From 1fdc0528db774db9dc53e337a1abe6c2180be869 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Maximilian=20Ke=C3=9Fler?= Date: Wed, 16 Feb 2022 01:13:05 +0100 Subject: [PATCH] replace fl --- 2021_Algebra_I.tex | 102 ++++++++++++++++++++++----------------------- algebra.sty | 1 - 2 files changed, 51 insertions(+), 52 deletions(-) diff --git a/2021_Algebra_I.tex b/2021_Algebra_I.tex index e37d036..42bbe9d 100644 --- a/2021_Algebra_I.tex +++ b/2021_Algebra_I.tex @@ -25,7 +25,7 @@ \newline \noindent $\mathfrak{k}$ is {\color{red} always} an algebraically closed field and $\mathfrak{k}^n$ is equipped with the Zariski-topology. -Fields which are not assumed to be algebraically closed have been renamed (usually to $\fl$). +Fields which are not assumed to be algebraically closed have been renamed (usually to $\mathfrak{l}$). \pagebreak @@ -357,19 +357,19 @@ Equivalent\footnote{used in a vague sense here} formulation: \end{remark} \begin{theorem}[Hilbert's Nullstellensatz (1b)] \label{hns1b} - 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 $\mathfrak{i}^m$. + 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 $\mathfrak{i}^m$. \end{theorem} \begin{proof} (HNS2 (\ref{hns2}) $\implies$ HNS1b (\ref{hns1b})) $I \se \fm$ for some maximal ideal. $R / \fm$ is a field, since $\fm$ is maximal. - $R / \fm$ is of finite type, since the images of the $X_i$ generate it as a $\fl$-algebra. - There are thus a field extension $\fri / \fl$ and an isomorphism $R / \fm \xrightarrow{\iota} \fri$ of $\fl$-algebras. - By HNS2 (\ref{hns2}), $\fri / \fl$ is a finite field extension. + $R / \fm$ is of finite type, since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra. + There are thus a field extension $\fri / \mathfrak{l}$ and an isomorphism $R / \fm \xrightarrow{\iota} \fri$ of $\mathfrak{l}$-algebras. + By HNS2 (\ref{hns2}), $\fri / \mathfrak{l}$ is a finite field extension. Let $x_i \coloneqq \iota (X_i \mod \fm)$. \[ P(x_1,\ldots,x_m) = \iota(P \mod \fm) \] - 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. + 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. Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \fm = 0$ for $P \in I \se \fm$). HNS1 (\ref{hns1}) can easily be derived from HNS1b. @@ -1101,39 +1101,39 @@ Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-satura % i = ic \begin{proposition}\label{trdegresfield} - Let $\fl$ be a %% ?? -field, $A$ a $\fl$-algebra of finite type and $\fp, \fq \in \Spec A$ with $\fp \subsetneq \fq$. + Let $\mathfrak{l}$ be a %% ?? +field, $A$ a $\mathfrak{l}$-algebra of finite type and $\fp, \fq \in \Spec A$ with $\fp \subsetneq \fq$. Then \[ - \trdeg(\mathfrak{k}(\fp) / \fl) > \trdeg(\mathfrak{k}(\fq) / \fl) + \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) \] \end{proposition} \begin{proof} Replacing $A$ by $A / \fp$, we may assume $\fp = \{0\} $ and $A$ to be a domain. Then $\mathfrak{k}(\fp) = Q(A / \fp) = Q(A)$. - 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}). - Thus, $\trdeg(\mathfrak{k}(\fq) / \fl) = 0$. - 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 $\fq = \fp \lightning$. + 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}). + Thus, $\trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$. + 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 $\fq = \fp \lightning$. This finishes the proof for $\fq \in \mSpec A$. We will use the following lemma to reduce the general case to this case: \begin{lemma}\label{ltrdegresfieldtrbase} - There are algebraically independent $a_1,\ldots,a_n \in A$ whose images in $A / \fq$ form a transcendence base for $\mathfrak{k}(\fq) / \fl$. + There are algebraically independent $a_1,\ldots,a_n \in A$ whose images in $A / \fq$ form a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{l}$. \end{lemma} \begin{subproof} - 