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Maximilian Keßler 2022-02-16 01:23:57 +01:00
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@ -86,11 +86,11 @@ Fields which are not assumed to be algebraically closed have been renamed (usual
\end{fact}
\begin{proof}
\begin{enumerate}
\item Consider a sequence $M_0'' \subset M_1'' \subset \ldots \subset M''$. Then $p\inv M_i''$ yields a strictly ascending sequence.
\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 $N' \coloneqq f\inv(N)$ and $N''\coloneqq p(N)$ are finitely generated. Since $0 \to N' \to N \to N'' \to 0$ is exact, $N$ is finitely generated.
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}
@ -195,8 +195,8 @@ This generalizes some facts about matrices to matrices with elements from commut
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 $\fA$ be the matrix with the $\rho_i$ as columns.
Then for all $v \in R^n: q(\fA \cdot v) = a \cdot q(v)$. By induction it follows that $q(P(\fA) \cdot v) = P(a)q(v)$ for all $P \in R[T]$. Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \fA)$ 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.
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 $a^d = \sum_{i=0}^{d-1} r_ia^i$.
@ -248,8 +248,8 @@ This generalizes some facts about matrices to matrices with elements from commut
\end{fact}
}
\begin{proof}
Let $B$ be a field and $a \in A \sm \{0\} $. Then $a\inv \in B$ 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
$a\inv = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$.
Let $B$ be a field and $a \in A \sm \{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 $b \in B \sm \{0\}$. 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}
@ -460,7 +460,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an algebraically
\subsubsection{Definition of the Zariski topology}
Let $\mathfrak{k}$ be algebraically closed, $R = \mathfrak{k}[X_1,\ldots,X_n]$.
\begin{corollary} (of \ref{fvop})
There is a topology on $\mathfrak{k}^n$ for which the set of closed sets coincides with the set $\fA$ of subsets of the form $\Va\left(I \right) $ for ideals $I \subseteq R$.
There is a topology on $\mathfrak{k}^n$ for which the set of closed sets coincides with the set $\mathfrak{A}$ of subsets of the form $\Va\left(I \right) $ for ideals $I \subseteq R$.
This topology is called the \vocab{Zariski-Topology}
\end{corollary}
@ -658,7 +658,7 @@ Let $X$ be a topological space.
\{f \in R | A \subseteq V(f)\} &\longmapsfrom A
\end{align}
Under this correspondence $A \subseteq \mathfrak{k}^n$ is irreducible iff $I \coloneqq f\inv(A)$ is a prime ideal.
Under this correspondence $A \subseteq \mathfrak{k}^n$ is irreducible iff $I \coloneqq f^{-1}(A)$ is a prime ideal.
Moreover, $\#A = 1$ iff $I$ is a maximal ideal.
\end{proposition}
\begin{proof}
@ -1004,7 +1004,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
\begin{notation}
Let $S \subseteq R$ be a multiplicative subset of $R$. We write $\frac{r}{s}$ for $[r,s]$.
The ring homomorphism $R \xrightarrow{i} R_S$ i given by $i(r) = \frac{r}{1}$.
For $X \subseteq R_S$ let $X \sqcap R$ denote $i\inv(X)$.
For $X \subseteq R_S$ let $X \sqcap R$ denote $i^{-1}(X)$.
\end{notation}
\begin{definition}[$S$-saturated ideal]
An ideal $I \subseteq R$ is called \vocab[Ideal!S-saturated]{$S$-saturated} if for all $s \in S, r \in R$
@ -1044,7 +1044,7 @@ Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-satura
But as $\frac{r}{1} = s \cdot \frac{r}{s}$ and $s \in R_S^{\times }$, we have $r \in I \iff \frac{r}{1} \in J \iff \frac{r}{s} \in J$ .
We have thus shown that the two maps between sets of ideals are well-defined and inverse to each other.
By \ref{invimgprimeideal}, $J \in \Spec R_S \implies f\inv(J) = J \cap R \in \Spec R_S$.
By \ref{invimgprimeideal}, $J \in \Spec R_S \implies f^{-1}(J) = J \cap R \in \Spec R_S$.
