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@ -33,10 +33,8 @@
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\section{Varieties}
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\section{Varieties}
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\input{inputs/varieties}
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\input{inputs/varieties}
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\iffalse
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% \section{Übersicht}
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\section{Übersicht}
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% \input{inputs/uebersicht}
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\input{inputs/uebersicht}
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\fi
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\cleardoublepage
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\cleardoublepage
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\printvocabindex
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\printvocabindex
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@ -5,9 +5,9 @@
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Then the following sets coincide
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Then the following sets coincide
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\begin{enumerate}
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\begin{enumerate}
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\item
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\item
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$\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{finite}, r_s \in R, \right\}$,
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\[\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{ finite}, r_s \in R\right\},\]
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\item
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\item
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$\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$,
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\[\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N,\]
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\item
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\item
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The $\subseteq$-smallest submodule of $M$ containing $S$.
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The $\subseteq$-smallest submodule of $M$ containing $S$.
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\end{enumerate}
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\end{enumerate}
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@ -366,7 +366,7 @@ For $f \in R$ let $V(f) = V(fR)$.
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\begin{proof}
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\begin{proof}
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Suppose $f$ vanishes on all zeros of $I$.
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Suppose $f$ vanishes on all zeros of $I$.
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Let $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n,T]$,
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Let $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n,T]$,
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$g(X_1,\ldots,X_n,T) \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)$
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\[g(X_1,\ldots,X_n,T) \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)\]
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and $J \subseteq R'$ the ideal generated by $g$ and the elements of $I$
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and $J \subseteq R'$ the ideal generated by $g$ and the elements of $I$
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(viewed as elements of $R'$ which are constant in the $T$-direction).
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(viewed as elements of $R'$ which are constant in the $T$-direction).
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@ -600,7 +600,7 @@ Let $X$ be a topological space.
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irreducible.
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irreducible.
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The remaining assertion follows from the fact,
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The remaining assertion follows from the fact,
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that the bijection is $\subseteq$-antimonotonic
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that the bijection is $\subseteq$-anti\-monotonic
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and thus maximal ideals correspond to minimal irreducible closed subsets,
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and thus maximal ideals correspond to minimal irreducible closed subsets,
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which are the one-point subsets as $\mathfrak{k}^n$ is T${}_1$.
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which are the one-point subsets as $\mathfrak{k}^n$ is T${}_1$.
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\end{proof}
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\end{proof}
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@ -609,7 +609,7 @@ Let $X$ be a topological space.
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Let $Z$ be an irreducible subset of the topological space $X$.
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Let $Z$ be an irreducible subset of the topological space $X$.
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Let $\codim(Z,X)$ be the maximum of the length $n$ of
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Let $\codim(Z,X)$ be the maximum of the length $n$ of
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strictly increasing chains
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strictly increasing chains
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$Z \subseteq Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n$
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\[Z \subseteq Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n\]
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of irreducible closed subsets of $X$ containing $Z$ or $\infty$
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of irreducible closed subsets of $X$ containing $Z$ or $\infty$
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if such chains can be found for arbitrary $n$.
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if such chains can be found for arbitrary $n$.
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Let
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Let
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@ -617,7 +617,7 @@ Let $X$ be a topological space.
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\dim X \coloneqq
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\dim X \coloneqq
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\begin{cases}
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\begin{cases}
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- \infty & \text{if } X = \emptyset,\\
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- \infty & \text{if } X = \emptyset,\\
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\sup_{\substack{Z \subseteq X \\ Z \text{ irreducible}}} \codim(Z,X) & \text{otherwise}.
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\sup\limits_{\substack{Z \subseteq X \\ Z \text{ irreducible}}} \codim(Z,X) & \text{otherwise}.
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\end{cases}
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\end{cases}
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\]
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\]
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\end{definition}
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\end{definition}
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@ -730,7 +730,7 @@ In general, these inequalities may be strict.
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it
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it
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is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$.
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is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$.
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Translation by $x \in \mathfrak{k}^n$ gives us
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Translation by $x \in \mathfrak{k}^n$ gives us
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$\codim(\{x\}, \mathfrak{k}^n) \ge n$.
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\[\codim(\{x\}, \mathfrak{k}^n) \ge n.\]
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The opposite inequality follows from \ref{upperbounddim}
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The opposite inequality follows from \ref{upperbounddim}
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($Z = \mathfrak{k}^n$,
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($Z = \mathfrak{k}^n$,
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@ -805,13 +805,13 @@ In general, these inequalities may be strict.
