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@ -25,7 +25,7 @@
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\newline
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\noindent $\mathfrak{k}$ is {\color{red} always} an algebraically closed field and $\mathfrak{k}^n$ is equipped with the Zariski-topology.
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Fields which are not assumed to be algebraically closed have been renamed (usually to $\fl$).
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Fields which are not assumed to be algebraically closed have been renamed (usually to $\mathfrak{l}$).
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\pagebreak
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@ -357,19 +357,19 @@ Equivalent\footnote{used in a vague sense here} formulation:
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\end{remark}
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\begin{theorem}[Hilbert's Nullstellensatz (1b)] \label{hns1b}
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Let $\fl$ be a field and $I \subset R = \fl[X_1,\ldots,X_m]$ a proper ideal. Then there are a finite field extension $\mathfrak{i}$ of $\fl$ and a zero of $I$ in $\mathfrak{i}^m$.
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Let $\mathfrak{l}$ be a field and $I \subset R = \mathfrak{l}[X_1,\ldots,X_m]$ a proper ideal. Then there are a finite field extension $\mathfrak{i}$ of $\mathfrak{l}$ and a zero of $I$ in $\mathfrak{i}^m$.
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\end{theorem}
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\begin{proof} (HNS2 (\ref{hns2}) $\implies$ HNS1b (\ref{hns1b}))
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$I \se \fm$ for some maximal ideal. $R / \fm$ is a field, since $\fm$ is maximal.
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$R / \fm$ is of finite type, since the images of the $X_i$ generate it as a $\fl$-algebra.
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There are thus a field extension $\fri / \fl$ and an isomorphism $R / \fm \xrightarrow{\iota} \fri$ of $\fl$-algebras.
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By HNS2 (\ref{hns2}), $\fri / \fl$ is a finite field extension.
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$R / \fm$ is of finite type, since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra.
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There are thus a field extension $\fri / \mathfrak{l}$ and an isomorphism $R / \fm \xrightarrow{\iota} \fri$ of $\mathfrak{l}$-algebras.
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By HNS2 (\ref{hns2}), $\fri / \mathfrak{l}$ is a finite field extension.
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Let $x_i \coloneqq \iota (X_i \mod \fm)$.
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\[
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P(x_1,\ldots,x_m) = \iota(P \mod \fm)
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\]
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Both sides are morphisms $R \to \fri$ of $\fl$-algebras. For for $P = X_i$ the equality is trivial. It follows in general, since the $X_i$ generate $R$ as a $\fl$-algebra.
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Both sides are morphisms $R \to \fri$ of $\mathfrak{l}$-algebras. For for $P = X_i$ the equality is trivial. It follows in general, since the $X_i$ generate $R$ as a $\mathfrak{l}$-algebra.
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Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \fm = 0$ for $P \in I \se \fm$).
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HNS1 (\ref{hns1}) can easily be derived from HNS1b.
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@ -1101,39 +1101,39 @@ Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-satura
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% i = ic
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\begin{proposition}\label{trdegresfield}
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Let $\fl$ be a %% ??
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field, $A$ a $\fl$-algebra of finite type and $\fp, \fq \in \Spec A$ with $\fp \subsetneq \fq$.
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Let $\mathfrak{l}$ be a %% ??
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field, $A$ a $\mathfrak{l}$-algebra of finite type and $\fp, \fq \in \Spec A$ with $\fp \subsetneq \fq$.
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Then \[
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\trdeg(\mathfrak{k}(\fp) / \fl) > \trdeg(\mathfrak{k}(\fq) / \fl)
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\trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) / \mathfrak{l})
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\]
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\end{proposition}
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\begin{proof}
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Replacing $A$ by $A / \fp$, we may assume $\fp = \{0\} $ and $A$ to be a domain. Then $\mathfrak{k}(\fp) = Q(A / \fp) = Q(A)$.
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If $\fq$ is a maximal ideal, $\mathfrak{k}(\fq) = A / \fq$ is of finite type over $\fl$, hence a finite field extension of $\fl$ by the Nullstellensatz (\ref{hns2}).
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Thus, $\trdeg(\mathfrak{k}(\fq) / \fl) = 0$.
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If $\trdeg(Q(A) / \fl) = 0$, $A$ would be integral over $\fl$, hence a field (fact about integrality and fields, \ref{fintaf}). But if $A$ is a field, $\fp = \{0\}$ is a maximal ideal of $A$, hence $\fq = \fp \lightning$.
