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% 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] Let $I \subseteq \mathfrak{l}[X_1,\ldots,X_n]$. $I \subseteq \mathfrak{m}$ maximal. $R / \mathfrak{m}$ 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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Thus $L / K$ algebraic $\implies$ integral $\implies$ finite.
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\item[HNS3] ($V(I) \subseteq V(f) \iff f \in \sqrt{I} $). Suppose $V(I) \subseteq V(f)$. $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n, T], J \subseteq R'$ the ideal generated by $I$ 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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% TODO
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Technical lemma for Noether normalization: For $S \subseteq \N^n$ finite, there exists $k \in \N^n$ such that $k_1 = 1$ and $s_1 \neq s_2 \in S \implies \langle k, s_1 \rangle \neq \langle k, s_2 \rangle$:
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For $s_1 \neq s_2$, % TODO
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Noether normalization:
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$a_i \in A$ minimal such that $A$ is integral over the subalgebra genereted by the $a_i$.
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Suppose $\exists P \in K[X_1,\ldots,X_n] \setminus \{ 0\} ~ P(a_1,\ldots,a_n) = 0$. $P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in \N^n | p_\alpha \neq 0\}$.
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Choose $k$ as in the lemma.
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$b_i \coloneqq a_{i+1} - a_1^{k_{i+1}}, 1 \le i <n$. Claim: $A$ is integral ober subalgebra $B$ generated by the $b_i$ ($\lightning$ minimality)
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$Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n})$. $Q(a_1) = P(\vec a) = 0$.
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For suitable $\beta_{\alpha, l} in B$:
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\[
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T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_{i+1}})^{\alpha_{i+1}} = T^{w_k(\alpha)} + \sum_{l=0}^{w_k(\alpha) - 1} \beta_{\alpha,l} T^l
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\]
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Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where $\alpha \in S$ such that $w_k(\alpha)$ is maximal. 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:
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$A \mathfrak{l}$-algebra of finite type, $\fp, \fq \in \Spec A, \fp \subsetneq \fq$. Then $\trdeg(\mathfrak{k}(\fp) /\mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) / \mathfrak{l})$:
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Without loss of generality loss of generality $\fp = \{0\}$ 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 finite (HNS) $\implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$.
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$\trdeg(Q(A) / \mathfrak{l}) = 0 \implies A$ integral over $\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$.
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If $\fq \not\in \MaxSpec A$, let $a_1,\ldots,a_n \in A$ alg. independent such that the $\overline{a_i}$ are a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{k}$
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Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. 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 ($\fp \subsetneq \fq \implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})$. Thus $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{l}) - \trdeg(\mathfrak{K}(X) / \mathfrak{l})$.
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TODO
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% List of proofs of HNS
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% Going up
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% TODO proof of dim Y = trdeg(K(Y) / k)
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$\dim Y \ge \trdeg(\mathfrak{k}(Y) / \mathfrak{k})$: Noether normalization. Subalgebra $\cong \mathfrak{k}[X_1,\ldots,X_d]$. 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. $A$ normal domain, $L / Q(A)$ normal field extension, $B$ int closure of $A$ in $L$, $\fp \in \Spec A$.
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Then $\Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$ :
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\begin{itemize}
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\item $\fq, \fr \in \Spec B$ lying over $\fp$.
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\item only need to show $\fq \subseteq \sigma(\fr)$ for some $\sigma \in G$ (Krull going-up, no inclusions)
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\item Suppose not. Then $x \in \fq \setminus \bigcup_{\sigma \in G} \sigma(\fr)$ (prime aviodance)
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\item $y = \prod_{\sigma \in G} \sigma(x) \in \fq \setminus \fr$ ($\fr$ prime ideal)
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\item $\exists k \in \N$ s.t. $y^k \in K$ ($y \in L^G$)
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\item $y^k \in K \cap B = A $ ($A$ normal). Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp$.
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\item $L / K$ infinite: Apply Zorn to pairs $(M, \sigma)$ where $K \subseteq M \subseteq L$ and $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap M$.
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\end{itemize}
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Going down Krull %TODO
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The ht p and trdeg
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==================
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% TODO % TODO % TODO %
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% Definitions
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Zariski-Topology, Spec, $\mathfrak{k}^n$
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Residue field $\mathfrak{k}(\fp) \coloneqq Q(A / \fp), \mathfrak{K}(V(\fp)) \coloneqq \mathfrak{k}(\fp)$. 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)
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Artin-Tate
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Uncountable fields
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\begin{landscape}
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\section{Übersicht}
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{\rowcolors{2}{gray!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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