188 lines
5.8 KiB
TeX
188 lines
5.8 KiB
TeX
% TODO REMARK ABOUT ZORNS LEMMA (LECTURE 1)
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% TODO REMARK ABOUT FIN PRESENTED MODULES (LECTURE 2)
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% TODO: LECTURE 9 LEMMA
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% ÜBERSICHT %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% List of forms of HNS
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\begin{itemize}
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\item[HNS2 $\implies$ HNS1b]
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Let $I \subseteq \mathfrak{l}[X_1,\ldots,X_n]$. $I \subseteq \mathfrak{m}$
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maximal. $R / \mathfrak{m}$ is isomorphic to a field extension of
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$\mathfrak{l}$.
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Finite by HNS2.
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\item[NNT $\implies$ HNS2]
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Apply NNT to $L / K$ $\leadsto$ alg. independent $a_i$ such that $L$ is finite
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over the image of $K[X_1,\ldots,X_n] \xrightarrow{\ev_a}
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L$ ($\ev_a : P \to P(a_1,\ldots,a_n)$).
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$\ev_a(K[X_1,\ldots,X_n])$ is a field by fact about integrality and fields
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\ref{fintaf}.
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Hence $n = 0$ and $L / K$ is finite.
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\item[UNCHNS2]
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$K$ uncountable, $L / K$ fin. type.
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Then $\dim_K L$ is countable.
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Suppose $l \in L$ is not integral.
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Then $K(l) \cong K(T)$ and $\dim_K L \ge \dim_K K(T) \ge \aleph_1$.
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Thus $L / K$ algebraic $\implies$ integral $\implies$ finite.
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\item[HNS3]
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($V(I) \subseteq V(f) \iff f \in \sqrt{I} $).
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Suppose $V(I) \subseteq V(f)$.
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$R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n, T], J \subseteq R'$ the ideal generated by $I$
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and $g(X_1,\ldots,X_n,T) \coloneqq 1 - Tf(X_1,\ldots,X_n)$.
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\end{itemize}
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% Proofs
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Def of integrality (<=>)
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Fact about integrality and field:
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Technical lemma for Noether normalization: For $S \subseteq \N^n$ finite, there
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exists $k \in \N^n$ such that $k_1 = 1$ and $s_1 \neq s_2 \in S \implies
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\langle k, s_1 \rangle \neq \langle k, s_2 \rangle$: For $s_1 \neq s_2$,
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Noether normalization: $a_i \in A$ minimal such that $A$ is integral over the
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subalgebra genereted by the $a_i$.
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% TODO% TODO
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Suppose $\exists P \in K[X_1,\ldots,X_n] \setminus \{ 0\} ~
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P(a_1,\ldots,a_n) =
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0$.
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$P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in \N^n
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| p_\alpha \neq 0\}$.
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Choose $k$ as in the lemma.
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$b_i \coloneqq a_{i+1} -
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a_1^{k_{i+1}}, 1 \le i <n$.
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Claim: $A$ is integral ober subalgebra $B$ generated by the $b_i$ ($\lightning$
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minimality) $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots,
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b_{n-1} + T^{k_n})$.
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$Q(a_1) = P(\vec a) = 0$.
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For suitable $\beta_{\alpha, l} in B$:
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\[
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T^{\alpha_1} \prod_{i=1}^{n-1} (b_i +
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T^{k_{i+1}})^{\alpha_{i+1}} =
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T^{w_k(\alpha)} + \sum_{l=0}^{w_k(\alpha) - 1}
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\beta_{\alpha,l} T^l
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\]
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Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where
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$\alpha \in S$ such that $w_k(\alpha)$ is maximal.
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Thus, $Q$ is normed.
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% TODO Artin-Tate
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%
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A first result of dimension theory: $A \mathfrak{l}$-algebra of
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finite type,
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$\fp, \fq \in \Spec A, \fp \subsetneq \fq$.
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Then $\trdeg(\mathfrak{k}(\fp) /\mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) /
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\mathfrak{l})$: Without loss of
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generality loss of generality $\fp = \{0\}$
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and $A$ a domain ($A' \coloneqq A / \fp$).
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For $\fq \in \MaxSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence
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finite (HNS) $\implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$.
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$\trdeg(Q(A) / \mathfrak{l}) = 0 \implies A$ integral over
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$\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq
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\lightning$.
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If $\fq \not\in \MaxSpec A$, let $a_1,\ldots,a_n \in A$ alg.
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independent such that the $\overline{a_i}$ are a transcendence base for
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$\mathfrak{k}(\fq) / \mathfrak{k}$
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Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$.
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Localize with respect to $S \coloneqq R \setminus \{0\}$.
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%TODO
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% TODO: LERNEN
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% Dim k^n
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$\dim(\mathfrak{k}^n)$
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$ \ge n$ build chian
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$\le n$ a first result in dim T ($\fp \subsetneq \fq \implies
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\trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})$.
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Thus $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{l}) -
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\trdeg(\mathfrak{K}(X) / \mathfrak{l})$.
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TODO
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$\dim Y \ge \trdeg(\mathfrak{k}(Y) / \mathfrak{k})$: Noether normalization.
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% List of proofs of HNS% Going up% TODO proof of dim Y = trdeg(K(Y) / k)
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Subalgebra $\cong \mathfrak{k}[X_1,\ldots,X_d]$.
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Lift chain of prime ideals using going up.
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% TODO prime avoidance
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Action of $\Aut(L/K)$ on prime ideals of a normal ring extension.
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$A$ normal domain, $L / Q(A)$ normal field extension, $B$ int closure of $A$ in
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$L$, $\fp \in \Spec A$.
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Then $\Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq
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\cap A =
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\fp\}$ :
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\begin{itemize}
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\item
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$\fq, \fr \in \Spec B$ lying over $\fp$.
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\item
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only need to show $\fq \subseteq \sigma(\fr)$ for some $\sigma \in G$ (Krull
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going-up, no inclusions)
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\item
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Suppose not.
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Then $x \in \fq \setminus \bigcup_{\sigma \in G} \sigma(\fr)$
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(prime
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aviodance)
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\item
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$y = \prod_{\sigma \in G} \sigma(x) \in \fq \setminus \fr$
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($\fr$ prime ideal)
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\item
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$\exists k \in \N$ s.t.
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$y^k \in K$ ($y \in L^G$)
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\item
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$y^k \in K \cap B = A $ ($A$ normal).
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Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp$.
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\item
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$L / K$ infinite: Apply Zorn to pairs $(M, \sigma)$ where $K \subseteq M
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\subseteq L$ and $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap
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M$.
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\end{itemize}
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Going down Krull
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The ht p and trdeg ==================
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Zariski-Topology, Spec, $\mathfrak{k}^n$ Residue field
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$\mathfrak{k}(\fp)
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\coloneqq Q(A / \fp), \mathfrak{K}(V(\fp)) \coloneqq \mathfrak{k}(\fp)$.
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%TODO% TODO % TODO % TODO %% Definitions
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TODO?
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% Counterexamples
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no going-up
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% list of definitions of codim, dim, trdeg, ht
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Original (Noether normalization) Artin-Tate Uncountable fields
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\begin{landscape}
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\section{Übersicht} {\rowcolors{2}{gray!
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10}{white}
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\begin{longtable}{lll}
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\end{longtable}
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}
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\end{landscape}
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