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\begin{enumerate}
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\begin{enumerate}
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\item[M]
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\item[M]
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If $m,n \in X$ and $A \subseteq X$ then
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If $m,n \in X$ and $A \subseteq X$ then
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$m \in \mathcal{H}( \{n\} \cup A) \setminus \mathcal{H}(A) %
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\[m \in \mathcal{H}( \{n\} \cup A) \setminus \mathcal{H}(A) %
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\iff n \in \mathcal{H}(\{m\} \cup A) \setminus \mathcal{H}(A)$,
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\iff n \in \mathcal{H}(\{m\} \cup A) \setminus \mathcal{H}(A),\]
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\item[F]
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\item[F]
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$\mathcal{H}(A) = %
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$\mathcal{H}(A) = %
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\bigcup_{F \subseteq A \text{ finite}} \mathcal{H}(F)$.
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\bigcup_{F \subseteq A \text{ finite}} \mathcal{H}(F)$.
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\end{enumerate}
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\end{enumerate}
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In this case, $S \subseteq X$ is called \vocab{Independent subset},
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In this case, $S \subseteq X$ is called \vocab{independent subset},
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if $s \not\in \mathcal{H}(S \setminus \{s\})$ for all $s \in S$ and
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if $s \not\in \mathcal{H}(S \setminus \{s\})$ for all $s \in S$ and
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\vocab[Generating subset]{generating} if $X = \mathcal{H}(S)$.
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\vocab[Generating subset]{generating} if $X = \mathcal{H}(S)$.
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$S$ is called a \vocab{base}, if it is both generating and independent.
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$S$ is called a \vocab{base}, if it is both generating and independent.
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@ -1944,9 +1944,9 @@ extensions:
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let $L$ be a field extension of $Q(A)$.
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let $L$ be a field extension of $Q(A)$.
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By \ref{intclosure} the set of elements of $L$ integral over $A$ is a
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By \ref{intclosure} the set of elements of $L$ integral over $A$ is a
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subring of $L$, the \vocab{integral closure} of $A$ in $L$.
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subring of $L$, the \vocab{integral closure} of $A$ in $L$.
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$A$ is \vocab{Domain!integrally closed} in $L$
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$A$ is \vocab[Domain!integrally closed]{integrally closed} in $L$
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if the integral closure of $A$ in $L$ equals $A$.
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if the integral closure of $A$ in $L$ equals $A$.
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$A$ is \vocab{Domain!normal} if it is integrally closed in $Q(A)$.
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$A$ is \vocab[Domain!normal]{normal} if it is integrally closed in $Q(A)$.
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\end{definition}
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\end{definition}
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\begin{proposition}
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\begin{proposition}
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@ -2029,7 +2029,8 @@ extensions:
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(\ref{characfixnormalfe}),
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(\ref{characfixnormalfe}),
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there is a positive integer $k$ with $y^k \in K$.
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there is a positive integer $k$ with $y^k \in K$.
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As $A$ is normal, we have $y^k \in K \cap B = A$.
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As $A$ is normal, we have $y^k \in K \cap B = A$.
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Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp = \emptyset \lightning$.
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Thus
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\[y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp = \emptyset \lightning.\]
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If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs
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If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs
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$(M, \sigma)$ where $M$ is an intermediate field
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$(M, \sigma)$ where $M$ is an intermediate field
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@ -2116,7 +2117,7 @@ extensions:
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Japanese.
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Japanese.
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\end{remark}
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\end{remark}
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\begin{align*}+[Counterexample to going down]
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\begin{example}+[Counterexample to going down]
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Let $R = \mathfrak{k}[X,Y]$ and $A = \mathfrak{k}[X,Y, \frac{X}{Y}]$.
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Let $R = \mathfrak{k}[X,Y]$ and $A = \mathfrak{k}[X,Y, \frac{X}{Y}]$.
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Then going down does not hold for $A / R$:
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Then going down does not hold for $A / R$:
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@ -2127,7 +2128,7 @@ extensions:
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and thus proper, we have $(X,Y)_R = \fq \cap R$.
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and thus proper, we have $(X,Y)_R = \fq \cap R$.