Suppose $I \in \Spec R$, $\frac{a}{s} \cdot \frac{b}{t} \in I_S$ for some $a,b \in R, s,t \in S$.
By \ref{ssatiis} $ab \in I$. Thus $a \in I \lor b \in I$, hence $\frac{a}{s} \in I_S \lor \frac{b}{t} \in I_S$ and we have $I_S \in \Spec R_S$.
@ -1294,16 +1294,16 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
Hence $a^n = \sum_{i=0}^{n-1} r_ia^i$ for some $r_i \in R$. Thus $x^n = \sum_{i=0}^{n-1} \rho_i x^i$ with $\rho_i \coloneqq s^{i-n} r_i \in R_\fp$.
\end{subproof}
As $i$ is injective and $R_\fp \neq \{0\} $ ($R_\fp$ is local!) $A_\fp \neq \{0\}$, there is $\mathfrak{m} \in \mSpec A_\fp$.
D has already been shown and applies to $A_\fp / R_\fp$, hence $i\inv(\mathfrak{m}) = \fp_\fp$ is the only maximal ideal of the local ring $R_\fp$. Hence $\fq = \alpha\inv(\mathfrak{m})$ satisfies
D has already been shown and applies to $A_\fp / R_\fp$, hence $i^{-1}(\mathfrak{m}) = \fp_\fp$ is the only maximal ideal of the local ring $R_\fp$. Hence $\fq = \alpha^{-1}(\mathfrak{m})$ satisfies
\[
\fq \cap R = \alpha\inv(\mathfrak{m}) \cap R = \rho\inv(i\inv (\mathfrak{m})) = \rho\inv(\fp_\fp) = \fp
\fq \cap R = \alpha^{-1}(\mathfrak{m}) \cap R = \rho^{-1}(i^{-1} (\mathfrak{m})) = \rho^{-1}(\fp_\fp) = \fp
\]
\item[B] The map $\Spec A_\fp \xrightarrow{\alpha\inv} \Spec A$ is injective with image equal to $\{\fq \in \Spec A | \fq \cap R \subseteq \fp\}$. In particular, it contains the set of all $\fq$ lying over $\fp$.
If $\fq = \alpha\inv(\fr)$ lies over $\fp$, then \[\rho\inv(i\inv(\fr)) = (\alpha\inv(\fr)) \cap R = \fq \cap R = \fp = \rho\inv(\fp_\fp)\]
hence $i\inv(\fr) = \fp_\fp$ by the injectivity of $\Spec R_\fp \xrightarrow{\rho\inv} \Spec R$.
\item[B] The map $\Spec A_\fp \xrightarrow{\alpha^{-1}} \Spec A$ is injective with image equal to $\{\fq \in \Spec A | \fq \cap R \subseteq \fp\}$. In particular, it contains the set of all $\fq$ lying over $\fp$.
If $\fq = \alpha^{-1}(\fr)$ lies over $\fp$, then \[\rho^{-1}(i^{-1}(\fr)) = (\alpha^{-1}(\fr)) \cap R = \fq \cap R = \fp = \rho^{-1}(\fp_\fp)\]
hence $i^{-1}(\fr) = \fp_\fp$ by the injectivity of $\Spec R_\fp \xrightarrow{\rho^{-1}} \Spec R$.
Because D applies to the integral ring extension $A_\fp / R_\fp$ and $\fp_\fp \in \mSpec R_\fp$, $\fr$ is a maximal ideal.
There are thus no inclusions between different such $\fr$. Because $\Spec A_\fp \xrightarrow{\alpha\inv} \Spec A$ is $\subseteq$-monotonic and injective, there are no inclusions between different $\fp \in \Spec A$ lying over $\fp$.
There are thus no inclusions between different such $\fr$. Because $\Spec A_\fp \xrightarrow{\alpha^{-1}} \Spec A$ is $\subseteq$-monotonic and injective, there are no inclusions between different $\fp \in \Spec A$ lying over $\fp$.
\item[C] Let $\fp \subseteq \tilde \fp$ be prime ideals of $R$ and $\fq \in \Spec A$ such that $\fq \cap R = \fp$.
