s21-algebra-1/inputs/uebersicht.tex

% 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
\begin{landscape}
    \section{Übersicht} {\rowcolors{2}{gray!
          10}{white}
      \begin{longtable}{lll}
      \end{longtable}
     }

\end{landscape}