\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}
        Without loss of generality loss of generality 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}
        Without loss of generality loss of generality 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}
