add Noether normalization proof

commutative-algebra-notes.tex

@@ -945,7 +945,7 @@ As in with rings, it is equivalent to say that
   $$J_m = (a_{m,1}, \ldots a_{m, r_m})$$
 
   By definition of $J_m$, for each $a_{m,j}$ with $1 \leq j \leq r_m$,\\
-  $\exissts$ a polynomial $f_{m, j} \in I$ of degree $m$ having the leading coefficient $a_{m, j}$.
+  $\exists$ a polynomial $f_{m, j} \in I$ of degree $m$ having the leading coefficient $a_{m, j}$.
 
   $$\Longrightarrow~~ \{ f_{m,j} \}_{m<n; 1 \leq j \leq r_m}$$
   the set of elements of $I$.
@@ -990,9 +990,13 @@ As in with rings, it is equivalent to say that
   $A[x_1, \ldots, x_n]$ being Noetherian implies that $A[x_1, \ldots, x_n]/I$ is Noetherian.
 \end{proof}
 
+
+
 \vspace{1cm}
 \section{Finite ring extensions and Noether normalisation}
 
+\subsection{A-algebras and integral domains}
+
 \begin{defn}{}{A-algebra.}
   An $A$-algebra is a ring $B$ with a ring homomorphism $\psi: A \longrightarrow B$.
 
@@ -1053,8 +1057,8 @@ As in with rings, it is equivalent to say that
 \end{proof}
 
 
-\begin{prop}{R.4.3}{Tower Laws}\label{R.4.3}
-  Let $B$ be an $A$-algebra.
+\begin{prop}{R.4.3}{Tower Laws.}\label{R.4.3}
+  \\Let $B$ be an $A$-algebra.
   \begin{enumerate}[a.]
     \item Transitivity of finiteness: if $A \subset B \subset C$ are extension rings such that $C$ is a finite $B$-algebra and $B$ a finite $A$-algebra,\\
       then $C$ is finite over $A$.
@@ -1072,7 +1076,7 @@ As in with rings, it is equivalent to say that
 \begin{proof}.\\
   \begin{enumerate}[a.]
     \item if $\{ \beta_1, \ldots, \beta_n \}$ generate $B$ as an $A$-module and $\{ \gamma_1, \ldots, \gamma_n \}$ generate $C$ as an $B$-module,\\
-      then the set of products $\{ \betA_i \gamma_j \}$ generates $C$ as an $A$-module.
+      then the set of products $\{ \beta_i \gamma_j \}$ generates $C$ as an $A$-module.
 
       Since there are $n \times m$ generators (ie. finite), $C$ is finite over $A$.
 
@@ -1141,6 +1145,77 @@ As in with rings, it is equivalent to say that
   For any integral domain $A$, the integral closure of $A$ in its field of fractions $K=Frac(A)$ is also called the \emph{normalization} of $A$.
 \end{defn}
 