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If $\fq$ is a maximal ideal, $\mathfrak{k}(\fq) = A / \fq$ is of finite type over $\mathfrak{l}$, hence a finite field extension of $\mathfrak{l}$ by the Nullstellensatz (\ref{hns2}).
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Thus, $\trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$.
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If $\trdeg(Q(A) / \mathfrak{l}) = 0$, $A$ would be integral over $\mathfrak{l}$, hence a field (fact about integrality and fields, \ref{fintaf}). But if $A$ is a field, $\fp = \{0\}$ is a maximal ideal of $A$, hence $\fq = \fp \lightning$.
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This finishes the proof for $\fq \in \mSpec A$.
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We will use the following lemma to reduce the general case to this case:
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\begin{lemma}\label{ltrdegresfieldtrbase}
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There are algebraically independent $a_1,\ldots,a_n \in A$ whose images in $A / \fq$ form a transcendence base for $\mathfrak{k}(\fq) / \fl$.
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There are algebraically independent $a_1,\ldots,a_n \in A$ whose images in $A / \fq$ form a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{l}$.
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\end{lemma}
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\begin{subproof}
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There exist $a_1,\ldots,a_n \in A$ such that $\mathfrak{k}(\fq)$ is algebraic over the subfield generated by $\fl$ and their images $\overline{a_i}$ (for instance generators of $A$ as a $\fl$-algebra).
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We may assume that $n$ is minimal. If the $a_i$ are $\fl$-algebraically dependent, then w.l.o.g. $\overline{a_n}$ can be assumed to be algebraic over the subfield generated by $\fl$ and the $\overline{a_i}, 1\le i <n$. Thus, $a_n$ could be removed, contradicting the minimality.
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There exist $a_1,\ldots,a_n \in A$ such that $\mathfrak{k}(\fq)$ is algebraic over the subfield generated by $\mathfrak{l}$ and their images $\overline{a_i}$ (for instance generators of $A$ as a $\mathfrak{l}$-algebra).
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We may assume that $n$ is minimal. If the $a_i$ are $\mathfrak{l}$-algebraically dependent, then w.l.o.g. $\overline{a_n}$ can be assumed to be algebraic over the subfield generated by $\mathfrak{l}$ and the $\overline{a_i}, 1\le i <n$. Thus, $a_n$ could be removed, contradicting the minimality.
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\end{subproof}
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Let $\fq$ be any prime ideal.
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Take $a_1,\ldots,a_n \in A$ as in the lemma. As the $a_i \mod \fq$ are $\fl$-algebraically independent, the same holds for the $a_i$ themselves.
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Thus $\trdeg(Q(A) / \fl) \ge n$ and the inequality is strict, if it can be shown that the $a_i$ fail to be a transcendence base of $Q(A) / \fl$.
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Let $R \se A$ denote the $\fl$-subalgebra generated by $a_1,\ldots,a_n$ and $S \coloneqq R \sm \{0\} $.
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We must show, that $Q(A)$ fails to be algebraic over $\fl_1 \coloneqq R_S = Q(R)$.
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Take $a_1,\ldots,a_n \in A$ as in the lemma. As the $a_i \mod \fq$ are $\mathfrak{l}$-algebraically independent, the same holds for the $a_i$ themselves.
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Thus $\trdeg(Q(A) / \mathfrak{l}) \ge n$ and the inequality is strict, if it can be shown that the $a_i$ fail to be a transcendence base of $Q(A) / \mathfrak{l}$.
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Let $R \se A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and $S \coloneqq R \sm \{0\} $.
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We must show, that $Q(A)$ fails to be algebraic over $\mathfrak{l}_1 \coloneqq R_S = Q(R)$.
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Let $A_1 \coloneqq A_S$ and $\fq_S$ the prime ideal corresponding to $\fq$ as in \ref{idealslocbij}.
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We have $\fq_S \neq \{0\} $ as $\{0_{A}\}_S = \{0_{A_S}\}$.
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$A_1$ is a domain with $Q(A_1) \cong Q(A)$ (\ref{locandquot}) and $A_1 / \fq_S$ is isomorphic to the localization of $A / \fq$ with respect to the image of $S$ in $A/\fq$ (\ref{locandfactor}).