By applying A to the ring extension $A / \fq$ of $R / \fp$, there is $\fr \in \Spec A /\fq$ such that $\fr \sqcap R / \fp = \tilde \fp / \fp$.
@ -1341,7 +1341,7 @@ This is part of the proof of \ref{trdegandkdim}.
is an isomorphism and in $\mathfrak{k}[X_1,\ldots,X_d]$ there is a strictly ascending chain of prime ideals corresponding to $\mathfrak{k}^d \supsetneq \{0\} \times \mathfrak{k}^{d-1} \supsetneq \ldots \supsetneq \{0\}$. Thus there is a strictly ascending chain $\{0\} = \fp_0 \subsetneq \fp_1 \subsetneq \ldots \subsetneq \fp_d$ of elements of $\Spec R$.
Let $\fq_0 = \{0\} \in \Spec A$. If $0 < i \le d$ and a chain $\fq_0 \subsetneq \ldots \subsetneq \fq_{i-1}$ in $\Spec A$ with $\fq_j \cap R = \fp_j$ for $0 \le j < i$ has been selected, we may apply going-up (\ref{cohenseidenberg}) to $A / R$ to extend this chain by a $\fq_i \in \Spec A$ with $\fq_{i-1} \subseteq \fq_i$ and $\fq_i \cap R = \fp_i$ (thus $\fq_{i-1} \subsetneq \fq_i$ as $\fp-i \neq \fp_{i-1})$.
Thus, we have a chain $\fq_0 = \{0\} \subsetneq \ldots \subsetneq \fq_d$ in $\Spec A$.
Let $\tilde \fq_i \coloneqq \pi_{B,\fp}\inv(\fq_i), Y_i \coloneqq V(\tilde \fq_i)$.
Let $\tilde \fq_i \coloneqq \pi_{B,\fp}^{-1}(\fq_i), Y_i \coloneqq V(\tilde \fq_i)$.
This is a chain $Y = Y_0 \supsetneq Y_1 \supsetneq \ldots \supsetneq Y_d$ of irreducible subsets of $\mathfrak{k}^n$.
Hence $\dim(Y) \ge \trdeg(\fK(Y) / \mathfrak{k})$.
@ -1482,7 +1482,7 @@ Recall the definition of a normal field extension in the case of finite field ex
This is equivalent to $\fq \subseteq \sigma(\fr)$, since the Krull going-up theorem (\ref{cohenseidenberg}) applies to the integral ring extension $B / A$, showing that there are no inclusions between different elements of $\Spec B$ lying above $\fp \in \Spec A$.
If $L / K$ is finite and there is no such $\sigma$, then by prime avoidance (\ref{primeavoidance}) there is $ x \in \fq \sm \bigcup_{\sigma \in G} \sigma(\fr)$.
As $\fr$ is a prime ideal, $y = \prod_{\sigma \in G} \sigma(x) \in \fq \sm \fr$.\footnote{$\prod_{\sigma \in G} \sigma(x) = \prod_{\sigma \in G} \sigma\inv(x)$}
As $\fr$ is a prime ideal, $y = \prod_{\sigma \in G} \sigma(x) \in \fq \sm \fr$.\footnote{$\prod_{\sigma \in G} \sigma(x) = \prod_{\sigma \in G} \sigma^{-1}(x)$}
By the characterization of $L^G$ for normal field extensions (\ref{characfixnormalfe}), there is a positive integer $k$ with $y^k \in K$.
As $A$ is normal, we have $y^k \in K \cap B = A$.
Thus $y^k \in (A \cap \fq) \sm (A \cap \fr) = \fp \sm \fp = \emptyset \lightning$.
@ -1558,7 +1558,7 @@ This is part of the proof of \ref{trdegandkdim}. %TODO: reorder
\codim(\{y\}, Y) \le d = \trdeg(\fK(Y) \sm \mathfrak{k})
\]
(see \ref{upperboundcodim})
equality holds for at least one pint $y \in F\inv(\{0\})$ but cannot rule out that there are other $y \in F\inv(\{0\})$ for which the inequality becomes strict.
equality holds for at least one pint $y \in F^{-1}(\{0\})$ but cannot rule out that there are other $y \in F^{-1}(\{0\})$ for which the inequality becomes strict.