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The prime ideal $(\frac{X}{Y},Y)_A = (\frac{X}{Y}, X,Y)_A$ is lying over
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The prime ideal $(\frac{X}{Y},Y)_A = (\frac{X}{Y}, X,Y)_A$ is lying over
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$(X,Y)_R$, so going down is violated.
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$(X,Y)_R$, so going down is violated.
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\end{align*}
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\end{example}
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\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}}
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\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}}
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@ -2280,7 +2281,7 @@ We will use the following
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$(a_i)_{i = 1}^m$ for $K / \mathfrak{l}$
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$(a_i)_{i = 1}^m$ for $K / \mathfrak{l}$
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with $a_i \in A$ for $1 \le i \le m$.
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with $a_i \in A$ for $1 \le i \le m$.
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\end{lemma}
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\end{lemma}
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\begin{align*}
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\begin{proof}
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The proof is similar to the proof of \ref{ltrdegresfieldtrbase}.
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The proof is similar to the proof of \ref{ltrdegresfieldtrbase}.
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There are a natural number $m \ge n$ and elements
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There are a natural number $m \ge n$ and elements
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$(a_i)_{i = n+1}^m \in A^{m-n}$ which generate $K$
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$(a_i)_{i = n+1}^m \in A^{m-n}$ which generate $K$
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@ -2296,7 +2297,7 @@ We will use the following
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But then $K$ is algebraic over its subfield generated by $\mathfrak{l}$
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But then $K$ is algebraic over its subfield generated by $\mathfrak{l}$
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and the $(a_i)_{i=1}^{m-1}$,
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and the $(a_i)_{i=1}^{m-1}$,
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contradicting the minimality of $m$.
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contradicting the minimality of $m$.
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\end{align*}
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\end{proof}
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\begin{theorem}
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\begin{theorem}
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\label{htandtrdeg}
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\label{htandtrdeg}
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@ -2724,13 +2725,15 @@ this more general theorem.
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defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$.
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defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$.
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Thus, $C$ is homeomorphic to an irreducible component $C'$ of
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Thus, $C$ is homeomorphic to an irreducible component $C'$ of
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$(A \times B) \cap \Delta$ and
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$(A \times B) \cap \Delta$ and
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\begin{align*}
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\begin{IEEEeqnarray*}{rCl}
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\codim(C, \mathfrak{k}^n) = n - \dim(C) = n - \dim(C')
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\codim(C, \mathfrak{k}^n)
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= n - \dim(A \times B) + \codim(C', A \times B) \\
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&=& n - \dim(C)\\
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\overset{\text{\ref{corpithm}}}{\le }2n - \dim(A \times B)
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&=& n - \dim(C')\\
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\overset{\text{\ref{dimprod}}}{=} 2n - \dim(A) - \dim(B)
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&=& n - \dim(A \times B) + \codim(C', A \times B) \\
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= \codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)
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&\overset{\text{\ref{corpithm}}}{\le }&2n - \dim(A \times B)\\
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\end{align*}
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&\overset{\text{\ref{dimprod}}}{=}& 2n - \dim(A) - \dim(B)\\
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&=& \codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)
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\end{IEEEeqnarray*}
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by the general
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by the general
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properties of dimension and codimension,
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properties of dimension and codimension,
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\ref{corpithm} applied to $(X_i - Y_i)_{i=1}^n$,
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\ref{corpithm} applied to $(X_i - Y_i)_{i=1}^n$,
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\subsection{The Jacobson radical}
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\subsection{The Jacobson radical}
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\begin{proposition}
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\begin{proposition}
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For a ring $A$,
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For a ring $A$,
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$\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m} = \{a \in A | \forall x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$,
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\[\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m} = \{a \in A | \forall x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A).\]
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the \vocab{Jacobson radical} of $A$.
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the \vocab{Jacobson radical} of $A$.
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\end{proposition}
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\end{proposition}
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\begin{proof}
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\begin{proof}
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We call the $r_d$ the \vocab{homogeneous components} of $r$.
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We call the $r_d$ the \vocab{homogeneous components} of $r$.
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An ideal $I \subseteq A$ is called \vocab{homogeneous} if
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An ideal $I \subseteq A$ is called \vocab{homogeneous} if
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$r \in I \implies \forall d \in \mathbb{I} ~ r_d \in I_d$
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\[r \in I \implies \forall d \in \mathbb{I} ~ r_d \in I_d\]
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where $I_d \coloneqq I \cap A_d$.