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$\mathfrak{k}(\fq_S)$ is algebraic over $\fl_1$ because the image of $\fl_1$ in $\mathfrak{k}(\fq_S)$ contains the images of $\fl$ and the $a_i$, and the images of the $a_i$ form a transcendence base for $\mathfrak{k}(\fq) / \fl$.
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$\mathfrak{k}(\fq_S)$ is algebraic over $\mathfrak{l}_1$ because the image of $\mathfrak{l}_1$ in $\mathfrak{k}(\fq_S)$ contains the images of $\mathfrak{l}$ and the $a_i$, and the images of the $a_i$ form a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{l}$.
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By the fact about integrality and fields (\ref{fintaf}) it follows that $A_1 / \fq_S$ is a field, hence $\fq_S \in \mSpec(A_1)$ and the special case of $\fq \in \mSpec(A)$
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can be applied to $\fq_S$ and $A_1 / \fl_1$ showing that $Q(A)$ cannot be algebraic over $\fl_1$.
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can be applied to $\fq_S$ and $A_1 / \mathfrak{l}_1$ showing that $Q(A)$ cannot be algebraic over $\mathfrak{l}_1$.
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\end{proof}
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\begin{corollary}\label{upperboundcodim}
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@ -1617,56 +1617,56 @@ To formulate a result which still applies in this context, we need the following
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\subsubsection{The relation between \texorpdfstring{$\hght(\fp)$}{ht(p)} and \texorpdfstring{$\trdeg$}{trdeg}}
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We will use the following
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\begin{lemma}\label{extendtotrbase}
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Let $\fl$ be an arbitrary field, $A$ a $\fl$-algebra of finite type which is a domain, $K \coloneqq Q(A)$ the field of quotients and let $(a_i)_{i=1}^n$ be $\fl$-algebraically independent elements of $A$. Then there exist a natural number $m \ge n$ and a transcendence base $(a_i)_{i = 1}^m$ for $K / \fl$ with $a_i \in A$ for $1 \le i \le m$.
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Let $\mathfrak{l}$ be an arbitrary field, $A$ a $\mathfrak{l}$-algebra of finite type which is a domain, $K \coloneqq Q(A)$ the field of quotients and let $(a_i)_{i=1}^n$ be $\mathfrak{l}$-algebraically independent elements of $A$. Then there exist a natural number $m \ge n$ and a transcendence base $(a_i)_{i = 1}^m$ for $K / \mathfrak{l}$ with $a_i \in A$ for $1 \le i \le m$.
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\end{lemma}
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\begin{proof}
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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 $(a_i)_{i = n+1}^m \in A^{m-n}$ which generate $K$ in the sense of a matroid used in the definition of $\trdeg$.
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For instance, one can use generators of the $\fl$-algebra $A$. We assume $m$ to be minimal and claim that $(a_i)_{i=1}^m$ are $\fl$-algebraically independent.
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Otherwise there is $j \in \N$, $1 \le j \le m$ such that $a_j$ is algebraic over the subfield of $K$ generated by $\fl$ and the $(a_i)_{i=1}^{j-1}$. We have $j > n$ by the algebraic independence of $(a_i)_{i=1}^n$.
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Exchanging $x_j$ and $x_m$, we may assume $j = m$. But then $K$ is algebraic over its subfield generated by $\fl$ and the $(a_i)_{i=1}^{m-1} $, contradicting the minimality of $m$.
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For instance, one can use generators of the $\mathfrak{l}$-algebra $A$. We assume $m$ to be minimal and claim that $(a_i)_{i=1}^m$ are $\mathfrak{l}$-algebraically independent.
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Otherwise there is $j \in \N$, $1 \le j \le m$ such that $a_j$ is algebraic over the subfield of $K$ generated by $\mathfrak{l}$ and the $(a_i)_{i=1}^{j-1}$. We have $j > n$ by the algebraic independence of $(a_i)_{i=1}^n$.
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Exchanging $x_j$ and $x_m$, we may assume $j = m$. But then $K$ is algebraic over its subfield generated by $\mathfrak{l}$ and the $(a_i)_{i=1}^{m-1} $, contradicting the minimality of $m$.
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\end{proof}
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\begin{theorem}\label{htandtrdeg}
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Let $\fl$ be an arbitrary field, $A$ a $\fl$-algebra of finite type which is a domain, and $\fp \in \Spec A$.
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Let $\mathfrak{l}$ be an arbitrary field, $A$ a $\mathfrak{l}$-algebra of finite type which is a domain, and $\fp \in \Spec A$.
