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Maximilian Keßler 2022-02-16 02:13:07 +01:00
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@ -2501,7 +2501,7 @@ The following is somewhat harder than in the affine case:
\begin{example} \begin{example}
\begin{itemize} \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 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 $W\st \xrightarrow{f\st} V\st$. \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. 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. \item The embedding of a subcategory is a faithful functor. In the case of a full subcategory it is also full.
\end{itemize} \end{itemize}
@ -2512,12 +2512,12 @@ The following is somewhat harder than in the affine case:
\subsection{The category of varieties} \subsection{The category of varieties}
\begin{definition}[Algebraic variety]\label{defvariety} \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 $f \in \mathcal{O}_X(V) \iff \iota\st_x(f) \in \mathcal{O}_{Y_x}(\iota_x^{-1}(V))$, 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} $\iota_x\st(f)$ of $f$ is defined by $(\iota_x\st(f))(\xi) \coloneqq f(\iota_x(\xi))$. 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 $f \in \mathcal{O}_Y(U)$, $\phi\st(f) \in \mathcal{O}_X(\phi^{-1}(U))$. 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. An isomorphism is a morphism such that $\phi$ is bijective and $\phi^{-1}$ also is a morphism of varieties.
\end{definition} \end{definition}
\begin{example} \begin{example}
@ -2564,14 +2564,14 @@ The following is somewhat harder than in the affine case:
\item Let $X,Y$ be varieties over $\mathfrak{k}$. Then the map \item Let $X,Y$ be varieties over $\mathfrak{k}$. Then the map
\begin{align} \begin{align}
\phi: \Hom_{\Var_\mathfrak{k}}(X,Y) &\longrightarrow \Hom_{\Alg_\mathfrak{k}}(\mathcal{O}_Y(Y), \mathcal{O}_X(X)) \\ \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\st} \mathcal{O}_X(X)) (X \xrightarrow{f} Y) &\longmapsto (\mathcal{O}_Y(Y) \xrightarrow{f^{\ast}} \mathcal{O}_X(X))
\end{align} \end{align}
is injective when $Y$ is quasi-affine and bijective when $Y$ is affine. is injective when $Y$ is quasi-affine and bijective when $Y$ is affine.
\item The contravariant functor \item The contravariant functor
\begin{align} \begin{align}
F: \Var_\mathfrak{k} &\longrightarrow \Alg_\mathfrak{k} \\ F: \Var_\mathfrak{k} &\longrightarrow \Alg_\mathfrak{k} \\
X &\longmapsto \mathcal{O}_X(X)\\ X &\longmapsto \mathcal{O}_X(X)\\
(X\xrightarrow{f} Y) &\longmapsto (\mathcal{O}_X(X) \xrightarrow{f\st} \mathcal{O}_Y(Y)) (X\xrightarrow{f} Y) &\longmapsto (\mathcal{O}_X(X) \xrightarrow{f^{\ast}} \mathcal{O}_Y(Y))
\end{align} \end{align}
restricts to an equivalence of categories between the category of affine varieties over $\mathfrak{k}$ and the full subcategory $\mathcal{A}$ of $\Alg_\mathfrak{k}$, 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. having the $\mathfrak{k}$-algebras $A$ of finite type with $\nil A = \{0\} $ as objects.
@ -2593,7 +2593,7 @@ The following is somewhat harder than in the affine case:
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. 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. 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\st(\xi_i) $. Thus $f$ is uniquely determined by $\mathcal{O}_Y(Y) \xrightarrow{f\st} \mathcal{O}_X(X)$. 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 $X \xrightarrow{f = (f_1,\ldots,f_n)} Y\subseteq \mathfrak{k}^n$. 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} \begin{claim}
$f$ has image contained in $Y$. $f$ has image contained in $Y$.
@ -2609,10 +2609,10 @@ The following is somewhat harder than in the affine case:
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 $Y \setminus \Omega = V(J)$. 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 $v \in V$. 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 $\alpha \coloneqq \phi(a)\defon{W} \in \mathcal{O}_X(W)$, $\beta \coloneqq \phi(b)\defon{W} \in \mathcal{O}_X(W)$. 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 $f\st(\lambda)\defon{W} = \frac{\alpha}{\beta} \in \mathcal{O}_X(W)$. 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 $f\st(\lambda) \in \mathcal{O}_X(U)$. The first part of \ref{localinverse} shows that $f^{\ast}(\lambda) \in \mathcal{O}_X(U)$.
\end{subproof} \end{subproof}
By definition of $f$, we have $f\st = \phi$. This finished the proof of the first point. By definition of $f$, we have $f^{\ast} = \phi$. This finished the proof of the first point.