcal{O}_X(V_x)$, then $\lambda \in \mathcal{O}_X(U)$.
\item If $\vartheta \in \mathcal{O}_X(U)$ and $\vartheta(x) \neq 0$ for all $x \in U$, then $\vartheta \in \mathcal{O}_X(U)^{\times }$.
\end{enumerate}
\end{lemma}
\begin{proof}
\begin{enumerate}[i)]
\item The property is local on $U$, hence it is sufficient to show it in the quasi-affine case. This was done in \ref{structuresheafcontinuous}.
\item For the second part, let $\lambda_x \coloneqq \lambda \defon{V_x} $.
We have $\lambda_x\defon{V_x \cap V_y} = \lambda \defon{V_x \cap V_y} = \lambda_y \defon{V_x \cap V_y} $.
The $V_x$ cover $U$. By the sheaf axiom for $\mathcal{O}_X$ there is $\ell \in \mathcal{O}_X(U)$ with $\ell\defon{V_x} =\lambda_x$. It follows that $\ell=\lambda$.
\item By the definition of variety, every $x \in U$ has a quasi-affine neighbourhood $V \subseteq U$. We can assume $U$ to be quasi-affine and $X = V(I) \subseteq \mathfrak{k}^n$, as the general assertion follows by an application of ii).
If $x \in U$ there are a neighbourhood $x \in W \subseteq U$ and $a,b \in R = \mathfrak{k}[X_1,\ldots,X_n]$ such that $\vartheta(y) = \frac{a(y)}{b(y)}$ for $y \in W$, with $b(y) \neq 0$.
Then $a(x) \neq 0$ as $\vartheta(x) \neq 0$. Replacing $W$ by $W \setminus V(a)$, we may assume that $a$ has no zeroes on $W$.
Then $\lambda(y) = \frac{b(y)}{a(y)}$ for $y \in W$ has a non-vanishing denominator and $\lambda \in \mathcal{O}_X(U)$.
We have $\lambda \cdot \vartheta = 1$, thus $\vartheta \in \mathcal{O}_X(U)^{\times}$.
\end{enumerate}
\end{proof}
\begin{proposition}[About affine varieties]
\label{propaffvar}
\begin{itemize}
\item Let $X,Y$ be varieties over $\mathfrak{k}$. Then the map
\begin{align}
\phi: \Hom_{\Var_\mathfrak{k}}(X,Y) &\longrightarrow \Hom_{\Alg_\mathfrak{k}}(\mathcal{O}_Y(Y), \mathcal{O}_X(X)) \\
(X \xrightarrow{f} Y) &\longmapsto (\mathcal{O}_Y(Y) \xrightarrow{f^{\ast}} \mathcal{O}_X(X))
\end{align}
is injective when $Y$ is quasi-affine and bijective when $Y$ is affine.
\item The contravariant functor
\begin{align}
F: \Var_\mathfrak{k} &\longrightarrow \Alg_\mathfrak{k} \\
X &\longmapsto \mathcal{O}_X(X)\\
(X\xrightarrow{f} Y) &\longmapsto (\mathcal{O}_X(X) \xrightarrow{f^{\ast}} \mathcal{O}_Y(Y))
\end{align}
restricts to an equivalence of categories between the category of affine varieties over $\mathfrak{k}$ and the full subcategory $\mathcal{A}$ of $\Alg_\mathfrak{k}$,
having the $\mathfrak{k}$-algebras $A$ of finite type with $\nil A = \{0\} $ as objects.
\end{itemize}
\end{proposition}
\begin{remark}
It is clear that $\nil(\mathcal{O}_X(X)) = \{0\}$ for arbitrary varieties. For general varieties it is however not true that $\mathcal{O}_X(X)$ is a $\mathfrak{k}$-algebra of finite type.
There are counterexamples even for quasi-affine $X$. %TODO
If, however, $X$ is affine, we may assume w.l.o.g. that $X = V(I)$ where $I = \sqrt{I} \subseteq R$ is an ideal with $R = \mathfrak{k}[X_1,\ldots,X_n]$.