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Let $K \coloneqq Q(A)$ be the field of quotients of $A$. Then
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\[
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\hght(\fp) = \trdeg(K /\fl) - \trdeg(\mathfrak{k}(\fp) / \fl)
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\hght(\fp) = \trdeg(K /\mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})
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\]
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\end{theorem}
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\begin{remark}
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By example \ref{htandcodim}, theorem \ref{trdegandkdim} is a special case of this theorem. %(\ref{htandtrdeg}).
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\end{remark}
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\begin{proof}
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If $\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_k$ is a chain of prime ideals in $A$, we have $\trdeg(\mathfrak{k}(\fp_i) / \fl) < \trdeg(\mathfrak{k}(\fp_{i+1}) / \fl)$ by \ref{trdegresfield} (``A first result of dimension theory'').
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If $\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_k$ is a chain of prime ideals in $A$, we have $\trdeg(\mathfrak{k}(\fp_i) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp_{i+1}) / \mathfrak{l})$ by \ref{trdegresfield} (``A first result of dimension theory'').
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Thus
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\[
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k \le \trdeg(\mathfrak{k}(\fp_k) / \fl) - \trdeg(\mathfrak{k}(\fp) / \fl) \le \trdeg(K / \fl) - \trdeg(\mathfrak{k}(\fp) / \fl)
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k \le \trdeg(\mathfrak{k}(\fp_k) / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) \le \trdeg(K / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})
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\]
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where the last inequality is another application of \ref{trdegresfield} (using $K = Q(A) = Q(A / \{0\}) = \mathfrak{k}(\{0\})$ and the fact that $\{0\} \se \fp_k$ is a prime ideal).
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Hence \[
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\hght(\fp) \le \trdeg( K / \fl) - \trdeg(\mathfrak{k}(\fp) / \fl)
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\hght(\fp) \le \trdeg( K / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})
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\]
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and it remains to show the opposite inequality.
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\begin{claim}
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For any maximal ideal $\fp \in \mSpec A$ \[
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\hght(\fm) \ge \trdeg(K / \fl)
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\hght(\fm) \ge \trdeg(K / \mathfrak{l})
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\]
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\end{claim}
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\begin{subproof}
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By the Noether normalization theorem (\ref{noenort}), there are $(x_i)_{i=1}^d \in A^d$ which are algebraically independent over $\fl$ such that $A$ is finite over the subalgebra $S$ generated by the $x_i$. We have $d = \trdeg(K / \fl)$ as the $x_i$ form a transcendence base of $K / \fl$.
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By the Noether normalization theorem (\ref{noenort}), there are $(x_i)_{i=1}^d \in A^d$ which are algebraically independent over $\mathfrak{l}$ such that $A$ is finite over the subalgebra $S$ generated by the $x_i$. We have $d = \trdeg(K / \mathfrak{l})$ as the $x_i$ form a transcendence base of $K / \mathfrak{l}$.
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\begin{claim}
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We can choose $x_i \in \fm$
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\end{claim}
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\begin{subproof}
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By the Nullstellensatz (\ref{hns2}), $\mathfrak{k}(\fm) = A / \fm$ is a finite field extension of $\fl$. Hence there exists a normed polynomial $P_i \in \fl[T]$ with $P_i(x_i \mod \fm) = 0$ in $\mathfrak{k}(\fm)$.
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By the Nullstellensatz (\ref{hns2}), $\mathfrak{k}(\fm) = A / \fm$ is a finite field extension of $\mathfrak{l}$. Hence there exists a normed polynomial $P_i \in \mathfrak{l}[T]$ with $P_i(x_i \mod \fm) = 0$ in $\mathfrak{k}(\fm)$.
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Let $\tilde x_i \coloneqq P_i(x_i) \in \fm$ and $\tilde S$ the subalgebra generated by the $\tilde x_i$. As $P_i(x_i) - \tilde x_i = 0$, $x_i$ is integral over $\tilde S$ and so is $S / \tilde S$. It follows that $A / \tilde S$ is integral, hence finite by \ref{ftaiimplf}. Replacing $x_i$ by $\tilde x_i$, we may thus assume that $x_i \in \fm$.