However using going-down (\ref{gdkrull}) for $F$, we can use a similar argument, but start lifting of the chain at the right end for the point $y \in Y$ for which we would like to show equality.
From this $\codim(X,Y) \ge \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$ can be derived similarly to \ref{upperboundcodim}.
Thus
@ -1592,11 +1592,11 @@ To formulate a result which still applies in this context, we need the following
\begin{example}\label{htandcodim}
Let $B = \mathfrak{k}[X_1,\ldots,X_n], \fq \in \Spec B$ and let $A \coloneqq B / \fp$.
Let $Y \coloneqq V(\fq) \subseteq \mathfrak{k}^n$, $\tilde \fp \coloneqq \pi_{B, \fq}\inv(\fp)$, where $B \xrightarrow{\pi_{B, \fp}} A $ is the projection to the ring of residue classes, and let $X = V(\tilde \fp)$.
Let $Y \coloneqq V(\fq) \subseteq \mathfrak{k}^n$, $\tilde \fp \coloneqq \pi_{B, \fq}^{-1}(\fp)$, where $B \xrightarrow{\pi_{B, \fp}} A $ is the projection to the ring of residue classes, and let $X = V(\tilde \fp)$.
By \ref{idealslocbij} we have a bijection between the prime ideals $\fr \subseteq \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime ideals $\tilde \fr \in \Spec B$ with $\fq \subseteq \tilde \fr \subseteq \tilde \fp$:
\begin{align}
f: \{\fr \in \Spec A | \fr \subseteq \fp \} &\longrightarrow \{\tilde \fr \in \Spec B | \fq \subseteq \tilde \fr \subseteq \tilde \fp\} \\
\fr &\longmapsto \pi_{B, \fq}\inv(\fr)\\
\fr &\longmapsto \pi_{B, \fq}^{-1}(\fr)\\
\tilde \fr / \fq &\longmapsfrom \tilde \fr
\end{align}
By \ref{bijiredprim}, the $\tilde \fr$ are in canonical bijection with the irreducible subsets $Z$ of $Y$ containing $X$.
@ -2132,7 +2132,7 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
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 $f\inv(\Vp\left( I \right)) = \sqrt{I} = I$.
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 $X \subseteq \bP^n$ 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.
@ -2360,7 +2360,7 @@ Let $X \subseteq \mathfrak{k}^n$ be open. Let $R = \mathfrak{k}[X_1,\ldots,X_n]$
\begin{remark}\label{structuresheafcontinuous}
$\cO_X$ is a subsheaf (of rings) of the sheaf of $\mathfrak{k}$-valued functions on $X$.
The elements of $\cO_X(U)$ are continuous:
Let $M \subseteq \mathfrak{k}$ be closed. We must show the closedness of $N \coloneqq \phi\inv(M)$ 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 $V_x$.
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}
@ -2456,13 +2456,13 @@ The following is somewhat harder than in the affine case:
\item For every $A \in \Ob(\cA)$, there is an $\Id_A \in \Hom_{\cA}(A,A)$ 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 $\cA $)} if there is a morphism $Y \xrightarrow{g} X$ (called the \vocab[Inverse morphism]{inverse $f\inv$ of $f$)} such that $g \circ f = \Id_X$ and $f \circ g = \Id_Y$.
A morphism $X \xrightarrow{f} Y$ is called an \vocab[Isomorphism]{isomorphism (in $\cA $)} 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 $f\inv$ of an isomorphism $f$ is uniquely determined.
\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}
@ -2518,13 +2518,13 @@ The following is somewhat harder than in the affine case:
\subsection{The category of varieties}
\begin{definition}[Algebraic variety]\label{defvariety}
An \vocab{algebraic variety} or \vocab{prevariety} over $\mathfrak{k}$ is a pair $(X, \cO_X)$, where $X$ is a topological space and $\cO_X$ 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 $f \in \cO_X(V) \iff \iota\st_x(f) \in \cO_{Y_x}(\iota_x\inv(V))$,
An \vocab{algebraic variety} or \vocab{prevariety} over $\mathfrak{k}$ is a pair $(X, \cO_X)$, where $X$ is a topological space and $\cO_X$ 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 $f \in \cO_X(V) \iff \iota\st_x(f) \in \cO_{Y_x}(\iota_x^{-1}(V))$,
In this, the \vocab{pull-back} $\iota_x\st(f)$ of $f$ is defined by $(\iota_x\st(f))(\xi) \coloneqq f(\iota_x(\xi))$.