Then $\mathcal{O}_X(X) \cong R / I$ (see \ref{structuresheafri}) is a $\mathfrak{k}$-algebra of finite type.
\end{remark}
\begin{proof}
It suffices to investigate $\phi$ when $Y$ is an open subset of $V(I) \subseteq \mathfrak{k}^n$, where $I = \sqrt{I} \subseteq R$ is an ideal and $Y = V(I)$ when $Y$ is affine.
Let $(f_1,\ldots,f_n)$ be the components of $X \xrightarrow{f} Y \subseteq \mathfrak{k}^n$. Let $Y \xrightarrow{\xi_i} \mathfrak{k}$ be the $i$-th coordinate.
By definition $f_i = f^{\ast}(\xi_i) $. Thus $f$ is uniquely determined by $\mathcal{O}_Y(Y) \xrightarrow{f^{\ast}} \mathcal{O}_X(X)$.
Conversely, let $Y = V(I)$ and $\mathcal{O}_Y(Y) \xrightarrow{\phi} \mathcal{O}_X(X)$ be a morphism of $\mathfrak{k}$-algebras. Define $f_i \coloneqq \phi(\xi_i)$ and consider $X \xrightarrow{f = (f_1,\ldots,f_n)} Y\subseteq \mathfrak{k}^n$.
\begin{claim}
$f$ has image contained in $Y$.
\end{claim}
\begin{subproof}
For $x \in X, \lambda \in I$ we have $\lambda(f(x)) = (\phi(\lambda \mod I))(x) = 0$ as $\phi$ is a morphism of $\mathfrak{k}$-algebras.
Thus $f(x) \in V(I) = Y$.
\end{subproof}
\begin{claim}
$f$ is a morphism in $\Var_\mathfrak{k}$
\end{claim}
\begin{subproof}
For open $\Omega \subseteq Y, U = f^{-1}(\Omega) = \{x \in X | \forall \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y \setminus \Omega = V(J)$.
If $\lambda \in \mathcal{O}_Y(\Omega)$ and $x \in U$, then $f(x)$ has a neighbourhood $V$ such that there are $a,b \in R$ with $\lambda(v) = \frac{a(v)}{b(v)}$ and $b(v) \neq 0$ for all $v \in V$.
Let $W \coloneqq f^{-1}(V)$. Then $\alpha \coloneqq \phi(a)\defon{W} \in \mathcal{O}_X(W)$, $\beta \coloneqq \phi(b)\defon{W} \in \mathcal{O}_X(W)$.
By the second part of \ref{localinverse} $\beta \in \mathcal{O}_X(W)^{\times}$ and $f^{\ast}(\lambda)\defon{W} = \frac{\alpha}{\beta} \in \mathcal{O}_X(W)$.
The first part of \ref{localinverse} shows that $f^{\ast}(\lambda) \in \mathcal{O}_X(U)$.
\end{subproof}
By definition of $f$, we have $f^{\ast} = \phi$. This finished the proof of the first point.
\begin{claim}
The functor in the second part maps affine varieties to objects of $\mathcal{A}$ and is essentially surjective.
\end{claim}
\begin{subproof}
It follows from the remark that the functor maps affine varieties to objects of $\mathcal{A}$.
If $A \in \Ob(\mathcal{A})$ then $ A /\mathfrak{k}$ is of finite type, thus $A \cong R / I$ for some $n$.
Since $\nil(A) = \{0\}$ we have $I = \sqrt{I}$, as for $x \in \sqrt{I}$, $x \mod I \in \nil(R / I) \cong \nil(A) = \{0\}$.
Thus $A \cong\mathcal{O}_X(X)$ where $X = V(I)$.
\end{subproof}
Fullness and faithfulness of the functor follow from the first point.
\end{proof}
\begin{remark}
Note that giving a contravariant functor $\mathcal{C} \to \mathcal{D}$ is equivalent to giving a functor $\mathcal{C} \to \mathcal{D}\op$. We have thus shown that the category of affine varieties is equivalent to $\mathcal{A}\op$, where $\mathcal{A} \subsetneq \Alg_\mathfrak{k}$ is the full subcategory of $\mathfrak{k}$-algebras $A$ of finite type with $\nil(A) = \{0\}$.