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). - 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 $\overline{a_i}, 1\le i n$ by the algebraic independence of $(a_i)_{i=1}^n$. - 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 $m$. + 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. + 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 $(a_i)_{i=1}^n$. + 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 $m$. \end{proof} \begin{theorem}\label{htandtrdeg} - Let $\fl$ be an arbitrary field, $A$ a $\fl$-algebra of finite type which is a domain, and $\fp \in \Spec A$. + Let $\mathfrak{l}$ be an arbitrary field, $A$ a $\mathfrak{l}$-algebra of finite type which is a domain, and $\fp \in \Spec A$. Let $K \coloneqq Q(A)$ be the field of quotients of $A$. Then \[ - \hght(\fp) = \trdeg(K /\fl) - \trdeg(\mathfrak{k}(\fp) / \fl) + \hght(\fp) = \trdeg(K /\mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) \] \end{theorem} \begin{remark} By example \ref{htandcodim}, theorem \ref{trdegandkdim} is a special case of this theorem. %(\ref{htandtrdeg}). \end{remark} \begin{proof} - 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''). + 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''). Thus \[ - k \le \trdeg(\mathfrak{k}(\fp_k) / \fl) - \trdeg(\mathfrak{k}(\fp) / \fl) \le \trdeg(K / \fl) - \trdeg(\mathfrak{k}(\fp) / \fl) + k \le \trdeg(\mathfrak{k}(\fp_k) / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) \le \trdeg(K / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) \] 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). Hence \[ - \hght(\fp) \le \trdeg( K / \fl) - \trdeg(\mathfrak{k}(\fp) / \fl) + \hght(\fp) \le \trdeg( K / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) \] and it remains to show the opposite inequality. \begin{claim} For any maximal ideal $\fp \in \mSpec A$ \[ - \hght(\fm) \ge \trdeg(K / \fl) + \hght(\fm) \ge \trdeg(K / \mathfrak{l}) \] \end{claim} \begin{subproof} - 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 $K / \fl$. + 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 $K / \mathfrak{l}$. \begin{claim} We can choose $x_i \in \fm$ \end{claim} \begin{subproof} - 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 $\mathfrak{k}(\fm)$. + 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 $\mathfrak{k}(\fm)$. 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 $x_i \in \fm$. \end{subproof} % TODO: fix names A_1 = A_S, k_1 = R_S - 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 $A$. + 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 $A$. 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. 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 $\fq_i \sqcap R = \fp_i$. It follows that $\hght(\fm) \ge d$. @@ -1674,15 +1674,15 @@ and it remains to show the opposite inequality. This finishes the proof in the case of $\fp \in \mSpec A$. To reduce the general case to that special case, we proceed as in \ref{trdegresfield}: -By lemma \ref{ltrdegresfieldtrbase} there are $a_1,\ldots,a_n \in A$ whose images in $A / \fp$ form a transcendence base for $\mathfrak{k}(\fp) / \fl$. -As these images are $\fl$-algebraically independent, the same holds for the $a_i$ themselves. +By lemma \ref{ltrdegresfieldtrbase} there are $a_1,\ldots,a_n \in A$ whose images in $A / \fp$ form a transcendence base for $\mathfrak{k}(\fp) / \mathfrak{l}$. +As these images are $\mathfrak{l}$-algebraically independent, the same holds for the $a_i$ themselves. -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 $K / \fl$. -Let $R \se A$ denote the $\fl$-subalgebra generated by $a_1,\ldots,a_n$ and let $S \coloneqq R \sm \{0\}$. +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 $K / \mathfrak{l}$. +Let $R \se A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and let $S \coloneqq R \sm \{0\}$. 