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where $I_d \coloneqq I \cap A_d$.
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By a \vocab{graded ring} we understand an $\N$-graded ring.
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By a \vocab{graded ring} we understand an $\N$-graded ring.
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Tin this case,
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In this case,
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$A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\} $
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\[A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\}\]
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is called the \vocab{augmentation ideal} of $A$.
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is called the \vocab{augmentation ideal} of $A$.
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\end{definition}
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\end{definition}
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\begin{remark}[Decomposition of $1$]
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\begin{remark}[Decomposition of $1$]
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\subsubsection{The Zariski topology on $\mathbb{P}^n$}
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\subsubsection{The Zariski topology on $\mathbb{P}^n$}
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\begin{notation}
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\begin{notation}
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Recall that for
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Recall that for
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$\alpha \in \N^{n+1}$ $|\alpha| = \sum_{i=0}^{n} \alpha_i$
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$\alpha \in \N^{n+1}$
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and $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$.
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\[|\alpha| = \sum_{i=0}^{n} \alpha_i \text{ and }
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x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}.\]
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\end{notation}
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\end{notation}
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\begin{definition}[Homogeneous polynomials]
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\begin{definition}[Homogeneous polynomials]
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Let $R$ be any ring and
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Let $R$ be any ring and
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$f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$.
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\[f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n].\]
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We say that $f$ is \vocab{homogeneous of degree $d$}
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We say that $f$ is \vocab{homogeneous of degree $d$}
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if $|\alpha| \neq d \implies f_\alpha = 0$ .
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if
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\[|\alpha| \neq d \implies f_\alpha = 0.\]
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We denote the subset of homogeneous polynomials of degree $d$ by
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We denote the subset of homogeneous polynomials of degree $d$ by
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$R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n]$.
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\[R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n].\]
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\end{definition}
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\end{definition}
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\begin{remark}
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\begin{remark}
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This definition gives $R$ the structure of a graded ring.
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This definition gives $R$ the structure of a graded ring.
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@ -151,8 +153,8 @@ Let $\mathfrak{l}$ be any field.
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\label{ztoppn}
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\label{ztoppn}
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Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.%
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Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.%
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\footnote{As always, $\mathfrak{k}$ is algebraically closed}
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\footnote{As always, $\mathfrak{k}$ is algebraically closed}
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For $f \in A_d = \mathfrak{k}[X_0,\ldots,
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For $f \in A_d = \mathfrak{k}[X_0,\ldots,X_n]_d$,
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_n]_d$, the validity of the equation $f(x_0,\ldots,x_{n}) = 0$
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the validity of the equation $f(x_0,\ldots,x_{n}) = 0$
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does not depend on the choice of homogeneous coordinates, as
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does not depend on the choice of homogeneous coordinates, as
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\[
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\[
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f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d f(x_0,\ldots,x_n).
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f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d f(x_0,\ldots,x_n).
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\begin{proof}
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\begin{proof}
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$\impliedby$ is clear.
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$\impliedby$ is clear.
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Let $\Vp(I) \subseteq \Vp(f)$.
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Let $\Vp(I) \subseteq \Vp(f)$.
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If $x = (x_0,
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If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in
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ldots,x_n) \in \Va(I)$, then either $x = 0$ in
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which case $f(x) = 0$ since $d > 0$
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which case $f(x) = 0$ since $d > 0$
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or the point $[x_0,\ldots,x_n] \in \mathbb{P}^n$ is
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or the point $[x_0,\ldots,x_n] \in \mathbb{P}^n$ is
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well-defined and belongs to $\Vp(I) \subseteq \Vp(f)$,
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well-defined and belongs to $\Vp(I) \subseteq \Vp(f)$,
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@ -330,8 +331,8 @@ Let $\mathfrak{l}$ be any field.
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(\ref{hns3}).
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(\ref{hns3}).