\end{remark}
\subsubsection{Affine open subsets are a topology base}
\begin{definition}
A set $\mathcal{B}$ of open subsets of a topological space $X$ is called a \vocab{topology base} for $X$ if every open subset of $X$ can be written as a (possibly empty) union of elements of $\mathcal{B}$.
\end{definition}
\begin{fact}
If $X$ is a set, then $\mathcal{B} \subseteq \mathcal{P}(X)$ is a base for some topology on $X$ iff $X = \bigcup_{U \in \mathcal{B}} U$ and for arbitrary $U, V \in \mathcal{B}, U \cap V$ is a union of elements of $\mathcal{B}$.
\end{fact}
\begin{definition}
Let $X$ be a variety.
An \vocab{affine open subset} of $X$ is a subset which is an affine variety.
\end{definition}
\begin{proposition}\label{oxulocaf}
Let $X$ be an affine variety over $\mathfrak{k}$, $\lambda \in \mathcal{O}_X(X)$ and $U = X \setminus V(\lambda)$.
Then $U$ is an affine variety and the morphism $\phi: \mathcal{O}_X(X)_\lambda \to \mathcal{O}_X(U)$ defined by the restriction $\mathcal{O}_X(X) \xrightarrow{\cdot |_U } \mathcal{O}_X(U)$ and the universal property of the localization is an isomorphism.
\end{proposition}
\begin{proof}
Let $X$ be an affine variety over $\mathfrak{k}, \lambda \in \mathcal{O}_X(X)$ and $U = X \setminus V(\lambda)$. The fact that $\lambda\defon{U} \in \mathcal{O}_x(U)^{\times}$ follows from \ref{localinverse}.
Thus the universal property of the localization $\mathcal{O}_X(X)_\lambda$ can be applied to $\mathcal{O}_X(X) \xrightarrow{\cdot |_U} \mathcal{O}_X(U)$.
\[
\begin{tikzcd}
\mathcal{O}_X(X) \arrow{d}{\cdot |_U}\arrow{r}{x \mapsto \frac{x}{1}} & \mathcal{O}_X(X)_\lambda \arrow[dotted, bend left]{dl}{\existsone \phi} \\
\mathcal{O}_X(U) &
\end{tikzcd}
\]
\[
\begin{tikzcd}
&Y \arrow[bend right, swap]{ld}{\pi_0} \arrow[bend right, swap]{d}{\pi}&\mathcal{O}_Y(Y) \cong A_\lambda \arrow{d}{\mathfrak{s}}& \\
X \arrow[hookrightarrow]{r}{}& U \arrow[swap]{u}{\sigma} & \mathcal{O}_X(U)
\end{tikzcd}
\]
For the rest of the proof, we may assume $X = V(I) \subseteq \mathfrak{k}^n$ where $I = \sqrt{I} \subseteq R \coloneqq\mathfrak{k}[X_1,\ldots,X_n]$ is an ideal.
Then $A \coloneqq \mathcal{O}_X(X) \cong R / I$ and there is $\ell \in R$ such that $\ell\defon{X} = \lambda$.
Let $Y = V(J) \subseteq \mathfrak{k}^{n+1}$ where $J \subseteq \mathfrak{k}[Z,X_1,\ldots,X_n]$ is generated by the elements of $I$ and $1 - Z\ell(X_1,\ldots,X_n)$.
Then $\mathcal{O}_Y(Y) \cong \mathfrak{k}[Z,X_1,\ldots,X_n] / J \cong A[Z] / (1 -\lambda Z) \cong A_\lambda$.
By the proposition about affine varieties (\ref{propaffvar}), the morphism $\mathfrak{s}: \mathcal{O}_Y(Y) \cong A_\lambda \to \mathcal{O}_X(U)$ corresponds to a morphism $U \xrightarrow{\sigma} Y$.