+
+\vspace{0.5cm}
+
+\subsection{Noether normalization}
+
+\begin{defn}{4.6}{Algebraically independent.}
+  $y_1, \ldots, y_n \in A$ are \emph{algebraically independent} over $K$ if the natural surjection $K[Y_1, \ldots, Y_n] \longrightarrow K[y_1, \ldots, y_n]$ is an isomorphism.
+
+  $\Longrightarrow~~ \nexists~ F(y_1, \ldots, y_n)=0$ ($F$ nonzero) with coefficients in $K$.
+\end{defn}
+
+Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some finite set $y_1, \ldots, y_n$.
+
+\begin{lemma}{R.4.6.L} \label{R.4.6.L}
+  Let $A = K[y_1, \ldots, y_n]$ and $0 \neq F \in K[Y_1, \ldots, Y_n]$ such that $F(y_1, \ldots, y_n)=0$.
+
+  Then $\exists~ y^*_1, \ldots, y^*_{n-1} \in A$ such that $y_n$ is integral over
+  $$A^* = K[y^*_1, \ldots, y^*_{n-1}] ~~\text{and}~~ A=A^*[y_n]$$
+\end{lemma}
+\begin{proof}
+  (todo)
+\end{proof}
+
+\begin{thm}{R.4.6}{Noether normalization lemma.} \label{noether-normalization}
+  Let $K$ a field, $A$ a fingen $K$-algebra.
+
+  Then $\exists~ z_1, \ldots, z_m \in A$ such that
+  \begin{enumerate}[i.]
+    \item $z_1, \ldots, z_m$ are algebraically independent over $K$
+    \item $A$ is finite over $B=K[z_1, \ldots, z_m]$
+  \end{enumerate}
+
+  That is, a fingen extension $K \subset A$ can be written as a composite
+  $$K \subset B = K[z_1, \ldots, z_m] \subset A$$
+  where $K \subset B$ is a polynomial extension, and $B \subset A$ is finite.
+\end{thm}
+\begin{proof}
+  induction on $n$.
+
+  if $n=0$, nothing to prove since $A$ is generated by $0$ elements $~\Longrightarrow~ A=K$, and $K$ is finite.
+
+  if $n>0$ we have two cases:
+  \begin{itemize}
+    \item $y_1, \ldots, y_n$ are algebraically independent over $K$, then $A \cong K[y_1, \ldots, y_n]$, so that $A$ is a finite module over itself.
+    \item $y_1, \ldots, y_n$ are algebraically dependent over $K$,
+      $$\exists 0 \neq f \in K[y_1, \ldots, y_n] ~\text{s.th}~ f(y_1, \ldots, y_n)=0$$
+  \end{itemize}
+
+  Goal: is to change variables so that $f$ becomes monic in one of the variables; this allows to express one generator as an integral element over the others.
+
+  Following from Lemma \ref{R.4.6.L}, define new variables $y^*_1, \ldots, y^*_{n-1} \in A$ such that $y_n$ is integral over
+  $$A^* = K[y^*_1, \ldots, y^*_{n-1}] ~\text{and}~ A=A^*[y_n]$$
+
+  By inductive hypothesis on $A^*,~~ \exists~ z_1, \ldots, z_m \in A^*$ algebraically independent over $K$ and with $A^*$ finite over $B=K[z_1, \ldots, z_m]$.
+
+  Since $y_n$ integral over $A^* ~~\Longrightarrow~ A^*[y_n]$ is finite over $A^*$.\\
+  Therefore, each step of $B \subset A^* \subset A^*[y_n]=A$ is finite, and $A$ is finite over $B$ as required.
+\end{proof}
+
+
+\begin{eg}{ }
+  $A = K[X,Y]/(XY-1)$. $Y$ is algebraic over $K[X]$, but not integral over $K[Y]$.
+
+  This corresponds to the fact that the hyperbola $XY=1$ has the line $X=0$ as an asymptotic line (so that its projection to the $X$-axis misses a root over $X=0$).
+
+  Take $X' = X- \epsilon Y$ as the element of $A$ instead of $X$; then the relation becomes $(X' + \epsilon Y) Y=1$, monic in $Y$ if $\epsilon \neq 0$.
+
+  This corresponds geometrically to tilting the hyperbola a little before projecting, so that no longer has a vertical asymtotic line.
+\end{eg}
+
+
 \newpage
 
 \section{Exercises}

galois-theory-notes.tex

@@ -517,12 +517,12 @@
       Note that $HN = \{ hn : h\in H, n\in N \}$. Let $h_1 n_1, h_2 n_2 \in HN$.
 
       Since $N$ normal $\Longrightarrow~ h_2^{-1} n_1 h_2 \in N$, so
-      $$(h_1 n_1)(h_2 n_2) = h_1 h_2 (h_2^{-1} n_1 h_2) \in HN$$
+      $$(h_1 n_1)(h_2 n_2) = h_1 h_2 (h_2^{-1} n_1 h_2) \cdot n_2 \in HN$$
 
       [Recall: since $N \triangleleft G$, $gN=Ng ~\forall g \in G$ $\Longrightarrow gn=n'g$ for some $n' \in N$.]
 
       To see that $(hn)^{-1} \in HN$:\\
-      since $(hn)^{-1} = n^{-1} h^{-1} = h^{-1} (h n^{-1} h^{-1})$, thus $(hn)^{-1} \in HN$.
+      since $(hn)^{-1} = h^{-1} n^{-1} = h^{-1} (h n^{-1} h^{-1})$, thus $(hn)^{-1} \in HN$.
 
       Thus $HN \subseteq G$.