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\end{subproof}
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% TODO: fix names A_1 = A_S, k_1 = R_S
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The ring homomorphism $\ev_x : R = \fl[X_1,\ldots,X_d] \xrightarrow{P \mapsto P(x_1,\ldots,x_d)} A$ is injective. Because $R$ is a UFD, $R$ is normal (\ref{ufdnormal}). Thus the going-down theorem (\ref{gdkrull}) applies to the integral $R$-algebra $A$.
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The ring homomorphism $\ev_x : R = \mathfrak{l}[X_1,\ldots,X_d] \xrightarrow{P \mapsto P(x_1,\ldots,x_d)} A$ is injective. Because $R$ is a UFD, $R$ is normal (\ref{ufdnormal}). Thus the going-down theorem (\ref{gdkrull}) applies to the integral $R$-algebra $A$.
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For $0 \le i \le d$, let $\fp_i \se R$ be the ideal generated by $(X_j)_{j=i+1}^d$. We have $\fm \sqcap R = \fp_0$ as all $X_i \in \fm$, hence $X_i \in \fm \sqcap R$ and $\fp_0$ is a maximal ideal.
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By applying going-down and induction on $i$, there is a chain $\fm = \fq_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_d$ of elements of $\Spec A$ such that $\fq_i \sqcap R = \fp_i$.
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It follows that $\hght(\fm) \ge d$.
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@ -1674,15 +1674,15 @@ and it remains to show the opposite inequality.
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This finishes the proof in the case of $\fp \in \mSpec A$.
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To reduce the general case to that special case, we proceed as in \ref{trdegresfield}:
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By lemma \ref{ltrdegresfieldtrbase} there are $a_1,\ldots,a_n \in A$ whose images in $A / \fp$ form a transcendence base for $\mathfrak{k}(\fp) / \fl$.
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As these images are $\fl$-algebraically independent, the same holds for the $a_i$ themselves.
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By lemma \ref{ltrdegresfieldtrbase} there are $a_1,\ldots,a_n \in A$ whose images in $A / \fp$ form a transcendence base for $\mathfrak{k}(\fp) / \mathfrak{l}$.
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As these images are $\mathfrak{l}$-algebraically independent, the same holds for the $a_i$ themselves.
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By lemma \ref{extendtotrbase} we can extend $(a_{i})_{i=1}^n$ to a transcendence base $(a_i)_{i=1}^m \in A^m$ of $K / \fl$.
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Let $R \se A$ denote the $\fl$-subalgebra generated by $a_1,\ldots,a_n$ and let $S \coloneqq R \sm \{0\}$.
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By lemma \ref{extendtotrbase} we can extend $(a_{i})_{i=1}^n$ to a transcendence base $(a_i)_{i=1}^m \in A^m$ of $K / \mathfrak{l}$.
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Let $R \se A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and let $S \coloneqq R \sm \{0\}$.
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Let $A_1 \coloneqq A_S$ and $\fp_S$ the prime ideal corresponding to $\fp$ under $\Spec(A_1) \cong \{\fr \in \Spec A | \fr \cap S = \emptyset\}$ (\ref{idealslocbij}).
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As in \ref{locandquot}, $A_1$ is a domain with $Q(A_1) \cong K = Q(A)$ and by \ref{locandfactor} $A_1 / \fp_S \cong (A / \fp)_{\overline{S}}$, where $\overline{S}$ denotes the image of $S$ in $A / \fp$.
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As in \ref{trdegresfield}, $\mathfrak{k}(\fp_S) \cong \mathfrak{k}(\fp)$ is integral over $A_1 / \fp_S$.
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From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 / \fp_S$ is a field. Hence $\fp_S \in \mSpec(A_1)$ and the special case can be applied to $\fp_S$ and $A_1 / \fl_1$, showing that $\hght(\fp_S) \ge e = \trdeg(K / \fl_1)$. We have $\trdeg(K / \fl_1) = m - n$, as $(a_i)_{i = n+1}^m$ is a transcendence base for $K / \fl_1$. By the description of $\Spec A_S$ (\ref{idealslocbij}), a chain $\fp_S = \fq_0 \supsetneq \ldots \supsetneq \fp_e$ of prime ideals in $A_S$ defines a similar chain $\fp_i \coloneqq \fq_i \sqcap A$ in $A$ with $\fp_0 = \fp$. Thus $\hght(\fp) \ge e$.