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\end{proof}
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\end{proof}
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||||||
|
|
||||||
\begin{definition}
|
\begin{definition}%
|
||||||
\footnote{This definition is not too important, the characterization in the following remark suffices.}.
|
\footnote{This definition is not too important, the characterization in the following remark suffices.}
|
||||||
For a graded ring $R_\bullet$,
|
For a graded ring $R_\bullet$,
|
||||||
let $\Proj(R_\bullet)$ be the set of $\fp \in \Spec R$
|
let $\Proj(R_\bullet)$ be the set of $\fp \in \Spec R$
|
||||||
such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$.
|
such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$.
|
||||||
|
@ -432,17 +433,18 @@ Let $\mathfrak{l}$ be any field.
|
||||||
\end{proof}
|
\end{proof}
|
||||||
|
|
||||||
\begin{remark}
|
\begin{remark}
|
||||||
If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$,
|
If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, then
|
||||||
then $\fp \oplus R_+ = \{r \in R | r_0 \in \fp\} $
|
\[\fp \oplus R_+ = \{r \in R | r_0 \in \fp\}\]
|
||||||
is a homogeneous prime ideal of $R$.
|
is a homogeneous prime ideal of $R$.
|
||||||
\[
|
\begin{IEEEeqnarray*}{rCl}
|
||||||
\{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } R_\bullet\}
|
&&\{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } R_\bullet\}\\
|
||||||
= \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec R_0\}.
|
&=&\Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec R_0\}.
|
||||||
\]
|
\end{IEEEeqnarray*}
|
||||||
\end{remark}
|
\end{remark}
|
||||||
|
|
||||||
|
\pagebreak
|
||||||
\subsection{Dimension of $\mathbb{P}^n$}
|
\subsection{Dimension of $\mathbb{P}^n$}
|
||||||
\begin{proposition}
|
\begin{proposition}\,
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
\item
|
\item
|
||||||
$\mathbb{P}^n$ is catenary.
|
$\mathbb{P}^n$ is catenary.
|
||||||
|
@ -450,11 +452,11 @@ Let $\mathfrak{l}$ be any field.
|
||||||
$\dim(\mathbb{P}^n) = n$.
|
$\dim(\mathbb{P}^n) = n$.
|
||||||
Moreover, $\codim(\{x\} ,\mathbb{P}^n) = n$ for every $x \in \mathbb{P}^n$.
|
Moreover, $\codim(\{x\} ,\mathbb{P}^n) = n$ for every $x \in \mathbb{P}^n$.
|
||||||
\item
|
\item
|
||||||
If $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$,
|
If $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$, then
|
||||||
then $\codim(\{x\}, X) = \dim(X) = n - \codim(X, \mathbb{P}^n)$.
|
\[\codim(\{x\}, X) = \dim(X) = n - \codim(X, \mathbb{P}^n).\]
|
||||||
\item
|
\item
|
||||||
If $X \subseteq Y \subseteq \mathbb{P}^n$ are irreducible subsets,
|
If $X \subseteq Y \subseteq \mathbb{P}^n$ are irreducible subsets,
|
||||||
then $\codim(X,Y) = \dim(Y) - \dim(X)$.
|
then \[\codim(X,Y) = \dim(Y) - \dim(X).\]
|
||||||
\end{itemize}
|
\end{itemize}
|
||||||
\end{proposition}
|
\end{proposition}
|
||||||
\begin{proof}
|
\begin{proof}
|
||||||
|
@ -472,12 +474,13 @@ Let $\mathfrak{l}$ be any field.
|
||||||
$\codim(X,Z) = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$
|
$\codim(X,Z) = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$
|
||||||
and $\codim(Y,Z) = \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
|
and $\codim(Y,Z) = \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
|
||||||
Thus
|
Thus
|
||||||
\begin{align*}
|
\begin{IEEEeqnarray*}{rCl}
|
||||||
\codim(X,Y) + \codim(Y,Z)
|
\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, Y \cap \mathbb{A}^n)\\
|
||||||
& = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)\\
|
& & + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\
|
||||||
& = \codim(X, Z)
|
& = &\codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)\\
|
||||||
\end{align*}
|
& = &\codim(X, Z)
|
||||||
|
\end{IEEEeqnarray*}
|
||||||
because $\mathfrak{k}^n$ is catenary and the first point follows.
|
because $\mathfrak{k}^n$ is catenary and the first point follows.
|
||||||
The remaining assertions can easily be derived from the first two.
|
The remaining assertions can easily be derived from the first two.
|
||||||
\end{proof}
|
\end{proof}
|
||||||
|
@ -492,17 +495,15 @@ Let $\mathfrak{l}$ be any field.
