s21-algebra-1/inputs/uebersicht.tex
2023-07-22 21:52:48 +02:00

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