A morphism $(X, \cO_X) \to (Y, \cO_Y)$ of varieties is a continuous map $X \xrightarrow{\phi} Y$ such that for all open $U \subseteq Y$ and $f \in \cO_Y(U)$, $\phi\st(f) \in \cO_X(\phi\inv(U))$.
An isomorphism is a morphism such that $\phi$ is bijective and $\phi\inv$ also is a morphism of varieties.
A morphism $(X, \cO_X) \to (Y, \cO_Y)$ of varieties is a continuous map $X \xrightarrow{\phi} Y$ such that for all open $U \subseteq Y$ and $f \in \cO_Y(U)$, $\phi\st(f) \in \cO_X(\phi^{-1}(U))$.
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}
@ -2612,9 +2612,9 @@ The following is somewhat harder than in the affine case:
$f$ is a morphism in $\Var_\mathfrak{k}$
\end{claim}
\begin{subproof}
For open $\Omega \subseteq Y, U = f\inv(\Omega) = \{x \in X | \A \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y \sm \Omega = V(J)$.
For open $\Omega \subseteq Y, U = f^{-1}(\Omega) = \{x \in X | \A \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y \sm \Omega = V(J)$.
If $\lambda \in \cO_Y(\Omega)$ 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 $W \coloneqq f\inv(V)$. Then $\alpha \coloneqq \phi(a)\defon{W} \in \cO_X(W)$, $\beta \coloneqq \phi(b)\defon{W} \in \cO_X(W)$.
Let $W \coloneqq f^{-1}(V)$. Then $\alpha \coloneqq \phi(a)\defon{W} \in \cO_X(W)$, $\beta \coloneqq \phi(b)\defon{W} \in \cO_X(W)$.
By the second part of \ref{localinverse} $\beta \in \cO_X(W)^{\times}$ and $f\st(\lambda)\defon{W} = \frac{\alpha}{\beta} \in \cO_X(W)$.
The first part of \ref{localinverse} shows that $f\st(\lambda) \in \cO_X(U)$.
\end{subproof}
@ -2676,8 +2676,8 @@ If $X$ is a set, then $\cB \subseteq \cP(X)$ is a base for some topology on $X$
Then $\cO_Y(Y) \cong \mathfrak{k}[Z,X_1,\ldots,X_n] / J \cong A[Z] / (1 -\lambda Z) \cong A_\lambda$.
By the proposition about affine varieties (\ref{propaffvar}), the morphism $\mathfrak{s}: \cO_Y(Y) \cong A_\lambda \to \cO_X(U)$ corresponds to a morphism $U \xrightarrow{\sigma} Y$.
We have $\mathfrak{s}(Z \mod J) = \lambda\inv$ and $\mathfrak{s}(X_i \mod J) = X_i \mod I$.
Thus $\sigma(x) = (\lambda(x)\inv, x)$ for $x \in U$.
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)
@ -2757,7 +2757,7 @@ If $X$ is a set, then $\cB \subseteq \cP(X)$ is a base for some topology on $X$
Let $g = (U, \gamma)/\sim \in \cO_{X,x}$ and $g(x) \neq 0$.
$\gamma$ is Zariski continuous (first point of \ref{localinverse}). Thus $V(\gamma)$ is closed. By replacing $U$ by $U \sm V(\gamma)$ we may assume that $\gamma$ vanishes nowhere on $U$.
By the third point of \ref{localinverse} we have $\gamma \in \cO_X(U)^{\times}$.
$(\gamma\inv)_x$ is an inverse to $g$.
$(\gamma^{-1})_x$ is an inverse to $g$.
\end{proof}
\begin{proposition}\label{proplocalring}