|
||||||
If $X = \Vp(I)$ where $I \subseteq A_+ = \mathfrak{k}[X_0,\ldots,X_n]_+$
|
If $X = \Vp(I)$ where $I \subseteq A_+ = \mathfrak{k}[X_0,\ldots,X_n]_+$
|
||||||
is homogeneous, then $C(X) = \Va(I)$.
|
is homogeneous, then $C(X) = \Va(I)$.
|
||||||
\end{definition}
|
\end{definition}
|
||||||
\begin{proposition}
|
\begin{proposition}\,
|
||||||
\label{conedim}
|
\label{conedim}
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
\item
|
\item
|
||||||
$C(X)$ is irreducible iff $X$ is irreducible or $X = \emptyset$.
|
$C(X)$ is irreducible iff $X$ is irreducible or $X = \emptyset$.
|
||||||
\item
|
\item
|
||||||
If $X$ is irreducible, then
|
If $X$ is irreducible, then
|
||||||
|
\[\dim(C(X)) = \dim(X) + 1\] and
|
||||||
$\dim(C(X)) = \dim(X) + 1$ and
|
\[\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \mathbb{P}^n).\]
|
||||||
|
|
||||||
$\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \mathbb{P}^n)$
|
|
||||||
\end{itemize}
|
\end{itemize}
|
||||||
\end{proposition}
|
\end{proposition}
|
||||||
\begin{proof}
|
\begin{proof}
|
||||||
|
@ -525,7 +526,7 @@ Let $\mathfrak{l}$ be any field.
|
||||||
Hence $\dim(C(X)) \ge 1 + d$ and
|
Hence $\dim(C(X)) \ge 1 + d$ and
|
||||||
$\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$.
|
$\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$.
|
||||||
Since
|
Since
|
||||||
$\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) = n+1$,
|
\[\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) = n+1,\]
|
||||||
the two inequalities must be equalities.
|
the two inequalities must be equalities.
|
||||||
\end{proof}
|
\end{proof}
|
||||||
\subsubsection{Application to hypersurfaces in $\mathbb{P}^n$}
|
\subsubsection{Application to hypersurfaces in $\mathbb{P}^n$}
|
||||||
|
|
|
@ -315,7 +315,7 @@ The following is somewhat harder than in the affine case:
|
||||||
\end{itemize}
|
\end{itemize}
|
||||||
\end{remark}
|
\end{remark}
|
||||||
\subsubsection{Examples of categories}
|
\subsubsection{Examples of categories}
|
||||||
\begin{example}
|
\begin{example}\,
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
\item
|
\item
|
||||||
The category of sets.
|
The category of sets.
|
||||||
|
@ -357,7 +357,7 @@ The following is somewhat harder than in the affine case:
|
||||||
addition $\Hom_\mathcal{B}(X,Y) = \Hom_\mathcal{A}(X,Y)$ for
|
addition $\Hom_\mathcal{B}(X,Y) = \Hom_\mathcal{A}(X,Y)$ for
|
||||||
arbitrary $X,Y \in \Ob(\mathcal{B})$.
|
arbitrary $X,Y \in \Ob(\mathcal{B})$.
|
||||||
\end{definition}
|
\end{definition}
|
||||||
\begin{example}
|
\begin{example}\,
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
\item
|
\item
|
||||||
The category of abelian groups is a full subcategory of the
|
The category of abelian groups is a full subcategory of the
|
||||||
|
@ -407,7 +407,7 @@ The following is somewhat harder than in the affine case:
|
||||||
It is called an \vocab{equivalence of categories} if it is full,
|
It is called an \vocab{equivalence of categories} if it is full,
|
||||||
faithful and essentially surjective.
|
faithful and essentially surjective.
|
||||||
\end{definition}
|
\end{definition}
|
||||||
\begin{example}
|
\begin{example}\,
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
\item
|
\item
|
||||||
There are \vocab[Functor!forgetful]{forgetful functors}
|
There are \vocab[Functor!forgetful]{forgetful functors}
|
||||||
|
@ -552,7 +552,7 @@ The following is somewhat harder than in the affine case:
|
||||||
|
|
||||||
|
|
||||||
\end{proof}
|
\end{proof}
|
||||||
\begin{proposition}[About affine varieties]
|
\begin{proposition}[About affine varieties]\,
|
||||||
\label{propaffvar}
|
\label{propaffvar}
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
\item
|
\item
|
||||||
|
|
Loading…
Reference in a new issue