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From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 / \fp_S$ is a field. Hence $\fp_S \in \mSpec(A_1)$ and the special case can be applied to $\fp_S$ and $A_1 / \mathfrak{l}_1$, showing that $\hght(\fp_S) \ge e = \trdeg(K / \mathfrak{l}_1)$. We have $\trdeg(K / \mathfrak{l}_1) = m - n$, as $(a_i)_{i = n+1}^m$ is a transcendence base for $K / \mathfrak{l}_1$. By the description of $\Spec A_S$ (\ref{idealslocbij}), a chain $\fp_S = \fq_0 \supsetneq \ldots \supsetneq \fp_e$ of prime ideals in $A_S$ defines a similar chain $\fp_i \coloneqq \fq_i \sqcap A$ in $A$ with $\fp_0 = \fp$. Thus $\hght(\fp) \ge e$.
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\end{proof}
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\begin{remark}
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@ -1918,12 +1918,12 @@ $\rad(A) = f A$ where $f = \prod_{i=1}^{n} p_i$.
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% Lecture 11
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\section{Projective spaces}
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Let $\fl$ be any field.
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Let $\mathfrak{l}$ be any field.
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\begin{definition}
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For a $\fl$-vector space $V$, let $\bP(V)$ be the set of one-dimensional subspaces of $V$.
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Let $\bP^n(\fl) \coloneqq \bP(\fl^{n+1})$, the \vocab[Projective space]{$n$-dimensional projective space over $\fl$}.
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For a $\mathfrak{l}$-vector space $V$, let $\bP(V)$ be the set of one-dimensional subspaces of $V$.
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Let $\bP^n(\mathfrak{l}) \coloneqq \bP(\mathfrak{l}^{n+1})$, the \vocab[Projective space]{$n$-dimensional projective space over $\mathfrak{l}$}.
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If $\fl$ is kept fixed, we will often write $\bP^n$ for $\bP^n(\fl)$.
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If $\mathfrak{l}$ is kept fixed, we will often write $\bP^n$ for $\bP^n(\mathfrak{l})$.
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When dealing with $\bP^n$, the usual convention is to use $0$ as the index of the first coordinate.
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@ -1936,10 +1936,10 @@ Let $\fl$ be any field.
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\end{remark}
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\begin{definition}[Infinite hyperplane]
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For $0 \le i \le n$ let $U_i \se \bP^n$ denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$.
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This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the same point $x \in \bP^n$ differ by scaling with a $\lambda \in \fl^{\times}$, $x_i = \lambda \xi_i$. Since not all $x_i$ may be $0$, $\bP^n = \bigcup_{i=0}^n U_i$. We identify $\bA^n = \bA^n(\fl) = \fl^n$ with $U_0$ by identifying $(x_1,\ldots,x_n) \in \bA^n$ with $[1,x_1,\ldots,x_n] \in \bP^n$.
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This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the same point $x \in \bP^n$ differ by scaling with a $\lambda \in \mathfrak{l}^{\times}$, $x_i = \lambda \xi_i$. Since not all $x_i$ may be $0$, $\bP^n = \bigcup_{i=0}^n U_i$. We identify $\bA^n = \bA^n(\mathfrak{l}) = \mathfrak{l}^n$ with $U_0$ by identifying $(x_1,\ldots,x_n) \in \bA^n$ with $[1,x_1,\ldots,x_n] \in \bP^n$.
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Then $\bP^1 = \bA^1 \cup \{\infty\} $ where $\infty=[0,1]$. More generally, when $n > 0$ $\bP^n \sm \bA^n$ can be identified with $\bP^{n-1}$ identifying $[0,x_1,\ldots,x_n] \in \bP^n \sm \bA^n$ with $[x_1,\ldots,x_n] \in \bP^{n-1}$.
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Thus $\bP^n$ is $\bA^n \cong \fl^n$ with a copy of $\bP^{n-1}$ added as an \vocab{infinite hyperplane} .
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Thus $\bP^n$ is $\bA^n \cong \mathfrak{l}^n$ with a copy of $\bP^{n-1}$ added as an \vocab{infinite hyperplane} .
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\end{definition}
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\subsubsection{Graded rings and homogeneous ideals}
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@ -2834,7 +2834,7 @@ Let $I_x(G,H) \se \cO_{\bP^2,x}$ denote the ideal generated by $\gamma_x$ and $\
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\begin{itemize}
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\item[HNS2 $\implies$ HNS1b] Let $I \se \fl[X_1,\ldots,X_n]$. $I \se \fm$ maximal. $R / \fm$ is isomorphic to a field extension of $\fl$. Finite by HNS2.