We have $\mathfrak{s}(Z \mod J) = \lambda^{-1}$ and $\mathfrak{s}(X_i \mod J) = X_i \mod I$.
Thus $\sigma(x) = (\lambda(x)^{-1}, x)$ for $x \in U$.
Moreover, the projection $Y \xrightarrow{\pi_0} X$ dropping the $Z$-coordinate has image contained in $U$, as for $(z,x) \in Y$ the equation
\[
1 = z\lambda(x)
\]
implies $\lambda(x) \neq 0$. It thus defines a morphism $Y \xrightarrow{\pi} U$ and by the description of $\sigma$ it follows that $\sigma \pi = \Id_U$.
Similarly it follows that $\sigma \pi = \Id_Y$. Thus, $\sigma$ and $\pi$ are inverse to each other.
\end{proof}
\begin{corollary}\label{affopensubtopbase}
The affine open subsets of a variety $X$ are a topology base on $X$.
\end{corollary}
\begin{proof}
Let $X = V(I) \subseteq \mathfrak{k}^n$ with $I = \sqrt{I}$. If $U \subseteq X$ is open then $X \setminus U = V(J)$ with $J \supseteq I$ and $U = \bigcup_{f \in J} (X \setminus V(f))$.
Thus $U$ is a union of affine open subsets. The same then holds for arbitrary quasi-affine varieties.
Let $X$ be any variety, $U \subseteq X$ open and $x \in U$.
By the definition of variety, $x$ has a neighbourhood $V_x$ which is quasi-affine, and replacing $V_x$ by $U \cap V_x$ which is also quasi-affine we may assume $V_x \subseteq U$.
$V_x$ is a union of its affine open subsets. Because $U$ is the union of the $V_x$, $U$ as well is a union of affine open subsets.
\end{proof}
% Lecture 14A TODO?
% Lecture 15
% CRTPROG
\subsection{Stalks of sheaves}
\begin{definition}[Stalk]
Let $\mathcal{G}$ be a presheaf of sets on the topological space $X$, and let $x \in X$.
The \vocab{stalk} (\vocab[Stalk]{Halm}) of $\mathcal{G}$ at $x$ is the set of equivalence classes of pairs $(U, \gamma)$, where $U$ is an open neighbourhood of $x$ and $\gamma \in \mathcal{G}(U)$
and the equivalence relation $\sim $ is defined as follows:
$( U , \gamma) \sim (V, \delta)$ iff there exists an open neighbourhood $W \subseteq U \cap V$ of $x$ such that $\gamma \defon{W} = \delta \defon{W}$.
If $\mathcal{G}$ is a presheaf of groups, one can define a groups structure on $\mathcal{G}_x$ by
\[
((U, \gamma) / \sim ) \cdot \left( (V,\delta) / \sim \right) = (U \cap V, \gamma \defon{U \cap V} \cdot \delta\defon{U \cap V}) / \sim
\]
If $\mathcal{G}$ is a presheaf of rings, one can similarly define a ring structure on $\mathcal{G}_x$.
If $U$ is an open neighbourhood of $x \in X$, then we have a map (resp. homomorphism)
\begin{align}
\cdot_x : \mathcal{G}(U) &\longrightarrow \mathcal{G}_x \\
\gamma &\longmapsto \gamma_x \coloneqq (U, \gamma) / \sim
\end{align}
\end{definition}
\begin{fact}
Let $\gamma,\delta \in \mathcal{G}(U)$. If $\mathcal{G}$ is a sheaf\footnote{or, more generally, a separated presheaf} and if for all $x \in U$, we have $\gamma_x = \delta_x$, then $\gamma = \delta$.
In the case of a sheaf, the image of the injective map $\mathcal{G}(U) \xrightarrow{\gamma \mapsto (\gamma_x)_{x \in U}} \prod_{x \in U} \mathcal{G}_x$
is the set of all $(g_x)_{x \in U} \in \prod_{x \in U} \mathcal{G}_x $ satisfying the following \vocab{coherence condition}:
For every $x \in U$, there are an open neighbourhood $W_x \subseteq U$ of $x$ and $g^{(x)} \in \mathcal{G}(W_x)$ with $g_y^{(x)} = g_y$ for all $y \in W_x$.