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}). 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 $A / \fp$. As in \ref{trdegresfield}, $\mathfrak{k}(\fp_S) \cong \mathfrak{k}(\fp)$ is integral over $A_1 / \fp_S$. -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 $\hght(\fp) \ge e$. +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 $\hght(\fp) \ge e$. \end{proof} \begin{remark} @@ -1918,12 +1918,12 @@ $\rad(A) = f A$ where $f = \prod_{i=1}^{n} p_i$. % Lecture 11 \section{Projective spaces} -Let $\fl$ be any field. +Let $\mathfrak{l}$ be any field. \begin{definition} - For a $\fl$-vector space $V$, let $\bP(V)$ be the set of one-dimensional subspaces of $V$. - Let $\bP^n(\fl) \coloneqq \bP(\fl^{n+1})$, the \vocab[Projective space]{$n$-dimensional projective space over $\fl$}. + For a $\mathfrak{l}$-vector space $V$, let $\bP(V)$ be the set of one-dimensional subspaces of $V$. + Let $\bP^n(\mathfrak{l}) \coloneqq \bP(\mathfrak{l}^{n+1})$, the \vocab[Projective space]{$n$-dimensional projective space over $\mathfrak{l}$}. - If $\fl$ is kept fixed, we will often write $\bP^n$ for $\bP^n(\fl)$. + If $\mathfrak{l}$ is kept fixed, we will often write $\bP^n$ for $\bP^n(\mathfrak{l})$. When dealing with $\bP^n$, the usual convention is to use $0$ as the index of the first coordinate. @@ -1936,10 +1936,10 @@ Let $\fl$ be any field. \end{remark} \begin{definition}[Infinite hyperplane] For $0 \le i \le n$ let $U_i \se \bP^n$ denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$. - 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 $[1,x_1,\ldots,x_n] \in \bP^n$. + 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 $[1,x_1,\ldots,x_n] \in \bP^n$. 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 $[x_1,\ldots,x_n] \in \bP^{n-1}$. - Thus $\bP^n$ is $\bA^n \cong \fl^n$ with a copy of $\bP^{n-1}$ added as an \vocab{infinite hyperplane} . + Thus $\bP^n$ is $\bA^n \cong \mathfrak{l}^n$ with a copy of $\bP^{n-1}$ added as an \vocab{infinite hyperplane} . \end{definition} \subsubsection{Graded rings and homogeneous ideals} @@ -2834,7 +2834,7 @@ Let $I_x(G,H) \se \cO_{\bP^2,x}$ denote the ideal generated by $\gamma_x$ and $\ \begin{itemize} - \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. + \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. \item[NNT $\implies$ HNS2] Apply NNT to $L / K$ $\leadsto$ alg. independent $a_i$ such that $L$ is finite over the image of $K[X_1,\ldots,X_n] \xrightarrow{\ev_a} L$ ($\ev_a : P \to P(a_1,\ldots,a_n)$). $\ev_a(K[X_1,\ldots,X_n])$ is a field by fact about integrality and fields \ref{fintaf}. Hence $n = 0$ and $L / K$ is finite. \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 $\dim_K L \ge \dim_K K(T) \ge \aleph_1$. @@ -2873,13 +2873,13 @@ Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where $\alpha \in S$ such that $ % A first result of dimension theory: -$A \fl$-algebra of finite type, $\fp, \fq \in \Spec A, \fp \subsetneq \fq$. Then $\trdeg(\mathfrak{k}(\fp) /\fl) > \trdeg(\mathfrak{k}(\fq) / \fl)$: +$A \mathfrak{l}$-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 ($A' \coloneqq A / \fp$). -For $\fq \in \mSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence finite (HNS) $\implies \trdeg(\mathfrak{k}(\fq) / \fl) = 0$. -$\trdeg(Q(A) / \fl) = 0 \implies A$ integral over $\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$. +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 $S \coloneqq R \sm \{0\}$. +Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. Localize with respect to $S \coloneqq R \sm \{0\}$. %TODO % TODO: LERNEN @@ -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 diff --git a/algebra.sty b/algebra.sty index fc03324..161de40 100644 --- a/algebra.sty +++ b/algebra.sty @@ -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}}