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\item[HNS2 $\implies$ HNS1b] Let $I \se \mathfrak{l}[X_1,\ldots,X_n]$. $I \se \fm$ maximal. $R / \fm$ is isomorphic to a field extension of $\mathfrak{l}$. Finite by HNS2.
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\item[NNT $\implies$ HNS2] Apply NNT to $L / K$ $\leadsto$ alg. independent $a_i$ such that $L$ is finite over the image of $K[X_1,\ldots,X_n] \xrightarrow{\ev_a} L$ ($\ev_a : P \to P(a_1,\ldots,a_n)$).
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$\ev_a(K[X_1,\ldots,X_n])$ is a field by fact about integrality and fields \ref{fintaf}. Hence $n = 0$ and $L / K$ is finite.
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\item[UNCHNS2] $K$ uncountable, $L / K$ fin. type. Then $\dim_K L$ is countable. Suppose $l \in L$ is not integral. Then $K(l) \cong K(T)$ and $\dim_K L \ge \dim_K K(T) \ge \aleph_1$.
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|
@ -2873,13 +2873,13 @@ Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where $\alpha \in S$ such that $
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%
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||||
A first result of dimension theory:
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||||
$A \fl$-algebra of finite type, $\fp, \fq \in \Spec A, \fp \subsetneq \fq$. Then $\trdeg(\mathfrak{k}(\fp) /\fl) > \trdeg(\mathfrak{k}(\fq) / \fl)$:
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||||
$A \mathfrak{l}$-algebra of finite type, $\fp, \fq \in \Spec A, \fp \subsetneq \fq$. Then $\trdeg(\mathfrak{k}(\fp) /\mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) / \mathfrak{l})$:
|
||||
\Wlog $\fp = \{0\}$ and $A$ a domain ($A' \coloneqq A / \fp$).
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For $\fq \in \mSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence finite (HNS) $\implies \trdeg(\mathfrak{k}(\fq) / \fl) = 0$.
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||||
$\trdeg(Q(A) / \fl) = 0 \implies A$ integral over $\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$.
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||||
For $\fq \in \mSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence finite (HNS) $\implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$.
|
||||
$\trdeg(Q(A) / \mathfrak{l}) = 0 \implies A$ integral over $\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$.
|
||||
|
||||
If $\fq \not\in \mSpec A$, let $a_1,\ldots,a_n \in A$ alg. independent such that the $\overline{a_i}$ are a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{k}$
|
||||
Let $R$ be the ring generated by $\fl$ and the $a_i$. Localize with respect to $S \coloneqq R \sm \{0\}$.
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Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. Localize with respect to $S \coloneqq R \sm \{0\}$.
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||||
%TODO
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||||
% TODO: LERNEN
|
||||
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||||
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|
@ -2887,7 +2887,7 @@ Let $R$ be the ring generated by $\fl$ and the $a_i$. Localize with respect to $
|
|||
% Dim k^n
|
||||
$\dim(\mathfrak{k}^n)$
|
||||
$ \ge n$ build chian
|
||||
$\le n$ a first result in dim T ($\fp \subsetneq \fq \implies \trdeg(\mathfrak{k}(\fq) / \fl) < \trdeg(\mathfrak{k}(\fp) / \fl)$. Thus $\codim(X,Y) \le \trdeg(\fK(Y) / \fl) - \trdeg(\fK(X) / \fl)$.
|
||||
$\le n$ a first result in dim T ($\fp \subsetneq \fq \implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})$. Thus $\codim(X,Y) \le \trdeg(\fK(Y) / \mathfrak{l}) - \trdeg(\fK(X) / \mathfrak{l})$.
|
||||
|
||||
TODO
|
||||
% List of proofs of HNS
|
||||
|
|
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|
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@ -36,7 +36,6 @@
|
|||
\DeclareMathOperator{\hght}{ht}
|
||||
\newcommand{\Wlog}{W.l.o.g. }
|
||||
|
||||
\newcommand{\fk}{\ensuremath\mathfrak{k}}
|
||||
\newcommand{\fl}{\ensuremath\mathfrak{l}}
|
||||
\newcommand{\fs}{\ensuremath\mathfrak{s}}
|
||||
\newcommand{\fri}{\ensuremath\mathfrak{i}}
|
||||
|
|
|
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Loading…
Reference in a new issue