\end{fact}
\begin{proof}
Because of $\gamma_x = \delta_x$, there is $x \in W_x \subseteq U$ open such that $\gamma\defon{W_x} = \delta\defon{W_x}$. As the $W_x$ cover $U$, $\gamma = \delta$ by the sheaf axiom.
\end{proof}
\begin{definition}
Let $\mathcal{G}$ be a sheaf of functions.
Then $\gamma_x$ is called the \vocab{germ} of the function $\gamma$ at $x$.
The \vocab[Germ!value at $x$]{value at $x$ } of $g = (U, \gamma) / \sim \in \mathcal{G}_x$ defined as $g(x) \coloneqq \gamma(x)$, which is independent of the choice of the representative $\gamma$.
\end{definition}
\begin{remark}
If $\mathcal{G}$ is a sheaf of $C^{\infty}$-functions (resp. holomorphic functions), then $\mathcal{G}_x$ is called the ring of germs of $C^\infty$-functions (resp. of holomorphic functions) at $x$.
\end{remark}
\subsubsection{The local ring of an affine variety}
\begin{definition}
If $X$ is a variety, the stalk $\mathcal{O}_{X,x}$ of the structure sheaf at $x$ is called the \vocab{local ring} of $X$ at $x$.
This is indeed a local ring, with maximal ideal $\mathfrak{m}_x = \{f \in \mathcal{O}_{X,x} | f(x) = 0\}$.
\end{definition}
\begin{proof}
By \ref{localring} it suffices to show that $\mathfrak{m}_x$ is a proper ideal, which is trivial, and that the elements of $\mathcal{O}_{X,x} \setminus \mathfrak{m}_x$ are units in $\mathcal{O}_{X,x}$.
Let $g = (U, \gamma)/\sim \in \mathcal{O}_{X,x}$ and $g(x) \neq 0$.
$\gamma$ is Zariski continuous (first point of \ref{localinverse}). Thus $V(\gamma)$ is closed. By replacing $U$ by $U \setminus V(\gamma)$ we may assume that $\gamma$ vanishes nowhere on $U$.
By the third point of \ref{localinverse} we have $\gamma \in \mathcal{O}_X(U)^{\times}$.
$(\gamma^{-1})_x$ is an inverse to $g$.
\end{proof}
\begin{proposition}\label{proplocalring}
Let $X = \Va(I) \subseteq \mathfrak{k}^n$ be equipped with its usual structure sheaf, where $I = \sqrt{I} \subseteq R = \mathfrak{k}[X_1,\ldots,X_n]$ . Let $x \in X$ and $A = \mathcal{O}_X(X) \cong R / I$.
$\{P \in R | P(x) = 0\} \text{\reflectbox{$\coloneqq$}} \fn_x \subseteq R$ is maximal, $I \subseteq \fn_x$ and $\mathfrak{m}_x \coloneqq \fn_x / I$ is the maximal ideal of elements of $A$ vanishing at $x$.
If $\lambda \in A \setminus \mathfrak{m}_x$, we have $\lambda_x \in \mathcal{O}_{X,x}^{\times}$, where $\lambda_x$ denotes the image under $A \cong \mathcal{O}_X(X) \to \mathcal{O}_{X,x}$.
By the universal property of the localization, there exists a unique ring homomorphism $A_{\mathfrak{m}_x} \xrightarrow{\iota} \mathcal{O}_{X,x}$
such that
\[
\begin{tikzcd}
A \arrow{r}{} \arrow{d}{\lambda \mapsto \lambda_x} & A_{\mathfrak{m}_x} \arrow[dotted, bend left]{ld}{\existsone \iota} \\
\mathcal{O}_{X,x}
\end{tikzcd}
\]
commutes.
The morphism $A_{\mathfrak{m}_x}\xrightarrow{\iota} \mathcal{O}_{X,x}$ is an isomorphism.
\end{proposition}
\begin{proof}
To show surjectivity, let $\ell = (U, \lambda) / \sim \in \mathcal{O}_{X,x}$, where $U$ is an open neighbourhood of $x$ in $X$.
We have $X \setminus U = V(J)$ where $J \subseteq A$ is an ideal. As $x \in U$ there is $f \in J$ with $f(x) \neq 0$. Replacing $U $ by $X \setminus V(f)$ we may assume $U = X \setminus V(f)$.
By \ref{oxulocaf}, $\mathcal{O}_X(U) \cong A_f$, and $\lambda = f^{-n}\vartheta$ for some $n \in \N$ and $\vartheta \in A$.
Then $\ell = \iota(f^{-n} \vartheta)$ where the last fraction is taken in $A_{\mathfrak{m}_x}$.
Let $\lambda = \frac{\vartheta}{g} \in A_{\mathfrak{m}_x}$ with $\iota(\lambda) = 0$.
It is easy to see that $\iota(\lambda) = (X \setminus V(g), \frac{\vartheta}{g}) / \sim $.
Thus there is an open neighbourhood $U$ of $x$ in $X \setminus V(g)$ such that $\vartheta$ vanishes on $U$.
Similar as before there is $h \in A$ with $h(x) \neq 0$ and $W = X \setminus V(h) \subseteq U$.
By the isomorphism $\mathcal{O}_X(W) \cong A_h$, there is $n \in \N$ with $h^{n}\vartheta = 0$ in $A$. Since $h \not\in \mathfrak{m}_x$, $h$ is a unit and the image of $\vartheta$ in $A_{\mathfrak{m}_x}$ vanishes, implying $\lambda = 0$.
\end{proof}
\subsubsection{Intersection multiplicities and Bezout's theorem}
\begin{definition}
Let $R = \mathfrak{k}[X_0,X_1,X_2]$ equipped with its usual grading and let $x \in \mathbb{P}^{2}$.
Let $G \in R_g, H \in R_h$ be homogeneous polynomials with $x \in V(G) \cap V(h)$.
Let $\ell\in R_1$ such that $\ell(x) \neq 0$. Then $x \in U = \mathbb{P}^2 \setminus V(\ell)$ and the rational functions $\gamma = \ell^{-g}G, \eta = \ell^{-h}H$ are elements of $\mathcal{O}_{\mathbb{P}^2}(U)$.
Let $I_x(G,H) \subseteq \mathcal{O}_{\mathbb{P}^2,x}$ denote the ideal generated by $\gamma_x$ and $\eta_x$.
\noindent The dimension $\dim_{\mathfrak{k}}(\mathcal{O}_{X,x} / I_x(G,H)) \text{\reflectbox{$\coloneqq$}} i_x(G,H)$ is called the \vocab{intersection multiplicity} of $G$ and $H$ at $x$.
\end{definition}
\begin{remark}
If $\tilde \ell \in R_1$ also satisfies $\tilde \ell(x) \neq 0$, then the image of $\tilde \ell / \ell$ under $\mathcal{O}_{\mathbb{P}^2}(U) \to \mathcal{O}_{\mathbb{P}^2,x}$ is a unit, showing that the image of $\tilde \gamma = \tilde \ell^{-g} G$ in $\mathcal{O}_{\mathbb{P}^2,x}$ is multiplicatively equivalent to $\gamma_x$, and similarly for $\eta_x$.
Thus $I_x(G,H)$ does not depend on the choice of $\ell \in R_1$ with $\ell(x) \neq 0$.
\end{remark}
\begin{theorem}[Bezout's theorem]
In the above situation, assume that $V(H)$ and $V(G)$ intersect properly in the sense that $V(G) \cap V(H) \subseteq \mathbb{P}^2$ has no irreducible component of dimension $\ge 1$.
Then
\[
\sum_{x \in V(G) \cap V(H)} i_x(G,H) = gh
\]
Thus, $V(G) \cap V(H)$ has $gh$ elements counted by multiplicity.
\end{theorem}
\printvocabindex
\end{document}